HOMOMORPHIC ENCRYPTION INVESTIGATE ENCRYPTION SCHEMES THAT ALLOW COM-
PUTATIONS TO BE PERFORMED ON CIPHERTEXTS
1. Question: In a fully homomorphic encryption scheme, if the ciphertext represents the integer 7 and the
encryption function is additive homomorphic, and the plaintext message is an integer 3, what would be the
result of decrypting the ciphertext after performing an addition operation on the ciphertext and plaintext?
Solution: Given ciphertext represents the integer 7. Plaintext message is an integer 3.
In a homomorphic encryption scheme, the encryption function is additive homomorphic. Therefore,
performing an addition operation on the ciphertext and plaintext will yield the same result as if the operation
was performed on the plaintexts before encryption.
So, the addition operation would result in: 7 (ciphertext) + 3 (plaintext) = 10
Therefore, after decrypting the modified ciphertext, the result would be the integer 10.
Final numerical answer: 10
2. Question: In a fully homomorphic encryption scheme, if the encryption function E adds noise ∆to
the ciphertext with plaintext m, and the noise grows with each homomorphic operation by a factor of α,
then after nhomomorphic operations, what is the total noise added to the ciphertext?
Solution: The total noise added to the ciphertext after nhomomorphic operations can be calculated
using the formula:
Total Noise = ∆ + α+α2+... +αn
This is a geometric series, and the sum of a geometric series can be expressed as:
Sum = a(1−αn)
1−α, where ais the first term of the series.
Since the noise grows with each operation by a factor of α, the first term of the series is α.
Therefore, Total Noise = α(1−αn)
1−α=α(1 −αn)
So, the total noise added to the ciphertext after nhomomorphic operations is α(1 −αn).
3. Question: In a fully homomorphic encryption scheme, if the encryption of integer 7 is represented as
25 and the encryption of integer 5 is represented as 16, what is the encryption of the product of 7 and 5?
Solution: In a fully homomorphic encryption scheme, the homomorphic property allows computations
to be performed on ciphertexts without decrypting them.
Given: Encryption of integer 7 = 25 Encryption of integer 5 = 16
Let’s denote the encryption function as Enc().
To find the encryption of the product of 7 and 5: Enc(7) = 25 Enc(5) = 16
Since fully homomorphic encryption schemes support multiplication, we can compute the product of
the ciphertexts to get the encryption of the product:
Enc(7) * Enc(5) = 25 * 16 = 400
Therefore, the encryption of the product of 7 and 5 is 400.
4. Question: In a Fully Homomorphic Encryption scheme, if a specific scheme requires 100 times the
computational resources to evaluate a function on encrypted data compared to evaluating it on plaintext data,
and the function takes 10 milliseconds to evaluate on plaintext data, how long would it take to evaluate the
same function on encrypted data in this scheme?
Solution: Let’s denote the time it takes to evaluate the function on plaintext data as "t", which is 10
milliseconds. In this case, the function takes 100 times the resources to evaluate on encrypted data compared
to plaintext data. Therefore, the time it would take to evaluate the function on encrypted data is 100 * t
milliseconds.
Substitute the value of t: 100 * 10 milliseconds = 1000 milliseconds
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
So, it would take 1000 milliseconds (or 1 second) to evaluate the function on encrypted data in this Fully
Homomorphic Encryption scheme.
5. Question: In a Fully Homomorphic Encryption (FHE) scheme, Bob wants to compute the multi-
plication of two encrypted integers: 5and 7which are encrypted as E(5) and E(7). If the multiplication
operation in this FHE scheme generates noise equivalent to 3, what would be the result of E(5) ×E(7) after
decryption?
Solution: In Fully Homomorphic Encryption (FHE) schemes, the homomorphic properties allow com-
putations to be performed on encrypted data. When we multiply two ciphertexts in an FHE scheme, the
noise grows due to the homomorphic properties.
