Decidability and Undecidability of Marked PCP

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Vesa Halava¹, Mika Hirvensalo²^;¹^?, and Ronald de Wolf³^;⁴

1 Turku Centre for Computer Science, Lemminkaisenkatu 14 A, 4th oor, FIN-20520, Turku, Finland,vehalava@cs.utu.fi

2 Department of Mathematics, University of Turku, FIN-20014, Turku, Finland.

mikhirve@utu.fi

3 CWI, P.O. Box 94079, Amsterdam, The Netherlands,rdewolf@cwi.nl

4 University of Amsterdam

Abstract. We show that themarkedversion of the Post Correspondence Problem, where the words on a list are required to dier in the rst letter, is decidable. On the other hand, PCP remains undecidable if we only require the words to dier in the rsttwoletters. Thus we locate the decidability/undecidability-boundary between marked and 2-marked PCP.

1 Introduction: PCP and Marked PCP

The Post Correspondence Problem (PCP) [6] is one of the most useful undecidable problems, because it can be simply described and many other problems can easily be reduced to it, particularly problems in formal language theory. The general form of the problem is as follows. An instance of PCP is a four-tuple I = (;;g;h), consisting of a nite source alphabet =^fa¹;:::;an^g, a nite target alphabet and two homomorphisms g;h : ^! (g(ab) = g(a)g(b) andh(ab) =h(a)h(b) whenevera;b²). It is enough to deneg;h:^!, the extension is just concatenation. PCP is the following decision problem:

GivenI = (;;g;h), is there anx²⁺ such thatg(x) =h(x)?

In other words, we have two lists of wordsg(a¹);:::;g(an) andh(a¹);:::;h(an) and we want to decide if there is a correspondence between them: are there ai¹;:::;ai^k ²such thatg(ai¹):::g(ai^k) =h(ai¹):::h(ai^k)?

The general form of this problem is undecidable [6], the reason being that the two morphisms together can simulate the computation of a Turing machine on a specic input. Examining restricted versions of PCP allows one to determine the exact boundary between decidability and undecidability. For instance, the problem becomes trivially decidable (but NP-complete) if we ask for the existence of a solution x of length at most some xed k [2, p. 228]. If we restrict to g;h which have to be injective (g is injective if x ⁶= y ⁾ g(x) ⁶=g(y)), the problem remains undecidable [4]. Also PCP(7), where we restrict to n = 7, is

?Supported by the Academy of Finland under grant 14047.

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still undecidable [5], but PCP(2) is decidable [1]. As far as we know, decidability or undecidability is still open for 2< n <7.

A further restriction which we will examine in this paper is to haveg andh marked, which we formally dene as follows. Ifzis a string, we usePrefk(z) to denote the prex of lengthkofz (Prefk(z) =z if^jz^jk). A homomorphismg isk-markedifg(a) andg(b) are nonempty and havePrefk(g(a))⁶=Prefk(g(b)) whenevera⁶=b². An instanceI = (;;g;h) of PCP isk-markedif bothg andharek-marked, andk-marked PCP is the PCP decision problem restricted to k-marked instances. We will abbreviate 1-marked to marked. If I is marked theng(a) andg(b) start with a dierent letter whenevera⁶=b², which implies that ^j^j^j^j. Without loss of generality we may assume . Markedness clearly implies injectivity: supposegis marked andx⁶=y²⁺, letx=zax⁰and y=zby⁰,aandbbeing the rst letter wherexandydier. Because of markedness we have g(a) ⁶=g(b), hence g(x) = g(z)g(a)g(x⁰)⁶= g(z)g(b)g(y⁰) = g(y), so g is injective. The converse does not hold. Consider for instance==^f1;2^g, g(1) = 11,g(2) = 12, thengis injective but not marked.

The proof of decidability of PCP(2) in [1] is based on a reduction from arbitrary instances of PCP(2) to marked instances of generalized PCP(2). [1]

then prove by means of extensive case analysis that marked generalized PCP(2) is decidable. In particular marked PCP(2) is decidable. Here we prove that marked PCP is decidable for any alphabet size. We will in fact show that marked PCP is in

EXPTIME

(the class of languages that can be recognized in time upper bounded by 2^p⁽^N⁾for some polynomialpof the input sizeN).

