On One-Rule Grid Semi-Thue Systems

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Michel Latteux, Yves Roos

To cite this version:

Michel Latteux, Yves Roos. On One-Rule Grid Semi-Thue Systems. Fundamenta Informaticae, Pol- skie Towarzystwo Matematyczne, 2012, Words, Graphs, Automata and Languages. Special Issue Honoring the 60th Birthday of Professor Tero Harju, 116 (1-4), pp.189-204. �10.3233/FI-2012-678�.

�hal-00749289�

On One-Rule Grid Semi-Thue Systems

Michel Latteux

Laboratoire d’Informatique Fondamentale de Lille Universit´e Lille 1

[email protected]

Yves Roos

Laboratoire d’Informatique Fondamentale de Lille Universit´e Lille 1

[email protected]

Abstract. The family of one-rule grid semi-Thue systems, introduced by Alfons Geser, is the family of one-rule semi-Thue systems such that there exists a lettercthat occurs as often in the left-hand side as the right-hand side of the rewriting rule. We prove that for any one-rule grid semi-Thue systemS, the setS(w)of all words obtainable fromwusing repeatedly the rewriting rule ofSis a constructible context-free language. We also prove the regularity of the setLoop(S)of all words that start a loop in a one-rule grid semi-Thue systemsS.

Keywords: one-rule semi-Thue system, termination, grid semi-Thue system.

1. Introduction

Semi-Thue systems, that are the non-symmetrical version of Thue systems introduced by Axel Thue in 1914, serve as a model for rewriting systems. Thus they are of primordial interest for computa- tional problems. For several years they have been intensively studied and several deep results have been obtained [10, 3, 2, 21, 19, 22]. However some intriguing decidability problems remain open.

With a semi-Thue systemS, one associates the setS1of words that start an infinite derivation in S. The recursiveness ofS1is called the termination problem forSand the emptiness ofS1is the uniform termination problem forS. The best result on the setS1is that the termination problem and the uniform termination problem are undecidable for 3-rules semi-Thue systems ([14]). Clearly the termination problem is decidable for length-preserving semi-Thue systems, contrarily to the uni- form termination problem that has been shown undecidable for length-preserving semi-Thue systems

([1]). This result remains true for 9-rules semi-Thue systems ([19]) and for length-two semi-Thue systems ([18]).

Another set naturally associated with semi-Thue systems is the setLoop(S)that brings together the wordswthat start a non-null derivation inStoward a word containingwas a factor. For length- preserving semi-Thue systems, the emptiness ofLoop(S)is equivalent to the emptiness ofS1but this equivalence does not hold for arbitrary semi-Thue systems, as shown in [7] where a 2-rules semi-Thue systemS is presented withLoop(S) = ;and S1 6= ;. Moreover, for an arbritary semi-Thue systemS, the setLoop(S)need not be recursive.

One-rule semi-Thue systems are the simplest rewriting systems. Indeed, they are defined by two words l, r,notedS = {l −! r}. For a wordw,S(w)is the set of words obtainable fromw by replacing repeatedlylbyr. It is clear that, except the particular casel=r,w2S1if and only if the setS(w)is infinite. However, despite numerous efforts for more than twenty years, these decidability problems remain open and have become challenging problems. We observe that these problems can be explained in a few minutes to non-scientific people and surely they point out a deep lack of understanding of the rewriting notion. To get new ideas in order to solve these difficult problems, it seems natural to examine closely particular one-rule semi-Thue systems. Several interesting partial results have been obtained in that direction([20, 23, 16, 5, 6, 17]). Here we continue the study of a natural subclass of one-rule semi-Thue systems, introduced by Alfons Geser in [4], and called one-rule grid semi-Thue systems.

It is obvious that the systemS={l−!r}withl6=ris uniformly terminating if|r|  |l|or if there is a letterxsuch that|r|x <|l|x. So, when studying the termination, we can assume that there is a letterxsuch that|r|x >|l|xand there is no letterysuch that|r|y<|l|y. Then a borderline case is when there is a unique letterxsuch that|r|x>|l|xand|r|y=|l|yfory6=x. The family of one-rule grid semi-Thue system satisfies a slightly more general condition. This family, that we denoteSgrid, is composed of all one-rule semi-Thue systems having a lettercsuch that|r|c = |l|c = k > 0¹. In [4], Alfons Geser has given a nice decidable characterization of one-rule grid semi-Thue systems that are uniformly terminating by proving that such a system is non-uniformly terminating if and only if it has a loop of length 1 or 2. To know whether a one-rule semi-Thue system has a loop of length 1 or 2 was previously shown decidable by Winfried Kurth in [11].

First, we give a new formulation of this decidable characterization of non-uniformly terminating systems inSgrid: there exist wordsx, ysuch thatlyis a left factor ofxrandxlis a right factor ofry.

We show that the wordsxandyare unique and give a simple way to compute these two words. This permits, when considering a nonterminating systemS = {l−! r} 2 S_grid, to give a very precise form of the two wordslandrand to get as a consequence that if|r|c=|l|c=kis oddlneeds to be a factor ofr.

Then we prove that, for anySinSgrid, the setLoop(S)is a constructible regular language by giving a simple rational expression. This property can not be generalized to arbitrary 1-rule semi-Thue system: for instanceLoop({c−!caca})is not regular. Concerning the link betweenLoop(S)and S1, it is proved in [4] that the familyS_gridsatisfies the equalityS1=S⁻¹(A^⇤Loop(S)A^⇤), that is w2S1if and only if there is a derivation fromwto a word in the regular languageA^⇤Loop(S)A^⇤ whereAis the alphabet oflr. Unfortunately, up to now, this relation does not imply thatS1is a regular language as shown in [20] whenl2a^⇤b^⇤. Then we prove that for anySinSgridand for any

1In the original definition of Alfons Geser, a one-rule grid semi-Thue system is a one rule semi-Thue systemS={l−!r} having a lettercsuch that|r|^c |l|^c. In this paper we do not consider the case|r|^c<|l|^cfor whichSis trivially uniformly terminating.

wordw, the setS(w)is a constructible context-free language. Note that this result does not hold for arbitrary one-rule semi-Thue systems ([12]). As a matter of fact,S(w)is a bounded context-free language. So we get both the decidability of the termination problem²inSgridand the decidability of the common descendant problem inSgrid. In the last section, we give an argument in favour of the regularity ofS1by proving that it is a regular language in the case|r|c=|l|c= 2.

