Many-valued truth functions, Cerny's conjecture and road. Lemminkaisenkatu 14 A, Turku, Finland and

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1 Mnyvlued truth functions, Cerny's conjecture nd rod coloring Alexndru Mteescu Turku Centre for Computer Science, Lemminkisenktu 4 A, 050 Turku, Finlnd nd Fculty of Mthemtics, University of Buchrest, Romni emil: mteescu@utu. Arto Slom Turku Centre for Computer Science nd Deprtment of Mthemtics, University of Turku, 004 Turku, Finlnd emil: slom@utu. Turku Centre for Computer Science TUCS Technicl Report No 74 My 999 ISBN ISSN 3989

2 Astrct We investigte interconnections etween mnyvlued truth functions nd functions dened y nite deterministic utomt. Propositions in one of the theory hve often their nturl counterprt in the other theory. Such is the cse with prolems out completeness, s well s with Cerny's conjecture in utomt theory. Sometimes results in one theory, such s those concerning selfconjugcy of truth functions, cn e pplied in the other theory. TUCS Reserch Group Mthemticl Structures of Computer Science

3 Sheer functions nd complete sets Consider functions whose vriles, nite in numer, rnge over nite set N = f; ; : : :; ng; n ; nd whose vlues re elements of N. There re n nm distinct mplce functions. If N is chosen to e the set of n truth vlues, then the functions considered re oviously truth functions in nvlued logic. A function g is sid to e generted y set F of functions if g cn e expressed s composition of functions in F. For instnce, ech of the functions f (x) = f(x; f(x; x)); f (x; y) = f(f(x; f(x; y)); f(x; y)); f 3 (x; y; z) = f(f(x; x); f(y; z)) is generted y the set F consisting of the plce function f(x; y). A set F of functions is complete or Sheer set if F genertes every function, no mtter wht the numer m of vriles is. (We of course men here functions of the kind considered.) A function f is Sheer function if its unit set is Sheer set. It ws shown y Post in [] tht the set of ll plce functions is Sheer set, for ny n. The following result ws estlished y the present columnist in 958, [4]. Theorem. Assume tht n 5 nd F is set of functions contining ll permuttions of the numers ; ; : : :; n nd, in ddition, n ritrry function f(x; y) which is nondegenertely inry nd ssumes ll of the numers ; ; : : :; n s vlues. Then F is Sheer set. Theorem. does not hold true for vlues n 4. Counter exmples for n = nd n = 3 re provided y the functions dened y the tles It is esy to verify tht neither of these functions, ugmented with ll permuttions, genertes ll functions.

4 Of specil interest for our considertions lter is the cse n = 4. To see tht the sttement corresponding to Theorem. does not hold, consider the function f 4 (x; y) dened y the tle Then the set F 4 consisting of f 4 nd the 4 permuttions is not complete. Indeed, it is not dicult to verify tht the set H of oneplce functions generted y F 4 consists of the 4 permuttions, the 4 constnt functions nd the 3 functions ssuming two vlues, ech of them twice. Thus, H misses the 44 functions ssuming exctly three vlues, s well s the 48 functions ssuming two vlues in the rtio 3:. We will discuss elow lso the set H 0 of oneplce functions generted y f 4 lone. For instnce, f 4 (x; y) is the circulr permuttion (34) nd f 4 (x; f 4 (x; x)) is the function whose vlue sequence is 33 (tht is, the sequence of the four vlues listed in the numericl order of the rguments). Thus, these two functions re in H 0. However, contrry to H, no constnt function is in H 0. This follows ecuse f 4 is selfconjugte under the permuttion p(x), dened s the product of two trnspositions: (3)(4). Selfconjugcy under p mens tht f stises the eqution f(x; y) = p(f(p (x); p (y))): By considering composition sequences, it cn e veried tht function selfconjugte under p genertes only functions selfconjugte under p. Since none of the constnts is selfconjugte under (3)(4), they re not in H 0. For n 3, ny function f(x; y) generting ll permuttions in the symmetric group S n is Sheer function. This is n immedite consequence of Theorem., for n 5. The sttetement holds lso for n = 3 nd n = 4, [5]. The sttement does not hold for n = ecuse S is cyclic. Oneplce functions cn never e Sheer functions. In the sequel we re mostly concerned with the (monoid of) n n oneplce functions. How mny oneplce functions re needed to generte ll oneplce functions nd, in prticulr, which oneplce functions re such genertors If n = then two functions suce, nmely, the trnsposition () nd one of the two constnts. The solution for n 3 is presented in the following theorem, where y sis of the symmetric group S n we men ny two permuttions generting S n. Theorem. Assume tht n 3. Then three oneplce functions generte ll oneplce functions i two of them constitute sis of the symmetric

