BACHELOR THESIS Star height

Transcription

1 BACHELOR THESIS Tomáš Svood Str height Deprtment of Alger Supervisor of the chelor thesis: Study progrmme: Study rnch: doc. Štěpán Holu, Ph.D. Mthemtics Mthemticl Methods of Informtion Security Prgue 217

2 I declre tht I crried out this chelor thesis independently, nd only with the cited sources, literture nd other professionl sources. I understnd tht my work reltes to the rights nd oligtions under the Act No. 121/2 S., the Copyright Act, s mended, in prticulr the fct tht the Chrles University hs the right to conclude license greement on the use of this work s school work pursunt to Section 6 susection 1 of the Copyright Act. Prgue, 21 July, 217 Tomáš Svood i

3 Title: Str height Author: Tomáš Svood Deprtment: Deprtment of Alger Supervisor: doc. Štěpán Holu, Ph.D., Deprtment of Alger Astrct: We present certin fmily of lnguges nd show tht for those lnguges infinite hierrchy of str heights exists. The proof ws first devised y Dejen nd Schützenerger [1]. More recently it ws reformulted y Skrovitch [2], who left some of the prts of the proof the the reder for more creful considertion. This thesis expnds on those prts nd provides more detiled proofs. We minly focus on construction of rtionl expression with the str height of the given lnguge. We lso compre the str height nd generlised str height nd the difference in chieved results for those two similr concepts. Keywords: str height, infinite hierrchy, utomt, rtionl expressions ii

4 Contents Introduction 2 1 Bsic concepts Words nd lnguges Rtionl expressions Str height Automt Eggn s question Automton recognising W q Rtionl expression denoting W q Witness words Infinite hierrchy Generlistion Generlised str height Open questions Conclusion 24 Biliogrphy 25 1

5 Introduction Kleene [3], in result known s Kleene s Theorem, shows tht utomt nd expressions correspond to ech other nd chrcterise the sme clss of lnguges. Eggn [4] refines this result y defining mesure of complexity for oth of them: loop complexity for utomt nd str height for expressions, nd y showing they correspond to ech other y chrcterising the sme clsses of lnguges. In this thesis we limit ourselves to the presenttion of str height, nd to the proof, due to Dejen nd Schützenerger [1], which sttes tht the str height hierrchy is infinite. They show tht for ny integer k, there exists rtionl lnguge of str height k over two letter lphet. We end y touching on nother notion, the generlised str height, which my divide the fmily of rtionl lnguges into only two prts. The determintion of the str height of lnguge turns out to e one of the most difficult prolems in utomt theory. McNughton [5] presented first notle result, n lgorithm for finding the str height of certin fmily of lnguges, so clled pure-group lnguges. Hshiguchi first [6] provided n lgorithm for deciding whether or not n ritrry rtionl lnguge is of str height one nd then [7], fter six yers, n lgorithm to determine the str height of ny rtionl lnguge. The lgorithm for the generl cse ws not prcticl, eing of non-elementary complexity clss. Kirsten [8] devised more efficient lgorithm thn Hshiguchi s, decidle in 2 2O(n) spce. 2

6 1. Bsic concepts In this chpter we introduce some stndrd concepts such s words, lnguges, rtionl expressions, nd utomt, ut we lso mke few key modifictions. For exmple, insted of the clssicl letters of n lphet, we use rtionl expressions s the lels of trnsitions in utomt. 1.1 Words nd lnguges An lphet is non-empty, usully finite, set of symols. Let A e n lphet. The elements of A, the symols, re clled letters, nd finite sequences of letters re clled words. The set of words, sequences of letters of A, is written A for resons tht should lter ecome ovious. Word f cn therefore e written with i in A, 1 i n. Product f = ( 1, 2,..., n ), Product on the set of words is equivlent to the opertion of conctention: ( 1,..., n ) ( 1,..., m ) = ( 1,..., n, 1,..., m ). This opertion hs neutrl element: the empty sequence or empty word, which is written 1, or 1 A if there could e some miguity. Note, tht conctention is ssocitive, ut it is not commuttive. The definition of product implies tht ech word is product of its letters nd therefore cn e written f = ( 1, 2,..., n ) = ( 1 ) ( 2 ) ( n ). By identifying the sequence () with the letter nd omitting the explicit symol for product, we re le to write the word f s f = 1 2 n. Similrly for the product of two words f nd g we write fg. Length The length of word is the length of the sequence of letters the word contins. Let 1, 2,..., n e letters. Then the length of word f = 1 2 n is n. It is written f. For the numer of occurrences of the letter in f we write f. If f is word in A we then hve f = f. A 3