2. Preliminaries and Notations

In the following,Awill denote a finite alphabet,A^⇤the free monoid overAandεthe empty word inA^⇤. For a wordw2A^⇤and a lettera2A,|w|denotes the length of the wordwand|w|adenotes the number of occurrences of the letterain the wordw.

Two wordsuandvareconjugateif there exist wordsxandysuch thatu=xyandv=yx. It is well known that two wordsuandvare conjugate if and only if there exists a wordzsuch thatuz=zv.

A worduis afactorof a wordvif there exist two wordsw1andw2such thatv=w1uw2and we denote byF(v)the set of the factors of the wordv. We denote byRF(w)(respectivelyLF(w)) the set ofright factors(respectivelyleft factors) of the wordw, that is:

RF(w) ={v2A^⇤| 9u2A^⇤, w=uv}, LF(w) ={u2A^⇤| 9v2A^⇤, w=uv}.

A semi-Thue system over an alphabetAis a subsetS✓A^⇤⇥A^⇤. Members ofSare denotedl−!

S r (orl −! rif there is no ambiguity). One-step derivation, denoted−!

S (−!if no ambiguity), is the binary relation over words defined by :8u, v2A^⇤, u−!viff there existsl−!r2Sandα, β2A^⇤ such thatu=αlβandv=αrβ. The relation−!^⇤ (resp. −⁺!) is the reflexive and transitive closure (resp. transitive closure) of the relation−!and, for any wordw2A^⇤, we shall denoteS(w)the set S(w) ={w⁰2A^⇤ | w−!^⇤

S w⁰}andS⁻¹(w) ={w⁰2A^⇤ | w⁰−!^⇤

S w}. We extend these notations to languages: for any languageL✓ A^⇤,S(L) = [w2LS(w)andS⁻¹(L) = [w2LS⁻¹(L). For a derivationw = w0 ! w1· · · ! wn = w⁰,nis called the length of the derivation that will be denoted byw−!ⁿ w⁰.

We notew −!¹

S iff there is an infinite derivation starting onwand we denote byS1the setS1 = {w2A^⇤ | w−!¹

S }.

For a semi-Thue systemS, for any positive integern, we denoteLoop_n(S) ={w2A^⇤ | 9x, y2 A^⇤, w−!ⁿ xwy}andLoop(S) =[n>0Loop_n(S).

The termination problemfor the alphabet Aand the semi-Thue system S ✓ A^⇤ ⇥ A^⇤ is the following:

instance:a wordw2A^⇤

2This result already appears implicitly inside proofs used by Alfons Geser in [4] to solve the uniform termination problem in Sgrid.

question: Does every derivation (moduloS) starting onwhave finite length? (that is doesw 62 S1?)

Theuniform termination problemfor a classSof semi-Thue systems is the following : instance:an alphabetAand a finite semi-Thue systemS✓A^⇤ ⇥ A^⇤which belongs toS question:Do all derivations (moduloS) starting from all wordsw2A^⇤have finite length? (that is

doesS1=;?)

To get shorter, we say that a systemS is nonterminatingif the uniform termination problem has a negative answer forS. In the sequel, we focus onSgrid, the family of one-rule grid semi-Thue systems.

Definition 2.1. A one-rule grid semi-Thue systemS ={u! v}is a one-rule semi-Thue system such that there exists a lettercwith|u|c=|v|c=k >0.

3. Uniform termination in S

In this section, we state some consequences of the following Geser’s result:

Proposition 3.1. ([4]) A one-rule grid semi-Thue systemS={u!v}is non-uniformly terminat- ing iff one of the following properties is satisfied³:

1. uis a factor ofv.

2. there exist wordsg, h, ksuch thatu=gh,v=hkandgghis a factor ofhkk.

Example 3.1. The simplest example of one-rule grid semi-Thue system satisfying the property 2 of proposition 3.1 but not the property 1 isS ={cac−!acca}withg =c,h= ac,k =caand ggh=ccacis a factor ofhkk=accaca.

LetS={u!v} 2 Sgrid. The wordsuandvbelong toA^⇤whereAis an alphabet that contains a lettercwith|u|c=|v|c=k >0. If we denoteB=A\ {c}, then

• u=lcu1. . . uk−1cr

• v=l⁰cv1. . . vk−1cr⁰

for some wordsl, r, l⁰, r⁰, u₁, . . . , uk−1, v₁, . . . , vk−12B^⇤. We get, as a consequence of the propo- sition 3.1, the following corollary that is also directly proved in [13]:

Corollary 3.1. LetS={u!v} 2 Sgridwithu6=v. ThenSis nonterminating iff one of the two following properties is satisfied:

(i) uis a factor ofv

(ii) there existx, y, s, e2A^⇤such thatxv=uyeandvy=sxu.

3More precisely, Alfons Geser proved in [4] that a one-rule grid semi-Thue systemS={u!v}is non-uniformly terminating if and only ifLoop₁(S)[Loop₂(S)6=;that is proved decidable by Winfried Kurth in [11].

Proof: Clearly, if (ii) is satisfied,Sis nonterminating sincexu −! xv = uye −! vye = sxue.

Now, ifu 62 F(v)and if the property 2 of the proposition 3.1 is satisfied, thenhkk = sgghefor somes, e. Since|g|c = |k|c, it followss, e 2B^⇤. Suppose thatg, k2 B^⇤then, sinceh 2l⁰cA^⇤ and hkk = sgghe, we getl⁰ = sggl⁰ that impliesg = εandu = h 2 F(v), a contradiction.

Thus|g|c = |k|c > 0andk = yefor somey 2 A^⇤. Takingx =g, we getxv = ghk = uye, vye=vk=hkk=sgghe=sxue, hencevy=sxu. ut

The following lemma states that ifu6=vthe conditions (i) and (ii) of the corollary 3.1 are mutually exclusive. This lemma also clarifies the relationship between the words x, y, sand ewhen the property (ii) of the corollary 3.1 is satisfied.

