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4.7 �Post's Correspondence Problem

������The Undecidability of Post's Correspondence Problem
������Applications of Post's Correspondence Problem

The Post's Correspondence Problem, or PCP for short, consists of the following domain and question.

**Domain:

{ <(x1, y1), . . . , (xk, yk)> | k 1 and x1,�. . . ,�xk, y1,�. . . ,�yk are strings over some alphabet. }

**Question:

Are there an integer n 1 and indices i1,�. . . ,�in for the given instance <(x1, y1), . . . , (xk, yk)> such that xi1 xin = yi1 yin? Each sequence i1,�. . . ,�in that provides a yes answer is said to be a witness for a positive solution to the given instance of PCP.

The problem can be formulated also as a "domino" problem of the following form.

**Domain:

{ | k 1, and each is a domino card with the string xi on its top and the string yi on its bottom, 1 i k. }

**Question:

Given k 1 piles of cards

with infinitely many cards in each pile, can one draw a sequence of n 1 cards

from these piles, so that the string xi1 xin formed on the top of the cards will equal the string yi1 yin formed on the bottom?

Example�4.7.1 �� PCP has the solution of yes for the instance <(01, 0), (110010, 0), (1, 1111), (11, 01)> or, equivalently, for the following instance in the case of the domino problem.

The tuple (i1, i2, i3, i4, i5, i6) = (1, 3, 2, 4, 4, 3) is a witness for a positive solution because x1x3x2x4x4x3 = y1y3y2y4y4y3 = 01111001011111. The positive solution has also the witnesses (1, 3, 2, 4, 4, 3, 1, 3, 2, 4, 4, 3), (1, 3, 2, 4, 4,3,1,3,2,4,4,3,1,3,2, 4, 4, 3), etc. On the other hand, the PCP has the solution no for <(0, 10), (01, 1)>.

�The Undecidability of Post's Correspondence Problem

Post's correspondence problem is very useful for showing the undecidability of many other problems by means of reducibility. Its undecidability follows from its capacity for simulating the computations of Turing machines, as exhibited indirectly in the following proof through derivations in Type 0 grammars.

Theorem�4.7.1 �� The PCP is an undecidable problem.

Proof� �� By Corollary�4.6.1 the membership problem is undecidable for Type 0 grammars. Thus, it is sufficient to show how from each instance (G, w) of the membership problem for Type 0 grammars, an instance I can be constructed, such that the PCP has a positive solution at I if and only if w is in L(G).

For the purpose of the proof consider any Type 0 grammar G = <N, , P, S> and any string w in *. With no loss of generality assume that #, �, and $ are new symbols not in N . Then let the corresponding instance I = <(x1, y1), . . . , (xk, yk)> of PCP be of the following form.

PCP has a positive solution at I if and only if I can trace a derivation that starts at S and ends at w.

For each derivation in G of the form S 1 m w, the instance�I has a witness (i1, . . . , in) of a positive solution such that either

or

depending on whether m is even or odd, respectively.

On the other hand, each witness (i1, . . . , in) of a positive solution for PCP at I has a smallest integer t 1 such that xi1 xit = yi1 yit. In such a case, xi1 xit = yi1 yit = �S#2#4 m for some derivation S * 1 * 2 * * m * w.

The instance I consists of pairs of the following form

1. �A pair of the form (�S, �).
2. �A pair of the form ($, $).
3. �A pair of the form (X, ), and a pair of the form (, X), for each symbol X in N {#}.
4. �A pair of the form (, ), and a pair of the form (, ), for each production rule in G that satisfies .
5. �A pair of the form (X, ), and a pair of the form (, X), for each production rule in G and for each X in N {#}. The underlined symbols are introduced to allow only (�S, �) as a first pair, and ($, $) as a last pair, for each witness of a positive solution. The pair (�S, �) in (a) is used to start the tracing of a derivation at S. The pair ($, $) in (b) is used to end the tracing of a derivation at w.

The other pairs are used to force the tracing to go from each given sentential form to a sentential form ', such that * '. The tracing is possible because each of the pairs (xi, yi) is defined so that yi provides a "window" into , whereas xi provides an appropriate replacement for yi in '.

The pairs of the form (X, ) and (, X) in (c) are used for copying substrings from to '. The pairs of the form (, ) and (, ), , in (d) are used for replacing substrings in by substrings in '. The pairs of the form (X, ) and (, X) in (e) are used for replacing substrings in by the empty string in'.

The window is provided because for each 1 i1, . . . , ij k, the strings x = xi1 xij and y = yi1 yij satisfy the following properties.