7 Fctors Let f, g, h nd u e words in A. Word g is left fctor or prefix of f if there exists h such tht f = gh nd g is proper left fctor or proper prefix if h is other thn the empty word. Word h is right fctor or suffix of f if there exists g such tht f = gh nd h is proper right fctor or proper suffix if g is other thn the empty word. Word u is fctor of f if there exist g nd h such tht f = guh nd u is proper fctor if g nd h re not oth equl to the empty word. Lnguges A lnguge over A, or lnguge of A, is ny set of words written in the lphet A. In other words, lnguge of A is suset of A, therefore n element of P(A ), the set of ll the susets of A. We cn thus nturlly define for lnguges ll the usul opertions on the susets of set: union, intersection, complement nd difference, with the usul nottion. Word f is fctor of lnguge L, if it is fctor of some word g in L. The product of words extends to product of lnguges, which is denoted, s in the cse of words, y dot, X Y, or y simple conctention, XY : X Y = {f g f X, g Y }. From this we otin, y induction on n, the definition of the nth power of X, for ll X in P(A ): X = {1 A }, X n+1 = X X n = X n X. Note tht even for P(A ) we get = {1 A }. The str of lnguge X is the union of ll the powers of X, nd is written s X : X = X n. n N Note tht the nottion X is somewht improper ecuse in this context X does not denote the set of ll words over X ut the set of products of elements of X. Nevertheless, if one tkes X = A, the impropriety vnishes: the set of products of elements of n lphet A is indeed equl to the set of ll words generted y A. 1.2 Rtionl expressions Let A e n lphet nd let {, 1, +,, } e five function symols. Definition. A rtionl expression over A is formul otined inductively from the letters of A nd the symols {, 1, +,, } y the following process: (i), 1, nd, for in A, re rtionl expressions, (ii) if E nd F re rtionl expressions, then (E+F), (E F), nd (E ) re rtionl expressions. 4

8 We write RtEA for the set of rtionl expressions over A. To e le to simplify the nottion, we cn think of the function symols s symols representing opertions nd we cn omit the prentheses in the rtionl expressions if we specify the order of precedence for +,, nd. We let tke precedence over, which in turn tkes precedence over +. As with letters in words, we omit the symol. With this convention we write, for exmple, + for (( ) + ( )), or ( + ( ) ) for (( + ( (((( ( )) ) ) ))) ). Definition. To ech rtionl expression E in RtEA we ssign lnguge of A, written L[E], nd defined inductively: L[] =, L[1] = {1 A }, nd L[] = {} for ll in A for the tomic rtionl expressions nd then for the composite rtionl expressions: L[E + F] = L[E] L[F], L[E F] = L[E] L[E], L[E ] = (L[E]). We sy tht E denotes the lnguge L[E]. Note tht the symol is used in two different contexts. Either s shorthnd for the infinite union of ll the powers of lnguge, or, here, s symol in formul of rtionl expression, in which it, however, directly represents the str of the lnguge the rtionl expression denotes. This direct trnsltion nd the fct tht it is stndrd nottion, we feel, justifies the impropriety of the somewht excessive usge of the symol. We sy tht lnguge of A is rtionl if nd only if it is denoted y rtionl expression over A. Two rtionl expressions re equivlent if they denote the sme lnguge. Tht mens E nd F re equivlent if L[E] = L[F] nd we write it s E F. Identities Let us stte few identities of rtionl expressions tht will e often used without explicit mention. Lemm 1. (E + F) + G E + (F + G), nd (E F) G E (F G), E + F F + E, E + + E E, E E, E 1 1 E E, E (F + G) E F + E G, nd (E + F) G E G + F G. 5

9 Proof. L[(E + F) + G] = L[E + F] L[G] = L[E] L[F] L[G] = L[E] L[F + G] = L[E + (F + G)], L[(E F) G] = L[E F] L[G] = L[E] L[F] L[G] = L[E] L[F G] = L[E (F G)], L[E + F] = L[E] L[F] = L[F] L[E] = L[F + E], L[E + ] = L[E] L[] = L[E] = L[] L[E] = L[ + E], L[E ] = L[E] L[] = L[] = L[] L[E] = L[ E], L[E 1] = L[E] L[1] = L[E] = L[1] L[E] = L[1 E], L[E (F + G)] = L[E] L[F + G] = L[E] (L[F] L[G]) = L[E] L[F] L[E] L[G] = L[E F] L[E G] = L[E F + E G], L[(E + F) G] = L[E + F] L[G] = (L[E] L[F]) L[G] = L[E] L[G] L[F] L[G] = L[E G] L[F G] = L[E G + F G]. since We lso hve 1, L[ ] = = {1} = L[1]. 1.3 Str height The symol defined for rtionl expressions is the only one tht tkes rtionl expression E denoting finite lnguge nd gives rtionl expression denoting n infinite lnguge, E. Hence the ide of mesuring the complexity of expressions y counting the numer of nested uses of this symol, numer which is clled the str height, which we will write h[e], for E RtEA, nd which is defined y induction: if E =, E = 1 or E =, for A, h[e] =, if E = F + G or E = F G, h[e] = mx (h[f], h[g]), if E = F, h[e] = 1 + h[f]. The str height of rtionl lnguge L over A, written h[l], is the minimum of the str heights of the rtionl expressions tht denote L: h[l] = min{h[e] E RtEA : L = L[E]}. 6