As stated above, PCP can be used for establishing the boundaries between decidability and undecidability. The main result of this paper is decidability of marked PCP. How much can we weaken the markedness condition before we lose decidability? We will show in Section 3 that 2-marked PCP is undecidable, thus locating the decidability/undecidability-boundary between 1-markedness and 2-markedness.

In another direction, we can weaken the markedness condition by only re- quiring g and h to be prex morphisms (g is prex if no g(ai) is a prex or another g(aj)) or even biprex (g is biprex if no g(ai) is a prex or sux of anotherg(aj)). It turns out that biprex PCP is undecidable [8].¹

2 Marked PCP Is Decidable

2.1 A Simpler Decision Problem

We would like to give a decision method for marked PCP. First we give an algorithm for the following simpler problem, which also occurs in [1, Section 6]:

Given markedI = (;;g;h) anda², are therex;y ²⁺ such that g(x) =h(y) andg(x) starts witha?

1 Clearly, a marked morphism is prex. Both marked and biprex PCP are special cases of injective PCP, but 2-marked PCP is not. See also at the end of Section 3.

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We do not look forg(x) =h(x) here but only forg(x) =h(y), and we additionally require thatg(x) starts with some specica². For example, if I has

g(a¹) =a¹ g(a²) =a² g(a³) =a³a⁴ g(a⁴) =a⁴ h(a¹) =a¹a³ h(a²) =a⁴a² h(a³) =a³a³ h(a⁴) =a²a² then fora=a¹, a solution would bex=a¹a³a² andy=a¹a².

The next algorithm decides the problem.

1. SetG=H =^;,i=j= 1.

2. If there are x¹;y¹ ² such that g(x¹) and h(y¹) start with a, then set x=x¹,y=y¹

else goto 4.

3. (a) If g(x) =h(y), then print \solutionx=x¹:::xi andy =y¹:::yj" and terminate.

(b) Ifg(x) is not a prex ofh(y) nor vice versa, then goto 4.

(c) Ifg(x)s=h(y), then do the following.

Ifs²Gthen goto 4; else seti=i+ 1 andG=G^[^fs^g.

If there is anxi such thatg(xi) and s start with the same letter, then setx=xxi and goto 3; else goto 4.

(d) Ifg(x) =h(y)s, then analogous to previous step.

4. Print \no solution" and terminate.

Informally, we are buildingx=x¹:::xi andy =y¹:::yj, trying to achieve g(x) = h(y). We add on a new xi⁺¹ as long asg(x) is a proper prex of h(y) (i.e.,g(x)s=h(y) for some suxs), and add on a newyj⁺¹ ifh(y) is a proper prex of g(x). Note that at each point such x_i⁺¹ ory_j⁺¹ are unique (if they exist) because of markedness; if they do not exist we know there is no solution.

We keep track of the suxes we have seen so far in the sets Gand H. Because the number of possible suxes is nite, either the process terminates with a solution, or at some point a sux is encountered for the second time, in which case we know the process will cycle forever and there is no solution.

The solutions produced by this algorithm are of minimal length. Note care- fully that the whole procedure is deterministic, because g and h are marked.

Furthermore, ifN is the length of the instance I given as input (i.e., the number of bits needed to describe the instance), then this procedure runs in time polynomial inN. Namely, eachg(ai) andh(ai) can have length at mostN, and hence can have at mostN ^?1 proper suxes. Since there are only 2n=O(N) dierentg(ai) andh(ai), there are onlyO(N²) dierent suxes, hence the loop of the algorithm can be repeated at most O(N²) times. This loop itself takes O(N) steps, because (1) to check ifg(x) =h(y) org(x)s=h(y) org(x) =h(y)s, we only need to check the wayg(x) andh(y) have been changed by the addition of the previousxi oryj, and (2) searching for a newxi(in step c) oryj (in step d) can be done in O(n) = O(N) steps. Therefore the whole procedure runs in O(N³) steps.

