Half a Century of the Cerny-Starke Conjecture: Synchronization in Automata

1. SYNCHRONIZATION. MORE THAN HALF A CENTURYOF THE CERNY-STARKE CONJECTURETrahtman A.N. Let any pair of outgoing edges of any vertex of the directed graph obtain different colors (letters) The considered directed graph has constant outdegree – the number of outgoing edges for any vertex is constant • So we obtain from a graph the complete deterministic finite automaton (DFA) q ά ά p For edge q → p suppose p= q ά For a set of states Q and mapping (a letter) ά consider a map Qά and Qs for the word s=ά1ά2… άi . Γs presents a map of Γ.

2. 2 A word w is called synchronizing (magic, reset) word of an automatonif w sends all states on an unique state. Jan Cerny in 1964 found n-state complete DFA with shortest synchronizing word of length (n-1)2. The hypothesis, well known today as theCerny conjecture states that it is an upper bound for the length of the shortest synchronizing word for any n-state automaton. (Starke 1966) known bounds Lower bound (n-1)2 Cerny 1964 : Upper bound (n3-n)/6 Frankl, 1982, Kljachko, Rystsov, Spivak, 1987, ,

3. 3 Several hundreds of articles considerthis problem from different points of view The conjecture is true for a lot of automata. Two conferences "Workshop on Synchronizing Automata“ (Turku, 2004) and "Around the Cerny conjecture“ (Wroclaw, 2008) were dedicated to this conjecture. The problem is discussed in "Wikipedia“ and other sites of Internet. Some automata with minimal synchronizing word of length (n-1)2 found by Kari,2001 n=6, size of alphabet 2, Roman, 2004, n=5, size of alphabet 3, 27 automata for n=3 and 4 (Cerny, Piricka, Rosenauerova, Trahtman, Don, Zantema. Together with the Road Coloring problem this simple-looking conjecture is arguably the most longstanding and famous open combinatorial problems in the theory of finite automata.

4. 4 Notions and some properties for the proof We connect a mapping of the set of states qi made by a word u of a matrix with element mi,j=1 if qi=qj and 0 otherwise. Matrix of word has in every row one unit and rest zeros. Let R(u) be the set of nonzero columns of Mu . The rank of the matrix Mt is equal to | R(t)|. cu is a set of states by mapping Mu Remark 1The columns of the matrix MuMa are obtained by permutation of columns Mu. Some of them can be merged. • The rows of the matrix MaMu are obtained by permutation of rows of matrix Mu. Some of these rows may disappear and replaced by another rows of Mu.

5. 5 Lemma 1. Remark 2 The rows of the matrix MaMu are obtained by permutation of rows of the matrix Mu Some of these rows may disappear and replaced by another rows of Mu. The columns of the matrix Mu Ma are obtained by permutation of columns Mu. Some of them can be merged with |R(ua)|<|R(u)|. The set R(u) of nonzero columns of Mu = set of states. Let Γ be the transition graph of a DFA. Then Γus  Γs for any words u and s. For every w R(u)ﬤR(v) implies R(uw)ﬤ R(vw). Rank of Muis equal to R(u) Lemma 2 • For every words a and u|R(ua)|<= |R(u) | • and R(au) is a subset of R(u)

6. 6 Space of matrices of word Lemma 3 The dimension of the space V of all n x k matrices of words is n(k-1)+1. Corollary. The dimension of the space generated by nX(n-1)-matrices of words is at most (n-1)2. Lemma 4Suppose that for the matrix Muof nonempty word u and for matrices Mujof words uj Mu=∑j muj. Then ∑ j =1 for every coordinate of Mu. ∑ j =1 or 0 for zero matrix.

7. 7 Distribution from left • Lemma 5 For every words b, and xi • Mb ∑ λi Mxi= ∑ λi Mb Mxi . • If ∑λiMxi is a matrix of word then also ∑ λiMb Mxi • is a matrix of word. RATIONAL SERIES Definition. Let P denote the subset of states with the characteristic column vector Pt with units in coordinates corresponding to the states of P and zeros everywhere else. Let C be a row of units. We denote by S the rational series depending on the set P defined as (S,u) = C MuPt-C Pt= C(Mu-E)Pt.

8. 8 Lemma 6 Let S be a rational series depending on the set P. Let Mu=∑ i Mui. Then (S,u)=∑ i (S,ui). If (S,ui)=i for every i then for nonzero Mu also (S,ui)=i. PROOF. By Lemma 3 ∑ j (Mu)j -E= ∑ j (Mu)j - ∑ jE= ∑ j((Mu)j - E). So (S,u)= C(∑ j Mu –E)PT= C(∑ j((Mu)j - E))PT = ∑ C( j (Mu)J–E)PT= ∑ j C (Mu–E)PT= ∑ j (S,uj) . Corollary 3.For the set of words u € U with constant rational series (S,u)=i the corresponding matrices Mugenerate a space V such that for every nontrivial matrix Mt € V of word t (S, t)=i.

9. 9The equivalence for the state q • Definition 1. For two matrices Mu and Mv ofword Mu~qMv ifcolumns of the state q of both matrices are equal. Lemma 7 Let Sq be a rational series depending on q. Then for matrices Mβ, Mu , Mu of words Mu~qMvimplies MβMu~q MβMv • Corollary. For As=q and Mu \sim_q Mv • Ms \sim_q MtMv -> Ms = MtMu = MtMv • Definition 2. The matrix Lx ~qMxofword x has (Sq,x)+1 units incolumn q and remaining units in next n-(Sq,x) columns

10. 10 The equation with unknown matrix LxMuLx=Ms • Lemma 8 Every equation Ms=Mu Lx has a solutions Lxwith (S,x) ≥0 and nonzero column q and units in next columns, one in column. • The units in the column q of of minimal Lx correspond nonzero columns of Mu, and |R(u)|-1=(S,x). • Every matrix Ly satisfies the equation iff Ly has in columnq units of qfrom minimalLx . • There exists one-to-one correspondence between the class cuand column q of minimal solution Lx.