Let’s assume the noise level of the decryption function is denoted as α. When we multiply two cipher-
texts, the noise in the result is the sum of the noises in the input ciphertexts plus some additional noise
related to the homomorphic operation:
αresult =αinput1+αinput2+additional noise
Given that the noise introduced by the multiplication operation is 3, we can set up the following equation:
α(E(5) ×E(7)) = α(E(5)) + α(E(7)) + 3
Since the ciphertexts for 5and 7are encrypted as E(5) and E(7), we assume they have negligible noise
initially. Hence,
α(E(5) ×E(7)) = 0 + 0 + 3 = 3
Therefore, after decryption, the result of E(5) ×E(7) would have noise equivalent to 3.
6. Question: In a Fully Homomorphic Encryption (FHE) scheme, if the noise budget is initially set to
1000 and each homomorphic multiplication operation increases the noise by 20, how many homomorphic
multiplications can be performed before the noise level exceeds the threshold of 2000?
Solution: Given: Initial noise budget = 1000 Noise increase per homomorphic multiplication = 20
Threshold noise level = 2000
Let’s denote the number of homomorphic multiplications as ’n’ that can be performed before the noise
level exceeds 2000.
The total noise after ’n’ homomorphic multiplications can be represented as: Total Noise = Initial Noise
+ Noise Increase per Multiplication * Number of Multiplications 2000 = 1000 + 20n 20n = 1000 n = 50
Therefore, before the noise level exceeds 2000, 50 homomorphic multiplications can be performed.
7. Question: In the context of Homomorphic Encryption, consider two encryption schemes: Scheme A
is a fully homomorphic encryption scheme, and Scheme B is a partially homomorphic encryption scheme.
If Scheme A allows for an unlimited number of homomorphic operations on ciphertexts, while Scheme B
allows for only two homomorphic operations before decryption, what is the maximum number of computa-
tions that can be performed on ciphertexts in Scheme B compared to Scheme A?
Solution: In a fully homomorphic encryption scheme such as Scheme A, computations can be performed
on encrypted data an unlimited number of times without needing to decrypt it. This means that Scheme A
offers the ability to perform an unlimited number of homomorphic operations on ciphertexts.
In contrast, in a partially homomorphic encryption scheme like Scheme B, only a limited number of
homomorphic operations can be performed before decryption is required to maintain the homomorphic
property. Let’s assume Scheme B can perform two homomorphic operations before decryption is needed.
Let’s denote the maximum number of computations as n. For Scheme A: n=∞(unlimited) For
Scheme B: Since Scheme B allows for only two homomorphic operations, the maximum number of compu-
tations would be 2.
Therefore, the maximum number of computations that can be performed on ciphertexts in Scheme B
compared to Scheme A is: 2(Scheme B) / ∞(Scheme A) = 0
So, the maximum number of computations that can be performed on ciphertexts in Scheme B compared
to Scheme A is 0.
8. Question: In a secure multiparty computation (MPC) protocol utilizing homomorphic encryption,
Alice encrypts a value x with a homomorphic encryption scheme and sends the ciphertext to Bob. If Bob
multiplies the ciphertext by 5 using the homomorphic properties and sends the result back to Alice, what
value does Alice decrypt after receiving the result?
Solution: Let’s denote the original value as x and the homomorphic encryption of x as Enc(x). When
Bob multiplies the ciphertext by 5, this operation is represented as 5 * Enc(x).
Since homomorphic encryption schemes support multiplication operation, Bob’s computation can be
done using the homomorphic properties. Therefore, Alice will receive the ciphertext corresponding to the
result of multiplying the original plaintext by 5, which we denote as Enc(5x).
When Alice decrypts the received ciphertext (ciphertext * 5), she will obtain the result of decrypting the
product, which corresponds to 5x.
Therefore, the value that Alice will decrypt after receiving the result is 5x.
9. Question: In a fully homomorphic encryption scheme, if the security parameter is increased from 80
bits to 128 bits, how much additional computational overhead (in percentage) would be typically added to
perform a single homomorphic operation on encrypted data?