Lemma 3.1. LetS = {u ! v} 2 Sgrid. If the property (ii) of the corollary 3.1 is satisfied and u6=v, we have

(i) |x|_c=|y|_c>0ands, e2B⁺, (ii) u62F(v),

(iii) u=xu⁰⁰=u⁰ywhereu⁰⁰is the longest word inRF(u)\LF(v)andu⁰is the longest word in LF(u)\RF(v),

(iv) xandyare conjugate;sandeare conjugate.

Proof: If the property (ii) of the corollary 3.1 is satisfied, there areeandsinA^⇤such thatxv=uye (1) andvy=sxu(2). From (1), we get|x|_c=|ye|_cand from (2),|y|_c=|sx|_c. So|x|_c=|y|_cand

|e|c=|s|c= 0that ise, s2B^⇤. Moreover|x|c=|y|c >0otherwisexandybelong toB^⇤and, from (1), we obtainxl⁰ =land from (2),l⁰ =sxl. It followsx=s=εand we similarly obtain y =e=ε. This leads to the contradictionu=v. Hencex2lcA^⇤andy2A^⇤crsovbelongs to slcA^⇤\A^⇤creandl⁰=sl,r⁰=re.

Let us now suppose thatuis a factor ofv. We prove that this leads to a contradiction. Indeed, if v=s⁰ue⁰for somee⁰, s⁰ 2B^⇤, then the equality (1) givesxs⁰ue⁰ =uye2A^⇤cre⁰\A^⇤creand it follows thate⁰=e. Similarly equality (2) impliess⁰=sand we obtainxs⁰u=uyandue⁰y=xu.

Thereforee⁰=s⁰=εandu=v; this contradiction proves (ii).

We can now finish the proof of (i): from (1) and (2), we getxsxu=xvy=uyeythat impliesu2 LF((xsx)^⇤). Supposes=εthenu2LF((x)^⇤)andvy =xu2 LF(x^⇤). It followsv2LF(x^⇤) that impliesu2F(v), a contradiction. Similarly we can showe6=ε.

For (iii), let us first observe that|x|<|u|and|y|<|u|: otherwise, we may assume that|y|  |x|;

then|x| ≥ |u|andx=uzfor some wordz. It follows thatxu=uzu2RF(vy)with|y|  |x|=

|zu|anduis a factor ofvwhich leads to a contradiction. Then we haveu =xu⁰⁰ = u⁰ywithu⁰⁰ inRF(u)\LF(v)andu⁰ inLF(u)\RF(v). We shall now prove thatu⁰⁰ is the longest word in RF(u)\LF(v)and we could similarly show thatu⁰is the longest word inLF(u)\RF(v).

We havex=lcu1. . . uj−1cu⁰_jwithu⁰_j2LF(uj)and, if we denotepthe first index such thatup6=

vp(pexists sinceu62F(v)), we havej p. Assume now thatj < pand setz =lcu1. . . up−jc, then we havexsz 2 LF(u)andsxz 2 LF(v). But this impliess = εandu_p = u_p−j = v_p, a contradiction. Thusp =j and it followsu_j = u⁰_jslandv_j = u⁰_jlwiths6= ε. Moreover, since ye2RF(ue)\RF(v)with|x|_c=|y|_c, we obtaink≥2pandxsx⁰2LF(u),sxx⁰2LF(v), with x⁰=lcu₁. . . up−1c.

In order to show thatu⁰⁰ is the longest word inRF(u)\LF(v), assume that there exists somez such that|z|<|x|andu2LF(zv). Then|z|c <|x|c,zsx2 LF(xsx⁰)andsxsx =szsxtwith t6=ε. Hencexsis not a primitive word andxs= (xrs)^q+1wherexrsis the primitive root ofxs.

That impliesx= (xrs)^qx_randx⁰= (xrs)x⁰_rwithx⁰_r2LF(xr),|x⁰_r|c=|xr|c>0,z= (xrs)ⁱx_r withi < qandzsx = xzs. Sinceu 2 LF(zv), we havexsx⁰ 2 LF(zsxx⁰) = LF(xszx⁰)and x⁰ = (xrs)^qx⁰_r 2 LF(zx⁰) = LF((xrs)ⁱxr(xrs)^qx⁰_r). Thus(xrs)^q−ix⁰_r 2 LF(xr(xrs)^qx⁰_r)and sx⁰_r2LF(xrs). But that impliess=ε, a contradiction that finishes the proof of (iii). Observe that it implies the unicity ofxandy. Moreover, if there are severalcsuch that|u|c=|v|c, the value of xandydoes not depend on the choice of a particularc.

Let us now prove (iv). Since|x|c = |y|c, we can writex =x₀cx₁. . . cx_tandy = y₀cy₁. . . cy_t with80  i  t, x_i, y_i 2 B^⇤. From (1) and (2), we can deducexvy = xsxu = uyey and it follows thatyeyandxsxare conjugate. Thuscy_tey₀c . . . cy_ty₀c 2F((xsx)^⇤). Sinces, e 2B⁺, it is easy to verify that we must haveytey₀ =xtsx₀andyty₀ = xtx₀. Suppose that|yt| ≥ |xt| (the case|yt|  |xt|is symmetric) thenyt = xtzfor somezand it followszy0 = x0. Then we getxtzey0=xtszy0sosandeare conjugate. Now from the equalityyteyy0=xtsxx0we obtain xtzeyy0=xtsxzy0that implieszey=szy=sxzthereforexandyare conjugate. ut

Since the equalities vy = sxuandxv = uyeimplyxvy = xsxu = uyey, it follows thatu 2 LF((xsx)^⇤), u 2 RF((yey)^⇤), v 2 LF((sxx)^⇤) and v 2 RF((yye)^⇤). Then we can state the precise form ofuandvfor a nonterminating one-rule grid semi-Thue system:

Proposition 3.2. S2 S_gridis nonterminating iff one of the following three conditions is satisfied 1. u2F(v)

2. u= (xsx)ⁿx⁰andv= (sxx)ⁿsx⁰withn >0andx⁰2LF(x)\LF(sx) 3. u= (xsx)ⁿxsx⁰andv= (sxx)ⁿsxx⁰ewithn≥0, x=x⁰x⁰⁰andx⁰⁰s=ex⁰⁰.