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2.2 Reducing to Simpler Instances

Consider an instance I = (;;g;h) of marked PCP: we have two marked homomorphismsg;h : ⁺ ^!⁺, where =^fa¹;:::;an^g, and we want to decide if there is an x ² ⁺ such that g(x) = h(x). Below we describe an approach to decideI by reducing it to an equivalent but simpler instanceI⁰ of marked PCP (\equivalent" meaning thatI has a solution iI⁰ has one).

Suppose = ^fa¹;:::;al^g, l n. We can run the procedure of the previous section for every ai ² , yielding pairs of (minimal-length) solutions (u¹;v¹);:::;(ul;vl) where ui;vi ²⁺ andg(ui) =h(vi) starts withai, or non- existence of solutions for certain i. At most n of the ai can have a solution.

Without loss of generality assume 1;:::;m n are the i that have a solution. We can turn this into a new instance I⁰ = (⁰;;g⁰;h⁰) of PCP, where

⁰=^fa¹;:::;am^g,g⁰(ai) =uiandh⁰(ai) =vi. Note thatg⁰ andh⁰ are marked, soI⁰ is an instance of marked PCP. Also, since the procedure of the previous section runs inO(N³) steps and has to be runntimes here,I⁰ can be built from I inO(N⁴) steps. The reduction fromI toI⁰ preserves equivalence:

Lemma 1.

Îf Î ândÎ⁰ are as above, thenI andI⁰ are equivalent.

Proof. Note that every solutionx to I must be built up fromui and vi: there must be i¹;:::;ik such that x = ui¹:::ui^k = vi¹:::vi^k. This is easy to see from the example in Figure 1. Here u¹ = a⁵a³a¹ and v¹ = a⁵a³ is a solution to the simpler problem for a¹, similarly (a²a⁴;a¹a²) is a solution for a⁶ and (a⁶a³;a⁴a⁶a³) is a solution for a². Herex =a⁵a³a¹a²a⁴a⁶a³ is a solution toI, x⁰=a¹a⁶a² is a solution toI⁰, related byx=g⁰(x⁰).

g(^x) =

g⁰⁽a¹⁾⁼u¹⁼a⁵a³a¹

z }| {

(a

1

) g(â⁵) ^g(â³) ^g(â¹)

g⁰⁽a⁶⁾⁼u⁶⁼a²a⁴

z }| {

(a

6

) g(^a²) ^g(^a⁴)

g⁰⁽a²⁾⁼u²⁼a⁶a³

z }| {

(a

2

) g(^a⁶) ^g(^a³)

h(^x) = ^(a¹⁾ ^h(^a⁵) ^h(^a³)

| {z }

h⁰⁽a¹⁾⁼v¹⁼a⁵a³

(a

6

) h(^a¹) ^h(^a²)

| {z }

h⁰⁽a⁶⁾⁼v⁶⁼a¹a²

(a

2

)h(â⁴) ^h(â⁶) ^h(â³)

| {z }

h⁰⁽a²⁾⁼v²⁼a⁴a⁶a³

Fig.1.How a solution to^I translates to^I⁰ and vice versa

In general, by construction, ifx⁰ is a solution to I⁰ thenx =g⁰(x⁰) =h⁰(x⁰) is a solution toI. And conversely, for every solutionxtoI there is a solutionx⁰ toI⁰ such thatx=g⁰(x⁰) =h⁰(x⁰). ThusI andI⁰ are equivalent. ² If we could prove thatI⁰is somehow simpler thanI, then we could repeat the procedure, reduce to simpler and simpler equivalent instances I⁰⁰, I⁰⁰⁰,..., and eventually decideI. There are at least two ways in whichI⁰can be simpler than I:^j⁰^j<^j^j(m < n) or(I⁰)< (I), wheremeasures the \sux complexity"

of an instanceI = (;;g;h) [1]:

(I) =^j^[a²^fx^jx is a proper sux ofg(a)^gj +^j^[a²^fx^jx is a proper sux ofh(a)^gj

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Ifn=m, we would likeI⁰ to be simpler thanI in the sense that (I⁰)< (I).

The following lemma shows thatI⁰ at least cannot be more complex than I:

Lemma 2.

Îf Î ândÎ⁰ are as above, then(I⁰)(I).