Solution: Fully homomorphic encryption (FHE) schemes allow for computations to be performed on
ciphertexts without requiring decryption. The security parameter in FHE determines the level of security
provided by the encryption scheme, typically measured in bits.
When the security parameter increases from 80 bits to 128 bits, the additional computational overhead
in terms of the number of operations needed to perform a single homomorphic operation on encrypted data
is usually calculated as a percentage increase. This overhead is incurred due to the increased number of
cryptographic operations required to maintain the higher security level.
Let’s say for an FHE scheme, increasing the security parameter from 80 bits to 128 bits leads to a 20
Therefore, the additional computational overhead when increasing the security parameter from 80 bits
to 128 bits in this FHE scheme is 20
10. Question: In a homomorphic encryption scheme, if the encryption key size is 2048 bits and the
plaintext message is 128 bits long, how many ciphertext bits will be produced after encryption?
Solution: In a homomorphic encryption scheme, the ciphertext size depends on factors such as the
security parameter and the encryption algorithm used. Assuming a simple scenario where the encryption
key size is 2048 bits and the plaintext message is 128 bits long, we need to calculate the size of the resulting
ciphertext.
To calculate the size of the ciphertext, we can use the formula: Ciphertext size = Encryption key size +
Plaintext size
Given: Encryption key size = 2048 bits Plaintext size = 128 bits
Ciphertext size = 2048 bits + 128 bits Ciphertext size = 2176 bits
Therefore, after encryption using the given parameters, the resulting ciphertext will be 2176 bits long.
11. Question: In a fully homomorphic encryption scheme, if the encryption function is represented as
E(m) = m + r, where m is the message and r is a random number, and the decryption function is D(c) = c,
what is the result of encrypting and then decrypting the sum of two plaintext messages, x and y?
Solution: 1. Encrypting x: E(x) = x + r 2. Encrypting y: E(y) = y + r 3. Adding the encrypted messages:
E(x) + E(y) = (x + r) + (y + r) = x + y + 2r 4. Decrypting the sum of the encrypted messages: D(x + y + 2r)
=x+y+2r
Therefore, the result of encrypting and then decrypting the sum of plaintext messages x and y in this
fully homomorphic encryption scheme will be x + y + 2r.
12. Question: In a comparative analysis of two fully homomorphic encryption schemes, Scheme A
supports up to 20 homomorphic additions and 15 homomorphic multiplications per ciphertext before de-
cryption, while Scheme B allows up to 30 homomorphic additions and 10 homomorphic multiplications
per ciphertext before decryption. If a computation requires 3 homomorphic additions and 5 homomorphic
multiplications, which scheme would be more suitable for this specific computation?
Solution: To solve this, we need to calculate the total cost in terms of homomorphic operations for the
given computation in each scheme.
For Scheme A: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
For Scheme B: Total cost = 3 homomorphic additions + 5 homomorphic multiplications = 8 homomor-
phic operations
Comparing the two schemes, both Scheme A and Scheme B have the same total cost of 8 homomorphic
operations for this specific computation. However, if we have to choose based on these numbers, Scheme
A would be more suitable as it has a higher limit on homomorphic multiplications which might be useful in
future computations requiring more multiplication operations.
Therefore, for this specific computation, both Scheme A and Scheme B are equally suitable, but Scheme
A might provide more flexibility for future computations requiring more homomorphic multiplications.
13. Question: In a secure multi-party computation scenario using a homomorphic encryption scheme,
Alice encrypts the value 5 using the encryption function, and Bob encrypts the value 3. They then perform
an addition operation on their respective ciphertexts. If the result of the addition operation is 18, what
encryption scheme are they likely using?
Solution: Let’s denote the ciphertext produced by Alice as CaandtheciphertextproducedbyBobasCb.T heencryptionfunctionforAliceisEnc(a)andforBobisEnc(b).
