aa r X i v : . [ m a t h . G R ] O c t A NOTE ON THE DEFINITION OFSMALL OVERLAP MONOIDS
Mark KambitesSchool of Mathematics, University of Manchester,Manchester M13 9PL, England.
Abstract.
Small overlap conditions are simple and natural combina-torial conditions on semigroup and monoid presentations, which serve tolimit the complexity of derivation sequences between equivalent wordsin the generators. They were introduced by J. H. Remmers, and morerecently have been extensively studied by the present author. However,the definition of small overlap conditions hitherto used by the authorwas slightly more restrictive than that introduced by Remmers; thisnote eliminates this discrepancy by extending the recent methods andresults of the author to apply to Remmers’ small overlap monoids in fullgenerality.
Small overlap conditions are simple and natural combinatorial conditionson semigroup and monoid presentations, which serve to limit the complexityof derivation sequences between equivalent words in the generators. Intro-duced by J. H. Remmers [2, 7, 8], and more recently studied by the presentauthor [3, 4, 5], they are the natural semigroup-theoretic analogue of thesmall cancellation conditions widely used in combinatorial group theory [6].The definitions of small overlap conditions originally introduced by Rem-mers are slightly more general than those used by the present author. Theaims of this note are to clarify this distinction, and then to extend the meth-ods and results introduced in [4, 5] to the full generality of small overlapmonoids as studied by Remmers.In addition to this introduction, this article comprises three sections. InSection 1 we briefly recall the definitions of small overlap conditions, andalso discuss the distinction between Remmers’ and the author’s definitions.In Section 2 we show how to extend the key technical results from [4], fromthe slightly restricted setting considered there to Remmers’ small overlapconditions in their more general form. Finally, Section 3 applies the resultsof the previous section to extend the main results of [4, 5] to the more generalcase.The proofs for certain of the results in this paper are very similar (in somecases identical) to arguments used in previous papers [4, 5]. In the inter-ests of brevity we refrain from repeating these, instead providing detailedreferences. Hence, while the results of this paper may be read in isolation,
Mathematics Subject Classification.
Key words and phrases. monoid, semigroup, word problem, finite presentation, smalloverlap, small cancellation. the reader wishing to fully understand the proofs is advised to read it inconjunction with [4, 5].1.
Small Overlap Monoids
We assume familiarity with basic notions of combinatorial semigroup the-ory, including free semigroups and monoids, and semigroup and monoid pre-sentations. Except where stated otherwise, we assume we have a fixed finitepresentation for a monoid (or semigroup, the difference being unimportant).Words are assumed to be drawn from the free monoid on the generating al-phabet unless otherwise stated. We write u = v to indicate that two wordsare equal in the free monoid or semigroup, and u ≡ v to indicate that theyrepresent the same element of the monoid or semigroup presented. We saythat a word p is a possible prefix of u if there exists a (possibly empty) word w with pw ≡ u , that is, if the element represented by u lies in the right idealgenerated by the element represented by p . The empty word is denoted ǫ .A relation word is a word which occurs as one side of a relation in thepresentation. A piece is a word in the generators which occurs as a factor insides of two distinct relation words, or in two different (possibly overlapping)places within one side of a relation word. Note that this definition differsslightly from that used in [4, 5] in the presence of the word “distinct”; weshall discuss the significance of this shortly. By convention, the empty wordis always a piece. We say that a presentation is weakly C ( n ), where n is apositive integer, if no relation word can be written as the product of strictlyfewer than n pieces. Thus for each n , being weakly C ( n + 1) is a strongercondition than being weakly C ( n ).In [4, 5] we used a slightly more general definition of a piece, followingthrough with which led to slightly more restrictive conditions C ( n ); theauthor is grateful to Uri Weiss for pointing out this discrepancy. Specifically,in [4, 5] we defined a piece to be a word which occurs more than once as afactor of words in the sequence of relation words. Under this definition, if thesame relation word appears twice in a presentation then it is considered to bea piece, and so the presentation fails to satisfy C (2). By contrast, Remmersdefined a piece to be a word which appears more than once as a factorof words in the set of relation words. The effect of this