Cs133 lec12

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CS 133 : Automata Theory and Computability Turing Machines Henry N. Adorna Nestine Hope S. Hernandez Algorithms and Complexity Laboratory Department of Computer Science University of the Philippines, Diliman {hnadorna,nshernandez}@dcs.upd.edu.ph

LEC 12

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Let M be a Turing machine and w an input string. An accepting computation history for M on w is a sequence of configurations, C1 , C2 , . . . , Ck , where C1 is the start configuration of M on w, Ck is an accepting configuration of M , and each Ci legally follows from Ci−1 according to the rules of M . A rejecting computation history for M on w is defined similarly, except that Ck is a rejecting configuration.

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A linear bounded automaton is a restricted type of Turing machine wherein the tape head isn’t permitted to move off the portion of the tape containing the input. If the machine tries to move its head off either end of the input, the head stays where it is, in the same way that the head will not move off the left-hand end of an ordinary Turing machine’s tape.

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A linear bounded automaton is a restricted type of Turing machine wherein the tape head isn’t permitted to move off the portion of the tape containing the input. If the machine tries to move its head off either end of the input, the head stays where it is, in the same way that the head will not move off the left-hand end of an ordinary Turing machine’s tape.

Despite their memory constraint, LBAs are quite powerful.

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A linear bounded automaton is a restricted type of Turing machine wherein the tape head isn’t permitted to move off the portion of the tape containing the input. If the machine tries to move its head off either end of the input, the head stays where it is, in the same way that the head will not move off the left-hand end of an ordinary Turing machine’s tape.

Despite their memory constraint, LBAs are quite powerful.

For example, the deciders for ADF A , ACF G , EDF A , ECF G are LBAs. Every CFL can be decided by an LBA.

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ALBA is the problem of determining whether an LBA accepts its input.

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ALBA is the problem of determining whether an LBA accepts its input.

ALBA = {< M, w > | M is an LBA that accepts string w}

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ALBA is the problem of determining whether an LBA accepts its input.

ALBA = {< M, w > | M is an LBA that accepts string w}

Lemma Let M be an LBA with q states and g symbols in the tape alphabet.

There are exactly qng n distinct configurations of M for a tape of length n.

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ALBA is the problem of determining whether an LBA accepts its input.

ALBA = {< M, w > | M is an LBA that accepts string w}

Lemma Let M be an LBA with q states and g symbols in the tape alphabet.

There are exactly qng n distinct configurations of M for a tape of length n.

Theorem ALBA is decidable.

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ALBA is the problem of determining whether an LBA accepts its input.

ALBA = {< M, w > | M is an LBA that accepts string w}

Lemma Let M be an LBA with q states and g symbols in the tape alphabet.

There are exactly qng n distinct configurations of M for a tape of length n.

Theorem ALBA is decidable.

• The algorithm that decides ALBA is as follows:

L = “On input < M, w >, where M is an LBA and w is a string: 1. Simulate M on w for qng n steps or until it halts.

2. If M has halted, accept if it has accepted and reject if it has rejected. If it has not halted, reject. ”

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ALBA is the problem of determining whether an LBA accepts its input.

ALBA = {< M, w > | M is an LBA that accepts string w}

Lemma Let M be an LBA with q states and g symbols in the tape alphabet.

There are exactly qng n distinct configurations of M for a tape of length n.

Theorem ALBA is decidable.

• The algorithm that decides ALBA is as follows:

L = “On input < M, w >, where M is an LBA and w is a string: 1. Simulate M on w for qng n steps or until it halts.

2. If M has halted, accept if it has accepted and reject if it has rejected. If it has not halted, reject. ”

If M on w has not halted within qng n steps, it must be repeating a configuration and therefore looping.

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ELBA = {< M > | M is an LBA where L(M ) = â&#x2C6;&#x2026;}

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ELBA = {< M > | M is an LBA where L(M ) = â&#x2C6;&#x2026;}

Theorem ELBA is undecidable.

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ELBA = {< M > | M is an LBA where L(M ) = â&#x2C6;&#x2026;}

Theorem ELBA is undecidable. â&#x20AC;˘ For a TM M and an input w we can determine whether M

accepts w by constructing a certain LBA B and then testing whether L(B) is empty.

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ELBA = {< M > | M is an LBA where L(M ) = ∅}

Theorem ELBA is undecidable.

• For a TM M and an input w we can determine whether M

accepts w by constructing a certain LBA B and then testing whether L(B) is empty.