10 Lemm 2. Every lnguge L over A with str height h is denoted y finite sum of rtionl expressions: where ech E j is product of the form: L = L[E E n ], E j = u F 1u 1 u m 1 F mu m, where ech u k, k m, is word in A nd ech F k, 1 k m, is rtionl expression over A of str height less then or equl to h 1. Proof. By induction on h, follows from the fct tht product distriutes over union. 1.4 Automt An utomton is directed grph which is lelled with rtionl expressions over n lphet, nd in which two susets of vertices re distinguished. Definition. An utomton A is specified y giving the following elements: (i) non-empty set Q, clled the set of sttes of A, (ii) set A, lso non-empty, clled the (input) lphet of A, (iii) suset I of Q, clled the set of initil sttes, of A, (iv) suset T of Q, clled the set of finl sttes of A, (v) suset E of Q RtEA Q, clled the set of trnsitions of A. We write A = Q, A, E, I, T nd we sy tht A is n utomton over A. Let A e finite lphet. We cll n utomton over A finite if set its sttes is finite. If e = (p, E, q) is trnsition of A, tht is, if e is in E, we sy tht E is the lel of e nd we will write p E q, or p E A q where it might e miguous which utomton we re considering. We lso sy tht p is the source nd q the destintion of the trnsition e. Trnsition is loop, if its source nd destintion re the sme stte. When trnsition hs lel 1, it is clled spontneous. Note tht we cn lwys ssume tht there is, etween ech pir of sttes of n utomton, t most one trnsition. This is ecuse we chose rtionl expressions s the lels of trnsitions, rther thn the usul letters of A. We mke this generlistion to e le to represent ny utomton y single trnsition, s ws shown to y possile y Kleene [3]. Similr generlistion ws first introduced y Brzozowski nd McCluskey [9]. When pproprite, we cn even ssume tht there is, etween ech pir of sttes of n utomton, exctly one trnsition, since if for some pir of sttes, there is no trnsition in the utomton, we cn dd trnsition lelled with. 7

11 A computtion in A is sequence of trnsitions where the source of ech trnsition is the destintion of the previous one, which cn e written s: or E p 1 E 2 E p1 3 p2 p E 1 E 2 E n pn. E n 1 pn 1 E n pn, We sy tht computtion is in A if every trnsition of the computtion is in A. The stte p is the source of the computtion c, nd p n its destintion. The length of the computtion c is n, the numer of trnsitions which mke up c. The lel of c is the product of the lels of the trnsitions of c. In the ove cse, the lel of c is E 1 E 2 E n. A computtion in A is successful if its source is n initil stte nd its destintion is finl stte. A word in A is clled ccepted or recognised y A if it is in lnguge denoted y lel of successful computtion in A. Definition. The lnguge ccepted, or recognised y A, written L(A), is the set of words ccepted (or recognised) y A: L(A) = {f A i I, t T : i E A t f L[E]}. Two utomt re equivlent if they recognise the sme lnguge. If L is lnguge, nd finite utomton A exists such tht L = L(A), we cll L recognisle. Lemm 3. Let L e recognisle lnguge over A. There exists N N such tht for every word f in L nd every fctoristion of the form f = uv 1 v 2 v N w, where every v i is non-empty word, there is pir (j, k) of indices, j < k N, tht uv 1 v 2 v j (v j+1 v k ) v k+1 v N w L. Proof. Let A = Q, A, E, I, T e n utomton tht recognises L. Set N = Q, mening N is size of the set Q. A computtion tht ccepts f cn e written i E q F 1 q1 F 2 q2 F 3 F N 1 qn 1 F N qn G t, where u L[E], v i L[F i ], for 1 i N, nd w L[G]. The N + 1 sttes q i cnnot ll e distinct, nd t lest two, sy q j nd q k, re equl to the sme stte p. The computtion cn therefore e written i E q F 1 F 2 F j p F j+1 F k p F k+1 F N q N G t, where L[F j+1 F k ] contins the non-empty word v j+1 v k. Hence, for every non-negtive integer n, i E q F 1 F 2 F j p (F j+1 F k ) n p F k+1 F N q N G t, is successful computtion of A nd uv 1 v 2 v j (v j+1 v k ) n v k+1 v N w is ccepted y A. 8

12 Let Z n e finite cyclic group of order n. When using Z n s set of sttes of n utomton, we consider integers {, 1, 2,..., n 1} to e the underlying set of the group nd we utilise the group opertions modulo n when enumerting the sttes. Definition. Automton Z n, {, }, E, {}, {} is clled ring utomton R(n), if for ech stte z Z n there re exctly two trnsitions with z s source nd they re (z,, z + 1) nd (z,, z 1). Note tht if stte is memer of Z n we often denote it y some ddition. For exmple, let s, r y sttes of R(n). Then s E s+r denotes trnsition (s, E, s+r), nd s E s+r F s+2 denotes computtion consisting of trnsitions (s, E, s+r) nd (s + r, F, s + 2). Exmple. Figure 1.1 shows ring utomton R(8) Figure 1.1: Automton R(8) Lemm 4. Let R(n) e ring utomton. For ny m, r, nd t in Z n, if the computtion r H t is in R(n) then r + m H t + m is lso in R(n). Proof. By induction on m. Let m =. Since stte r + is the sme s r nd s + is the sme s s, the proposition holds. Given the induction hypothesis for some m 1, let the computtion r H t e written s sequence of trnsitions: X s 1 X 2 s1 X k 1 sk 1 X k s k, where s = r, s k = t, nd ech X i is either letter or. Then computtion s + m 1 X 1 s 1 + m 1 X 2 X k 1 sk 1 + m 1 X k s k + m 1 is in R(n). For ny stte p R(n), trnsitions (p,, p + 1) nd (p,, p 1) re in R(n), thus, if (s i + m 1, X i, s i + m 1) is in R(n), so is (s i + m, X i, s i + m), for i k. Therefore s + m X 1 s 1 + m X 2 X k 1 sk 1 + m X k s k + m, which is the computtion r + m H t + m, is lso in R(n). 9