5 group S n nd the third ssumes exctly n vlues. No less thn three functions generte ll oneplce functions. See [5] for proof of Theorem.. [] is comprehensive study out the ses of the symmetric group (lso out ses consisting of more thn two elements). For instnce, the circulr permuttion (... n) nd the trnsposition () constitute sis, for ny n. A circulr permuttion nd n ritrry trnsposition form lwys sis if n is prime. Functions dened y nite utomt Insted of truth vlues, the elements of our sic set N = f; ; : : :; ng cn eqully well e viewed s sttes of nite deterministic utomton A = (f; ; : : :; ng; ; ), where is the lphet nd the trnsition function, everywhere dened. The utomt we consider re without initil nd nl sttes. (In the erly dys of utomt theory, such utomt were often referred to s Medvedev utomt, fter Ju. T. Medvedev. He trnslted the seminl 95 volume \Automt Studies" into Russin, nd included his own rticle in the trnsltion. The ltter remined lrgely unknown in the West.) All utomt considered in this pper re s indicted ove. Clerly, ech letter of the lphet cn e ssocited to the oneplce function g (x), dened y the condition g (x) = (x; ). This ssocition is extended in the nturl wy to concern words over : ctention of letters corresponds to the composition of functions. (Thus, the ssocition is morphism of the semigroup + into the semigroup of functions mpping N to N. We prefer here semigroups to monoids since we wnt to represent functions with nonempty words over.) The following prolem res now rise quite nturlly. (i) Wht is the set F (A) (susemigroup of the whole semigroup) of functions ssocited to given utomt A Does this set contin functions of prticulr type, for instnce, constnts Is the utomton functionlly complete in the sense tht its set of functions contins ll functions (Of course, we re concerned here with oneplce functions only.) (ii) Assuming tht F (A) contins previously specied function or one in specied clss of functions, wht is the \simplest" representtion of such function s word over A nturl mesure of complexity here is the word length. Prolems (ii) cn e restricted to concern only functionlly complete utomt. Consider rst the twostte utomton 3

6 ; Figure with the lphet f; g. The vlue sequences of the two functions g nd g re nd, respectively. All 4 functions, with vlue sequences,,,, re represented y the words ; ; ;, respectively. Thus, the utomton is functionlly complete nd ll functions re represented y words of length. Consider next the threestte utomton AK A ; c A A A A ; 3 A A A c c Figure with the lphet f; ; cg. By Theorem., the utomton is functonlly complete. In the following tle we give shortest possile representtion for ech of the 7 functions. The functions re numered ccording to the \lpheticl" order of their vlue sequences. 4

7 numer vlue sequence representtion cc c 3 3 c 4 c 5 c c c 0 c c 3 3 c 4 c c 5 3 c c 8 33 c 9 3 c c c 4 33 c 5 33 c 33 c cc Thus, ltogether 0 dierent functions re represented y words of length. Additionlly, functions re represented y words of length 3, nd 7 further functions y words of length 4. The remining exceptionl functions, 5, 0, 9 require word of length 5 for their representtion. We oserve tht one of these exceptionl functions is constnt, wheres the other two constnts re represented y words of length 4 ut y no shorter words. Since there re ltogether 0 (nonempty) words of length 4, some functions possess mny representtions using such words. The gretest numer is possessed y function 5 which hs no less thn 7 representtions using words of length 4: c; c ; c; c; c ; c 3 ; 3 c; c; c ; c ; c 3 ; cc; cc; c c; cc; c ; c 4 : Further detils re contined in [7]. By Theorem., functionlly complete nstte utomt over the lphet f; ; cg cn e constructed, for ny n. 5