Proof. Dene the following four sets:

G =^[a²^fx^jxis a proper sux ofg(a)^g G⁰ =^[a²⁰^fx^jxis a proper sux ofg⁰(a)^g H =^[a²^fx^jxis a proper sux ofh(a)^g H⁰=^[a²⁰^fx^jxis a proper sux ofh⁰(a)^g

We will dene an injective function p: G⁰ ^!H. Letu ²G⁰, sou is a proper sux of some specicg⁰(ai) =ui=x¹:::xc generated by the procedure of the previous section. Let xr be the rst letter of u, ands be the shortest sux of some h(yt) due to which xr was added to ui in the procedure of the previous section, sosis a prex ofg(xr) (see Figure 2) or vice versa. Denepasp(u) =s.

g(^ui) = ^g(^x¹) ^:^:^:^:^:^: ^g(^xr^?1)

u⁼x^rx^{r +1}:::x^c

z }| {

g(^xr) ^g(^xr⁺¹) ^:^:^:^:^:^: ^g(^xc)

h(^vi) = ^h(^y1) ^:^:^:^:^:^: ^h(^yt)

|{z}s

h(^yt⁺¹) ^:^:^:^:^:^:^:^:^: ^h(^yd)

Fig.2.The sux^scorresponding to^u

We will showpis injective. If u;u⁰ ²G⁰ andp(u) =p(u⁰), then uandu⁰ are associated with the same sux s = p(u), hence u and u⁰ must start with the samexrand (by determinism of the procedure of the previous section) continue in the same way, givingu=u⁰. Thuspis injective, which implies^jG⁰^j^jH^j.

Similarly we can dene an injective function from H⁰ to G, which proves

jH⁰^j^jG^j. It now follows that(I⁰) =^jG⁰^j+^jH⁰^j^jG^j+^jH^j=(I). ²

2.3 The Algorithm

We will here give a method to decide if a given instance I = (;;g;h) of marked PCP has a solution. The idea is to make a sequence of equivalence- preserving reductions I⁰ = I;I¹;I²;:::, such that once in a while a reduction from Ii to Ii⁺¹ simplies the instance (makes the source alphabet or the sux complexity smaller). We will show that either this sequence of reductions reaches an Ij which has source alphabet of size 1 or equal to 0 (so Ij is decidable), or the sequence will repeat itself after a while and start cycling. Such cycles are detectable, and we will show that every I leading to such a cycle is easily decidable.

So suppose the sequence of reductions does not reach anIj with alphabet of size 1 or (Ij) = 0. Then it must get \stuck" at a certain source alphabet size

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and. That is, there exist ak,mandz such that allIi in the innite sequence Ik;Ik⁺¹;Ik⁺²;::: have source alphabet of sizem and have (Ii) = z. Now this sequence must repeat itself after a while, for otherwise there would be innitely many distinct instances with the same alphabet and -value, contradicting the next lemma.

Lemma 3.

Let=^fa¹;:::;am^g be nite sets andz be a positive natural number. There exist only nitely many distinct instancesI = (;;g;h) of PCP that satisfy(I)z.

Proof. An instance I = (;;g;h) is completely specied by giving the 2m wordsg(a¹);:::;g(a_m),h(a¹);:::;h(a_m)²⁺. Note that if one of those words has length> z+1, then this word has more thanzproper suxes and(I)> z. Accordingly, each of the 2m words can have length at most z+ 1. There are

Pzi⁼¹⁺¹^j^jⁱ ^j^j^z⁺² such words. Thus there are at most^j^j⁽^z⁺²⁾²^m choices for 2msuch words, and hence nitely many dierentI that satisfy(I)z. ² This lemma shows that if the procedure does not converge to very simple instances then it will cycle, and we can detect this by noting that some I_k and I_r (k < r) are equal. It remains to show how we can decide such \cycling"

instances of marked PCP. So suppose we have a cycle, assume without loss of generality that it already starts atI⁰:

I⁰^!I¹^!^!Ir^?1^!Ir=I⁰;

where Ii = (;;gi;hi). By the proof of Lemma 1, for every solution xi to someIi, there is a solutionxi⁺¹ toIi⁺¹ such thatxi =gi⁺¹(xi⁺¹) =hi⁺¹(xi⁺¹).