• The language that B recognizes comprises all accepting

computation histories for M on w.

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ELBA = {< M > | M is an LBA where L(M ) = ∅}

Theorem ELBA is undecidable.

• For a TM M and an input w we can determine whether M

accepts w by constructing a certain LBA B and then testing whether L(B) is empty.

• The language that B recognizes comprises all accepting

computation histories for M on w.

• If M accepts w, L(B) contains one string and so is nonempty.

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ELBA = {< M > | M is an LBA where L(M ) = ∅}

Theorem ELBA is undecidable.

• For a TM M and an input w we can determine whether M

accepts w by constructing a certain LBA B and then testing whether L(B) is empty.

• The language that B recognizes comprises all accepting

computation histories for M on w.

• If M accepts w, L(B) contains one string and so is nonempty.

• If M does not accept w, L(B) is empty.

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ELBA = {< M > | M is an LBA where L(M ) = ∅}

Theorem ELBA is undecidable.

• For a TM M and an input w we can determine whether M

accepts w by constructing a certain LBA B and then testing whether L(B) is empty.

• The language that B recognizes comprises all accepting

computation histories for M on w.

• If M accepts w, L(B) contains one string and so is nonempty.

• If M does not accept w, L(B) is empty.

• If we can determine whether B’s language is empty, clearly we

can determine whether M accepts w.

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ELBA = {< M > | M is an LBA where L(M ) = ∅}

Theorem ELBA is undecidable.

• For a TM M and an input w we can determine whether M

accepts w by constructing a certain LBA B and then testing whether L(B) is empty.

• The language that B recognizes comprises all accepting

computation histories for M on w.

• If M accepts w, L(B) contains one string and so is nonempty.

• If M does not accept w, L(B) is empty.

• If we can determine whether B’s language is empty, clearly we

can determine whether M accepts w.

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How can we obtain a description of B, given descriptions of M and w?

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How can we obtain a description of B, given descriptions of M and w? â&#x20AC;˘ We construct B to accept its input x if x is an accepting

computation history for M on w.

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How can we obtain a description of B, given descriptions of M and w? â&#x20AC;˘ We construct B to accept its input x if x is an accepting

computation history for M on w. â&#x20AC;˘ For the purposes of this proof, we assume that the accepting

computation history is presented as a single string, with the configurations separated from each other by the symbol # symbol.

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How can we obtain a description of B, given descriptions of M and w?

• We construct B to accept its input x if x is an accepting

computation history for M on w.

• For the purposes of this proof, we assume that the accepting

computation history is presented as a single string, with the configurations separated from each other by the symbol # symbol.

• When LBA B receives an input x, it is supposed to accept if x is an accepting computation for M on w.

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How can we obtain a description of B, given descriptions of M and w?

• We construct B to accept its input x if x is an accepting

computation history for M on w.

• For the purposes of this proof, we assume that the accepting

computation history is presented as a single string, with the configurations separated from each other by the symbol # symbol.

• When LBA B receives an input x, it is supposed to accept if x is an accepting computation for M on w.

• B breaks up x according to the delimiters into strings

C1 , C 2 , . . . , C k .

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How can we obtain a description of B, given descriptions of M and w?

• We construct B to accept its input x if x is an accepting

computation history for M on w.

• For the purposes of this proof, we assume that the accepting

computation history is presented as a single string, with the configurations separated from each other by the symbol # symbol.

• When LBA B receives an input x, it is supposed to accept if x is an accepting computation for M on w.

• B breaks up x according to the delimiters into strings

C1 , C 2 , . . . , C k .

• Then B determines whether the Ci satisfy the three conditions of

an accepting computation history. If satisfied, B accepts its input.

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How can we obtain a description of B, given descriptions of M and w?

• We construct B to accept its input x if x is an accepting

computation history for M on w.

• For the purposes of this proof, we assume that the accepting

computation history is presented as a single string, with the configurations separated from each other by the symbol # symbol.

• When LBA B receives an input x, it is supposed to accept if x is an accepting computation for M on w.

• B breaks up x according to the delimiters into strings

C1 , C 2 , . . . , C k .

• Then B determines whether the Ci satisfy the three conditions of

an accepting computation history. If satisfied, B accepts its input.

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â&#x20AC;˘ Suppose that R is a decider for ELBA .

We construct TM S to decide AT M as follows: S = â&#x20AC;&#x153;On input < M, w >, where M is a TM and w is a string: 1. Construct LBA B from M and w as described.

