L = { <M,t> : t ∈ L(M) and s ∉ L(M), t,s ∈ {a,b}*, where t is the string after s in a lexicographic ordering of {a,b}*}.
Rony
Chapter19TheUnsolvabilityoftheHaltingProblemFile.pptThe Unsolvability of the Halting Problem
Chapter 19
Languages and Machines
SD
D
Context-Free
Languages
Regular
Languages
reg exps
FSMs
cfgs
PDAs
unrestricted grammars
Turing Machines
D and SD
● A TM M with input alphabet decides a language L * iff,
for any string w *,
● if w L then M accepts w, and
● if w L then M rejects w.
A language L is decidable (in D) iff there is a Turing
machine M that decides it.
● A TM M with input alphabet semidecides L iff for any string
w *,
● if w L then M accepts w
● if w L then M does not accept w. M may reject or loop.
A language L is semidecidable (in SD) iff there is a Turing
machine that semidecides it.
Defining the Universe
What is the complement of:
- AnBn = {anbn : n 0}
- {<M, w> : TM M halts on input string w}.
Defining the Universe
L1 = {<M, w> : TM M halts on input string w}.
L2 = {<M> : M halts on nothing}.
L3 = {<Ma, Mb> : Ma and Mb halt on the same strings}.
For a string w to be in L1, it must:
● be syntactically well-formed.
● encode a machine M and a string w such that M halts
when started on w.
Define the universe from which we are drawing strings to contain only those strings that meet the syntactic requirements of the language definition.
This convention has no impact on the decidability of any of these languages since the set of syntactically valid strings is in D.
Our earlier definition:
L1 = {x: x is not a syntactically well formed <M, w> pair}
{<M, w> : TM M does not halt on input string w}.
We will use a different definition:
Define the complement of any language L whose member strings include at least one Turing machine description to be with respect to a universe of strings that are of the same syntactic form as L.
Now we have:
L1 = {<M, w> : TM M does not halt on input string w}.
A Different Definition of Complement
The Language H
H = {<M, w> : TM M halts on input string w}
Theorem: The language:
H = {<M, w> : TM M halts on input string w}
● is semidecidable, but
● is not decidable.
Does This Program Halt?
times3(x: positive integer) =
While x 1 do:
If x is even then x = x/2.
Else x = 3x + 1
25
Does This Program Halt?
times3(x: positive integer) =
While x 1 do:
If x is even then x = x/2.
Else x = 3x + 1
25
76
38
19
58
Does This Program Halt?
times3(x: positive integer) =
While x 1 do:
If x is even then x = x/2.
Else x = 3x + 1
25
76
38
19
58
29
88
44
22
11
Does This Program Halt?
times3(x: positive integer) =
While x 1 do:
If x is even then x = x/2.
Else x = 3x + 1
25
76
38
19
58
29
88
44
22
11
34
17
52
26
13
Does This Program Halt?
times3(x: positive integer) =
While x 1 do:
If x is even then x = x/2.
Else x = 3x + 1
25
76
38
19
58
40
20
10
5
16
29
88
44
22
11
34
17
52
26
13
Does This Program Halt?
times3(x: positive integer) =
While x 1 do:
If x is even then x = x/2.
Else x = 3x + 1
25
76
38
19
58
40
20
10
5
16
29
88
44
22
11
34
17
52
26
13
8
4
2
1
H is Semidecidable
Lemma: The language:
H = {<M, w> : TM M halts on input string w}
is semidecidable.
Proof:
H is Semidecidable
Lemma: The language:
H = {<M>, w : TM M halts on input string w}
is semidecidable.
Proof: The TM MH semidecides H:
MH(<M, w>) =
1. Run M on w.
MH halts iff M halts on w. Thus MH semidecides H.
The Unsolvability of the Halting Problem
Lemma: The language:
H = {<M, w> : TM M halts on input string w}
is not decidable.
Proof: If H were decidable, then some TM MH would decide it. MH would implement the specification:
halts(<M: string, w: string>) =
If <M> is a Turing machine description
and M halts on input w
then accept.
Else reject.
Trouble
Trouble(x: string) =
if halts(x, x) then loop forever, else halt.
If there exists an MH that computes the function halts, Trouble exists:
What is Trouble(<Trouble>)?
What is halts(<Trouble, Trouble>)?
● If halts reports that Trouble(<Trouble>) halts, Trouble loops.
● But if halts reports that Trouble(<Trouble>) does not halt, then
Trouble halts.
● Lexicographically enumerate Turing machines.
● Let 1 mean halting, blank mean non halting.
But Trouble behaves as:
Or maybe halts said that trouble(<trouble>) would halt. But then trouble would loop.
Viewing the Halting Problem as Diagonalization
| i1 | i2 | i3 | … | <Trouble> | … |
| machine1 | 1 | ||||
| machine2 | 1 | ||||
| machine3 | 1 | ||||
| … | 1 | ||||
| Trouble | 1 | 1 | |||
| … | 1 | 1 | 1 | ||
| … | 1 |
| Trouble | 1 | 1 | 1 |
If H were in D
Theorem: If H were in D then every SD language would be in D.
Proof: Let L be any SD language. There exists a TM ML that
semidecides it.
H = {<M>, w : TM M halts on input string w}
If H were also in D, then there would exist an O that decides it.
If H were in D
M'(w: string) =
1. Run O on <ML, w>.
2. If O accepts (i.e., ML will halt), then:
2.1. Run ML on w.
2.2. If it accepts, accept. Else reject.
3. Else reject.
So, if H were in D, all SD languages would be.
To decide whether w is in L(ML):
Back to the Entscheidungsproblem
Theorem: The Entscheidungsproblem is unsolvable.
Proof: (Due to Turing)
1. If we could solve the problem of determining whether a given Turing
machine ever prints the symbol 0, then we could solve the problem of
determining whether a given Turing machine halts.
2. But we can’t solve the problem of determining whether a given Turing
machine halts, so neither can we solve the problem of determining
whether it ever prints 0.
3. Given a Turing machine M, we can construct a logical formula F that is
true iff M ever prints the symbol 0.
4. If there were a solution to the Entscheidungsproblem, then we would
be able to determine the truth of any logical sentence, including F and
thus be able to decide whether M ever prints the symbol 0.
5. But we know that there is no procedure for determining whether M
ever prints 0.
6. So there is no solution to the Entscheidungsproblem.
IN SD OUT
Semideciding TM H
D
Deciding TM AnBnCn Diagonalize
Context-Free
CF grammar AnBn Pumping
PDA Closure
Closure
Regular
Regular Expression a*b* Pumping
FSM Closure
Language Summary