Supposex⁰is a solution toI⁰of minimal length. There must exist some solution xrtoIr such that

x⁰=g¹g²:::gr(xr) x⁰=h¹h²:::hr(xr)

Since thegiandhicannot be length-decreasing, we have^jx⁰^j^jxr^j. Butx⁰was chosen to be a minimal-length solution toI⁰andxris also a solution toIr=I⁰, hence ^jx⁰^j = ^jxr^j. This implies that g⁰(= gr) and h⁰(= hr) map the letters occurring in xr to letters. But then the rst letter ofxr is already a solution, hence^jx⁰^j=^jxr^j = 1. ThusI⁰ has a solution iI⁰ has a 1-letter solution (i.e., there is ana²⁰ such thatg⁰(a) =h⁰(a)), and this is trivially decidable.

Below we summarize this analysis in an algorithm and a theorem:

Decision procedure for marked PCP

1. Set^I =^;,i= 0,I⁰=I. 2. Seti=i+ 1.

3. ReduceIi^?1 toIi in the way stated above.

4. IfIihas source alphabet of size 1 or= 0, then decideIi, print the outcome and terminate.

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5. If Ii is simpler thanIi^?1 (smaller source alphabet or) then set^I=^;and goto 2.

6. If Ii ²^I then there is a cycle and we can decide Ii by checking if it has a 1-letter solution, print the outcome and terminate;

else set^I=^I^[^fIi^gand goto 2.

Theorem 1.

Marked PCP is decidable.

2.4 Complexity Analysis

Let us analyze the complexity of this algorithm. LetN be the length of the input instance I. Each reduction from Ii to Ii⁺¹ can be done in O(N⁴) steps. How many dierent reductions do we need to make? For a xed alphabet size^j^j

j^j = m and sux complexity z, we can make at most m⁽^z⁺²⁾²^m reductions before detecting a cycle (proof of Lemma 3). Since m=O(N) and z=O(N²), this gives an upper bound of 2Ô^(log^N^N³⁾ on the number of reductions for xed alphabet size and sux complexity. Alphabet size and sux complexity cannot increase during the process. There are at mostn=O(N) dierent alphabet sizes and at most(I) =O(N²) dierent sux complexities possible, so we have to make no more thanO(N³)2Ô^(log^N^N³⁾ reductions. Since the setÎ can contain at most 2Ô^(log^N^N³⁾ instances, the test I_i ² Î in step 6 can be performed in 2Ô^(log^N^N³⁾ steps. Thus the whole algorithm works in 2Ô^(log^N^N³⁾ steps, which means that marked PCP is in

EXPTIME

.

3 2-Marked PCP Is Undecidable

Here we will show that if we weaken the condition of markedness, by only re- quiring the morphisms to be 2-marked, then PCP becomes undecidable again.

Consider the following semi-groupS⁷with set of 5 generators? =^fa;b;c;d;e^g and 7 relations:

S⁷=^ha;b;c;d;e^jRⁱ

R=^fac=ca;ad=da;bc=cb;bd=db;eca=ce;edb=de;cca=ccae^g Tzeitin [10] (see also [7, p. 445]) proved that the following problem for this semi-group is undecidable:

Givenu;v²?⁺, isu=v²S⁷?

Note that the set of 7 left-hand-sides of R is 2-marked, and similarly for the set of 7 right-hand-sides of R. We will reduce this problem to 2-marked PCP.

We use a slight modication of the standard reduction from word problems to PCP, involving an alphabet with some underlined letters in order to ensure 2- markedness.

Dene the source alphabet as

=? ^[?^[^fB;E;#;#;r¹;r²;:::;r⁷;r¹;r²;:::;r⁷^g,

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where ? = ^fa;b;c;d;e^g, and r¹;:::;r⁷ are the 7 relations in R and r¹;:::;r⁷ are their underlined versions (considered as single letters), so r¹ = [ac = ca], r¹= [ac=ca] etc. Dene the target alphabet as

=?^[?^[^fB;E;#;#^g.

