1. For edge q → p suppose p= q ά
1 Comments on the proof (see ArXiv) with accent on more complicated points MINIMAL SYNCHRONIZING WORD AND MORE THAN 50 YEARS OF THE CERNY CONJECTURE Trahtman 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 q ά
So we obtain from a graph the complete deterministic finite automaton (DFA) 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. Upper bound (n3-n)/6 Frankl, 1982, Kljachko, Rystsov, Spivak, 1987,
2 A word w is called synchronizing (magic, reset) word of an automaton if 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 the Cerny 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,
Shykula 2017, declared also cubic, but less
Gap

3. 3 Several hundreds of articles consider this 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, 15 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 of word with element mi,j=1 if qi=qj and 0 otherwise
Lemma 1. Let Γ be the transition graph of a DFA.
Then Γus  Γs for any words u and s. For the state p outside Γs also p outside Γus , the column i in both matrices As and Aus consists of zeros. Nonzero columns of Aus have units in As.
Corollary 1 The matrix Ms of synchronizing word s has zeros in all columns except one and units in the residuary column.
Remark 1 The rank of the matrix Mt of word t is equal to the number of nonzero columns of Mt.

5. 5 Space of matrices of word
Lemma 2 The dimension of the space V generated by all n x k matrices of words is n(k-1)+1.
Corollary 2. The dimension of the space generated by nX(n-1)-matrices of words is at most (n-1)2.
Lemma 3 Suppose that for the matrix Mu of nonempty word u and for matrices Muj of words uj one has Mu =∑ j muj.
Then ∑ j =1.

6. We denote by S the rational series depending on the set P defined as
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.
Lemma 5 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 - ∑ j E= ∑ 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 Mu generate a space V such that for every nontrivial matrix Mt € V of word t (S, t)=i.

7. Definition Two matrices Mu and Mv of word are called q-equivalent (~q)
Q-EQUIVALENCE
Definition Two matrices Mu and Mv of word are called q-equivalent (~q)
if the columns of the state q of both matrices are equal. If units in
column q of Mv is a subset of the set of units of Mu then Mv Mu
Lemma 6. Let S be a rational series depending on the state q.
Then for matrices Ma, Mu, Mv of words
Mu ~q Mv  MaMu ~q MaMv
Mv Mu  Ma Mv Ma Mu
Corollary 7. For synchronizing word s, matrices Ma and Mu ~q Mv
Ms ~q MaMv  Ms=MaMu = MaMv. ≤
Remark. (S,u) = (S,v) if M_u ~q M_v, (S,v) ≤ (S,u) if M_v M_u

8. 8 Matrices Lx of word with Lx ~q Mx of word x.
Definition.The matrix of word Lu ~q Mu with (S,u)=n-i for 1<i≤ n has n-i+1 units in the column one of the state q and remaining i-1 units in the column i.
Lemma 7. MvLu ~q Lvu. If (S,u)=(S,vu) then MvLu = Lvu.
For invertible matrix Mv also MvLu = Lvu
The word u of every matrix Lu is not synchronizing.
Lemma 8. Let the space W be generated by matrices Lw with (S,w)=n-i for 1<i≤n.
The matrix Ms not in W for the word s with (S,s)=n If the sum ∑j=1k λwjLwj is a matrix Mx of word x then Mx=Lx for all (S,wj)=(S,x).
The number of linear independent matrices Lw with constant (S,w)=n-i is at most n and kn with k distinct values of (S,x).

9. Sketch of proof: Suppose that Mt =∑λwLw.
Let us divide the sum and let Wi be set of Liw with nonzero column i .
Mt = ∑i€I∑λiwLiw for nonzero columns I.
So in the sum Wi = ∑λiwLw every Lw has common value (S,w) = n-i and zeros in all columns except i and q. The number of units in the column i is a multiple of i-1 and is equal to (i-1) λiw and in the column q is equal to (n - i + 1) λiw . The number of all units of the matrix Mt is n. So
n= ∑ i € IλiwLiw (i-1)+∑ i € IλiwLiw (n-i+1) =∑i€In.
Therefore the size |I|=1, whence Mt=Lt (by Corollary 3 ).
The proof for the matrix Ms is similar of Mt, but from (i-1) λiw=0 and i-1≠0 follows that ∑ λiw=0, whence the matrix Ms must be zero.
The space W with k distinct values of (S, x) has dimension at most kn by Lemmas 2 and 5.

10. 9 The equation with unknown matrix Lx Mu Lx=Ms
The matrix of word Lx of rank two is not necessarily matrix of words in the alphabet ∑ unlike Mu and Ms.
Remark The columns of matrix are enumerated with number one for q.
A s = q for the irreducible synchronizing word s.
The 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 the matrix Mu.

11. Lemma 9
Every equation Ms=Mu Lx has a solutions Lx with
(S,x) ≥0 and nonzero column q .
The units in the column q of of minimal Lx correspond nonzero columns of Mu, the rest of cells in q are zero
cells and |N(u)|-1=(S,x).
Every solution Ly satisfies the equation if and only if
Lx Ly for Lx with minimal (S,x).
Corollary For matrix Mu with (S,x)=0 the word u is synchronizing.
The set G of nonzero columns of size |N(u)|=m of the
matrix Mu can be considered as the set of states of the automaton of the same size m or the set of m units in
the column q of the minimal solution Lx of the equation MuLx=Ms also of the same size. It is one-to-one correspondence.