1. �If x is a prefix of y, then x = y. Otherwise there would have been a least l such that xi1 xil is a proper prefix of yi1 yil. In which case (xil, yil) would be equal to (v, uvv') for some nonempty strings v and v'. However, by definition, no pair of such a form exists in I.
2. �If y is a proper prefix of x, then the sum of the number of appearances of the symbols # and in x is equal to one plus the sum of the number of appearances of # and in y. Otherwise, there would be a least l > 1 for which xi1 xil and yi1 yil do not satisfy the property. In such a case, because of the minimality of l, xil and yil would have to differ in the number of # and they contain. That is, by definition of I, (xil, yil) would have to equal either ($, $) or (�, �). However, ($, ) is an impossible choice because it implies that xi1 xil = yi1 yil, and (�, �) is an impossible choice because it implies that xi1 xil-1 = yi1 yil-1 (and hence that the property holds). The correctness of the construction can be shown by induction on the number of production rules used in the derivation under consideration or, equivalently, on the number of pairs of type (d) and (e) used in the given witness for a positive solution.

Example�4.7.2 �� If G is a grammar whose set of production rules is {S aSaSaS, aS }, then the instance of the PCP that corresponds to (G, ) as determined by the proof of Theorem�4.7.1, is <(�S, �), (, S), (aSaSaS, ), (, #aS), (#, ), (, aaS), (a, ), (, SaS), (S, ), (#, ), (, #), (a, ), (, a), (S, ), (, S), ($, $)>.

The instance has a positive solution with a witness that corresponds to the arrangement in Figure�4.7.1.

The witness also corresponds to the derivation S * S aSaSaS aSaS aS in G.

�Applications of Post's Correspondence Problem

The following corollary exhibits how Post's correspondence problem can be used to show the undecidability of some other problems by means of reducibility.

Corollary�4.7.1 �� The equivalence problem is undecidable for finite-state transducers.

Proof� �� Consider any instance <(x1, y1), . . . , (xk, yk)> of PCP. Let be the minimal alphabet such that x1,�. . . ,�xk,�y1,�. . . ,�yk are all in *. With no loss of generality assume that = {1, . . . , k} is an alphabet.

Let M1 = <Q1, , , 1, q0, F1> be a finite-state transducer that computes the relation * � *, that is, a finite-state transducer that accepts all inputs over�, and on each such input can output any string over .

Let M2 = <Q2, , , 2, q0, F2> be a finite-state transducer that on input i1 inoutputs some w such that either w xi1 xin or w yi1 yin. Thus, M2 on input i1 in can output any string in * if xi1 xin yi1 yin. On the other hand, if xi1 xin = yi1 yin, then M2 on such an input i1 in can output any string in *, except for xi1 xin.

It follows that M1 is equivalent to M2 if and only if the PCP has a negative answer at the given instance <(x1, y1), . . . , (xk, yk)>.

Example�4.7.3 �� Consider the instance <(x1, y1), (x2, y2)> = <(0, 10), (01, 1)> of PCP. Using the terminology in the proof of Corollary�4.7.1, = {0, 1} and = {1, 2}. The finite-state transducer M1 can be as in Figure�4.7.2(a),

� �

and the finite-state transducer M2 can be as in Figure�4.7.2(b).

M2 on a given input i1 in nondeterministically chooses between its components Mx and My. In Mx it outputs a prefix of xi1 xin, and in My it outputs a prefix of yi1 yin. Then M2 nondeterministically switches to M>, M<, or M.

M2 switches from Mx to M> to obtain an output that has xi1 xin as a proper prefix. M2 switches from Mx to M< to obtain an output that is proper prefix of xi1 xin. M2 switches from Mx to M to obtain an output that differs from xi1 xinwithin the first |xi1 xin| symbols.

M2 switches from My to M>, M<, M for similar reasons, respectively.

The following corollary has a proof similar to that given for the previous one.

Corollary�4.7.2 �� The equivalence problem is undecidable for pushdown automata.

Proof� �� Consider any instance <(x1, y1), . . . , (xk, yk)> of PCP. Let 1 be the minimal alphabet such that x1,�. . . ,�xk,�y1,�. . . ,�yk are all in 1*. With no loss of generality assume that2 = {1, . . . , k} is an alphabet, that 1 and 2 are mutually disjoint, and that Z0 is a new symbol not in 1.

Let M1 = <Q1, 1 2, 1 Z0, 1, q0, Z0, F1> be a pushdown automaton that accepts all the strings in (1 2)*. (In fact, M1 can also be a finite-state automaton.)

Let M2 = <Q2, 1 2, 1 Z0, 2, q0, Z0, F2> be a pushdown automaton that accepts an input w if and only if it is of the form in i1u, for some i1 in in 1* and some u in 2*, such that either u xi1 xin or u yi1 yin.

It follows that M1 and M2 are equivalent if and only if the PCP has a negative answer at the given instance.

The pushdown automaton M2 in the proof of Corollary�4.7.2 can be constructed to halt on a given input if and only if it accepts the input. The constructed pushdown automaton halts on all inputs if and only if the PCP has a negative solution at the given instance. Hence, the following corollary is also implied from the undecidability of PCP.

Corollary�4.7.3 �� The uniform halting problem is undecidable for pushdown automata.

PCP is a partially decidable problem because given an instance <(x1, y1), . . . , (xk, yk)> of the problem one can search exhaustively for a witness of a positive solution, for example, in {1, . . . , k}* in canonical order. With such an algorithm a witness will eventually be found if the instance has a positive solution. Alternatively, if the instance has a negative solution, then the search will never terminate.

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