Proof: Let us first observe that these three conditions are sufficient by verifying that for any system S satisfying 1, 2 or 3,Loop(S) 6= ;. It is clear for 1 since, in this case, u 2 Loop₁(S). IfS satisfies 2 then x(xsx)ⁿx⁰ 2 Loop₂(S): indeed, in this case we getx = x⁰x⁰⁰ andsx⁰ = x⁰x⁰⁰⁰ for some wordsx⁰⁰, x⁰⁰⁰andx(xsx)ⁿx⁰ −!x(sxx)ⁿsx⁰ = (xsx)ⁿx⁰x⁰⁰sx⁰ −! (sxx)ⁿsx⁰x⁰⁰sx⁰ = sx(xsx)ⁿx⁰x⁰⁰⁰. Finally, ifSsatisfies 3,x(xsx)ⁿxsx⁰ −!x(sxx)ⁿsxx⁰e= (xsx)ⁿxsx⁰x⁰⁰x⁰e−! (sxx)ⁿsxx⁰ex⁰⁰x⁰e= (sxx)ⁿsxx⁰x⁰⁰sx⁰e=sx(xsx)ⁿxsx⁰e, thereforex(xsx)ⁿxsx⁰2Loop₂(S).

It now remains to prove that the condition (1 or 2 or 3) is necessary.

Suppose that 1 is not satisfied, then we can consider three cases sinceu2LF((xsx)^⇤):

(i) u= (xsx)ⁿx⁰withx=x⁰x⁰⁰.

First we can observe thatv= (sxx)ⁿsx⁰sincev2LF((sxx)^⇤)and|v|=|u|+|s|. Moreover n >0since|x|c<|u|c. It remains to prove thatx⁰2LF(sx). The property (iii) of the lemma 3.1 givesy 2RF(u), thereforey =x⁰⁰x⁰ since|x|=|y|. From the equalityxv =uyewe obtainx(sxx)ⁿsx⁰= (xsx)ⁿx⁰x⁰⁰x⁰eandx⁰2LF(sx⁰)since|e|=|s|.

(ii) u= (xsx)ⁿxs⁰withs⁰2LF(s).

Since v 2 LF((sxx)^⇤) and|v| = |u|+|s|, it followsv = (sxx)ⁿsxx⁰ with|x⁰| = |s⁰|.

Thenx⁰ 2B^⇤since|u|_c =|v|_cands2 B^⇤. It follows thatx=x⁰x⁰⁰for somex⁰⁰. Let us now consider the factorizationx=zctwitht2 B^⇤(we know that|x|c >0) then we obtain

u = u1cts⁰and v = v1ctx⁰ with|ts⁰| = |tx⁰|andts⁰, tx⁰ 2 B^⇤,u1, v1 2 A^⇤. This is a contradiction sinceu=u1crandv=v1crewith|re|>|r|. The caseu= (xsx)ⁿxs⁰with s⁰2LF(s)is finally impossible.

(iii) u= (xsx)ⁿxsx⁰withx⁰2LF(x).

Sincev2 LF((sxx)^⇤)and, from the property (iv) of the lemma 3.1,|v|=|u|+|e|, we get v = (sxx)ⁿsxx⁰x⁰⁰⁰ with|x⁰⁰⁰| = |e|. Moreoverv 2 RF((yye)^⇤) thereforex⁰⁰⁰ = eand v = (sxx)ⁿsxx⁰e. Now, from the equalityxv = uye, we can deduce thaty = x⁰⁰x⁰ since

|x|=|y|. Let us consider now the equalityvy=sxu: we get(sxx)ⁿsxx⁰ey=sx(xsx)ⁿxsx⁰ and it followsx⁰ey=xsx⁰. Sincex=x⁰x⁰⁰andy=x⁰⁰x⁰, we finally obtainex⁰⁰=x⁰⁰s.

u t Example 3.2. LetS={a²ca⁵c−!a⁴ca³ca²}. In this case,s=e=a²and from the property (iii) of the lemma 3.1,x=a²ca. Then we are in the case 3 of the proposition 3.2 withn= 0,x⁰=a²c andx⁰⁰=a.

The condition 2 of the proposition 3.2 impliesx⁰ 2 LF(sx⁰)sox⁰ 2 B^⇤. It follows that|u|c = 2n|x|c is even. Similarly, the condition 3 of the proposition 3.2 implies that x⁰⁰ 2 B^⇤. Then

|x⁰|_c=|x|_cand|u|_c= 2n|x|_cis even. We obtain:

Corollary 3.2. LetS = {u −! v} such that|u|c = |v|c is odd. ThenS is nonterminating iff u2F(v).

4. The construction of Loop( S ) with S in S

The aim of this section is to prove that for any systemS= {u−!v}inSgrid, the setLoop(S)is a constructible regular language. Clearly, ifu=vthenLoop(S) =A^⇤uA^⇤and in the rest of this section, we suppose thatu6=v.

We shall use the following lemma:

Lemma 4.1. LetS = {u−! v}inSgrid that is nonterminating withu 62 F(v). Ifz0cw0cz₀⁰ −!⁴ z4cw4cz₄⁰ withz0, z⁰₀, z4, z⁰₄2B^⇤and|w0|=|w4|thenz0cw0cz₀⁰ 2B^⇤(xu+uy)B^⇤.

Proof: We consider the derivationz₀cw₀cz⁰₀−!z₁cw₁cz₁⁰ −!z₂cw₂cz⁰₂−!z₃cw₃cz₃⁰ −!z₄cw₄cz₄⁰ withzi, z_i⁰ 2 B^⇤. Observe first that|cw₀c|c > |u|cotherwise we can not apply a derivation step on z1cw1cz⁰₁sinceu 62 F(v). That implies that the first and the last occurrences of the letterc can not be both involved in a same step of the derivation. From the proposition 3.2,|wi+1|=|wi| if the derivation involves the first or the last occurrence of the letterc, else|wi+1| >|wi|. From the hypothesis, we are in the first case and8i,|wi| =|w0|. We can also suppose thatz0 = land z₀⁰ =r. Thenu2LF(lcw0cr)[RF(lcw0cr)and we consider the caselcw0cr=uy0−!vy0. Since su62LF(v), we necessarily haveu2RF(vy0). It follows thatvy0=sx1u−!sx1v=suy1e−! svy1e=s²x2ue. We can verify that|y0|=|y1|andy0, y12RF(u). Finallyy0=y1and we get vy₀=sx₁uandx₁v=uy₀e. Now, from the property (ii) of the corollary 3.1 and from the property (iii) of the lemma 3.1, we gety₀=y,x₁ =xthat implylcw₀cr =uy. Observe that ifz₀6=lor z₀⁰ 6=r we still havez₀cw₀cz₀⁰ 2 B^⇤uyB^⇤. Ifu2 RF(lcw0cr), thenz₀cw₀cz⁰₀2 B^⇤xuB^⇤that

proves the lemma. ut

Let us denote, for any wordw2A^⇤cA^⇤,int(w) = (B^⇤)⁻¹w(B^⇤)⁻¹\cA^⇤\A^⇤c.