2. Run R on input < B >.

3. If R rejects, accept; if R accepts, reject.

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• Suppose that R is a decider for ELBA .

We construct TM S to decide AT M as follows: S = “On input < M, w >, where M is a TM and w is a string: 1. Construct LBA B from M and w as described.

2. Run R on input < B >.

3. If R rejects, accept; if R accepts, reject.

• If R decides ELBA , then S decidesAT M .

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• Suppose that R is a decider for ELBA .

We construct TM S to decide AT M as follows: S = “On input < M, w >, where M is a TM and w is a string: 1. Construct LBA B from M and w as described.

2. Run R on input < B >.

3. If R rejects, accept; if R accepts, reject.

• If R decides ELBA , then S decidesAT M .

• But AT M is undecidable, so ELBA also must be undecidable

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Let ALLCF G be the problem of determining whether a context-free grammar generates all possible strings.

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Let ALLCF G be the problem of determining whether a context-free grammar generates all possible strings.

ALLCF G = {< G > | G is a CFG and L(G) = ÎŁâ&#x2C6;&#x2014; }

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Let ALLCF G be the problem of determining whether a context-free grammar generates all possible strings.

ALLCF G = {< G > | G is a CFG and L(G) = ÎŁâ&#x2C6;&#x2014; }

Theorem ALLCF G is undecidable.

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Let ALLCF G be the problem of determining whether a context-free grammar generates all possible strings.

ALLCF G = {< G > | G is a CFG and L(G) = Σ∗ }

Theorem ALLCF G is undecidable.

• (Proof by contradiction) Assume ALLCF G is decidable and use

this to show that AT M is decidable.

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Let ALLCF G be the problem of determining whether a context-free grammar generates all possible strings.

ALLCF G = {< G > | G is a CFG and L(G) = Σ∗ }

Theorem ALLCF G is undecidable.

• (Proof by contradiction) Assume ALLCF G is decidable and use

this to show that AT M is decidable.

• For a TM M and an input w we construct a CFG G that generates all strings if and only if M does not accept w.

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Let ALLCF G be the problem of determining whether a context-free grammar generates all possible strings.

ALLCF G = {< G > | G is a CFG and L(G) = Σ∗ }

Theorem ALLCF G is undecidable.

• (Proof by contradiction) Assume ALLCF G is decidable and use

this to show that AT M is decidable.

• For a TM M and an input w we construct a CFG G that generates all strings if and only if M does not accept w.

• If M does accept w, G does not generate the accepting

computation history for M on w.

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Let ALLCF G be the problem of determining whether a context-free grammar generates all possible strings.

ALLCF G = {< G > | G is a CFG and L(G) = Σ∗ }

Theorem ALLCF G is undecidable.

• (Proof by contradiction) Assume ALLCF G is decidable and use

this to show that AT M is decidable.

• For a TM M and an input w we construct a CFG G that generates all strings if and only if M does not accept w.

• If M does accept w, G does not generate the accepting

computation history for M on w.

• That is, G is designed to generate all strings that are not

accepting computation histories for M on w.

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Let ALLCF G be the problem of determining whether a context-free grammar generates all possible strings.

ALLCF G = {< G > | G is a CFG and L(G) = Σ∗ }

Theorem ALLCF G is undecidable.

• (Proof by contradiction) Assume ALLCF G is decidable and use

this to show that AT M is decidable.

• For a TM M and an input w we construct a CFG G that generates all strings if and only if M does not accept w.

• If M does accept w, G does not generate the accepting

computation history for M on w.

• That is, G is designed to generate all strings that are not

accepting computation histories for M on w.

• If we can determine whether G generates all strings, clearly we

can determine whether M accepts w.

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Let ALLCF G be the problem of determining whether a context-free grammar generates all possible strings.

ALLCF G = {< G > | G is a CFG and L(G) = Σ∗ }

Theorem ALLCF G is undecidable.

• (Proof by contradiction) Assume ALLCF G is decidable and use

this to show that AT M is decidable.

• For a TM M and an input w we construct a CFG G that generates all strings if and only if M does not accept w.

• If M does accept w, G does not generate the accepting

computation history for M on w.

• That is, G is designed to generate all strings that are not

accepting computation histories for M on w.

• If we can determine whether G generates all strings, clearly we

can determine whether M accepts w.

• But AT M is undecidable, so ALLCF G must also be undecidable.