13 Stte removl lgorithm Brzozowski nd McCluskey [9] used generlised forms of utomt to show tht there is simple lgorithm tht tkes finite utomton nd returns rtionl expression denoting the sme lnguge tht is ccepted y the utomton. This lgorithm is used in Lemm 8 to find rtionl expression denoting the lnguge recognised y ring utomton. Lemm 5. Let A = Q, A, E, I, T e n utomton with n sttes which hs t lest one stte tht is neither initil nor ccepting, tht mens Q \ (I T ). An utomton B with n 1 sttes exists such tht L(B) = L(A). Proof. Let q e stte in Q \ (I T ). Next, we consider the trnsitions (p 1, E 1, q),..., (p k, E k, q), (q, G 1, r 1 ),..., (q, G l, r l ), nd (q, F, q), where ech p h q, nd r j q. These re ll the different trnsitions of A with q s destintion, source, or oth respectively. Note tht F my e 1, since every stte hs t lest spontneous loop. To crete the utomton B, we remove ll the mentioned trnsitions nd the stte q, nd for ech pir of sttes (p h, r j ), 1 h k nd 1 j l, we replce the trnsition (p h, H, r j ) with (p h, H + E h F G j, r j ). Note tht H my e for some stte pirs. Now we verify tht L(B) = L(A). Let f e word ccepted y A. A computtion tht ccepts f cn e written s i A 1 A 2 A s1 3 s2 A N 1 sn 1 A N t. If some s M = q, there re non-negtive integers K nd L such tht s M K = s M K+1 = = s M 1 = s M = s M+1 = = s M+L 1 = s M+L = q, nd the computtion ccepting f is i A 1 A M K 1 ph E h q K+L times F F q G j rj A M+L+2 A N t, therefore fctoristion of f exists such tht f = u 1 u 2 v 1 v K+L w 1 w 2, where u 1 L[A 1 A M K 1 ], u 2 L[E h ], v 1,..., v K+L L[F], w 1 L[G j ], nd w 2 L[A M+L+2 A N ]. It follows tht, for every non-negtive integer J nd ny words g 1,..., g J L[F], the word u 1 u 2 g 1 g J w 1 w 2 is lso in L(A). Specificlly it is ccepted y computtions: i i i A 1 A 1 A 1 ut lso y i A M K 1 ph A M K 1 ph A M K 1 ph A 1 E h q E h q A M 1 ph J times F F q F q E h F G j rj G j rj G j rj A M+L+2 H+E h F G j rj 1 A M+L+2 A M+L+2 A N t, A M+L+2 A N t, A N t, A N t,

14 which is successful computtion is B nd therefore f is ccepted y B. Otherwise, if every stte s M q, ut there re some sttes s M = p h nd s M+1 = r j, then f is ccepted y successful computtion: i A 1 A M ph therefore f is lso ccepted y B, since i A 1 A M ph H r j A M+2 H+E h F G j rj A M+2 A N t, A N t is successful computtion in B. If neither of the ove cses re true, tht mens every s M s M = p h then s M+1 r j, the word f is lso ccepted y q nd if ny i A 1 A s 2 A 1 s 3 2 B B B A N 1 B s N 1 A N t, B which is successful computtion in B, therefore f L[B]. This shows tht every word ccepted y A is lso ccepted y B. The process of removing the stte descried in the Proof of Lemm 5 is clled stte removl lgorithm. By iterting on the sttes of utomton A = Q, A, E, I, T, pplying stte removl lgorithm ech step, we cn crete n utomton with only I T sttes tht ccepts L(A). It is importnt to note tht depending on the order of the sttes chosen to e removed, we my otin different utomt. 11

15 2. Eggn s question In this chpter we present n nswer to the question whether there re rtionl lnguges with ritrrily lrge str heights. The first exmple of such lnguges is given y Eggn [4]. Eggn s exmple uses n lphet of size 2 n 1 for the lnguge with str height n. He therefore sked whether there re some exmples of lnguges over inry lphets. Answer to tht question, which is ffirmtive, ws provided shortly fter y Dejen nd Schützenerger [1]. Their nswer is lso reformulted y Skrovitch [2], who phrsed the proof in the form we work with here. The fmily of lnguges provided y Dejen nd Schützenerger is the following: Definition. Let q e non-negtive integer, then W q is lnguge, such tht W q = {f {, } f f (mod 2 q )}. Theorem 6. The lnguge W q hs str height q. The proof hs three prts. First, we find n utomton tht recognises W q in Lemm 7. Second, we infer rtionl expression of str height q, denoting W q, in Lemm 8. And lstly, we show tht the str height of the lnguge hs to e t lest q. 2.1 Automton recognising W q Lemm 7. Lnguge W q is recognised y n utomton R(2 q ). Proof. We show tht ech word in W q is ccepted y successful computtion in R(2 q ). Let f e word in W q. Since f hs even length, let k e such tht 2k = f. Since f W q, f f = n2 q for some integer n. For n =, f = f. We use induction on k to show tht for every fctoristion f = uvw, if f is ccepted y R(2 q ), words uvw nd uvw re lso ccepted y R(2 q ). For k =, empty word is ccepted since the initil stte is lso finl nd oth computtions 1, nd 2 q 1 re successful in R(2 q ). Given the induction hypothesis for some k 1, every word f = uvw of length 2k 2 such tht f = f is ccepted y computtion E i F j G, where u L[E], v L[F], nd v L[G]. We need to show tht computtions E i E i i + 1 i 1 F j + 1 F j 1 G j, j G, nd re successful. For ny two sttes r nd t of R(2 q ), the computtion r H t cn e written s sequence of trnsitions: r X 1 X s 2 1 X m 1 sm 1 12 X m t,