8 Finlly, consider the following two fourstte utomt: ; c c ; 4 3 ; c c Figure 3 ; c ; c ; c 4 3 ; c Figure 4 In oth cses,, nd c re circulr permuttion, trnsposition nd nonpermuttion, respectively. However, the sitution is drsticly dierent in oth cses. While the former utomton is functionlly complete y Theorem., the ltter utomton A 0 genertes only functions in the set H 0 discussed in Section. Indeed, the vlue sequences of the functions g (x), g (x) nd g c (x) in the ltter utomton re 34, 43, 33, respectively. Considering the function f 4 (x; y) dening the set H 0, we see tht g (x) = f 4 (x; x); g c (x) = f 4 (x; f 4 (x; x)); g (x) = f 4 (x; f 4 (f 4 (x; x); f 4 (x; x))): This shows tht every function represented y the ltter utomton A 0 is in H 0 nd, consequently, selfconjugte under the permuttion (3)(4). Hence,

9 no constnts re represented. We leve it to the reder to verify tht the permuttions represented constitute the group consisting of the identity (), two circulr permuttions (34) nd (43), two trnspositions (3) nd (4), s well s three products ()(34), (3)(4) nd (4)(3). All these permuttions p stisfy the eqution p = gpg, where g is the permuttion (3)(4). 3 Trget functions We now discuss the prolem re (ii) referred to in the preceding section. Thus, we specify function f : f; ; : : :; ng! f; ; : : :; ng nd consider representtions of f y nstte deterministic utomt, in the sense descried ove. Of course, given f, we cn immeditely construct n utomton, where one of the letters directly gives rise to f. We wnt to consider ll nstte utomt nd, for ech utomton, look for the \simplest" representtion of f. It is lso possile tht f is not t ll represented y specic utomton. For instnce, if f is constnt, 3cycle or function ssuming exctly 3 vlues, then it is not represented y the utomton A 0 considered t the end of the preceding section. By denition, if n utomton A is functionlly complete, then ny f is represented. Given nite deterministic utomton A = (f; ; : : :; ng; ; ) nd function f : f; ; : : :; ng! f; ; : : :; ng, we denote y L(A; f) + the lnguge, consisting of ll words representing f. Thus w L(A; f) mens tht w tkes the utomton A from the stte i to the stte f(i), for ny i = ; ; : : :; n. Being n intersection of regulr lnguges, L(A; f) is lwys regulr. L(A; f) eing empty mens tht f is not represented y A. In choosing the \simplest" representtion, we indicte complexity y word length. This gives rise to the following denitions, [7]. For lnguge L, we denote y min(l) the length of the shortest word in L. By denition min(;) =. (Since we don't consider lnguges contining the empty word, min(l) is lwys positive integer or.) For given A nd f, the complexity C(A; f) of f with respect to A is dened y C(A; f) = min(l(a; f)): The complexity C(f) of function f : f; ; : : :; ng! f; ; : : :; ng is de ned y C(f) = mx C(A; f); where A rnges over nstte utomt such tht L(A; f) = ;. The complete complexity CC(f) of f is dened y CC(f) = mx C(A; f); where A now rnges over functionlly complete nstte utomt. 7

10 Some oservtions re immedite. For ny f, n utomton A f with C(A f ; f) =, s well s n utomton A 0 with f C(A0 ; f) =, cn e eectively constructed. Both of the complexities C(f) nd CC(f) re positive f integers nd, for ny f, CC(f) C(f). Also the upper ound C(f) n n is cler, in view of the totl numer of functions. The next theorem, [7], shows tht, for some functions f, C(f) hs n exponentil lower ound. Theorem 3. There re functions f : f; ; : : :; ng! f; ; : : :; ng for which no polynomil P (n) with the property C(f) P (n) exists. Proof outline. We let f e the identity function nd consider vlues of n of the form n = p + p + : : : + p k, where the p's re distinct primes. (For instnce, we my tke the rst k primes.) Consider now the utomton A n = (f; ; : : :; ng; fg; ), where the trnsitions re dened s follows. The stte set is divided into k pirwise disjoint susets, consisting of p ; p ; : : :; p k elements, respectively. In ech of the susets, the only letter ects circulr permuttion. It is now cler tht C(A n ; f) = p p : : :p k : Since there exists no polynomil upper ound for the product of k primes in terms of their sum, our theorem follows. The proof is not vlid for the complete complexity CC(f). We return to the discussion of CC(f) in [7]. Clerly, there re functions f such tht CC(f) < C(f). 4 Cerny's conjecture We consider now constnt functions f c : f; ; : : :; ng! fcg, where c n. The following result holds for the complexity C(f c ). Theorem 4. C(f c ) n(n ): The proof of Theorem 4. is given in detil in [7]. The following outline should e quite helpful. We my ssume tht c = n; this cn lwys e chieved y renumering. Consider the following utomton B n = (f; ; : : :; ng; f; g; ), where is dened y the following grph: 8