Given that Alice encrypts 5 and Bob encrypts 3, we have: Ca=Enc(5)Cb=Enc(3)
If the addition operation on the ciphertexts results in 18, we can express this as: Dec(Ca) + Dec(Cb) =
18Dec(Enc(5)) + Dec(Enc(3)) = 185 + 3 = 188 = 18
Since 8 is not equal to 18, we can see that standard encryption schemes do not produce this result.
This indicates that Alice and Bob are likely using a homomorphic encryption scheme, specifically a
Partially Homomorphic Encryption (PHE) scheme. In a PHE scheme, operations can be performed on ci-
phertexts that correspond to specific mathematical operations, allowing the computation of a certain function
on encrypted values to match the result of the function on the plaintexts.
So, the encryption scheme they are likely using is a Partially Homomorphic Encryption (PHE) scheme.
14. Question: In a fully homomorphic encryption (FHE) scheme, if the original message is represented
by the integer 7 (m=7) and the encryption function is defined as E(m) = m + 3 mod 10, what would be the
ciphertext when m=7?
Solution: To find the ciphertext, we substitute m=7 in the encryption function E(m) = m + 3 mod 10:
E(7) = 7 + 3 mod 10 E(7) = 10 mod 10 E(7) = 0
Therefore, when the original message m=7 is encrypted using the given FHE scheme, the ciphertext
would be 0.
15. Question: In Fully Homomorphic Encryption (FHE) schemes, if Bob wants to multiply two cipher-
texts encrypted with the public key, shown as (Epk(m1), Epk(m2)), to compute the multiplication of the
plaintexts m1×m2, how many homomorphic operations would be required in a typical scenario?
Solution: In Fully Homomorphic Encryption (FHE) schemes, to compute the multiplication of plaintexts
m1×m2from their encrypted forms (Epk(m1), Epk(m2)), several homomorphic operations are required.
Typically, for fully homomorphic encryption schemes, to perform a multiplication operation, it would
require several basic operations equivalent to the multiplicative depth of the circuit representing the opera-
tion. This multiplicative depth depends on the encryption scheme and the specific operations required.
In a typical scenario, multiplying two ciphertexts encrypted with the public key involves multiple oper-
ations equivalent to the circuit depth, which could range from 102to 104or more depending on the security
parameters and efficiency of the encryption scheme.
Therefore, in a general sense for Fully Homomorphic Encryption, to multiply two ciphertexts together, it
would require a numerical range of operations typically falling within the scale of 102to 104homomorphic
operations.
16. Question: In the context of Performance Optimization in Fully Homomorphic Encryption Algo-
rithms, if a computation requires 5 homomorphic multiplications and 3 homomorphic additions on cipher-
texts, and each homomorphic operation takes 10 milliseconds to execute, how long will it take to complete
the entire computation?
Solution: Given that: - Number of homomorphic multiplications = 5 - Number of homomorphic addi-
tions = 3 - Time taken for each homomorphic operation = 10 milliseconds
Time taken for 5 homomorphic multiplications = 5 multiplications * 10 milliseconds/multiplication =
50 milliseconds Time taken for 3 homomorphic additions = 3 additions * 10 milliseconds/addition = 30
milliseconds
Total time taken to complete the computation = Time taken for multiplications + Time taken for additions
= 50 milliseconds + 30 milliseconds = 80 milliseconds
Therefore, it will take 80 milliseconds to complete the entire computation.
17. Question: In terms of computational complexity, how does Fully Homomorphic Encryption (FHE)
differ from Partially Homomorphic Encryption (PHE)?
Solution: Fully Homomorphic Encryption (FHE) schemes generally involve higher computational com-
plexity compared to Partially Homomorphic Encryption (PHE) schemes. The key difference lies in the
ability of FHE to carry out both addition and multiplication operations on ciphertexts, while PHE can only
support one of these operations.
For FHE, the computational complexity is often significantly higher due to the need to perform complex
operations on encrypted data. This typically involves a higher number of cryptographic operations per
homomorphic operation which can lead to increased computational overhead.
On the other hand, PHE schemes are less computationally intensive as they support only one type of
operation (either addition or multiplication) on ciphertexts, thus requiring fewer operations overall.