16 where ech X l is either letter or. Due to Lemm 4, computtions r + 1 X 1 s X 2 r 1 X 1 s 1 1 X 2 X m 1 sm X m 1 sm 1 1 Xm t + 1, Xm t 1 nd re lso in R(2 q ), nd so re r r r + 1 X 1 s X 2 r 1 X 1 s 1 1 X 2 X m 1 sm X m 1 sm 1 1 Xm t + 1 Xm t 1 t, t. nd F If we tke computtion i j in plce of r H t, the ssertion is proved. For n, f f = n2 q. Let g e word such tht g = g, therefore g is ccepted y R(2 q ). Computtion ccepting g cn e written s X 1 X s 2 1 X m 1 sm 1 X m, where ech X l is either letter or. Let Y 1,..., Y m+1 e expressions such tht f is in L[Y 1 X 1 Y m X m Y m+1 ]. Let e h e in L[Y h ] for 1 h m + 1. n2 q = f f = g + m+1 h=1 e h g m+1 h=1 e h = m+1 By m + 1 itertions of Lemm 4, it follows tht computtion Y 1X 1 Y p1 + ( e 1 e 1 ) 2 X 2 p2 + m 1 Y m 1 X m 1 p m 1 + ( e h e h ) h=1 h=1 2 ( e h e h ) h=1 m+1 Y mx my m+1 is in R(2 q ). It is successful computtion in R(2 q ) ecuse n2 q = m+1 h=1 ( e h e h ) (mod 2 q ). ( e h e h ). Y 3 X 3 h=1 ( e h e h ). 2.2 Rtionl expression denoting W q We hve just proved tht lnguge W q is recognised y the ring utomton R(2 q ). We hve lso shown tht the stte removl lgorithm tkes n utomton nd finds nother one with one less stte, tht recognises the sme lnguge. We comine these results to find n utomton with only one stte, tht recognises W q. Lemm 8. Lnguge W q is denoted y rtionl expression of str height q. 13

17 Proof. Due to Lemm 7, lnguge W q is recognised y n utomton R(2 q ). By itertive ppliction of the stte removl lgorithm, we will otin n utomton with only one stte. First, we set X =, Y =, nd Z =, then, for every non-negtive integer n, X n+1 = X n Z nx n, Y n+1 = Y n Z ny n, nd Z n+1 = Z n + X n Z ny n + Y n Z nx n. Second, we define utomton A k s A k = Z 2 q k, A, E, {}, {} such tht ny stte p in A k is source of exctly three trnsitions in A k : p X k p + 1, p Y k p 1, nd p Z k p. By induction on k, we show tht W q, ccepted y R(2 q ), is lso ccepted y A k. Let k =. Note tht for ech stte p in A, trnsitions p X p + 1, p Y p 1, nd p Z p re in R(2 q ). Similrly, for ny stte r in R(2 q ), trnsitions r r + 1, r r 1 re in A. Therefore R(2 q ) is equivlent to A. Given the induction hypothesis for some k 1, W q is ccepted y A k 1. By 2 q k itertions of the stte removl lgorithm n utomton with 2 q k sttes is otined. The set of sttes of A k 1 is Z 2 q k+1. Consecutively, in the incresing order, we remove ll the odd sttes of A k 1, tht is, sttes 1, 3, 5,..., 2 q k+1 1. Let m e the removed odd stte. The process of removing the stte is techniclly the sme for ll the sttes, ut in the interest of clrity, we will differentite etween m eing equl to 1, eing etween 3 nd 2 q k+1 3, nd eing equl 2 q k+1 1. For m = 1, the trnsitions tht hve 1 s source, destintion, or oth re: 1 X k 1 2, 1 Y k 1, X k 1 1, 2 Y k 1 1, nd 1 Z k 1 1. Those trnsitions re removed, in ccordnce with the stte removl lgorithm. For ech pir of sttes, (, ), (, 2), (2, ), (2, 2), these trnsitions re replced: Z k 1 replced y Z k 1 + X k 1 Z k 1 Y k 1, E 2 replced y E + X k 1Z k 1 X k 1 2, 2 E replced y 2 E + Y k 1Z k 1 Y k 1, 2 Z k 1 2 replced y 2 Z k 1 + Y k 1 Z k 1 X k 1 2. For 3 m 2 q k+1 3, the sttes 1, 3,..., m 2 re lredy removed. The trnsitions tht hve m s source, destintion, or oth re: m X k 1 m + 1, m Y k 1 m 1, m 1 X k 1 m, m + 1 Y k 1 m, nd m Z k 1 m. 14