Lemma 4.2. LetSbe nonterminating withu6=v.

1. Ifw−!^⇤ w⁰withu2F(v)or|w|c>|u|c, then|int(w⁰)| ≥ |int(w)|, 2. ifw−⁺!w⁰!−^⇤ zwz⁰then|int(w)|=|int(w⁰)|

Proof: For the property 1, it is sufficient to prove thatw−!w⁰implies|int(w⁰)| ≥ |int(w)|. That is clear ifu2 F(v)since|v|>|u|andint(v) = int(u). Ifu 62F(v)then the property (ii) of the corollary 3.1 is satisfied. Since|w|c>|u|cwe have to consider three cases:

1. w=zuw⁰⁰andw⁰=zvw⁰⁰withz2B^⇤,w⁰⁰2(A^⇤\B^⇤), 2. w=w⁰⁰uzandw⁰=w⁰⁰vzwithz2B^⇤,w⁰⁰2(A^⇤\B^⇤), 3. w=w⁰⁰uw⁰⁰⁰withw⁰⁰, w⁰⁰⁰2(A^⇤\B^⇤).

Let us consider the case 1. From the property (iv) of the lemma 3.1, |int(u)| = |int(v)|+|e|, moreoveru2A^⇤crandv2A^⇤creso we get|int(w⁰)|=|int(w)|. The second case can be proved similarly and for the third case, we clearly obtain |int(w⁰)| > |int(w)|since|u| < |v|. For the property 2, we haveint(zwz⁰) = int(w)andw2S1. So, ifu62F(v),|w|c>|u|cfrom the lemma 4.1. Then property 1 yields:|int(w)|  |int(w⁰)|  |int(zwz⁰)|=|int(w)|. ut

Lemma 4.3. If the property (ii) of the corollary 3.1 is satisfied andu6=vthen 1. v62F(xuB^⇤),

2. v62F(B^⇤uy), 3. xu62A^⇤uA⁺, 4. uy62A⁺uA^⇤.

Proof: From the property (iii) of the lemma 3.1,u=gywheregis the longest word inLF(u)\ RF(v). Assume first thatv2F(xu). Sincev2slcA^⇤andxu2lcA^⇤,v62LF(xu). Thenu=g⁰y⁰ withg⁰2LF(u)\RF(v)and|y⁰|<|x|=|y|. Thus|g|<|g⁰|, a contradiction. Assume now that v2F(xuB^⇤). Sincev2A^⇤creandu2A^⇤cr, we getv2RF(xue)andv=ue, a contradiction.

Assume now thatxu2A^⇤uA⁺. Consider the equalityvy=sxu. Ifu2LF(xu)thenv=su, a contradiction. Otherwise, u= g⁰y⁰ withg⁰ 2 LF(u)\RF(v)and|y⁰| < |y|. Thus|g| <|g⁰|, a

contradiction. Properties 2 and 4 can be proved similarly. ut

We obtain as a consequence:

Corollary 4.1. If the property (ii) of the corollary 3.1 is satisfied andu 6= vthenS⁻¹(B^⇤(xu+ uy)B^⇤) =B^⇤(xu+uy)B^⇤.

Proof: Let us prove thatS⁻¹(B^⇤xuB^⇤)✓B^⇤(xu+uy)B^⇤. It is clearly sufficient to consider a single step of derivation. Letw1uw2 −!w1vw2=w⁰₁xuw₂⁰ withw1, w22A^⇤andw⁰₁, w⁰₂2B^⇤. From the lemma 4.3,v62 F(xuB^⇤)therefore there existt2 B^⇤andz2A^⇤such thatw⁰₁=w1t, w₂=zw⁰₂andtxu=vz. Sincev2slcA^⇤andx2lcA^⇤it followst=sthat impliesz=yand w₁uw₂=w₁uyw⁰₂2B^⇤uyB^⇤. We can prove similarly thatS⁻¹(B^⇤uyB^⇤)✓B^⇤(xu+uy)B^⇤.

u t

Proposition 4.1.

1. Ifv=suwiths2B⁺thenLoop(S) = RF(s^⇤)uA^⇤, 2. Ifv=uewithe2B⁺thenLoop(S) =A^⇤uLF(e^⇤), 3. Ifv=suewiths, e2B⁺thenLoop(S) = RF(s^⇤)uLF(e^⇤).

Proof: LetS={u−!su}withs2B⁺andw2Loop(S). Thenw=αuβ−!αsuβ−!^⇤ zαuβz⁰ withz, z⁰ 2 B^⇤. From the lemma 4.2, we have|int(αsuβ)| = |int(αuβ)|and, sinces 2 B⁺, we get α 2 B^⇤. Moreover we obtain by induction on the lengthn > 0of the derivation that zαuβz⁰ = αsⁿuβ. It followsαsⁿ = zαandα 2 RF(αs^i⇥n)for any integeri >0. Therefore α2RF(s^⇤)andLoop(S)✓RF(s^⇤)uA^⇤. Since the converse inclusion is immediate we finally get Loop(S) = RF(s^⇤)uA^⇤. The other cases can be proved similarly. ut Lemma 4.4. If the property (ii) of the corollary 3.1 is satisfied andu6=vthen for any wordα, β2 B^⇤