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We need to construct the CFG G which generates all strings that fail to be an accepting computation history for M on w.

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We need to construct the CFG G which generates all strings that fail to be an accepting computation history for M on w.

That is, G generates all strings that 1. DO NOT start with the start configuration C1 , or 2. DO NOT end with an accepting configuration, or 3. where some Ci DOES NOT properly yield Ci+1 under the rules of M .

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In constructing the CFG G, we design a PDA D first.

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In constructing the CFG G, we design a PDA D first.

D will start by nondeterministically branching to guess which of the preceding 3 conditions to check.

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In constructing the CFG G, we design a PDA D first.

D will start by nondeterministically branching to guess which of the preceding 3 conditions to check.

1. First branch checks on whether the beginning of the input string is C1 and accepts if it isnâ&#x20AC;&#x2122;t.

2. Next branch checks on whether the input string ends with a configuration containing the accept state, qaccept , and accepts if it isnâ&#x20AC;&#x2122;t.

3. Last branch is supposed to accept if some Ci does not properly yield Ci+1 .

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In constructing the CFG G, we design a PDA D first.

D will start by nondeterministically branching to guess which of the preceding 3 conditions to check.

1. First branch checks on whether the beginning of the input string is C1 and accepts if it isn’t.

2. Next branch checks on whether the input string ends with a configuration containing the accept state, qaccept , and accepts if it isn’t.

3. Last branch is supposed to accept if some Ci does not properly yield Ci+1 .

• scans input till it has come to Ci • pushes Ci onto stack • pops stack to compare with Ci+1

B suppose to match except around head position where difference is dictated by transition function

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In constructing the CFG G, we design a PDA D first.

D will start by nondeterministically branching to guess which of the preceding 3 conditions to check.

1. First branch checks on whether the beginning of the input string is C1 and accepts if it isn’t.

2. Next branch checks on whether the input string ends with a configuration containing the accept state, qaccept , and accepts if it isn’t.

3. Last branch is supposed to accept if some Ci does not properly yield Ci+1 .

• scans input till it has come to Ci • pushes Ci onto stack • pops stack to compare with Ci+1

B suppose to match except around head position where difference is dictated by transition function B there is a problem here ... what?

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In constructing the CFG G, we design a PDA D first.

D will start by nondeterministically branching to guess which of the preceding 3 conditions to check.

1. First branch checks on whether the beginning of the input string is C1 and accepts if it isn’t.

2. Next branch checks on whether the input string ends with a configuration containing the accept state, qaccept , and accepts if it isn’t.

3. Last branch is supposed to accept if some Ci does not properly yield Ci+1 .

• scans input till it has come to Ci • pushes Ci onto stack • pops stack to compare with Ci+1

B suppose to match except around head position where difference is dictated by transition function B there is a problem here ... what?

• we must write the configuration history differently ... every other configuration appears in reverse order

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EQCF G = {< G, H > | G and H are CFGs and L(G) = L(H)} Show that EQCF G is undecidable.

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EQCF G = {< G, H > | G and H are CFGs and L(G) = L(H)} Show that EQCF G is undecidable. â&#x20AC;˘ Assume EQCF G is decidable and obtain a contradiction.

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EQCF G = {< G, H > | G and H are CFGs and L(G) = L(H)} Show that EQCF G is undecidable. â&#x20AC;˘ Assume EQCF G is decidable and obtain a contradiction. â&#x20AC;˘ Suppose that R is a decider for EQCF G .

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EQCF G = {< G, H > | G and H are CFGs and L(G) = L(H)} Show that EQCF G is undecidable.

• Assume EQCF G is decidable and obtain a contradiction.

• Suppose that R is a decider for EQCF G .

We construct TM S to decide ALLCF G as follows: S = “On input < G >, an encoding of a CFG G: 1. Run R on input < G, G1 > where G1 is a CFG that generates Σ∗ .

2. If R accepts, accept; if R rejects, reject.”

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EQCF G = {< G, H > | G and H are CFGs and L(G) = L(H)} Show that EQCF G is undecidable.

• Assume EQCF G is decidable and obtain a contradiction.

• Suppose that R is a decider for EQCF G .

We construct TM S to decide ALLCF G as follows: S = “On input < G >, an encoding of a CFG G: 1. Run R on input < G, G1 > where G1 is a CFG that generates Σ∗ .

2. If R accepts, accept; if R rejects, reject.”

• Clearly, if R decides EQCF G , then S decides ALLCF G .