B and E will mark the beginning and end of expressions, respectively. Given u;v²?⁺,gandhare dened by Table 1:

B E # # â ... ê â ... ê [^s=^t] [^s=^t]

g Bu# Ê # # â ... ê â ... ê ^t ^s

h B #^vE # # â ... ê â ... ê ^s ^t

Table1.Denition of^gand^h

Note that the constructed instanceI = (;;g;h) is an instance of 2-marked PCP. The following lemma shows that the reduction preserves equivalence with Tzeitin's problem:

Lemma 4.

Letu;v;I be as above. Then u=v²S⁷ iI has a solution.

Proof.

=⁾: Suppose u=v ²S⁷. Then there is a sequenceu=u¹^!u²^!^! uk =v, whereui =u⁰su⁰⁰ andui⁺¹ =u⁰tu⁰⁰, ands=t ²R ort =s ²R. We construct a solution toI by induction onk.

Ifk= 1, thenu=v²?⁺. Nowx=Bu#uE is a solution toI.

Now let I⁰ = (;;g⁰;h⁰) be the instance of 2-marked PCP corresponding to u = uk^?1 ² S⁷. By the induction hypothesis we can assume that I⁰ has a minimal-length solution x⁰. It is easy to see that every solution must begin with B and end with E, so x⁰ = ByE, and therefore g⁰(By) = w#uk^?1 and h⁰(By) =wfor some w. Note that sinceI andI⁰ only dier in the assignment h(E) andh⁰(E), andE cannot occur iny (becausex⁰ is minimal), we also have g(By) =w#uk^?1 and h(By) = w. We distinguish two cases. Firstly, uk^?1 = u⁰su⁰⁰ and v =u_k = u⁰tu⁰⁰, wherer = [s =t] is one of the 7 relations. Then it is easily veried that x =By#u⁰ru⁰⁰#u⁰tu⁰⁰E is a solution to I. Secondly, if u_k^?1 =u⁰tu⁰⁰ and v =u_k =u⁰su⁰⁰, thenx =By#u⁰tu⁰⁰#u⁰ru⁰⁰E is a solution.

This completes the induction step.

(=: Suppose I has a solution x. We can assume x is of minimal length.

Thisxmust be of the formBx¹x²:::xmE, wherexi², sog(Bx¹:::xmE) = Bu#g(x¹:::xm)E = h(Bx¹:::xmE) = Bh(x¹:::xm)#vE. Ignoring the underlining, g(x) = h(x) must be of the formBu¹#u²#:::#uk^?1#ukE, where ui ² ?, u¹ = u and uk = v. We will show that ui = ui⁺¹ ² S⁷ for every 1ik^?1, from whichu=v²S⁷follows.

Because # occurs inh(x¹:::xm), there must be some leastisuch thatxi=

#, and henceu=h(x¹:::xi^?1). Since there is no underlining inu, it follows that

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x¹;:::;xi^?1 must have been chosen from a;:::;e;r¹;:::;r⁷. Let x¹:::xi^?1 = w¹ri¹w²ri²:::wl, with wi ² ? and ri = [si = ti] ² ^fr¹;:::;r⁷^g. Then u = h(w¹ri¹w²ri²:::wl) =w¹si¹w²si²:::wl. See Figure 3 for illustration.

g(^Bx1^:^:^:^xi^:^:^:^xm^E) =

g⁽B⁾⁼Bu^#

z }| {

B w1si¹^w2si²^:^:^:^wl # ^g(^x1) ^g(^x2) ^:^:^:^:^:^:

g⁽E⁾

z}|{

E

h(^B^x¹^:^:^:^xi^:^:^:^xm^E) = ^B

|{z}h⁽B⁾

w

1

si¹^w²^si²^:^:^:^wl

| {z }

h⁽x¹:::x^i?1⁾⁼u⁼u¹

#

|{z}h⁽xⁱ⁾

h(^xi⁺¹) ^:^:^:^:^:^: #^vE

| {z }

h⁽E⁾

Fig.3.Picture leading to^u=^v

Note thatg(x¹:::xi^?1) =g(w¹ri¹w²ri²:::wl) =w¹ti¹w²ti²:::wl. But now, since we must haveg(x¹:::xmE) =h(xi⁺¹:::xmE), there must be a leastj > i such that xj ²^f#;#^gandh(xi⁺¹:::xj^?1) =g(x¹:::xi^?1) =w¹ti¹w²ti²:::wl. The latter string (without underlining) is u². Note that u¹ =u²²S⁷, because u¹(=u) andu² only dier byu² havingti whereu¹hassi.