1. S(αxuβ) ={αsⁿxueⁿβ|n≥0} [ {αsⁿ⁻¹uyeⁿβ|n >0}, 2. S(αuyβ) ={αsⁿuyeⁿβ|n≥0} [ {αsⁿ⁻¹xueⁿβ|n >0}.

Proof: We prove thatS(αxuβ) ={αsⁿxueⁿβ |n ≥ 0} [ {αsⁿ⁻¹uyeⁿβ | n >0}: from the lemma 4.3, xu 62 A^⇤uA⁺anduy 62 A⁺uA^⇤ therefore the first steps of a derivation fromw are w =αxuβ −!αxvβ =αuyeβ−!αvyeβ =αsxueβand the property is proved by induction on the length of the derivation. The second property of this lemma can be proved similarly. ut Proposition 4.2. If the property (ii) of the corollary 3.1 is satisfied andu6=vthen

Loop(S) = RF(s^⇤)(xu+uy)LF(e^⇤)

Proof: Letw 2Loop(S)thenw −!ⁿ zwz⁰withz, z⁰ 2B^⇤andn > 0. We can supposen ≥ 4 and there exists a derivationw −! w₁ −! w₂ −! w₃ !− w₄ −!^⇤ zwz⁰withz, z⁰ 2 B^⇤. From the lemma 4.2, we get |int(w4)| = |int(w)|and, from the lemma 4.1, we getw = αxuβ orw = αuyβwithα, β 2 B^⇤. Let us suppose thatw = αxuβthenS(w) = {αsⁿxueⁿβ | n ≥ 0} [ {αsⁿ⁻¹uyeⁿβ | n > 0}from the lemma 4.4. Moreover, from the lemma 4.3 we deduce that xu 6= uysinceu 62 LF(xu). On the other hand, from the property (iii) of the lemma 3.1, we have xu 2 lcA^⇤cr and uy 2 lcA^⇤cr then it follows that int(uy) 6= int(xu) = int(w). That implies that ifw =αxuβ −⁺! zwz⁰ = zαxuβz⁰we must havezαxuβz⁰ =αsⁿxueⁿβfor some n >0thereforeα2RF(s^⇤)andβ2LF(e^⇤). The casew=αuyβis symmetric and we finally get Loop(S)✓RF(s^⇤)(xu+uy)LF(e^⇤). We can easily verify the converse inclusion sincexu−!^⇤ sxue

anduy−!^⇤ suye. ut

We directly obtain from the two previous propositions:

Corollary 4.2. IfS2 S_gridthenLoop(S)is a constructible regular language.

The conditionS2 Sgridin the corollary 4.2 is necessary as shown by the following example:

Example 4.1. ConsiderS = {c −! caca}. We shall show thatLoop(S)\c(ca)⁺a^⇤ = L = c{(ca)ⁿa^p |n > p≥0}. Sincec −!^⇤ (ca)ⁿ⁻¹caⁿ⁻¹, it follows that forn > p≥0,c(ca)ⁿa^p −!^⇤ c(ca)ⁿa^pa^n−1−p(ca)ⁿ⁻¹a^pandL✓Loop(S). For the reverse inclusion, remark that(ca+a)^⇤isS- closed. So, ifcw=c(ca)ⁿa^p−!^⇤ αcwβfor someα, β2A^⇤, thenα=εandw−⁺!wβwith|β|c>

0. Thusβ=β⁰a^p,(ca)ⁿ−!^⇤ (ca)ⁿa^pβ⁰2LF(S(c))andp < nsince8z2LF(S(c)),2|z|c>|z|a.

5. Testing membership in S

with S in S

The aim of this section is to show that the termination problem of a systemS2 Sgridis decidable.

Recall that this problem is, givenS2 Sgridand a wordw2A^⇤to know whetherw2S1. Clearly, for a systemS= {u−!v}withu2 F(v),S1 =A^⇤uA^⇤and we conjecture that for any system S2 Sgridthe setS1is a constructible regular language. This is true when|u|c =|v|c = 2as it is proved in the next section but here we must find another way in order to prove thatS1is a recursive language for anyS 2 Sgrid. More precisely, we show that for anyS 2 Sgridand any wordw, the setS(w)is a constructible bounded context-free language.

Definition 5.1. ([9]) A languageL✓A^⇤is said to beboundedif there exist wordsw1, . . . , wn2A^⇤ such thatL✓w^⇤₁. . . w^⇤_n.

As a matter of fact, it would be sufficient to prove that for anyS 2 Sgridand any wordw, the set S(w)is a constructible context-free language in order to decide whetherS(w)is finite. Nevertheless bounded languages have nice structural properties that will also permit to solve in this section the common descendant problem for anyS2 Sgridthat is to decide, given two wordswandw⁰, whether S(w)\S(w⁰)is not empty. We also observe that the imageS(w)of a wordwby an arbitrary one-rule semi-Thue systemSneed not be context-free([12]).

Following Alfons Geser in [4], we first establish that if a derivation is applied to a word long enough, this word can then be factorized so that the derivation applies independently on each factor of the word. Let us denoteN⁰= max{|cui| |i < k}andN = max{N⁰,|l|,|r|}.

Lemma 5.1. Letw=z⁰cm⁰m⁰⁰cz⁰⁰withm⁰, m⁰⁰ 2B^⇤andz⁰, z⁰⁰ 2A^⇤. If|m⁰| ≥N and|m⁰⁰| ≥ N thenS(w) =S(z⁰cm⁰)S(m⁰⁰cz⁰⁰).

Proof: ClearlyS(z⁰cm⁰)S(m⁰⁰cz⁰⁰)✓S(w). Conversely, letw⁰2S(w), we prove by induction on the lengthnof the derivation fromwtow⁰thatw⁰ 2S(z⁰cm⁰)S(m⁰⁰cz⁰⁰). The base casew =w⁰ is immediate. Otherwisew −!w⁰⁰ −−−!ⁿ⁻¹ w⁰. Since|m⁰m⁰⁰| > N, the first step of the derivation applies on an occurrence ofuthat can only appear inz⁰cm⁰or inm⁰⁰cz⁰⁰. Suppose it is inz⁰cm⁰, thenz⁰cm⁰−!z₁⁰cm⁰₁withm⁰₁2B^⇤,|m₁⁰| ≥ |m⁰|andw⁰⁰=z₁⁰cm⁰₁m⁰⁰cz⁰⁰. Now, we can apply the induction hypothesis onw⁰⁰−−−!ⁿ⁻¹ w⁰:w⁰2S(z₁⁰cm⁰₁)S(m⁰⁰cz⁰⁰)✓S(z⁰cm⁰)S(m⁰⁰cz⁰⁰). The other

case can be proved similarly. ut

We directly deduce:

Lemma 5.2. Letzcwcz⁰withz, z⁰ 2B^⇤andw 2A^⇤. If|cwc| ≥2⇥(|w|c+ 2)⇥N then there exist two constructible wordsw⁰, w⁰⁰ 2 (A^⇤\B^⇤)such thatzcwcz⁰ = w⁰w⁰⁰ andS(zcwcz⁰) = S(w⁰)S(w⁰⁰).