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EQCF G = {< G, H > | G and H are CFGs and L(G) = L(H)} Show that EQCF G is undecidable.

• Assume EQCF G is decidable and obtain a contradiction.

• Suppose that R is a decider for EQCF G .

We construct TM S to decide ALLCF G as follows: S = “On input < G >, an encoding of a CFG G: 1. Run R on input < G, G1 > where G1 is a CFG that generates Σ∗ .

2. If R accepts, accept; if R rejects, reject.”

• Clearly, if R decides EQCF G , then S decides ALLCF G .

• Because ALLCF G is undecidable, EQCF G also must be

undecidable.

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A Simple Undecidable Problem We give an example of an undecidable problem concerning simple manipulation of strings.

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A Simple Undecidable Problem We give an example of an undecidable problem concerning simple manipulation of strings.

The Post correspondence problem is to determine whether a collection of dominos has a match.

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A Simple Undecidable Problem We give an example of an undecidable problem concerning simple manipulation of strings.

The Post correspondence problem is to determine whether a collection of dominos has a match.

An instance a collection P of dominos: nh i hof the i PCP h isio tk t1 t2 P = , , 路 路 路 , . b1 b2 bk

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A Simple Undecidable Problem We give an example of an undecidable problem concerning simple manipulation of strings.

The Post correspondence problem is to determine whether a collection of dominos has a match.

An instance a collection P of dominos: nh i hof the i PCP h isio tk t1 t2 P = , , · · · , . b1 b2 bk A match is a sequence i1 , i2 , . . . , im , where ti1 ti2 · · · tim = bi1 bi2 · · · bim .

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A Simple Undecidable Problem We give an example of an undecidable problem concerning simple manipulation of strings.

The Post correspondence problem is to determine whether a collection of dominos has a match.

An instance a collection P of dominos: nh i hof the i PCP h isio tk t1 t2 P = , , · · · , . b1 b2 bk A match is a sequence i1 , i2 , . . . , im , where ti1 ti2 · · · tim = bi1 bi2 · · · bim .

Example a b ca abc ab , b ab , ca , a , c

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Post Correspondence Problem

Let P CP = {< P > | P is an instance of the Post correspondence problem with a match} .

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Let P CP = {< P > | P is an instance of the Post correspondence problem with a match} .

Theorem P CP is undecidable.

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Let P CP = {< P > | P is an instance of the Post correspondence problem with a match} .

Theorem P CP is undecidable.

Suppose that R decide the P CP .

We construct TM S to decide AT M

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Post Correspondence Problem

Let P CP = {< P > | P is an instance of the Post correspondence problem with a match} .

Theorem P CP is undecidable.

Suppose that R decide the P CP .

We construct TM S to decide AT M That is, we show that from any TM M and input w, we can construct an instance P where a match is an accepting computation history for M on w.

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Post Correspondence Problem

Let P CP = {< P > | P is an instance of the Post correspondence problem with a match} .

Theorem P CP is undecidable.

Suppose that R decide the P CP .

We construct TM S to decide AT M That is, we show that from any TM M and input w, we can construct an instance P where a match is an accepting computation history for M on w.

If we could determine whether the instance has a match, we would be able to determine whether M accepts w.

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Post Correspondence Problem

Let P CP = {< P > | P is an instance of the Post correspondence problem with a match} .

Theorem P CP is undecidable.

Suppose that R decide the P CP .

We construct TM S to decide AT M That is, we show that from any TM M and input w, we can construct an instance P where a match is an accepting computation history for M on w.

If we could determine whether the instance has a match, we would be able to determine whether M accepts w.

Before getting into the construction, for convenience, we make the ff modifications: â&#x20AC;˘ we assume that M on w never attempts to move its head off the left-hand end of the tape â&#x20AC;˘ if w = , we use the string t in place of w â&#x20AC;˘ we modify h PCP i to require that a match starts with the first domino, bt11

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Let M = (Q, Σ, Γ, δ, q0 , qaccept , qreject ).

Post Correspondence Problem

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Let M = (Q, Σ, Γ, δ, q0 , qaccept , qreject ).

Part 1.

Part 2.

Part 3.

Part 4.

Part 5.

Post Correspondence Problem

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Let M = (Q, Σ, Γ, δ, q0 , qaccept , qreject ).

Part 6.

Part 7.

Post Correspondence Problem

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Questions?

See you next meeting!

Post Correspondence Problem