is that Remmers’definition permits C (2) (and higher) presentations to have relations of, forexample, the form ( u, v ) and ( u, v ) with v = v . (Equivalently, one couldchoose to define a piece in terms of the sequence of relation words but permit“ n -ary” relations of the form ( u, v , v ), to be interpreted as equivalent torelations ( u, v ) and ( u, v )). In this paper, we say that a presentation is strongly C ( n ) if it is weakly C ( n ) and has no repeated relation words, thatis, if it satisfies the condition which was called C(n) in [4, 5].In fact it transpires that the weakly C ( n ) conditions still suffice to es-tablish the main methods and results of [4, 5]. However, this fact is ratherobscured by the technical details and notation in [4, 5]. In particular, for arelation word R we defined R to be the (necessarily unique) word such that R = R or R = R is a relation in the presentation. The extensive use of thisnotation makes it difficult to convince oneself that the arguments in [4, 5] do NOTE ON SMALL OVERLAP MONOIDS 3 indeed apply in the more general case, so the aim of this paper is to providefull proofs of the results of those papers in the more general setting.For each relation word R , let X R and Z R denote respectively the longestprefix of R which is a piece, and the longest suffix of R which is a piece. Ifthe presentation is weakly C (3) then R cannot be written as a product of twopieces, so this prefix and suffix cannot meet; thus, R admits a factorisation X R Y R Z R for some non-empty word Y R . If moreover the presentation isweakly C (4), then the relation word R cannot be written as a product ofthree pieces, so Y R is not a piece. The converse also holds: a weakly C (3)presentation such that no Y R is a piece is a weakly C (4) presentation. Wecall X R , Y R and Z R the maximal piece prefix , the middle word and the maximal piece suffix respectively of R .Assuming now that the presentation is weakly C (3), we shall use theletters X , Y and Z (sometimes with adornments or subscripts) exclusivelyto represent maximal piece prefixes, middle words and maximal piece suffixesrespectively of relation words; two such letters with the same subscript oradornment (or with none) will be assumed to stand for the appropriatefactors of the same relation word.We say that a relation word R is a complement of a relation R if thereare relation words R = R , R , . . . , R n = R such that either ( R i , R i +1 ) or( R i +1 , R i ) is a relation in the presentation for 1 ≤ i < n . We say that R is a proper complement of R if, in addition, R = R . Abusing notationand terminology slightly, if R = X R Y R Z R and R = X R Y R Z R then we write X R = X R , X R Y R = X R Y R and so forth. We say that X R is a complementof X R , and X R Y R is a complement of X R Y R .A relation prefix of a word is a prefix which admits a (necessarily unique,as a consequence of the small overlap condition) factorisation of the form aXY where X and Y are the maximal piece prefix and middle word re-spectively of some relation word XY Z . An overlap prefix (of length n ) ofa word u is a relation prefix which admits an (again necessarily unique)factorisation of the form bX Y ′ X Y ′ . . . X n Y n where • n ≥ • bX Y ′ X Y ′ . . . X n Y n has no factor of the form X Y , where X and Y are the maximal piece prefix and middle word respectively of somerelation word, beginning before the end of the prefix b ; • for each 1 ≤ i ≤ n , R i = X i Y i Z i is a relation word with X i and Z i the maximal piece prefix and suffix respectively; and • for each 1 ≤ i < n , Y ′ i is a proper, non-empty prefix of Y i .Notice that if a word has a relation prefix, then the shortest such must bean overlap prefix. A relation prefix aXY of a word u is called clean if u does not have a prefix aXY ′ X Y where X and Y are the maximal piece prefix and middle word respectivelyof some relation word, and Y ′ is a proper, non-empty prefix of Y . As in [4],clean overlap prefixes will play a crucial role in what follows.If u is a word and p is a piece, we say that u is p -active if pu has a relationprefix aXY with | a | < | p | , and p -inactive otherwise. A NOTE ON SMALL OVERLAP MONOIDS Technical Results
In this section we show how some technical results and methods from[4] concerning strongly C (4) monoids can be extended to cover weakly C (4)monoids. We assume throughout initially a fixed monoid presentation whichis weakly C (4). The following three foundational statements are completelyunaffected by our revised definitions, and can still be proved exactly as in[4]. Proposition 1.
Let aX Y ′ X Y ′ . . . X n Y n be an overlap prefix of some word.Then this prefix contains no relation word as a factor, except possibly thesuffix X n Y n in the case that Z n = ǫ . Proposition 2.
Let u be a word. Every overlap prefix of u is contained ina clean overlap prefix of u . Corollary 1.
If a word u has no clean overlap prefix, then it contains norelation word as a factor, and so if u ≡ v then u = v . The following lemma is essentially a restatement of [4, Lemma 1] using ournew notation. The proof is essentially the same as in [4], with the additionof an obvious inductive argument to allow for the fact that several rewritesmay be needed to obtain
XY Z from
XY Z . Lemma 1.
Suppose u = wXY Zu ′ with wXY a clean overlap prefix and XY Z is a complement of
XY Z . Then wXY is a clean overlap prefix of wXY Zu ′ . From now on, we shall assume that our presentation is weakly C (4). Weare now ready to prove our first main technical result, which is an analogue of[4, Lemma 2], and is fundamental to our approach to weakly C (4) monoids. Lemma 2.
Suppose a word u has clean overlap prefix wXY . If u ≡ v then v has overlap prefix wXY for some complement XY Z of XY Z , andno relation word occurring as a factor of v overlaps this prefix, unless it is XY Z in the obvious place.Proof.
Since wXY is an overlap prefix of u , it has by definition a factorisa-tion wXY = aX Y ′ . . . X n Y ′ n XY for some n ≥
0. We use this fact to prove the claim by induction on thelength r of a rewrite sequence (using the defining relations) from u to v .In the case r = 0, we have u = v , so v certainly has (clean) overlap prefix wXY . By Proposition 1, no relation word factor can occur entirely withinthis prefix, unless it is the suffix XY and Z = ǫ . If a relation word factorof v overlaps the end of the given overlap prefix and entirely contains XY then, since XY is not a piece, that relation word must clearly be XY Z .Finally, a relation word cannot overlap the end of the given overlap prefixbut not contain the suffix XY , since this would clearly contradict either thefact that the given overlap prefix is clean, or the fact that Y is not a piece.Suppose now for induction that the lemma holds for all values less than r , and that there is a rewrite sequence from u to v of length r . Let u be NOTE ON SMALL OVERLAP MONOIDS 5 the second term in the sequence, so that u is obtained from u by a singlerewrite using the defining relations, and v from u by r − u which is to be rewritten in order to obtain u , and in particular its position in u . By Proposition 1, this relation wordcannot be contained in the clean overlap prefix wXY , unless it is XY where Z = ǫ .Suppose first that the relation word to be rewritten contains the finalfactor Y of the given clean overlap prefix. (Note that this covers in particularthe case that the relation word is XY and Z = ǫ .) From the weakly C (4)assumption we know that Y is not a piece, so we may deduce that therelation word is XY Z contained in the obvious place. In this case, applyingthe rewrite clearly leaves u with a prefix w ˆ X ˆ Y for some complement ˆ X ˆ Y ˆ Z of XY Z . By Lemma 1, this is a clean overlap prefix. Now v can be obtainedfrom u by r − v has overlap prefix wXY where XY Z is a complement of ˆ X ˆ Y ˆ Z andhence of XY . It follows also that no relation word occurring as a factor of v overlaps this prefix, unless it is XY Z ; this completes the proof in this case.Next, we consider the case in which the relation word factor in u to berewritten does not contain the final factor Y of the clean overlap prefix, butdoes overlap with the end of the clean overlap prefix. Then u has a factor ofthe form ˆ X ˆ Y , where ˆ X is the maximal piece prefix and ˆ Y the middle wordof a relation word, which overlaps XY , beginning after the start of Y . Thisclearly contradicts the assumption that the overlap prefix is clean.Finally, we consider the case in which the relation word factor in u whichis to be rewritten does not overlap the given clean overlap prefix at all. Thenobviously, the given clean overlap prefix of u remains an overlap prefix of u . If this overlap prefix is clean, then a simple application of the inductivehypothesis again suffices to prove that v has the required property.There remains, then, only the case in which the given overlap prefix is nolonger clean in u . Then by definition there exist words ˆ X and ˆ Y , being