11. 11Pseudoinversematrix • For Pseudoinverse matrix Ma_of the matrix Ma every nonzero column j of Ma with ai,j=1 the cell (j,i) of Ma- has unit. In still zero row_s of Ma- is added unit. • Eb=Mb Mb- Lemma 9 .If R(u)≠R(ub) then Mb merges columns Mu . The product Eb =MbMb– does not depend on arbitrary placing of units in MbMb–in empty rows. If R(u)=R(ub) then MbMb– returns all images of column Mb to its origin place.

12. 12Insert a matrix Eb=Mb Mb- Lemma 10For equation MuLx=Ms,some Mβ and every letter β MuβLy =Msfor minimal Lxand Ly. If R(u) is a subset of R(β) then for every matrix Mβ- forminimal solutions Lx one has Ly= Mβ- Lx and (Sq,y)<= (Sq,x). For R(uβ) ≠ R(u) then for minimal Lxand Ly=Ma-Lx for minimalsolutions (Sq,y) < (Sq,x). For R(u β) =R(u) or invertible Ma- for minimal Lx,Ly=Ma-Lxholds Sq,y)= (Sq,x).

13. 13Properties of pseudoinversematrix Lemma 11 The product Eb= Mb Mb- does not depend on arbitrary placing of units in Mb-in empty rows. In the case of |R(u)|=|R(ub) the product MuEbreturns columns of Mu to its origin place. Remark A set of solutions Ly=Ma-Lx with constant (Sq,y)=(Sq,x) can be created by help of invertible matrices Ma- (a maximal such set) .

14. 13The sequence of words of growing length • Let us consider the space Wjgenerated by j linear independent solutions Lx ofequation MuLx = Ms for a sequence of irreducible words u of growing length≤ j. So dim(Wj) = j. Let all matrices u have common zero column. Then lR(u) l ≤ n-1 and (S; x)≤ n -2 for minimal Lx by Lemma 11. • Assume u1 =ä for the right letter ä of minimal synchronizing word s. Theminimal solution Lx1 of the equation MuLx1 = Ms generates the space W1.Matrix Mu is singular with lR(u) l ≤ n - 1.We consider for Wj the set of solutions Ly of the equations MuLy = Ms withlul ≤ j + 1 and choose a matrix Lynot in Wj. Then Ly is added to the space ofwords Wj turning it into Wj+1. • The distinct linear independent solutions of theequationMuLx1 = Msare added consistently.lR(u) l > 1 • implies(S,x) > 0 for every solution Lxby Lemma 11.

15. 14 Main Lemma One can byhelp of invertible generalized inverse matrix to get maximal set of linear independent matrices Lxwith fixed rational series (Sq,x)by the expansion of the space Wj . Lemma 11 • Let proper subspace Wj of W be generated by Ms and a sequence of linear independent solutions Mx of words x of equationMuLx = Ms for |u)|≤ j and n>|R(u)|> 1 . • Thenthere exists outside Wj a solution Lzof the equation • MuLz = Ms for |u)|≤ j+ 1

16. 14 Proof of Lemma 11 • A set of solutions Lyof the equation MuaLy=Ms with constant (Sq,y)=(Sq,x) can be created by help of invertible matrices Ma- . • One can create so a maximal set of such linear independent matrices Ly with known (Sq,x) . • By Remark 2 the invertible matrix Ma- does not change the set of nonzero columns of Ly, doing only permutation of rows without changing number of units in every column. • On some step, Wjreaches maximal dimension for matrices with units in given columns. • Dim(Wj) with n-(Sq,x) nonzero columns is at most • n(n-(Sq, x) -1) +1.

17. 15 Continue the proof of Lemma 11 • Assume that contrary to Lemma every solution Ly of the equation MuβLy=Ms with |u|<=j belongs to Wj for any letter β . • Let us consider rational series (Si,x) that depends on the set of states corresponding first i nonzero columns. The generators Lx have units only in these columns. Then for all solutions Lx for • |u|< j+1 we have (Si,x) <=n-i. • Let Lw =∑λiLwi with generators Lwi ofWj and |wi|<j . So (Si,wi) =n-ix =(Si,w) for Lw. • Let Lt= Mβ- Lw= Mβ- ∑λiLwi= ∑λiMβ-Lwifor • Mβ-Lwi~Lβ-Lwi |u)|with(Si, β-wi) =n-i=(Si,wi).Hence • (Si, β-wi) =n-i=(Si,w).Thus Lt in Wj for every t with|t|≤ j+2. • Consequently Lt in Wj for every t. • Contradiction for proper subspace Wj .

18. 16 Theorem The deterministic finite complete n-state synchronizing automaton has synchronizing word of length at most (n-1)2. Theorem Let |Aw|<|A|-1 for a letter w of complete n-state DFA over alphabet W. Then the minimal synchronizing word has length less than(n-1)2.

19. 17Automata on the border (n-1)2(at most 3 letters) The vertices of automata for n=3 and n=4 can be united and present some new examples on the border (n-1)2 with more edges. (Don and Zantema)

20. 18 All automata of minimal reset word of length less than(n-1)2 The growing gap between (n-1)2 and Maxof minimal length inspires Conjecture The set of n-state complete DFA (n>2) with minimal reset word of length (n-1)2 contains only the sequence of Cerny, set of automata of size 3 and of size 4, one of size 5 and one of size 6.

21. THANKS !