# The Conjugate Post Correspondence Problem

We introduce a modification to the Post Correspondence Problem where (in the formulation using morphisms) we require the images to be conjugate words. This problem is then shown to be undecidable by reducing it to the word problem for a special type of semi-Thue systems.

## Authors

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12/04/2021

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## 1 Introduction

Two words and are conjugates if they can be written in the form and for some words and . The problem of deciding whether there exist two conjugate words with the same images under a pair of morphisms is knows as the circular Post Correspondence Problem, or CPCP for short. More formally CPCP asks given morphisms and , whether there exist words and such that when . The CPCP is known to be undecidable as was shown in [4], see [2] for a somewhat simpler proof. Here we give a new variant of the problem by requiring that the images are conjugate words with the same pre-image. We call this problem the conjugate-PCP and give it the following formal definition:

###### Problem 1.

Given two morphisms , decide whether or not there exists a word such that and for some words .

The behaviour of the instances of conjugate-PCP differ vastly from the more traditional variants of the PCP where a valid presolution (prefix of a possible solution) can be verified by a matching of the images. Working out a possible solution to a conjugate-PCP instance is much less intuitive.

For example let us have some morphisms and guess that a solution begins with the letter . Then the situation is the following:

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{}{}{}\pgfsys@endscope}}\pgfsys@endscopeto0.0pt\hss\pgfsys@discardpath\pgfsys@endscope\hss\endpgfpictureg(a)⋯to18.17pt\vboxto19.67pt\pgfpicture\makeatletterto0.0pt\pgfsys@beginscope\definecolorpgfstrokecolorrgb0,0,0\pgfsys@color@rgb@stroke000\pgfsys@color@rgb@fill000\pgfsys@setlinewidth0.4pt\nullfontto0.0pt\pgfsys@beginscope\hbox{{\pgfsys@beginscope{}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope{}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-3.749943pt% }{-2.499962pt}{}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}% \pgfsys@color@rgb@stroke{0}{0}{0}{}\pgfsys@color@rgb@fill{0}{0}{0}{}\hbox{} }}{}{}\pgfsys@endscope}}} {}{}{}\pgfsys@endscope}}\pgfsys@endscopeto0.0pt\hss\pgfsys@discardpath\pgfsys@endscope\hss\endpgfpicture g(w) = 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The validity of the presolution cannot be verified because there is no matching that needs to happen between and . Moreover the factorization of the images to and need not be unique even for minimal solutions:

###### Example.

Let be morphisms defined by

 h(a) =aba, g(a) =bab, h(b) =b, g(b) =a.

Now is a minimal solution for the conjugate-PCP instance having two factorizations: or .

In the next section we will prove that the conjugate-PCP is indeed also undecidable by reducing it to a word problem for a special type of semi-Thue systems.

## 2 The Proof of Undecidability

We recall the construction of the semi-Thue system in [3]

. The construction used the structure of a given Turing machine

. We can simplify our presentation by acknowledging the existence of such a system and declaring that our new system has the same properties.

Let be our semi-Thue system with the following properties:

1. with pairwise disjoint alphabets . Notably where are markers for the left and right border of the word, respectively.

2. is -deterministic in the following way:

1. .

2. If there is a rule in where all symbols are non-overlined, then the corresponding overlined rule , where all symbols are overlined is also in , and vice versa.

3. For all words , if there is a rule in giving then the rule is unique.

4. There is a single rule from and a single rule from , moreover these rules are such that they re-write everything between the and markers, namely if there are rules giving and for a then the rules are and , respectively.

3. has an undecidable circular word problem. In particular it is undecidable whether has a circular derivation where is the word appearing in rules of 2(iv). Note that and in case 2(iv) are fixed words from the construction of the semi-Thue system for a TM , and .

The special -determinism of

can be interpreted as derivations being in two different phases: the normal phase and the overlined phase. Transitioning between phases is via the unique rules from 2/(iv). It is straightforward to see that all derivations do not go through phase changes and that the phase is changed more than once if and only if

has a circular derivation. The system considered is now clear from the context and we write the derivations omitting the index simply as .

We now add a few additional rules to : we remove the unique rule and replace it with one extra step by introducing rules and where is a new symbol for the intermediate step. The corresponding overlined rules and are added also to replace the rule . These new rules are needed in identifying the border between words and , and adding them has no effect on the behaviour of .

By theorem LABEL:circular we have the following lemma.

###### Lemma 1.

Assume that the semi-Thue system is constructed as above. Then has an undecidable individual circular word problem for the word .

We now reduce the individual circular word problem of the system to the conjugate-PCP.

Let , where the rules are pairs . We denote by and the left and right desynchronizing morphisms defined by

 lx(a)=xa,rx(a)=ax

for all words . In the following we consider the elements of as letters. Denote by the alphabet where letters are given subscripts . Define morphisms according to the following table:

Here the re-writing rules are of the form for some . The following rules play important roles:
where is the unique word such that , and
.