18 As in the cse m = 1, those trnsitions re removed. For ech pir of sttes, (m 1, m 1), (m 1, m + 1), (m + 1, m 1), (m + 1, m + 1), these trnsitions re replced: m 1 Z k 1 + Y k 1 Z k 1 X k 1 m 1 replced y m 1 Z k m 1, since Z k = (Z k 1 + Y k 1 Z k 1X k 1 ) + X k 1 Z k 1Y k 1, m 1 E m + 1 replced y m 1 E + X k 1Z k 1 X k 1 m + 1, m+1 E m 1 replced y m+1 E + Y k 1Z k 1 Y k 1 m 1, m+1 Z k 1 m + 1 replced y m+1 Z k 1 + Y k 1 Z k 1 X k 1 m+1. Lstly for m = 2 q k+1 1, the sttes 1, 3,..., 2 q k+1 3 re lredy removed. Since m = 2 q k+1 1, the stte m+1 is 2 q k+1 (mod 2 q k+1 ). The trnsitions tht hve m s source, destintion, or oth re: m X k 1, m Y k 1 m 1, m 1 X k 1 m, Y k 1 m, nd m Z k 1 m. As for every m 2 q k+1 3, these trnsitions re removed. For ech pir of sttes, (m 1, m 1), (m 1, ), (, m 1), (, ), these trnsitions re replced: m 1 m 1 Z k 1 + Y k 1 Z k 1 X k 1 m 1 replced y m 1 Z k m 1, E replced y m 1 E + X k 1Z k 1 X k 1, E m 1 replced y E + Y k 1Z k 1 Y k 1 m 1, Z k 1 + X k 1 Z k 1 Y k 1 replced y Z k. Thus, fter 2 q k odd sttes removed, we hve utomton with 2 q k sttes, ll non-negtive even numers 2 q k+1 2, tht ccepts W q. To complete the inductive step, crete utomton A k, we renme ech stte 2t s t, t {, 1, 2,..., 2 q k 1}, mening we replce ech stte 2t with t, ech trnsition F F 2t s with t s, nd s F 2t with s F t, for every stte s. For k = q 2, utomton A 2 hs two sttes. Note tht, to conform with the definition of A 2, it should hve six distinct trnsitions, X k 1, 1 Y k, Z k, 1 X k, Y k 1, 1 Z k 1. We cn see tht the pir of trnsitions, 1 nd 1, cn e replced y X k+y k X 1, nd the pir of trnsitions, 1 k nd 1 Y k cn e replced y 1 X k+y k. Automton A 2 is reduced into one stte utomton y ppliction of the stte removl lgorithm on the stte 1. Thus creting loop lelled with Z q 1 + (X q 1 + Y q 1 )Z q 1(X q 1 + Y q 1 ) on the stte. 15 X k Y k

19 Note the following equivlence: Therefore Z q 1 + (X q 1 + Y q 1 )Z q 1(X q 1 + Y q 1 ) X q + Y q + Z q. W q = L(A q ) = L[(Z q 1 + (X q 1 + Y q 1 )Z q 1(X q 1 + Y q 1 )) ] = L[(X q +Y q +Z q ) ]. By induction on i, we show, tht h[x i ] = h[z i ] = h[z i ] = i 1. Let i = 1. h[ 2 ] = h[ 2 ] = h[ + ] =. Given the induction hypothesis for some i 1, h[z i 1 ] = i 1. Therefore h[z i 1] = i = h[x i 1 Z i 1X i 1 ] = h[x i ] = h[y i 1 Z i 1Y i 1 ] = h[y i ] = h[z i 1 + X i 1 Z i 1Y i 1 + Y i 1 Z i 1X i 1 ] = h[z i ]. We conclude tht h[(x q + Y q + Z q ) ] = q, which proves the lemm. Exmple. Lnguge W 3 is recognised y R(8). In Figure 2.1, we show how the rtionl expression X 3 + Y 3 + Z 3 is derived. Then W 3 = L[(X 3 + Y 3 + Z 3 ) ]. 16

20 () Automton R(8) () Automton A (c) (d) (e) (f) Automton A ( + ) 2 Z Y 2 X 2 X 2 + Y 2 X 2 + Y ( + ) 2 (g) Z 2 (h) Automton A 2 Z 2 + (X 2 + Y 2 )Z 2(X 2 + Y 2 ) (i) Figure 2.1: Steps of the stte removl lgorithm on R(8) 17

21 2.3 Witness words We hve found rtionl expression of str height q denoting the lnguge W q, which mens tht the str height of the lnguge will not e higher then q. In the following sections we proceed to show tht it lso hs to e t lest q. First, to simplify the proof, we define specil words, show tht they re in every lnguge W q, nd show some properties of fctors of those words. Definition. For ny non-negtive integers k nd n, witness word w k,n is word defined recursively s: w,n =, w 1,n = 2 () n 2 () n,. w k,n = 2k (w k 1,n ) n 2k (w k 1,n ) n. Lemm 9. For ny non-negtive integers k nd n, w k,n = w k,n. Proof. By induction on k. For k =, = regrdless of n. Given the induction hypothesis for some k 1, w k 1,n w k 1,n =. The word w k,n consists wholly of fctors 2k, 2k, nd two (w k 1,n ) n. Thus: w k,n w k,n = 2k + 2 (w k 1,n ) n + 2k 2k 2 (w k 1,n ) n 2k = 2 k + 2 n w k 1,n 2 n w k 1,n 2 k = 2 n ( w k 1,n w k 1,n ) =. From Lemm 9 it follows tht ny witness word w k,n is in ny lnguge W q. Lemm 1. Any prefix u nd suffix v of (w k,n ) n stisfy the inequlities: u u 2 k+1 1, v v 2 k+1 1. Proof. First, we prove the inequlities of the prefix u. Let u = (w k,n ) i u such tht u is proper prefix of w k,n. Since u u = (w k,n ) i + u (w k,n ) i u = i ( w k,n w k,n ) + u u = u u, we only need to prove tht the inequlities hold for prefix of w k,n. This lets us ssume, without loss of generlity, tht u is proper prefix of w k,n. We proceed y induction on k. For k =, u is prefix of nd we cn see tht the ssertion is true. Given the induction hypothesis for some k 1, prefix of w k 1,n hs etween nd 2 k 1 more s thn s. There re four possiilities: (i) u is prefix of 2k, 18