Then we can state:

Proposition 5.1. LetS2 S_gridsatisfying the property (ii) of the corollary 3.1.

1. For any wordw2A^⇤, the setS(w)is a bounded context-free language.

2. S1=S⁻¹(A^⇤(xu+uy)A^⇤)=S⁻¹(A^⇤Loop(S)A^⇤).

Proof: The proof is based on an induction over|w|c. For the property 2, we have only to prove S1✓S⁻¹(A^⇤(xu+uy)A^⇤)since the inclusionsS⁻¹(A^⇤(xu+uy)A^⇤)✓S⁻¹(A^⇤Loop(S)A^⇤)✓ S1are clear.

• If|w|c<|xu|cthenS(w)is finite (therefore bounded context-free) from the lemma 4.1,

• If|w|c=|xu|cthen

– ifw62 B^⇤(xu+uy)B^⇤thenw 62S⁻¹(B^⇤(xu+uy)B^⇤)from the corollary 4.1 and it follows thatS(w)is finite from the lemma 4.1,

– ifw 2 B^⇤(xu+uy)B^⇤then, from the lemma 4.4,S(w)is a (constructible) bounded context-free language andwis inS⁻¹(A^⇤(xu+uy)A^⇤)

• If|w|c>|xu|c, we make a new induction onKw= 2⇥(|w|c)⇥N− |int(w)|. IfKw 0 then it follows from the lemma 5.2 that there exist two constructible wordsw⁰, w⁰⁰2(A^⇤\B^⇤) such thatw=w⁰w⁰⁰andS(w) = S(w⁰)S(w⁰⁰). As|w⁰|c <|w|cand|w⁰⁰|c <|w|c, we can apply the induction hypothesis and it follows thatS(w)is a constructible bounded context-free language. IfKw>0, let us denoteS₁={w⁰2A^⇤|w−−!^<4 w⁰}andS₂={w⁰2A^⇤|w−!⁴ w⁰}. ClearlyS(w) =S1[(S

w⁰2S₂S(w⁰)). The family of bounded languages is closed by finite union, consequently it remains to prove that for any wordw⁰inS2,S(w⁰)is a bounded context-free language. Sincew 62 B^⇤(xu+uy)B^⇤it follows from the lemma 4.1 and the lemma 4.2 that, for any wordw⁰ 2 S2,|int(w⁰)| >|int(w)|. ThenKw⁰ < Kw and, by the induction hypothesis,S(w⁰)is a bounded context-free language, moreover ifw⁰ 2 S1then w⁰2S⁻¹(A^⇤(xu+uy)A^⇤)and it follows thatw2S⁻¹(A^⇤(xu+uy)A^⇤)which concludes the proof of the proposition.

u t

Whenu2F(v), a similar proof can be used to show that the property 1 is also satisfied in this case.

Then we get as a corollary:

Corollary 5.1. The termination problem is decidable for any systemS2 Sgrid.

Since it is proved in [8] that the non-emptiness of the intersection of two bounded context-free languages is decidable, we also obtain:

Corollary 5.2. The common descendant problem is decidable for any systemS2 Sgrid.

6. The special case | u |

= | v |

= 2

It has been shown in the section 4 thatLoop(S)is regular forSinSgridand that this result does not hold for arbitrary one-rule semi-Thue system. The status ofS1is different. G´eraud S´enizergues has proved in [20] thatS1is regular whenu2a^⇤b^⇤, but the regularity ofS1is an open problem for arbitrary one-rule semi-Thue system. Note that the regularity ofS1does not hold for two-rules grid semi-Thue system as shown by the exampleS={aca−!c, cc−!cca}since it is easy to see that S1 =A^⇤cA^⇤cA^⇤cA^⇤[ {aⁱca^pca^q |pi+q}. We conjecture thatS1is a constructible regular language for anySinSgrid. As we are unable to prove this fact, we give an argument in favour by proving it in the simpler case|u|c =|v|c= 2. The following proof shows also that the structure of S1can be involved and that the proof in the general caseSinSgridcould be tricky.

Clearly the question arises only whenSsatisfies the property (ii) of the corollary 3.1 since, when u2F(v),S1=A^⇤uA^⇤. Then we consider in the following a nonterminating systemS={u−!v}

withu=lcu₁cr,v=slcv₁cre,xv=uyeandvy=sxuforu₁, v₁, s, e2B^⇤andx, y2A^⇤. Then there exist wordsu⁰₁andu⁰⁰₁such thatu₁=u⁰₁sl=reu⁰⁰₁, v₁=u⁰₁l=ru⁰⁰₁, x=lcu⁰₁andy=u⁰⁰₁cr Let us denoteS1^min=S1\(A S1[S1A)and, for any wordw2A^⇤,Lw=S⁻¹(A^⇤w)\(S1[ A S⁻¹(A^⇤w))andRw=S⁻¹(wA^⇤)\(S1[S⁻¹(wA^⇤)A).

Lemma 6.1. S^min1 ✓Lxlcu1cRr[Llcu1cRry.