We begin by examining the forms of the images of and . The morphisms are a modification of the ones in [3] with slight alterations made such that it is possible to have (finite) solutions to the conjugate-PCP instance with easily identifiable borders between the factors and using special symbols and . The symbols and function as desynchronizing symbols. The desynchronizing symbols and make sure that in the solution the factors that will represent the configurations of the semi-Thue system are of the correct form, that is of the form where the determinism is kept intact. This follows from the forms of and : under all images are desynchronized by either (non-overlined letters) or (overlined letters). To get similarly desynchronized factors in the image under we note that in the pre-image the words between two -symbols (similarly for overlined symbols ) are of the form where , and (with end markers and omitted from and ). The symbol is not really used in desynchronizing but making sure that the change between phases is carried out correctly.

The following lemma is useful in our proof:

###### Lemma 2.

Words and are conjugates if and only if and are conjugates for all conjugates and of .

###### Proof.

If and are conjugates for all conjugates and of then and are conjugates.

Assume then that and are conjugates and let and be conjugates of . We may now write and . Denote and . Now is a conjugate of and is a conjugate of . By our assumption also and are conjugates. ∎

Next we will show that a circular derivation beginning from word exists in if and only if there is a solution to the conjugate-PCP instance . We will prove the claim in the following two lemmata.

###### Lemma 3.

If there is a circular derivation in beginning from , then there exists a non-empty word such that and for some words and .

###### Proof.

Assume that a circular derivation exists. The derivation is of the form . This derivation can be coded into a word

 w=Iw1#w2#w3#⋯#th−1th¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯w1#w2#w3#⋯#th−1th,

where for each , where we recall that is the unique rewriting rule used in each derivation step. The rules and appear right before the transition to overlined derivation as they correspond to the final and intermediate steps before the transition. Let us consider the images of under the morphisms and :

 h(w)=$ld2(w0#α1v1β1#α2v2β2#⋯#s)fff$\poundsle2(w0#α1v1β1⋯#s)fff\pounds)

and

 g(w)=re2(\poundsα1u1β1#α2u2β2#⋯#u#)sfff\poundsrd2($α1u1β1⋯#u#)sfff$).

These images are indeed very similar. The images match at all positions that do not contain a desynchronizing symbol ( or ) or a special symbol ( or ). Thus, if we erased all of these non-matching symbols we would have images that are equal (and of the form for a word ). Also the non-matching symbols are such that is always matched with and is always matched with . It is quite clear that the factors in both and beginning and ending in the same special symbol are the same, that is, the factors and appearing in both images are equal. It follows that and , which proves our claim. ∎

###### Lemma 4.

If there exists a non-empty word such that and for some words and , then there is a circular derivation in beginning from .

###### Proof.

Firstly we show that the factor must appear in and hence or is a factor in . Assume the contrary: there is no factor in .

From the construction of we know that also is desynchronized so that between each letter there is either a factor or . Conjugation of does not break this property except possibly for the beginning and the end of ( could start and end in a single desynchronizing symbol).

Take now the first letter of . We can assume that it is a non-overlined letter as the considerations are similar for the overlined case. The letter cannot be as it would have to be followed by : does not appear as a factor under without , and produces , which is uncoverable by . From the construction of we see that the letters following must also be non-overlined, otherwise the desynchronization would be broken. Thus the desynchronizing symbol is the same for all these following letters. But as we can see from the form of the morphisms and , we have a different desynchronizing symbols under for and its successors. It is clear that must contain both and and so must have both non-overlined and overlined letters. If there is a change in the desynchronizing symbol in then it contradicts the form of the images under . Hence we must have the factor in to make the transition without breaking the desynchronization.

The images of the factor are as follows:

 h(th−1th)=dsfff$\poundsle2(w0#)ee and  g(th−1th)=re2(u#)sfff$\poundsdd.

As we can see the desynchronizing symbols do not match. Hence we also must have the overlined copy of this factor in , that is the factor , the images of which are as follows (the letter is a forced continuation to the overlined factor to account for the special symbols and ):

 h(¯¯¯¯¯¯¯¯¯¯¯¯¯¯th−1thI)=sfff\pounds$\ld2(w0#)d and  g(¯¯¯¯¯¯¯¯¯¯¯¯¯¯th−1thI)=rd2(u#)sfff$\poundsee.

One of either of these factors has one swap between symbol and . From the above we concluded that we need an even number of these swaps as for every factor we must also have the factor and vice versa. It is possible that ends in the letter . In this case the swap happens "from the end to the beginning", i.e., the prefix of a factor doing the swap is at the end of and the remaining suffix is at the beginning of . The following proposition shows that we can in fact restrict ourselves to the case where the factors and are intact, that is, the swap does not happen from the end to the beginning of as a result of the conjugation between and .

###### Proposition.

It may be assumed that the first and last symbols of are and .

###### Proof of proposition.

If is not of the desired form then it has as a factor (by above the symbols from are in ). Images of the letters under do not have as a factor so there is a factorization such that ends in and begins with ( ends in and begins with ). By Lemma 2, and are conjugates if and only if and are, where now has as the first symbol and as the last symbol.

Now by the proposition we may assume that begins with and ends with . From this it also follows that when and the word has as the first and the last symbol and has as the first and the last symbol. It follows that , where the border between and is in the image :

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Here the border between and need not be in the image of the same instance of . Nevertheless we know by above that in the image under the word begins with . To get this image as a factor of we must have in , where is the first rewriting rule used and