amaximal piece prefix and middle word respectively of some relation word,such that u has the prefix aX Y ′ . . . X n Y ′ n XY ′ ˆ X ˆ Y for some proper, non-empty prefix Y ′ of Y . Now certainly this is not a prefixof u , since this would contradict the assumption that aX Y ′ . . . X n Y ′ n XY isa clean overlap prefix of u . So we deduce that u can be transformed to u by rewriting a relation word overlapping the final ˆ X ˆ Y . This relationword factor cannot contain the entire of this factor ˆ X ˆ Y , since then it wouldoverlap with the prefix aX Y ′ . . . X n Y n XY , which would again contradictthe assumption that this prefix is a clean overlap prefix of u . Nor can therelation word contain the final factor ˆ Y , since ˆ Y is not a piece. Hence, u must have a prefix aX Y ′ . . . X n − Y ′ n − X n Y ′ n XY ′ ˆ X ˆ Y ′ R for some relation word and proper, non-empty prefix ˆ Y ′ of ˆ Y and somerelation word R . Suppose R = X R Y R Z R where X R and Z R are the maximal A NOTE ON SMALL OVERLAP MONOIDS piece prefix and suffix respectively. Then it is readily verified that aX Y ′ . . . X n − Y ′ n − X n Y ′ n XY ′ ˆ X ˆ Y ′ X R Y R is a clean overlap prefix of u . Indeed, the fact it is an overlap prefix isimmediate, and if it were not clean then some factor of u of the form ˜ X ˜ Y would have to overlap the end of the given prefix; but this factor wouldeither be contained in Y R Z R (contradicting the fact that ˜ X is a maximumpiece prefix of ˜ X ˜ Y ˜ Z ) or would contain a non-empty suffix of Y R followed by Z R (contradicting the fact that Z R is a maximum piece prefix of X R Y R Z R ).Now by the inductive hypothesis, v has prefix aX Y ′ . . . X n − Y ′ n − X n Y ′ n XY ′ ˆ X ˆ Y ′ X R Y R . (1)for some complement X R Y R of X R Y R . But now v has prefix aX Y ′ . . . X n − Y ′ n − X n Y ′ n XY ′ ˆ X ˆ Y ′ which in turn has prefix aX Y ′ . . . X n − Y ′ n − X n Y ′ n XY. (2)Moreover, by Proposition 1, the prefix (1) of v contains no relation word asa factor, unless it is the final factor X R Y R and Z R = ǫ , and it follows easilythat no relation word factor overlaps the prefix (2) of v . (cid:3) The following results are now proved exactly as their analogues in [4].
Corollary 2.
Suppose a word u has (not necessarily clean) overlap pre-fix wXY . If u ≡ v then v has a prefix w and contains no relation wordoverlapping this prefix. Proposition 3.
Suppose a word u has an overlap prefix aXY and that u = aXY u ′′ . Then u ≡ v if and only if v = av ′ where v ′ ≡ XY u ′′ . Proposition 4.
Let u be a word and p a piece. If u is p -inactive then pu ≡ v if and only if v = pw for some w with u ≡ w . Proposition 5.
Let p and p be pieces and suppose u is p -active and p -active. Then p and p have a common non-empty suffix, and if z is theirmaximal common suffix then (i) u is z -active; (ii) p u ≡ v if and only if v = z v ′ where z z = p and v ′ ≡ zu ; and (iii) p u ≡ v if and only if v = z v ′ where z z = p and v ′ ≡ zu . Corollary 3.
Let p and p be pieces. Suppose p u ≡ p v and u is p -active.Then p u ≡ p v . The following is a strengthening of the [4, Corollary 4]
Corollary 4.
Let u and v be words and p , p , . . . , p k be pieces. Supposethere exist words u = u , . . . , u n = v such that for ≤ i < n there exists ≤ j i ≤ k with p j i u i ≡ p j i u i +1 . Then p j u ≡ p j v for some j with ≤ j ≤ k .Proof. Fix u , v and p , . . . , p k , and suppose n is minimal such that a sequence u , . . . , u n with the hypothesized properties exists. Our aim is thus to showthat n ≤
2. Suppose for a contradiction that n > NOTE ON SMALL OVERLAP MONOIDS 7 If u was p j -inactive then by Proposition 4 we would have u ≡ u so that p j u ≡ p j u ≡ p j u which clearly contradicts the minimalityassumption on n . Thus, u is p j -active. But now since p j u ≡ p j u , weapply Corollary 3 to see that p j u ≡ p j u ≡ p j u , which again contradictsthe minimality of n . (cid:3) We now present a lemma which gives a set of mutually exclusive combi-natorial conditions, the disjunction of which is necessary and sufficient fortwo words of a certain form to represent the same element.
Lemma 3.