Proof: Let us considerw₀2S1^minand letnbe the length of the shortest derivation fromw₀to any word inA^⇤(xu+uy)A^⇤. Let us suppose that there exists a derivation fromw₀tow_n2A^⇤(xu)A^⇤. Thenw₀−!^⇤ w₁−!^⇤ wi

−!⇤ wnandwn=pncqncrnwithpn2A^⇤xl,qn=u₁andrn2rA^⇤. For any i2[0, n], let us consider the factorizationwi=picqicriwith|pi|c=|pn|candqi2B^⇤. We claim thatq0= u1, else letjbe the greatest index such thatqj 6=u1. Thenwj+1 =pj+1cu1crj+1and p_j+1−!^⇤ p_n, r_j+1−!^⇤ r_n. This will lead to the following contradiction: there exists a derivation from w₀to a word inA^⇤(xu+uy)A^⇤whose length is strictly less thann. It follows that for anyi,q_i=u₁ andp₀−!^⇤ p_n,r₀−!^⇤ r_nthereforep₀2S⁻¹(A^⇤xl), r02 S⁻¹(rA^⇤). Sincew₀2S1^min, it follows p₀2Lxl, r₀2Rrandw₀2Lxlcu₁cRr. Let us distinguish two cases for the stepwj!wj+1:

1. qj=u⁰₁l, rj=u1crzandrj+1=v1crez.

Thenpncu⁰₁2A^⇤lcu⁰₁=A^⇤xandw⁰=pncqjcrj=pncu₁⁰lcu1crz2A^⇤xuA^⇤. Clearly, the length of the derivationw0

−!⇤ wj=pjcqjcrj

−⇤

!w⁰=pncqjcrjis strictly smaller thann.

2. qj=ru⁰⁰₁, pj=zlcu1andpj+1=zslcv1.

Thenu⁰⁰₁cr_n2u₁⁰⁰crA^⇤=yA^⇤andw⁰=p_jcq_jcr_n=zlcu₁cru⁰⁰₁cr_n2A^⇤uyA^⇤. Clearly, the length of the derivationw0

−!⇤ wj=pjcqjcrj

−⇤

!w⁰=pjcqjcrnis strictly smaller thann.

u t

Lemma 6.2. S⁻¹(A^⇤u)\A^⇤cu1cr✓S1.

Proof: Letw2S⁻¹(A^⇤u)\A^⇤cu₁cr. Then there exists a derivation:w₀!w₁−!^⇤ w_i −!^⇤ w_n2 A^⇤u. Letjbe the greatest index such thatw_j 62 A^⇤cu₁cr. Thenw_j = p_juu⁰⁰₁cr 2 A^⇤uy, and it

followsw2S⁻¹(A^⇤uy)✓S1. ut

Let us denoteF = {cv1} [ {cu1crz | v1 =rez}. Remark thatF is finite andS⁻¹(A^⇤l)F ✓ S⁻¹(A^⇤xl).

Lemma 6.3. Lxl✓LlF.

Proof: Letw2Lxl. Thenw−!^⇤ zxl=zlcv1. Let us consider two cases:

1. Ifw=w⁰cv₁, we can deduce thatw⁰ −!^⇤ zl, sow⁰ 2S⁻¹(A^⇤l). Sincew=w⁰cv₁62S1, it followsw⁰ 62 S1and sincew 62AS⁻¹(A^⇤xl), it followsw⁰ 62AS⁻¹(A^⇤l). Thenw⁰ 2 L_l andw=w⁰cv₁2LlF.

2. Elsew=w⁰crzwithz2B^⇤andrz6=v1. Thenw−!^⇤ puz!pvz −!^⇤ p⁰cv1. We show that rez=v1. Indeed, if it is not the case,pslcv1cr2S⁻¹(A^⇤u)\A^⇤cu1cr✓S1which implies pvz 2 S1andw 2 S1. Sincew =w⁰crz −!^⇤ puz, it followsw⁰cr 2 S⁻¹(A^⇤u)\S1 ✓ A^⇤cu1cr, sow⁰crz =w⁰⁰cu1crz −!^⇤ pvz 2 A^⇤xl. Finallyw⁰⁰ 2 S⁻¹(A^⇤l), w⁰⁰ 62S1and w⁰⁰62AS⁻¹(A^⇤l)which impliesw⁰⁰2Llandw2LlF.

u t

Let us denote H = {z | l 2 RF(rez)\RF(rz)}. Remark thatHis finite andS⁻¹(A^⇤u)H ✓ S⁻¹(A^⇤l).

Lemma 6.4. Ll✓l+LuH.

Proof: Letw2 Ll. Ifw2 A^⇤lthenw=lotherwisew =w⁰crzwithl 62RF(rz). Then there exists a derivationw−!^⇤ puz−!pvzwithpvz2A^⇤crez\S⁻¹(A^⇤l). Suppose thatl62RF(rez), thenpslcv1cr 2 S⁻¹(A^⇤u)\A^⇤cu1crand it follows from the lemma 6.2 thatpvz, and thus also w, is inS1. This contradiction impliesl 2 RF(rez)thereforez 2 H andw⁰cr 2 S⁻¹(A^⇤u).

Moreoverw62S1andw62AS⁻¹(A^⇤l)imply thatw⁰cr62 S1andw⁰cr62AS⁻¹(A^⇤u). Finally

w⁰cr2Luandw2LuH. ut

Lemma 6.5. L_u✓L_lcu₁cr.

Proof: Letw 2 Lu ✓ S⁻¹(A^⇤u)\S1 ✓ A^⇤cu1cr. Then w = w⁰cu1cr −!^⇤ w⁰⁰cu1cr with w⁰⁰ 2 A^⇤l. It follows thatw⁰ 2 S⁻¹(A^⇤l). Moreoverw⁰ 62S1andw⁰ 62 AS⁻¹(A^⇤l)that imply

w⁰2Llandw2Llcu1cr. ut

Since, clearly,S⁻¹(A^⇤l)cu1crH✓S⁻¹(A^⇤l), it follows from the lemmata 6.4 and 6.5 thatLl ✓ l(cu1crH)^⇤✓S⁻¹(A^⇤l). Then, from the lemma 6.3, we getLxl✓l(cu1crH)^⇤F ✓S⁻¹(A^⇤l)F ✓ S⁻¹(A^⇤xl):

Lemma 6.6. Ll✓l(cu1crH)^⇤✓S⁻¹(A^⇤l)andLxl✓l(cu1crH)^⇤F ✓S⁻¹(A^⇤xl).

Symmetrically, denotingF⁰={v₁c} [ {zlcu₁c|v₁=zsl}andH⁰={z|r2LF(zsl)\LF(zl)}, we get: