SHAMIR’S AND BLAKLEY’S SECRET SHARING
SCHEMES IN ADVANCED COMPUTER SECURITY
1 SHAMIR’S SECRET SHARING SCHEME
Problem 1. In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The
polynomial used is f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x
= 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
2 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 9. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 10. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 11. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
3 BLAKLEY’S SECRET SHARING SCHEME
Problem 12. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret
point is (2,3,4). Generate equations for three shares.
Solution 12. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 13. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y
+ 3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 14. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x
- y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 15. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of
the space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 16. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is
the geometric representation of each share, and what is the maximum number of shares that
can be generated while maintaining security (i.e., any k-1 shares reveal no information about
the secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 17. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 18. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 19. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
4 BLAKLEY’S SECRET SHARING SCHEME
Problem 20. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret
point is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
5 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
6 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
7 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
8 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
9 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
10 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
11 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
12 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
13 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
14 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
15 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
16 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
17 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
18 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
19 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
20 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
21 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
22 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
23 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
24 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
25 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
26 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
27 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
28 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
29 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
30 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
31 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
32 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
33 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
34 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
35 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
36 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
37 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
38 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
39 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
40 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
In a (3,5) Shamir’s secret sharing scheme over GF(17), the secret is 10. The polynomial used is
f(x) = 10 + 7x + 4x². Calculate the shares for the first three participants (x = 1, 2, 3).
Solution 1. Step 1: Use the polynomial f(x) = 10 + 7x + 4x² to calculate shares.
For x = 1: f(1) = 10 + 7(1) + 4(1)² = 10 + 7 + 4 = 21 ≡ 4 (mod 17)
For x = 2: f(2) = 10 + 7(2) + 4(2)² = 10 + 14 + 16 = 40 ≡ 6 (mod 17)
For x = 3: f(3) = 10 + 7(3) + 4(3)² = 10 + 21 + 36 = 67 ≡ 16 (mod 17)
Therefore, the shares for the first three participants are (1,4), (2,6), and (3,16).
Problem 2. In a Shamir’s secret sharing scheme, three shares are given: (1,4), (2,7), and
(4,13). The scheme is over GF(17). Reconstruct the secret.
Solution 2. Step 1: Use Lagrange interpolation to reconstruct the polynomial.
L₁(x) = ((x-2)(x-4)) / ((1-2)(1-4)) = (x²-6x+8) / 3 ≡ 6x²-2x+3 (mod 17) L₂(x) = ((x-1)(x-4)) /
((2-1)(2-4)) = (x²-5x+4) / 2 ≡ 9x²-11x+2 (mod 17) L₄(x) = ((x-1)(x-2)) / ((4-1)(4-2)) = (x²-
3x+2) / 6 ≡ 3x²-9x+1 (mod 17)
f(x) = 4L₁(x) + 7L₂(x) + 13L₄(x) = 4(6x²-2x+3) + 7(9x²-11x+2) + 13(3x²-9x+1) ≡ 24x²-
8x+12 + 63x²-77x+14 + 39x²-117x+13 ≡ 126x²-202x+39 ≡ 7x²+13x+5 (mod 17)
Step 2: The secret is the constant term of the polynomial.
Therefore, the reconstructed secret is 5.
Problem 3. In a (4,7) Shamir’s secret sharing scheme over GF(23), what is the maximum
degree of the polynomial used to generate shares?
Solution 3. In a (k,n) Shamir’s secret sharing scheme, the degree of the polynomial is k-1.
For a (4,7) scheme, k = 4.
Therefore, the maximum degree of the polynomial is 4 - 1 = 3.
Problem 4. In a Shamir’s secret sharing scheme over GF(19), the polynomial used is f(x) = 7
+ 3x + 5x² + 2x³. What is the threshold (minimum number of shares required to reconstruct
the secret) for this scheme?
Solution 4. Step 1: Determine the degree of the polynomial. The polynomial f(x) = 7 + 3x +
5x² + 2x³ has degree 3.
Step 2: Calculate the threshold. In Shamir’s scheme, the threshold k is equal to the degree of
the polynomial plus 1.
