Problems That Are Not Recursive¶

Examples¶

1. The Halting Problem¶

Proof 1¶

Proof 2¶

\(A_{TM} = \{<M, w>|M\text{ is a TM and M accepts w}\}\)

\(A_{TM}\) is not recursive

Construct a TM \(M^*\ H_{TM}\) to solve \(A_{TM}\):

\(H_{TM} = {<M, w>|M\text{ is a TM and M halts on w}}\)

\(M^* =\) On input x:

run M on w
If M accepts w, accept
If M rejects w, reject
If M loops on w, \(M^*\) is looping on w

Suppose \(M^*\) is a recursive TM, then we can construct a TM \(M_{H}\) to decide \(H_{TM}\):

\(M_{A} =\) On input :

Construct a TM \(M^*\) as above using M
Run \(M_{H}\) on \(M^*\), w
If \(M_{H}\) accepts, accept
If \(M_{H}\) rejects, reject

If all this is true, then \(M_A\) can decide \(A_{TM}\)

But \(A_{TM}\) is not recursive
So \(M^*\) is not recursive
So the halting problem is not recursive

1. \(L = \{"M"|\text{M is a TM that accepts e}\}\) is not recursive¶

Use \(A_{TM}\) to prove this

A Turing machine \(M\) accepts \(w\) if and only if \(M^*\) accepts \(e\)

\(M^* =\) On input x:

Run M on w
If M accepts w, accept
If M rejects w, reject

If M accepts w, then \(M^*\) accepts all inputs, including e

If M rejects or loops on w, then \(M^*\) rejects all inputs, including e

Suppose \(L\) is recursive, then we can construct a TM \(M_{L}\) to decide \(L\):

\(M_{A} =\) On input :

Construct a TM \(M^*\) as above using M
Run \(M_{L}\) on \(M^*\), w
If \(M_{L}\) accepts, accept
If \(M_{L}\) rejects, reject

If all this is true, then \(M_A\) can decide \(A_{TM}\)

But \(A_{TM}\) is not recursive
So \(M_L\) is not recursive
So \(L\) is not recursive

2. \(L_2 = \{"M"|\text{M is a TM that accepts all strings}\}\) is not recursive¶

Same as above

A Turing Machine accepts w if and only if \(M^*\) accepts all strings

3. \(L_3 = \{"M1, M2"|\text{M1 and M2 accept the same set of strings}\}\) is not recursive¶

Suppose that \(L_3\) is recursive, \(\exist M_3\) that decides \(L_3\)

Use \(M_3\) to construct \(M_2\) that deides \(L_2\)

\(M_2\) = on input "M"

construct \(M_E\) as follows: \(M_E\) = on input x
1. Accept
Run \(M_3\) on "\(M\)","\(M_E\)" -- check whether \(M\) and \(M_E\) accepts the same set of strings
If \(M_3\) accepts, accepts \(M\)
If \(M_3\) rejects, rejects \(M\)

So \(M_2\) decides \(L_2\) but \(L_2\) is not recursive

4. \(L_4 = \{"M"|\text{M is a TM with L(M) is regular}\}\) is not recursive¶

Proof: \(L_4\) is not recursive

Suppose that \(L_4\) is recursive, \(\exist M_4\) that decides \(L_4\)

Construction of \( M^* \): - Define a modified Turing machine \( M^* \) which takes two inputs: a string \( w \) and a separate input \( x \). - \( M^* \) operates as follows: On input x
1. It first runs \( M \) on \( w \).
2. If \( M \) accepts on \( w \), \( M^* \) then runs a universal Turing machine \( U \) on \( x \).
  1. If U accepts,accept
  2. If U rejects, reject
3. If \( M \) does not halt on \( w \), \( M^* \) does nothing (or rejects all inputs).

Universal Turing Machine U : On input : 1. Run M on w 2. If M accepts w, accept 3. else reject

Language of \( M^* \): The language \( L(M^*) \) is constructed so that:

\( L(M^*) \) is non-empty if \( M \) accepts \( w \).
\( L(M^*) \) = \(L(U)\) = \(A_{TM}\) Not regular/Not Context-Free /Not Recursive
\( L(M^*) = \emptyset \) if \( M \) does not accept \( w \). -- Regular/Contex-Free/Recursive

Conclusion:

If \( L(M^*) \) is non-regular, it implies that \( M \) accepts \( w \).
If \( L(M^*) \) is regular, then \( M \) does not accept \( w \).
Since determining whether \( L(M^*) \) is regular would solve the Halting Problem, it follows that \( R_{TM} \) is undecidable.