Suppose u = XY u ′ where XY is a clean overlap prefix of u .Then u ≡ v if and only if one of the following mutually exclusive conditionsholds: (1) u = XY Zu ′′ and v = XY Zv ′′ and Zu ′′ ≡ Zv ′′ for some complement Z of Z ; (2) u = XY u ′ , v = XY v ′ , and Z fails to be a prefix of at least one of u ′ and v ′ , and u ′ ≡ v ′ ; (3) u = XY Zu ′′ , v = XY Zv ′′ for some uniquely determined propercomplement XY Z of XY Z , and ˆ Zu ′′ ≡ ˆ Zv ′′ for some complement ˆ Z of Z ; (4) u = XY u ′ , v = XY Zv ′′ for some uniquely determined proper com-plement XY Z of XY Z but Z is not a prefix of u ′ and u ′ ≡ Zv ′′ ; (5) u = XY Zu ′′ , v = XY v ′ for some uniquely determined proper com-plement XY Z of XY Z , but Z is not a prefix of v ′ and Zu ′′ ≡ v ′ ; (6) u = XY u ′ , v = XY v ′ for some uniquely determined proper comple-ment XY Z of XY Z , Z is not a prefix of u ′ and Z is not a prefix of v ′ , but Z = z z , Z = z z , u ′ = z u ′′ , v ′ = z v ′′ where u ′′ ≡ v ′′ and z is the maximal common suffix of Z and Z , z is non-empty, and z isa possible prefix of u ′′ .Proof. It follows easily from the definitions that no complement of XY isa prefix of another. Hence, v can have at most one of them as a prefix.Thus, conditions (1)-(2) are not consistent with conditions (3)-(6), and theprefixes of v in (3)-(6) are uniquely determined. The mutual exclusivity of(1) and (2) is self-evident from the definitions, and likewise that of (3)-(6).It is easily verified that each of the conditions (1)-(5) imply that u ≡ v .We show next that (6) implies that u ≡ v . Since z is a possible prefix of u ′′ and u ′′ ≡ v ′′ , we may write u ′′ ≡ zx ≡ v ′′ for some word x . Now we have u = XY u ′ = XY z u ′′ ≡ XY z zx = XY Zx ≡ XY Zx = XY z zx ≡ XY z v ′′ = XY v ′ = v. It remains to show that u ≡ v implies that one of the conditions (1)-(6)holds. To this end, suppose u ≡ v ; then there is a rewrite sequence taking u to v . By Lemma 2, every term in this sequence will have prefix which is acomplement of XY , and this prefix can only be modified by the applicationof a relation, both sides of which are complements of XY Z , in the obviousplace. We now prove the claim by case analysis.By Lemma 2, v begins either with XY or with some proper complement XY . Consider first the case in which v begins with XY ; we split this into A NOTE ON SMALL OVERLAP MONOIDS two further cases depending on whether u and v both begin with the fullrelation word XY Z ; these will correspond respectively to conditions (1) and(2) in the statement of the lemma.
Case (1).
Suppose u = XY Zu ′′ and v = XY Zv ′′ . Then clearly there isa rewrite sequence taking u to v which by Lemma 2 can be broken up as: u = XY Zu ′′ = X Y Z u ′′ → ∗ X Y Z u → X Y Z u → ∗ X Y Z u → X Y Z u → ∗ · · · → X n Y n Z n u n → ∗ X n Y n Z n v ′′ = XY Zv ′′ = v where each prefix X i Y i Z i is a complement of XY Z , and none of the steps inthe sequences indicated by → ∗ involves rewriting a relation word overlappingwith the prefix X i Y i . It follows that there are rewrite sequences. Zu ′′ → ∗ Zu , Z u → ∗ Z u , Z u → ∗ Z u , . . . , Z n u n → ∗ Z n v ′′ Now by Corollary 4, we have Z i u ′′ ≡ Z i v ′′ for some 1 ≤ i ≤ n , where Z i isa complement of Z as required to show that condition (1) holds. Case (2).