Therefore, the threshold for this scheme is 3 + 1 = 4.
41 BLAKLEY’S SECRET SHARING SCHEME
Problem 5. In a (3,5) Blakley’s secret sharing scheme in 3-dimensional space, the secret point
is (2,3,4). Generate equations for three shares.
Solution 5. In Blakley’s scheme, each share is a plane that passes through the secret point.
We’ll generate three random planes that pass through (2,3,4).
Step 1: Generate three planes (using random coefficients): Plane 1: 2x + y + z = 11 Plane 2: x
+ 3y - z = 8 Plane 3: -x + 2y + 2z = 13
Step 2: Verify that each plane passes through (2,3,4): Plane 1: 2(2) + 3 + 4 = 11 ✓ Plane 2: 2
+ 3(3) - 4 = 8 ✓ Plane 3: -2 + 2(3) + 2(4) = 13 ✓
Therefore, three possible share equations are: 1. 2x + y + z = 11 2. x + 3y - z = 8 3. -x + 2y +
2z = 13
Problem 6. In a Blakley’s secret sharing scheme, three shares (planes) are given: 1. x + 2y +
3z = 14 2. 2x - y + z = 6 3. 3x + y - z = 8 Find the secret point (the intersection of these
planes).
Solution 6. Step 1: Set up the system of linear equations: x + 2y + 3z = 14 (Equation 1) 2x -
y + z = 6 (Equation 2) 3x + y - z = 8 (Equation 3)
Step 2: Solve the system using elimination or substitution method. Multiply Equation 1 by 2 and
subtract Equation 2: 2x + 4y + 6z = 28 2x - y + z = 6 —————– 5y + 5z = 22 (Equation 4)
Multiply Equation 1 by 3 and subtract Equation 3: 3x + 6y + 9z = 42 3x + y - z = 8 —————–
5y + 10z = 34 (Equation 5)
From Equation 4: y + z = 22/5 From Equation 5: y + 2z = 34/5
Subtracting these equations: z = 12/5
Substituting back: y = 22/5 - 12/5 = 2
Substituting y and z into Equation 1: x + 2(2) + 3(12/5) = 14 x + 4 + 36/5 = 14 x = 14 - 4 -
36/5 = 10 - 36/5 = 50/5 - 36/5 = 14/5
Therefore, the secret point (intersection of the planes) is (14/5, 2, 12/5) or (2.8, 2, 2.4).
Problem 7. In a (4,6) Blakley’s secret sharing scheme, what is the minimum dimension of the
space in which the scheme operates?
Solution 7. In a (k,n) Blakley’s secret sharing scheme, the minimum dimension of the space is
k-1.
For a (4,6) scheme, k = 4.
Therefore, the minimum dimension of the space is 4 - 1 = 3.
The scheme operates in at least 3-dimensional space.
Problem 8. In a Blakley’s secret sharing scheme operating in 4-dimensional space, what is the
geometric representation of each share, and what is the maximum number of shares that can
be generated while maintaining security (i.e., any k-1 shares reveal no information about the
secret)?
Solution 8. Step 1: Determine the geometric representation of each share. In a 4-dimensional
space, each share is represented by a hyperplane (a 3-dimensional subspace).
Step 2: Calculate the threshold k. In an n-dimensional space, the threshold k for Blakley’s
scheme is n + 1. For a 4-dimensional space, k = 4 + 1 = 5.
Step 3: Determine the maximum number of shares. In Blakley’s scheme, there is no theoretical
upper limit to the number of shares that can be generated. However, in practice, the number of
shares is limited by the precision of the computations and the desire to maintain a certain level
of security.
A common guideline is to keep n significantly smaller than the size of the finite field being used
(if one is used) to prevent information leakage through statistical analysis.
Assuming we’re using a large finite field, a reasonable maximum number of shares could be in
the hundreds or even thousands, as long as it’s much smaller than the field size.
Therefore, in a 4-dimensional Blakley’s scheme: 1. Each share is represented by a 3-
dimensional hyperplane. 2. The threshold k is 5. 3. The maximum number of shares can be very
large (hundreds or thousands) depending on the implementation details and security
requirements.