22 (ii) 2k is prefix of u nd u is prefix of 2k (w k 1,n ) n, (iii) 2k (w k 1,n ) n is prefix of u nd u is prefix of 2k (w k 1,n ) n 2k, (iv) 2k (w k 1,n ) n 2k is prefix of u nd u is prefix of 2k (w k 1,n ) n 2k (w k 1,n ) n. For (i), let u = i, for i 2 k. Then u u = u 2 k < 2 k+1 1. For (ii), let u = 2k (w k 1,n ) i g, for i n, such tht word g is prefix of w k 1,n. Then < 2 k u u 2 k + 2 k 1 = 2 k+1 1. For (iii), let u = 2k (w k 1,n ) n i, for i 2 k. Then u u = 2k + (w k 1,n ) n + i 2k (w k 1,n ) n i = 2 k i. Therefore u u 2 k < 2 k+1 1. For (iv), let u = 2k (w k 1,n ) n 2k (w k 1,n ) i g, for i n, such tht word g is prefix of w k 1,n. Then u u = 2k + (w k 1,n ) n + 2k + (w k 1,n ) i + g 2k (w k 1,n ) n 2k (w k 1,n ) i g = 2 k + g + (n + i)( w k 1,n w k 1,n ) 2 k g = g g. Therefore u u 2 k 1 < 2 k+1 1. For the proof of inequlities of suffix v, we write w k,n s w k,n = gv. Since = w k,n w k,n = g + v g v, v v = g g. Word g is prefix of w k,n, therefore v v 2 k Infinite hierrchy We use the witness words from the previous section to define properties of lnguges. Ech of those properties specifies fmily of lnguges with common str height tht re ll susets of W q. We show tht for the lnguge W q, there re q different fmilies, ech with the common str height strictly higher then the previous one, proving tht W q itself hs to hve str height of t lest q. Definition. We sy tht lnguge L stisfies property P k if there exists n infinite numer of vlues of n such tht (w k,n ) n is fctor of t lest one word in L. Note tht if L stisfies P k, it lso stisfies P l for l < k since (w l,n ) n is fctor of (w k,n ) n. Proof of Theorem 6. Lemm 8 shows tht the str height of W q is t most q. Now we show tht it lso hs to e t lest q. By W k we denote fmily of lnguges L tht stisfy the following conditions: (i) L W q, 19

23 (ii) L stisfies P k, (iii) L hs minimum str height of ny lnguge stisfying the first two conditions. Let h k e the common vlue of the str height of the lnguges in W k. Lnguges in W re necessrily infinite, therefore < h. P k implies P k 1, which mens tht h k 1 h k. W q oviously stisfies (i). It lso stisfies P q 1 since (w q 1,n ) n = (w q 1,n ) n for every n. Becuse we hve found rtionl expression of str height q denoting W q, it follows tht h q 1 q. Therefore we hve < h h 1 h q 1 q. To prove the theorem it is enough to show tht h k 1 < h k for ech k, k = 1,..., q 1. Let L e in W k. Due to Lemm 2 L cn e written s finite union of lnguges E j, ech denoted y rtionl expression of the form E j = u (H 1 ) u 1 u m 1 (H m ) u m, where ech rtionl expression H i denotes lnguge H i with str height less then or equl to h k 1. Since L stisfies P k nd the union of the lnguges E j is finite, it follows tht t lest one of E j hs to stisfy P k. We cn therefore sfely ssume tht L itself is denoted y rtionl expression of the sme form s E j. For ech word g in H i, words u u 1 u m nd u u 1 u i 1 gu i u m re oth in lnguge L. Therefore, ecuse it hs to e true tht g g (mod 2 q ), H i W q. Since L is of miniml str height to stisfy P k, none of the H i stisfies P k. Now we need to show tht some H i stisfies P k 1. In fct, we will hve h k 1 h k 1. L stisfies P k, therefore words (w k,n ) n re fctors of L for ritrrily lrge n. Lemm 3 shows tht we cn find N lrge enough, such tht there is infinitely mny n N, tht (w k,n ) n is fctor of L, nd, for ech n, we hve infinitely mny l s, tht (w k,n ) l is fctor of L. Let Hi e the lnguge recognised y H i. Since m is finite nd fixed for L, there hs to e infinitely mny n s, tht (w k,n ) n is fctor of r n Hi for some i, 1 i m. This mens tht Hi stisfies P k. Next we show tht for these n s, (w k 1,n ) n is fctor of word in H i, mening H i stisfies P k 1. We write r n s fctoristion r n = g g 1 g l, where g j H i, for j l. If w k,n, from (w k,n ) n, is fctor of some g j, the condition is stisfied, since (w k 1,n ) n is fctor of w k,n. Otherwise, let us show how fctor (w k,n ) 2 is covered y the fctoristion of r n. Written explicitly, we hve (w k,n ) 2 = 2k (w k 1,n ) n 2k (w k 1,n ) n 2k (w k 1,n ) n 2k (w k 1,n ) n. Let us consider the g j tht covers, t lest prtilly, the fctor 2k. There re two possiilities: (i) 2k is fctor of g j, (ii) left fctor of 2k is right fctor of g j. 2