Suppose now that u = XY u ′ , v = XY v ′ and Z fails to be aprefix of at least one of u ′ and v ′ . We must show that u ′ ≡ v ′ ; suppose for acontradiction that this does not hold. We again consider rewrite sequencesfrom u = XY u ′ to v = XY v ′ . Again using Lemma 2, we see that there iseither (i) such a sequence taking u to v containing no rewrites of relationwords overlapping the prefix XY , or (ii) such a sequence taking u to v whichcan be broken up as: u = XY u ′ = X Y u ′′ → ∗ X Y Z u → X Y Z u → ∗ X Y Z u → X Y Z u → ∗ · · · → X n Y n Z n u n → ∗ X n Y n Z n v ′′ = X n Y n v ′ = XY v ′ = v where each prefix X i Y i Z i is a complement of XY Z , and none of the steps inthe sequences indicated by → ∗ involves rewriting a relation word overlappingwith the prefix X i Y i . In case (i) there is clearly a rewrite sequence taking u ′ to v ′ so that u ′ ≡ v ′ as required. In case (ii), there are rewrite sequences. u ′ → ∗ Zu , Z u → ∗ Z u , Z u → ∗ Z u , . . . , Z n u n = Zu n → ∗ v ′ Now if u ′ does not begin with Z , we can deduce from Proposition 4 that u is Z -active. By Corollary 4, we have ˆ Zu ≡ ˆ Zu n for some complement ˆ Z of Z . Since u is Z -active, Corollary 3 tells us that we also have Zu ≡ Zu n .But now u ′ ≡ Zu ≡ Zu n ≡ v ′ so condition (2) holds. A similar argument applies if v ′ does not begin with Z . Case (3).
Suppose u = XY Zu ′′ and v = XY Zv ′′ . Then u = XY Zu ′′ ≡ v ≡ XY Zv ′′ , so by the same argument as in case (1) we have either Zu ′′ ≡ Zv ′′ or Zu ′′ ≡ Zv ′′ as required to show that condition (3) holds. Case (4).
Suppose u = XY u ′ and v = XY Zv ′′ but Z is not a prefixof u ′ . Then u = XY u ′ ≡ v ≡ XY Zv ′′ . Now applying the same argumentas in case (2) (with XY Zv ′′ in place of v and setting v ′ = Zv ′′ ) we have u ′ ≡ v ′ = Zv ′′ so that condition (4) holds. Case (5).
Suppose u = XY Zu ′′ , v = XY v ′ but Z is not a prefix of v ′ . Then we have XY Zu ′′ ≡ u ≡ v = XY v ′ , and moreover, Lemma 1guarantees that XY is a clean overlap prefix of XY Zu ′′ . Now applying the NOTE ON SMALL OVERLAP MONOIDS 9 same argument as in case (1) (but with
XY Zu ′′ in place of u and setting u ′ = Zu ′′ ) we obtain u ′ ≡ v ′ = Zu ′′ so that condition (5) holds. Case (6).
Suppose u = XY u ′ , v = XY v ′ and that Z is not a prefix of u ′ and Z is not a prefix of v ′ . It follows this time that there is a rewritesequence taking u to v of the form u = XY u ′ = X Y u ′ → ∗ X Y Z u → X Y Z u → ∗ X Y Z u → X Y Z u → ∗ · · · → X n Y n Z n u n → ∗ X n Y n v ′ = XY v ′ = v where once more by Lemma 2 each prefix X i Y i Z i is a complement of XY Z ,and none of the steps in the sequences indicated by → ∗ involves rewritinga relation word overlapping with the prefix X i Y i . Now there are rewritesequences. u ′ → ∗ Zu , Z u → ∗ Z u , Z u → ∗ Z u , . . . , Z n u n = Zu n → ∗ v ′ Notice that, since u ′ does not begin with Z , we may deduce from Proposi-tion 4 that u is Z -active. By Corollary 4, we have ˆ Zu ≡ ˆ Zu n for somecomplement ˆ Z of Z . Now since u is Z -active, Corollary 3 tells us that wealso have Zu ≡ Zu n . But now u ′ ≡ Zu ≡ Zu n where u ′ does not begin with Z , and also v ′ ≡ Zu n were v ′ does not beginwith Z . By applying Proposition 4 twice, we deduce that u n is both Z -activeand Z -active.Let z be the maximal common suffix of Z and Z . Then applying Propo-sition 5 (with p = Z and p = Z ), we see that z is non-empty and • u ′ = z u ′′ where Z = z z and u ′′ ≡ zu n ; and • v ′ = z v ′′ where Z = z z and v ′′ ≡ zu n .But then we have u ′′ ≡ zu n ≡ v ′′ and also z is a possible prefix of u ′′ asrequired to show that condition (6) holds. (cid:3) Lemma 4.