Therefore, the key distinction in terms of computational complexity between FHE and PHE is that Fully
Homomorphic Encryption tends to have a higher computational overhead compared to Partially Homomor-
phic Encryption.
Final Answer: The numerical answer required is that Fully Homomorphic Encryption (FHE) generally
involves higher computational complexity than Partially Homomorphic Encryption (PHE).
18. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user can perform up to 5 homo-
morphic operations on ciphertexts, and a Partially Homomorphic Encryption (PHE) scheme allows compu-
tation on a ciphertext only for addition or multiplication but not both, how many homomorphic operations
can the user perform on ciphertexts using PHE?
Solution: In a PHE scheme, the user can perform either addition or multiplication homomorphic opera-
tion on ciphertexts but not both. Therefore, in one operation, the user can either add or multiply encrypted
values but not both simultaneously.
Since the PHE scheme allows computation for only one type of operation, the user can perform a total
of 1 homomorphic operation on ciphertexts using the PHE scheme.
Final numerical answer: 1
19. Question: In a homomorphic encryption scheme, if a computation on ciphertexts increases the noise
by 5 units and the initial noise level was 12, what is the final noise level after 8 computations?
Solution: Each computation on ciphertexts increases the noise by 5 units. If the initial noise level was
12, after 8 computations, the total increase in noise would be 5 * 8 = 40 units.
Therefore, the final noise level after 8 computations would be the initial noise level (12) plus the total
increase in noise (40), which equals 12 + 40 = 52 units.
Final Numerical Answer: The final noise level after 8 computations in the given homomorphic encryp-
tion scheme would be 52 units.
20. Question: In a fully homomorphic encryption scheme, if a plaintext value x is encrypted and then
multiplied by a ciphertext value y to get a new ciphertext z, what operation can be performed on z to obtain
the result of (x * y) * y?
Solution: - In a fully homomorphic encryption scheme, the encryption operation can be represented as
E(x) = cxandE(y) = cy, whereEistheencryptionfunction, xandyareplaintextvalues, andcxandcyaretheircorrespondingciphertexts.−
T hemultiplicationof aciphertextvaluecxwithanotherciphertextvaluecy, denotedascz=cx∗cy, yieldsanewciphertextczsuchthatwhendecrypted, theresultistheproductoftheplaintextvaluesencryptedbycxandcy, i.e., x∗
y.−T herefore, toobtaintheresultof (x∗y)∗yusinghomomorphicproperties, theoperationthatneedstobeperf ormedontheciphertextczisanothermultiplicationwiththeciphertextcy.−
Mathematically, wehave : (x∗y)∗y=x∗(y∗y) = E(x)∗E(y∗y) = cx∗c2
y−T hus, theoperationtobeperf ormedoncztoobtaintheresult(x∗
y)∗yissquaringtheciphertextcy.
Therefore, the numerical answer to the question is: squaring the ciphertext cy.
21. Question: Consider a fully homomorphic encryption scheme that supports additions and multiplica-
tions on encrypted data. If a user encrypts the numbers 5 and 3 separately using this scheme, what would be
the encrypted value of the result when performing the multiplication operation on these encrypted values?
Solution: Let’s assume the encryption of 5 is represented as Enc(5) and the encryption of 3 is represented
as Enc(3).
Given that the homomorphic encryption scheme supports both addition and multiplication, we can per-
form the multiplication operation on the encrypted values.
The multiplication operation on two encrypted values can be performed in a fully homomorphic en-
cryption scheme by multiplying the ciphertexts together. Therefore, the encrypted value of the result would
be:
Enc(5) * Enc(3)
When the encrypted values are multiplied, the resulting ciphertext will contain the encrypted value of 5
* 3 = 15.
Thus, the encrypted value of the result when performing the multiplication operation on the encrypted
values of 5 and 3 would be 15.