The undecidability of \( R_{TM} = \{ \langle M \rangle : M \text{ is a Turing machine with } L(M) \text{ regular} \} \) (i.e., determining if a Turing machine \( M \) has a regular language \( L(M) \)) stems from the fact that if we could decide \( R_{TM} \), we could also decide the Halting Problem, which is known to be undecidable.

Here’s a step-by-step explanation of why this leads to undecidability:¶

Assume \( R_{TM} \) is Decidable: Suppose there exists a Turing machine \( D \) that decides \( R_{TM} \), meaning that \( D \) can take as input any Turing machine \( M \) and decide if \( L(M) \) is regular.
Reduction from the Halting Problem: We know that the Halting Problem (determining whether a Turing machine \( M \) halts on an input \( w \)) is undecidable. To show that \( R_{TM} \) is also undecidable, we construct a new Turing machine \( M^* \) based on \( M \) and \( w \) and show that if we could decide whether \( L(M^*) \) is regular, we could solve the Halting Problem.
Construction of \( M^* \): Define \( M^* \) as a Turing machine that: - First simulates \( M \) on \( w \). - If \( M \) halts on \( w \), \( M^* \) then simulates a universal Turing machine \( U \) on any additional input \( x \). - If \( M \) does not halt on \( w \), \( M^* \) simply rejects all inputs or does not accept any string beyond a certain structure.
Behavior of \( L(M^*) \): - If \( M \) halts on \( w \), then \( L(M^*) \) becomes non-regular (since it depends on \( L(U) \), the language of a universal Turing machine, which is non-regular). - If \( M \) does not halt on \( w \), \( L(M^*) \) is regular (e.g., it could be empty or finite, depending on the construction).
Implication: If we had a way to determine whether \( L(M^*) \) is regular using the hypothetical decider \( D \) for \( R_{TM} \), we could answer whether \( M \) halts on \( w \): - If \( D \) tells us that \( L(M^*) \) is regular, then \( M \) does not halt on \( w \). - If \( D \) tells us that \( L(M^*) \) is non-regular, then \( M \) halts on \( w \).
Contradiction: Since this method would solve the Halting Problem (which is known to be undecidable), the assumption that \( R_{TM} \) is decidable must be incorrect. Therefore, \( R_{TM} \) is undecidable.

In short, if we could decide whether \( L(M) \) is regular for any Turing machine \( M \), we could also solve the Halting Problem by constructing \( M^* \) as described. This leads to a contradiction, proving that \( R_{TM} \) is undecidable.

4.1. \(L_5 = \{"M"|\text{M is a TM with L(M) is context-free}\}\) is not recursive¶

4.2. \(L_6 = \{"M"|\text{M is a TM with L(M) is recursive}\}\) is not recursive¶

Rice's Theorem¶

See Link

\(All_{PDA}\) is not recursive¶

\(All_{PDA} = \{<M>|M\text{ is a PDA and }L(M) = \Sigma^*\}\)

Turn into proving \(NOTAll_{PDA} = \{<M>|M\text{ is a PDA and }L(M) \neq \Sigma^*\}\) is not recursive

Construct a TM \(M\) to solve \(NOTAll_{PDA}\): * M halts on w if and only if \(L(P) \neq \Sigma^*\)

Turn the computing history of M that halts on w to a string

Construct P accepts all strings that are not computing history of M that halts on w.

Given a string \(C_1\#C_2\#...\#C_k\), where \(C_i\) is a configuration.
The string is not a computing history of M if: 1. \(C_1\) is not the start configuration of M 2. \(C_k\) is not the accept configuration of M 3. There exists \(C_i\) and \(C_{i+1}\) such that \(C_i\) does not lead to \(C_{i+1}\)

If \(NOTAll_{PDA}\) is recursive, then \(H_{TM}\) is recursive, which is not true.

Reduction¶

Let A, B be two languages, A reduces to B if there is a computable function f \(\Sigma^* \to \Sigma^*\) such that:

\(x \in A \Leftrightarrow f(x) \in B\)

Lemma 1¶

Suppose \(\exist f\) that reduces A to B:

If B is recursive, then A is recursive

B is recursive, then we can construct a TM \(M_B\) to decide B
\(\exist f\) that reduces A to B
\(M_A\) = On input x: 1. Compute f(x) 2. Run \(M_B\) on f(x) 3. If \(M_B\) accepts, accept 4. If \(M_B\) rejects, reject

If A is not recursive, then B is not recursive

If B is recursive, then A is recursive, which is not true(Clearly)

最后更新: 2024年11月11日 23:37:39
创建日期: 2024年11月11日 23:05:25