Continuing this reasoning, we can show that for every two wordsui;ui⁺¹²? occurring in g(x) =h(x) separated by #, ignoring underlining, we must have ui=ui⁺¹²S⁷(some of the wordsui andui⁺¹may actually already be equal in

⁺). Henceu=v²S⁷, sinceg(x) starts withu¹=uand ends with uk =v. ² Together with Tzeitin's result, the above lemma implies:

Theorem 2.

2-Marked PCP is undecidable.

To end this section, we note that 2-marked PCP is not a special case of injective PCP. For example, the morphism dened byg(1) = 23,g(2) = 2,g(3) = 3 is 2-marked but not injective. We can combinek-markedness and injectivity by calling a morphismgstronglyk-markedifgis bothk-marked and prex (i.e., no g(ai) is a prex of anotherg(aj)). This clearly implies injectivity. It follows from a construction of Ruohonen [8] that strongly 5-marked PCP is undecidable: the biprex instances of PCP constructed there to show undecidability of biprex PCP are also 5-marked. Decidability of stronglyk-marked PCP for 1< k <5 is still open.

4 Conclusion and Future Work

We can investigate the boundary between decidability and undecidability by examining which restrictions on the Post Correspondence Problem render the problem decidable. We have shown here that restricting PCP to marked morphisms gives us decidability. On the other hand, 2-marked PCP is still undecidable.

The following questions are left open by this research:

(10)

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Is exponential time the best we can do when deciding marked PCP, or is there a polynomial-time algorithm for the problem?

{

What about decidability of stronglyk-marked PCP for 1< k <5?

{

What about decidability of marked generalized PCP [1,3]?

{

The decidability status of PCP with elementary morphisms [9, pp. 72{

77] is still open. A morphism g is elementary if it cannot be written as a composition g²g¹ via a smaller alphabet. Marked PCP is a subcase of elementary PCP which we have shown here to be decidable. Can our results help to settle the decidability status of elementary PCP?

Acknowledgment

We would like to thank Tero Harju, Juhani Karhumaki, and John Tromp for reading and commenting this paper, and Harry Buhrman for some discussions.

The second author would like to thank the CWI for its hospitality during the summer of 1998, when part of this work was done.

References

1. A. Ehrenfeucht, J. Karhumaki, and G. Rozenberg. The (generalized) Post correspondence problem with lists consisting of two words is decidable. The- oretical Computer Science, 21(2):119{144, 1982.

2. M. Garey and D. Johnson. Computers and Intractability: A Guide to the Theory of NP-Completeness. W. H. Freeman, 1979.

3. T. Harju, J. Karhumaki, and D. Krob. Remarks on generalized Post correspondence problem. InProceedings of 13th STACS, volume 1046 of Lecture Notes in Computer Science, pages 39{48. Springer-Verlag, 1996.

4. Y. Lecerf. Recursive insolubilite de l'equation generale de diagonalisation de deux monomorphisms de monodes libres ^x=^x. Comptes Rendus Acad.

Sci. Paris, 257:2940{2943, 1963.

5. Y. Matiyasevich and G. Senizergues. Decision problems for semi-Thue sys- tems with a few rules. InProceedings of the 11th IEEE Symposium on Logic in Computer Science, pages 523{531, 1996.

6. E. L. Post. A variant of a recursively unsolvable problem. Bulletin of the American Mathematical Society, 52:264{268, 1946.

7. G. Rozenberg and A. Salomaa, editors. Handbook of Formal Languages, volume 1. Springer-Verlag, Berlin, 1997.

8. K. Ruohonen. Reversible machines and Post's correspondence problem for biprex morphisms. Journal of Information Processing and Cybernetics (EIK), 21(12):579{595, 1985.

9. A. Salomaa. Jewels of Formal Language Theory. Pitman, 1981.

10. G. C. Tzeitin. Associative calculus with an unsolvable equivalence problem.

Tr. Mat. Inst. Akad. Nauk, 52:172{189, 1958. In Russian.