22. Question: In a fully homomorphic encryption scheme, if the encryption of a plaintext requires x
multiplication operations and yaddition operations, how many multiplication and addition operations are
needed to evaluate the product of two ciphertexts encrypted under this scheme?
Solution: In a fully homomorphic encryption scheme, evaluating the product of two ciphertexts en-
crypted under this scheme involves multiplying the ciphertexts element-wise and performing addition oper-
ations on the resulting ciphertext.
To multiply the ciphertexts, we need to perform xmultiplication operations for the first ciphertext and
xmultiplication operations for the second ciphertext, resulting in a total of x+x= 2xmultiplication
operations.
To add the ciphertexts element-wise, we need to perform yaddition operations for the first ciphertext
and yaddition operations for the second ciphertext, resulting in a total of y+y= 2yaddition operations.
Therefore, to evaluate the product of two ciphertexts encrypted under a fully homomorphic encryption
scheme, we need a total of 2xmultiplication operations and 2yaddition operations.
23. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a ciphertext encrypts the num-
ber 5 and the encryption scheme supports addition and multiplication operations with depth 3, how many
homomorphic operations can be performed on the ciphertext before decrypting the result?
Solution: Fully Homomorphic Encryption (FHE) schemes allow computations to be carried out on
encrypted data without decrypting it. The depth of an FHE scheme restricts the number of homomorphic
operations that can be performed on the ciphertext before it needs to be decrypted. In this case, let’s assume
the FHE scheme supports addition and multiplication operations with a depth of 3.
For an FHE scheme with depth 3, it allows for a chain of three homomorphic operations to be carried out
on the ciphertext before it requires decryption. This means we can perform a sequence of three homomorphic
operations (e.g., two additions and one multiplication, or any other combination) on the encrypted data.
Therefore, the number of homomorphic operations that can be performed on the ciphertext with depth 3
is 3.
24. Question: In a Fully Homomorphic Encryption (FHE) scheme, if a user encrypts a value x= 7 and
the encryption scheme supports both addition and multiplication homomorphisms with public parameters,
then applies the following computations: 1. Compute x+ 3 on the ciphertext. 2. Compute (x+ 3) ×2on
the resulting ciphertext from the first computation. What is the final decrypted result if the user decrypts the
final ciphertext?
Solution: 1. Encrypting the value x= 7: - Let x= 7 - Encryption of xgives ciphertext C=Enc(x)
2. Compute x+ 3 on the ciphertext: - Add 3 to the ciphertext Cusing the homomorphic addition:
C+ 3 = Enc(x)+3- The computation x+ 3 results in the ciphertext C′where C′=Enc(x+ 3)
3. Compute (x+ 3) ×2on the resulting ciphertext from the first computation: - Multiply the resulting
ciphertext C′by 2 using the homomorphic multiplication: 2∗C′= 2 ∗Enc(x+ 3) - The computation
(x+ 3) ×2results in the ciphertext C′′ where C′′ =Enc((x+ 3) ×2)
4. Decryption of the final ciphertext: - Decrypt the final ciphertext C′′ to get the result of the computa-
tion: Dec(C′′) = (x+ 3) ×2 = (7 + 3) ×2 = 10 ×2 = 20
Therefore, the final decrypted result after performing the computations on the ciphertext is 20.
25. Question: In a secure multiparty computation using homomorphic encryption, Alice encrypts a
message x= 10 under a fully homomorphic encryption scheme. Bob encrypts a message y= 5 under the
same scheme. They perform the homomorphic addition operation on their ciphertexts. If the result is z= 25
after decryption, what is the decryption result of x×y?
Solution: 1. Let’s denote the encrypted messages as follows: - Enc(x) = a-Enc(y) = b-Enc(z) =
a+b
2. Given that after decryption the result is z= 25, it means: - Dec(Enc(z)) = Dec(a+b) = x+y=
10 + 5 = 15
AL
AL
3. To find the decryption result of x×y, we perform the corresponding operations on the plaintexts: -
x×y= 10 ×5 = 50
4. Therefore, the decryption result of x×yis 50.