11 @ q ; n Figure 5 n Thus, (i; ) = i +, for i n, (n; ) = nd (i; ) = i, for i n, (n; ) =. It is esily veried tht the constnt f n is represented y the word ( n ) n of length n(n ). It is somewht more complicted to show tht no shorter word suces. The result C(B n ; f n ) = n(n ) proves the theorem. Cerny's conjecture. C(f n ) = n(n ). In view of Theorem 4., Cerny's conjecture mounts to the following sttement. For ny utomton A with L(A; f c ) = ;, we hve C(A; f c ) n(n ). Cerny's conjecture is usully presented in terms of synchronizing words. By denition, w is synchronizing word for n utomton A if w tkes A from ny of its sttes to the sme stte. In this terminology, Cerny's conjecture sys tht, whenever n nstte utomton possesses synchronizing word, then it possesses synchronizing word of length (n ). Our version of the conjecture, presented ove, is slightly weker thn the \synchroversion". Indeed, ssume tht the synchroversion holds true nd A is n nstte utomton with L(A; f c ) = ;. Then some constnt function f d is represented y word of length (n ). This implies tht the previously chosen constnt function f c is represented y word of length (n ) + n = n(n ), ecuse the longest pth without loops in A is of length n. There is no direct rgument from our version to the synchroversion. However, it is very unlikely tht the synchroversion of the conjecture is flse nd our version true. Indeed, in view of the rich possiilities for functionl constructions (for instnce, see []), oth versions cn very well e wrong. We hve not seen Cerny's originl pper [] nd cnnot comment on his possile rguments supporting the conjecture. We discuss here riey two rguments tht could e used to support it. 9

12 Consider rst the following rgument y enumertion. The totl numer of ll words of length t most (n) is much lrger thn the totl numer of functions. Hence, it is likely tht some constnt gets n erly representtion mong such huge numer of words. The wekness of this rgument lies in the fct tht mny functions possess multiple representtions. This ws oserved lredy for the functionlly complete 3stte utomton discussed in Section. Another rgument possily supporting the conjecture concerns the longest possile descent to constnt. Consider the synchroversion of the conjecture. We know tht synchronizing word exists, nd wnt to nd short one. The function g ssocited to letter mps ech suset S of the stte set N into nother suset S 0. Therey the crdinlity of S 0 is t most tht of S. We egin with the whole set N nd im t set of crdinlity, using dierent functions g. In this process, there re descending steps when the crdinlity ecomes strictly smller, s well s permuting steps, prepring for the next descent. There re t most n descending steps. Between two descending steps, there re t most n permuting steps ecuse the longest pth without loops hs n edges. Hence, the totl numer of steps is t most (n ) + (n )(n ) = (n ). The wekness of this rgument lies in the estimte concerning permuting steps. Doule trnsitivity might e needed: we might hve to trnsfer the pir (i ; i ) to the pir (j ; j ) in order to descend further. For this trnsfer, n steps might not suce. (Actully, they lwys suce with n = 3, for which vlue the rgument is vlid.) The following exmple, originlly due to [3], illustrtes this point @ 4 3 Figure The utomton is smll enough for us to give lso its \suset chrt", indicting the suset trnsformtions ected y the two functions g nd g. Oserve tht, in generl, the suset chrt gives very little informtion 0

13 out the semigroup of functions generted y the utomton ut it gives ll necessry informtion out synchronizing words. For the ove utomton, the suset chrt looks s follows: 34 ' $ & % 3 3 & 4 3 ; 4 34 % 3 4 ' $ & % & & % Figure 7