24 In cse (i), we hve g j = v 2k u. If v covers the fctor (w k 1,n ) n to the left of 2k, or u covers the fctor (w k 1,n ) n to the right of 2k, the condition is stisfied. If not, u is left fctor of (w k 1,n ) n nd v is right fctor. Set Hence x = g j g j, y = u u, nd z = v v. x = 2 k (y z). We see tht 1 2 k y z 2 k 1, from Lemm 1. Tht gives us < x < 2 k+1 2 q. which contrdicts the fct tht g j H i W q. In cse (ii), we hve g j = v r, < r < 2 k. If (w k 1,n ) n is fctor of v, the condition is stisfied. Otherwise v is right fctor of (w k 1,n ) n nd, similrly s ove, we set x = g j g j, nd z = v v. Hence Therefore, due to Lemm 1, x = r + z. r x r + 2 k 1 < 2 k+1 2 q. which produces the sme contrdiction s the cse (i). 21

25 3. Generlistion The definition of str height cn e expnded into generlised str height, concept tht is quite similr, yet current understnding of its properties is much more limited. 3.1 Generlised str height Definition. A generlised rtionl expression over A is formul otined inductively from the letters of A nd the symols {, 1, +,,,,, } y the following process: (i), 1, nd, for in A, re generlised rtionl expressions, (ii) if E nd F re generlised rtionl expressions, then (E + F), (E F), (E ), (E F), (E F), (E ) re generlised rtionl expressions. We write GRtEA for the set of generlised rtionl expressions over A. The lnguges denoted y generlised rtionl expressions re the sme s for the rtionl expressions plus, to ccommodte the dded symols, we let: L[E F] = L[E] \ L[F], L[E F] = L[E] L[E], L[E ] = A L[E]. Definition. Let E GRtEA. Generlised str height, written gsh[e], is defined y induction on the complexity of expressions: if E =, E = 1 or E =, for A, gsh[e] =, if E = F + G or E = F G, gsh[e] = mx (gsh[f], gsh[g]), if E = F, gsh[e] = 1 + gsh[f], if F = E, gsh[e] = gsh[f]. For E = F G, or E = F G, due to De Morgn s lws, identities exist: F G F G, nd F G (F + G ), thus gsh[e] = mx (gsh[f], gsh[g]). The generlised str height of rtionl lnguge L over A, written gsh[l], is the minimum of the generlised str heights of the generlised rtionl expressions tht denote L: gsh[l] = min{gsh[e] E GRtEA : L = L[E]}. 22

26 3.2 Open questions The determintion of the str height of rtionl lnguge, however difficult prolem, hs een solved. The determintion of the generlised str height still remins to e solved. Two open questions, tht re rised y the definition of generlised str height of rtionl lnguge, re nmely the existence of n infinite hierrchy nd the computtion of generlised str height. Not only do we not know whether there re rtionl lnguges with ritrrily lrge generlised str height, ut even lnguge with generlised str height greter thn 1 hs not yet een shown. On the other hnd, for lnguges whose generlised str height is equl to, or so clled str-free lnguges, Schützenerger [1] provided n lgeric chrcteristion. 23

27 Conclusion This thesis proves, in greter detil thn the originl works, tht there is n infinite hierrchy of str heights of lnguges over inry lphet. In the first chpter, we defined n unusul modifiction of clssicl utomt with trnsitions lelled with rtionl expressions. This let us throughout the thesis consider only utomt with t most one trnsition etween ech pir of sttes. We devised precise definition of ring utomton nd proved Lemm 4 out the existence of computtions in ring utomt. For the stte removl lgorithm we compred the successful computtions in n nd n 1 stte utomt to prove their equivlence. In the second chpter, the im of this thesis ws to tke the proof of existence of lnguges with ritrrily high str height, s formulted y Skrovitch [2], nd in detil prove the prts tht were only outlined. In the Section 2.1 we used our Lemm 4 to show tht lnguge W q is recognised y the ring utomton with 2 q sttes. In the Section 2.2 we showed tht, with prticulr order of sttes chosen for stte removl lgorithm, lnguge W q is denoted y rtionl expression of str height q. In the Section 2.3 witness words re defined. In Lemm 9 we hve shown tht ech of them hs the sme numer of oth letters of the inry lphet over which they re defined. This equlity is used to simplify the proof of Lemm 1, where we show how much more letters of one type cn e in either prefix, or suffix of the witness words. This inequlity is used in the proof of Theorem 6. 24

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