Suppose u = XY u ′ where XY is a clean overlap prefix, andsuppose p is a piece. Then u ≡ v and p is a possible prefix of u if and onlyif one of the following mutually exclusive conditions holds: (1’) u = XY Zu ′′ and v = XY Zv ′′ and Zu ′′ ≡ Zv ′′ for some complement Z of Z , and also p is a prefix of some complement of X ; (2’) u = XY u ′ , v = XY v ′ , and Z fails to be a prefix of at least one of u ′ and v ′ , and u ′ ≡ v ′ , and also either – p is a prefix of X ; or – p is a prefix of some complement of X and Z is a possible prefixof u ′ . (3’) u = XY Zu ′′ , v = XY Zv ′′ for some uniquely determined propercomplement XY Z of XY Z , and ˆ Zu ′′ ≡ ˆ Zv ′′ for some complement ˆ Z of Z , and p is a prefix of some complement of X ; (4’) u = XY u ′ , v = XY Zv ′′ for some uniquely determined proper com-plement XY Z of XY Z , but Z is not a prefix of u ′ and u ′ ≡ Zv ′′ ,and also p is a prefix of some complement of X ; (5’) u = XY Zu ′′ , v = XY v ′ for some uniquely determined proper com-plement XY Z of X , but Z is not a prefix of v ′ and Zu ′′ ≡ v ′ , andalso p is a prefix of some complement of X ; (6’) u = XY u ′ , v = XY v ′ for some uniquely determined proper comple-ment XY Z of XY Z , Z is not a prefix of u ′ and Z is not a prefixof v ′ , but Z = z z , Z = z z , u ′ = z u ′′ , v ′ = z v ′′ where u ′′ ≡ v ′′ , z is the maximal common suffix of Z and Z , z in non-empty, z isa possible prefix of u ′′ , and also p is a prefix of some complement of X .Proof. Mutual exclusivity of the six conditions is proved exactly as forLemma 3. Suppose now that one of the six conditions above applies. Eachcondition clearly implies the corresponding condition from Lemma 3, so wededuce immediately that u ≡ v . We must show, using the fact that p is aprefix of a complement of X , that p is a possible prefix of u , or equivalentlyof v .In case (1’), p is clearly a possible prefix of u = XY Zu ′′ , and cases (3’),(4’) and (5’) are entirely similar. In case (2’), if p is a prefix of X then it isalready a prefix of u , while if p is a prefix of a proper complement X of X and Z is a possible prefix of u ′ , say u ′ ≡ Zw , then u = XY u ′ ≡ XY Zw ≡ XY Zw where the latter has p as a possible prefix. Finally, in case (6’) we know that z is a possible prefix of u ′′ , say u ′′ ≡ zx , so we have u = XY u ′ = XY z u ′′ = XY z zx = XY Zx and it is again clear that p is a possible prefix of u .Conversely, suppose u ≡ v and p is a possible prefix of u . Then exactlyone of the six conditions in Lemma 3 applies. By Lemma 2, every wordequivalent to u begins with a complement of XY , so p must be a prefix of aword beginning with some complement ˆ X ˆ Y . Since ˆ X is the maximal pieceprefix of ˆ X ˆ Y ˆ Z and ˆ Y is non-empty, it follows that p is a prefix of ˆ X . Ifany but condition (2) of Lemma 3 is satisfied, this suffices to show that thecorresponding condition from the statement of Lemma 4 holds.If condition (2) from Lemma 3 applies, we must show additionally thateither p is a prefix of X , or that Z is a possible prefix of u ′ . Suppose p is nota prefix of X . Then by the above, p is a prefix of some complement ˆ X . Itfollows from Lemma 2, that the only way the prefix XY of the word u canbe changed using the defining relations is by application of a relation of theform ( XY Z, XY Z ). In order for this to happen, one must clearly be ableto rewrite u = XY u ′ to a word of the form XY Zw ; consider the shortestpossible rewrite sequence which achieves this. By Lemma 2, no term in thesequence except for the last term will contain a relation word overlappingthe initial XY . It follows that the same rewriting steps rewrite u ′ to Zw ,so that Z is a possible prefix of u ′ , as required. (cid:3) Applications
The main application presented in [4] was for each strongly C (4) monoidpresentation, a linear time recursive algorithm to decide, given words u , v and a piece p , whether u ≡ v and p is a possible prefix of u . In particular,by fixing p = ǫ , we obtain an algorithm which solves the word problemfor the presentation in linear time. Figure 1 shows a modified version of NOTE ON SMALL OVERLAP MONOIDS 11 the algorithm which works for weakly C (4) presentations. The proofs ofcorrectness and termination are essentially the same as those in [4], butrelying on the more general results of Section 2. Thus, we establish thefollowing theorem. Theorem 1.
