Syntactic Complexity of Suffix-Free Languages. Marek Szykuła

Inroducion Upper Bound on Synacic Complexiy of Suffix-Free Language Univeriy of Wrocław, Poland Join work wih Januz Brzozowki Univeriy of Waerloo, Canada DCFS, 25.06.2015

Abrac Inroducion Sae and ynacic complexiy Suffix-free language We udy ynacic complexiy of he cla of uffix-free language: A language L i uffix-free if w = uv L, where u i non-empy, implie ha he uffix v L. Conribuion Brzozowki, Li, and Ye (TCS 2012) conjecured ha: For n 6, he ynacic complexiy of uffix-free language i (n 1) n 2 + n 2. For n 6, he ynacic complexiy of bifix-free language i (n 1) n 3 +(n 2) n 3 +(n 3)2 n 3. For n 6, he ynacic complexiy of facor-free language i (n 1) n 3 +(n 3)2 n 3 + 1. We prove he fir conjecure (abou uffix-free language).

Lef quoien Inroducion Sae and ynacic complexiy Suffix-free language The (lef) quoien of a regular language L by a word w i w 1 L = {x Σ wx L}. Analogouly, for a ae q of a minimal DFA A = (Q,Σ,δ,q 0,F) recognizing L: L q = {x Σ q x F}, where x i he ranformaion induced by word x. So L q i he e of word aking q o an acceping ae.

Sae complexiy Inroducion Sae and ynacic complexiy Suffix-free language Nerode righ congruence on Σ For a regular language L and word x,y Σ : x L y if and only if xv L yv L, for all v Σ Sae complexiy The ae complexiy or quoien complexiy κ(l) of a regular language L i: The number of equivalence clae of L. The number of lef quoien of L. The number of ae in a minimal DFA recognizing L.

Synacic complexiy Inroducion Sae and ynacic complexiy Suffix-free language Myhill congruence For a regular language L and word x,y Σ : x L y if and only if uxv L uyv L for all u,v Σ Synacic complexiy The ynacic complexiy σ(l) of a regular language L i: Σ + / L he number of equivalence clae of L. The ize of he ynacic emigroup of L. The ize of he raniion emigroup of a minimal DFA recognizing L.

Inroducion Sae and ynacic complexiy Suffix-free language Synacic complexiy of a cla of language The ynacic complexiy of a cla of language i: The ize of he large ynacic emigroup of language in ha cla. Expreed a a funcion of he ae complexiie n = κ(l) of he language. In oher word Suppoe we have an n-ae minimal DFA recognizing ome language from he given cla. We ak how many ranformaion (a mo) can be in he raniion emigroup of he DFA.

Inroducion Sae and ynacic complexiy Suffix-free language Propoiion n 1 σ(l) n n The bound are igh for n > 1 in he cla of all regular language.

Previou reul Inroducion Sae and ynacic complexiy Suffix-free language Gome, Howie 1992: (parially) monoonic emigroup. Krawez, Lawrence, Shalli 2003: unary and binary alphabe. Holzer, König 2004: unary and binary alphabe. Brzozowki, Ye 2010: ideal and cloed language. Beaudry, Holzer 2011: emigroup of reverible DFA. Brzozowki, Liu 2012: finie, cofinie, definie, revere definie language. Brzozowki, Li, Ye 2012: prefix-, uffix-, bifix-, facor-free language. Iván, Nagy-György 2013: (generalized) definie language. Brzozowki, Li 2013: J -rivial and R-rivial language. Brzozowki, Li, Liu 2013: aperiodic, nearly monoonic emigroup. Brzozowki, Szykuła 2014: aperiodic emigroup. Brzozowki, Szykuła 2014: lef and wo-ided ideal language.

Suffix-free language Inroducion Sae and ynacic complexiy Suffix-free language Suffix-free language A language L i uffix-free if w = uv L, where u i non-empy, implie ha he uffix v L. Baic properie L ha he empy quoien. For w,x Σ +, if w 1 L hen w 1 L (xw) 1 L. For w Σ, he chain of quoien w 1 L,(ww) 1 L,... end in.

Inroducion Sae and ynacic complexiy Suffix-free language Synacic complexiy of uffix-free language Brzozowki, Li, Ye 2012 Here Two maximal emigroup were inroduced: W 5 (n) and W 6 (n). For n 5 he large raniion emigroup i W 5 (n). For n = 6 a large raniion emigroup i W 6 (n). I wa conjecured ha W 6 (n) i alo large for all n 7. The ize of W 6 (n) i (n 1) n 2 +(n 2). We have proved ha W 6 (n) i he unique large raniion emigroup for n 7.

Inroducion Sae and ynacic complexiy Suffix-free language Properie of DFA of uffix-free language Le D = (Q,Σ,δ,0,F) be a minimal DFA of a uffix-free language. Q = {0,1,...,n 1}. 0 i he ar ae. n 1 i he empy ae. Le T n be he raniion emigroup of D. Baic propery For any ranformaion T n and any ae q Q \{0}, we have 0 q unle 0 = q = n 1.