14 In this exmple, (n ) = 9 nd n(n ) =. On the other hnd, the shortest pth from the whole set f; ; 3; 4g to some singleton set is of length 9, nd to the previously chosen singleton set f4g is of length. However, 4 > n permuting steps re needed t the level of twoelement susets. The conjecture is sved ecuse, \mirculously", < n permuting steps suce t the level of threeelement susets. It would e too lengthy to summrize here the literture round Cerny's conjecture. The contriutions of J.E. Pin hve een quite sustntil. (For instnce, see [8 0].) In fct, his recent inspiring visit to Turku, lrgely initited our interest in the conjecture. The references [4], [5], [], [3] should e mentioned s recent contriutions to the re. 5 Rod coloring The decidility of most (ut not ll!) of the prolems for functions introduced in Section is firly ovious. It is decidle whether or not given utomton is functionlly complete, or whether specic function elongs to the set of functions represented. It is decidle whether or not synchronizing word exists for given utomton nd, if it does, shortest synchronizing word cn e found. However, the lgorithms re mostly sed on n exhustive serch; nice necessry nd sucient conditions re lrgely missing. The rod coloring prolem, [], elongs to the re of synchronizing words. We re given nite directed grph G, where the edges re unleled. The tsk is to nd leling of the edges tht turns G into deterministic utomton possessing synchronizing word. The term \rod coloring" is due to the following interprettion. A trveler lost in the grph G cn lwys nd wy ck home regrdless of where he/she ctully strted, provided he/she follows sequence of lels (colors) constituing synchronizing word. In the setup of the prolem, the given grph G stis es the following requirements: (i) All vertices hve the sme outdegree ( condition necessry for mking G n utomton) nd (ii) G is strongly connected nd the gretest common divisor of the cycle lengths is ( condition necessry for synchronizing words). For instnce, consider the grph

15 4 Figure 8 3 The leling c c c 4 Figure 9 3 c constitutes wrong solution: s seen in Section, no synchronizing word cn exist. However, the chnge of the lels nd c in the rrows emnting from the stte leds to the correct solution 3

16 c c c 4 Figure 0 3 c Here we hve the shortest synchronizing word. References [] Adler, R., Goodwin, I. nd Weiss, B.; Equivlence of topologicl Mrkov shifts, Isrel. J. Mth. 7 (977) [] Cerny, J.; Poznmk k homogennym experimentom s konecnymi utomtmi, Mt. fyz. cs. SAV 4 (94), [3] Cerny, J., Pirick, A. nd Rosenuerov; On directle Automt, Kyernetik, 7, 4 (97) [4] Duuc, L.; Sur les utomtes circulires et l conjecture de Cerny, Informtique theorique et Applictions, 3, 3 (998) 34. [5] Gohring, W.; Miniml Initilizing Word: Contriution to Cerny's Conjecture, Journl of Automt, Lnguges nd Comintorics (997) 4, 09. [] Imreh, B. nd Steiny, M.; Directle nondeterministic utomt, Act Cyernetic, 4, (999) 05. [7] Mteescu, A. nd Slom, A.; Functionl constructions in nite utomt. In preprtion. 4

17 [8] Pin, J.E.; Le proleme de l synchronistion, Contriution l'etude de l conjecture de Cerny, These de 3 e cycle l'universite Pierre et Mrie Curie (Pris ), 978. [9] Pin, J.E.; Le proleme de l synchronistion et le conjecture de Cerny, Noncommuttive structures in lger nd geometric comintorics, De Luc, A., ed., Quderni de l Ricerc Scientic, CNR, Rom, 98, 09, [0] Pin, J.E.; On two comintoril prolems rising from utomt theory, Annls of Discrete Mthemtics, 7 (983) [] Piccrd, S.; Sur les ses du groupe symetrique et les couples de sustitutions qui engendrent un groupe regulier, Pris, Liririe Vuiert (94). [] Post, E.L.; Introduction to generl theory of elementry propositions, Amer. J. Mth. 43 (9) [3] Rystsov, I.K.; Qusioptiml Bound for Length Reset Words for Regulr Automt, Act Cyernetic, (995) [4] Slom, A.; A theorem concerning the composition of functions of severl vriles rnging over nite set, J. Symolic Logic, 5 (90) [5] Slom, A.; On the composition of functions of severl vriles rnging over nite set, Ann. Univ. Turkuensis, Ser. AI, 4 (90). [] Slom, A.; On sic groups for the set of functions over nite domin, Ann. Acd. Scient. Fennice, Ser. AI, 338 (93). 5

18 Turku Centre for Computer Science Lemminkisenktu 4 FIN050 Turku Finlnd University of Turku Deprtment of Mthemticl Sciences Ao Akdemi University Deprtment of Computer Science Institute for Advnced Mngement Systems Reserch Turku School of Economics nd Business Administrtion Institute of Informtion Systems Science