For every weakly C (4) finite monoid presentation, there existsa two-tape Turing machine which solves the corresponding word problem intime linear in the lengths the input words. The algorithms presented [4, Section 5] for finding the pieces of a presen-tation and hence testing strong small overlap conditions may clearly also beused to test the weak variants of those conditions, with the proviso that oneconsiders the set of relation words in the presentation, with any duplicatesdisregarded. In particular, we have:
Corollary 5.
There is a RAM algorithm which, given as input a finitepresentation h A | R i , decides in time O ( | R | ) whether the presentation isweakly C (4) . Theorem 2.
There is a RAM algorithm which, given as input a weakly C (4) finite presentation h A | R i and two words u, v ∈ A ∗ , decides whether u and v represent the same element of the semigroup presented in time O (cid:0) | R | min( | u | , | v | ) (cid:1) . Just as with the algorithm from [4], the algorithm in Figure 1 is essentiallya finite state process, and can be implemented on a 2-tape prefix-rewritingautomaton using a slight variation on the technique described in the proofof [5, Theorem 2]. It follows that we have:
Theorem 3.
Let h A | R i be a finite monoid presentation which is weakly C (4) . Then the relation { ( u, v ) ∈ A ∗ × A ∗ | u ≡ v } is deterministic rational and reverse deterministic rational. Moreover, onecan, starting from the presentation, effectively compute 2-tape deterministicautomata recognising this relation and its reverse. Just as in [5], we obtain as corollaries large number of other facts aboutweakly C (4) monoids. For brevity we refrain from explaining all terms, andinstead refer the reader to [5] for definitions. Corollary 6.
Every monoid admitting a weakly C (4) finite presentation • is rational (in the sense of Sakarovitch [9] ); • is word hyperbolic (in the sense of Duncan and Gilman [1] ); • is asynchronous automatic; • has a regular language of linear-time computable normal forms (namely,the set of words minimal in their equivalence class with respect to thelexicographical order induced by any total order on the generatingset); • has a boolean algebra of rational subsets; • has uniformly decidable rational subset membership problem; and • has rational subsets which coincide with its recognisable subsets. WP-Prefix ( u, v, p )1 if u = ǫ or v = ǫ then if u = ǫ and v = ǫ and p = ǫ then return Yes else return No elseif u does not have the form XY u ′ with XY a clean overlap prefix6 then if u and v begin with different letters7 then return No elseif p = ǫ and u and p begin with different letters9 then return No else u ← u with first letter deleted12 v ← v with first letter deleted13 if p = ǫ then p ← p with first letter deleted15 return WP-Prefix ( u, v, p )16 else let X, Y, u ′ be such that u = XY u ′ if p is not a prefix of a complement of X then return No elseif v does not begin with a complement of XY then return No elseif u = XY Zu ′′ and v = XY Zv ′′ for some complement XY Z of XY Z then if u ′′ is ˆ Z -active for some complement ˆ Z of Z then return WP-Prefix ( ˆ Zu ′′ , ˆ Zv ′′ , ǫ ) for some such ˆ Z else return WP-Prefix ( Zu ′′ , Zv ′′ , ǫ )26 elseif u = XY u ′ and v = XY v ′ then if p is a prefix of X then return WP-Prefix ( u ′ , v ′ , ǫ )29 else return WP-Prefix ( u ′ , v ′ , Z )30 elseif u = XY u ′ and v = XY Zv ′′ for some complement XY Z of XY Z then return WP-Prefix ( u ′ , Zv ′′ , ǫ )32 elseif u = XY Zu ′′ and v = XY v ′ for some complement XY Z of XY Z then return WP-Prefix ( Zu ′′ , v ′ , ǫ )34 elseif u = XY u ′ and v = XY v ′ for some complement XY of XY then let z be the maximal common suffix of Z and Z let z be such that Z = z z let z be such that Z = z z if u ′ does not begin with z or v ′ does not begin with z ;39 then return NO else let u ′′ be such that u ′ := z u ′′ let v ′′ be such that v ′ := z v ′′ ;42 return WP-Prefix ( u ′′ , v ′′ , z ) Figure 1.
Algorithm to solve the word problem for a fixedweakly C (4) presentation. NOTE ON SMALL OVERLAP MONOIDS 13
Acknowledgements
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