Colliding pair of ae Inroducion Sae and ynacic complexiy Suffix-free language Definiion Two ae p,q Q are colliding in T n, if here i a ranformaion T n uch ha: 0 = p, r = q for ome ae r Q \{0,n 1}. No ranformaion can map a pair of colliding ae o a ingle ae, excep o n 1. u r q 0 p... n 1

Colliding ae Inroducion Sae and ynacic complexiy Suffix-free language Remark In W 5 (n) all pair p,q Q \{0,n 1} are colliding. In W 6 (n) here are no colliding pair.

Inroducion The wine (lower bound) The proof of he upper bound Wine DFA wih he raniion emigroup W 6 (n) generaed by 5 leer. c c, d a, b 0 e 1 b 2 a... a n 2 a, c b, c, d b, d Σ\{e} d, e e e n 1 Σ

Inroducion The wine (lower bound) The proof of he upper bound Le S n = W 6 (n) be he raniion emigroup of he wine The raniion emigroup S n S n conain: All ranformaion ha map 0 o n 1. All ranformaion ha map 0 o a ae in {1,...,n 2}, and all oher ae o n 1. Thee are (n 1) n 2 +(n 2) ranformaion in oal.

Inroducion The wine (lower bound) The proof of he upper bound Idea We aume n 7. S n i he raniion emigroup of he wine. T n i he raniion emigroup of an arbirary DFA of a uffix-free language. We how ha T n S n = (n 1) n 2 +(n 2). I i poible ha T n S n. We conruc an injecive funcion: ϕ: T n S n (Injecive: for every we have ϕ() ϕ( ).)

Cae 1 Inroducion The wine (lower bound) The proof of he upper bound If S n, hen le ϕ() =. Thi i obviouly injecive, and ϕ() T n. From now, for S n we need o aign a ranformaion ϕ() T n.

Cae 2 Inroducion The wine (lower bound) The proof of he upper bound If S n and ha a cycle, hen we revere he chain p,p,... and map p o a minimal ae ha appear in cycle. : r r... 0 p... p k n 1 : r r... 0 p... p k n 1

Cae 3 Inroducion The wine (lower bound) The proof of he upper bound If S n and ha no cycle, hen we revere he chain and make p a fixed poin. : 0 p... p k n 1 : 0 p... p k n 1

Cae 11 wih p < r Inroducion The wine (lower bound) The proof of he upper bound : q r r 1 r 2 0 p n 1 : q r r 1 r 2 0 p n 1

Cae 11 wih p > r Inroducion The wine (lower bound) The proof of he upper bound : q r r 1 r 2 0 p n 1 : q r r 1 r 2 0 p n 1

The la cae 12 Inroducion The wine (lower bound) The proof of he upper bound : : f r 1 r 2... 0 p n 1 f r 1 r 2... 0 p n 1

We have 12 cae Inroducion The wine (lower bound) The proof of he upper bound Cae 1 S n. Cae 2 ha a cycle. Cae 3 p n 1. Cae 4 There i a fixed poin r Q \{0, n 1} wih in-degree 2. Cae 5 There i r wih in-degree 1 ha i no a fixed poin and r n 1. Cae 6 There i r Q \{0, n 1} wih in-degree 2. Cae 7 There are q 1,q 2 Q \{0, n 1} ha are no fixed poin and aify q 1 n 1 and q 2 n 1. Cae 8 There are wo fixed poin r 1 and r 2 in Q \{0, n 1} wih in-degree 1. Cae 9 There i q Q \{0, n 1} ha i no a fixed poin and aifie q n 1, p < q, and a fixed poin f n 1. Cae 10 There i q Q \{0, n 1} ha i no a fixed poin and aifie q n 1, and a fixed poin f n 1. Cae 11 There i q Q \{0, n 1} ha i no a fixed poin and aifie q n 1. Cae 12 Any oher ranformaion. In each cae we aume ha doe no belong o he previou cae.

Summary Inroducion The wine (lower bound) The proof of he upper bound The 12 cae cover all poibiliie for. Funcion ϕ i injecive and ϕ(t n ) S n, and T n = ϕ(t n ) S n.

Inroducion Uniquene of maximaliy The wine (lower bound) The proof of he upper bound Theorem For n 7, he raniion emigroup S n of he wine i he only one reaching he upper bound. Theorem Five leer are neceary o generae S n. So five leer are needed o reach he upper bound.

Fuure work Inroducion The wine (lower bound) The proof of he upper bound Oher problem The echnique of injecive funcion can be applied o imilar problem. Earlier, we olved he problem of ynacic complexiy of lef ideal (5 ubcae) and wo-ided ideal (8 ubcae) (DLT 2014). The problem of upper bound for ynacic complexiy of bifix- and facor-free language remain open. Thank you!