12. (S,z) ≤ (S,x) and |N(ua)| ≤ |N(u)|.
Lemma 10
For solutions Lx and Lz with minimal (S,x) and (S,z) of equations Ms=MuLx and Ms=MuaLz
(S,z) ≤ (S,x) and |N(ua)| ≤ |N(u)|.
Corollary: Let two equations MuLx=Ms and MuwLy=Ms have minimal solutions Lx and Ly.
Then |N(uw)| ≤ |N(u)| and (S,y) ≤ (S,x),
in particular,
|N(uw)| = |N(u)| implies (S,y) = (S,x).

13. 12 The directed acyclic graph (DAG)
The pairs (Mu, Lx) such that MuLx=Ms are vertices of DAG. The root (Mα,Lx) of the graph has singular matrix Mα of a letter α.
The path to the vertex (Mu, Lx) defines a space Vx generated by matrices Lx from vertices of the path.
If the matrix Lz from the vertex (Muβ, Lz) does not belong to the space Vx of a path to the vertex (Mu, Lx) then the edge β goes from (Mu, Lx) to (Muβ, Lz) .
The generators of Vx are linearly independent and dim(Vx) is not less than the length of the path from the root.

14. Vx is a union of subspaces Wj with fixed (S,x).
Main Lemma
For every nonnegative k<n-1 there is a word u of length at most kn+1 such that |N(u)|<n-k of the matrix Mu.
In Proof
We consider the space Vx of every path in DAG to (Mu,Lx) from the root with a singular matrix Lx of a letter. dim(V_x) grows together with the length of path to (Mu,Lx) and |u|.
Vx is a union of subspaces Wj with fixed (S,x).
Our goal is to prove that for some generator Lx of some Vx and some letter β the solution Ly of the equation MuβLy=Ms does not belong to Vx.

15. 14 Continue main Lemma (to contradiction in Wp)
For opposite case let the space Wp have minimal |N(u)| for generators Lx of vertex (MuL,x ) Equations MuLx=Ms and MuβLy=MuMβLy=MuMβy =Ms with solutions Lx and Ly with minimal (S,x)=(S,y) are studied.
Using almost all former lemmas and results we prove first that for every generator Lx in Wp Lx= Mβ Ly for suitable Ly in Wp and then we prove the same for arbitrary Lw in Wp. (Details on next pages).
Consequently for every matrix Lw in Wp Lw = Md Ly for every letter, and therefore for every word d, in particular, for s with |N(s)|= This contradicts to n-1>(S,w) of Lw.
So Wp and some Vx can be extended and dim(Vx) grows together with |u|. The dimension of Vx is restricted by kn in view of |N(u)|≥n-k (Lemma 7).

16. Muβ Ly=Mu Mβ Ly=Mu Mβy =Ms
15 Two equations for matrices Mu and Ms with both solutions in Wp
MuLx=Ms and
Muβ Ly=Mu Mβ Ly=Mu Mβy =Ms
Let us prove the equality of solutions Lx and Mβy
Mβy ~ L β y by definition. MβLy ~ Lβy by Lemma 7. (S, y) = (S, x) for minimal solutions. (S, y) = (S, x) and MβLy ~ Lβy imply MβMy = Lβy
by Lemma 7. So for every letter β and suitable Ly
Lx = MβLy = Lβy for generator Lx and Ly of Wp
Let‘s go to an arbitrary matrix in the space Wp
Lz=∑λm Lxm = ∑λmMβLym=Mβ ∑λm Lym for suitable generators Lxm
All Lym have two common nonzero columns due to com mon (S,ym)=(S,z). As well as Lz by Corollary 3.

17. 16 Continuation - to closed Wp in main lemma
Unit in Lz is a product of two units, one from Mβ and one from ∑. So every nonzero column of Mβ has corresponding unit in ∑. The second column has in such row zero in view of ∑λm=1.
Let now j be zero column of Mβ of a letter β. Hence j is also zero column of MumMβ. Then in the equation MumMβ=Ms every minimal solution Lym must have zero in the cell (j,q) in view of minimality of the solution.
Thererore the sum ∑Lym has zero in this cell.
So the matrix ∑Lym has unit in second nonzero column due to ∑λm=1
Thus the ∑Lym is a matrix of word t in the space Wp.
For arbitrary Lz in Wp
Lz=MβLt = Lβt .

18. 17 Theorem The deterministic finite complete
n-state synchronizing automaton has synchronizing word of length at most (n-1)2.
Corollary For every k<n in deterministic complete n-state strongly connected DFA B there exists a word t of length at most (n-i-1)n+1 such that |Bt|≤k.
Theorem Let |Aw|<|A|-1 for a letter w of complete n-state DFA over alphabet W.
Then the minimal length of synchronizing word is less than (n-1)2.

19. 18 Automata on the border (n-1)2
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. (found by Don and Zantema)

20. 19 All automata of minimal reset word of length less than (n-1)2
Size
n=4
n=5
n=6
n=7
n=8
n=9 q<3
n=10 q<3
n=11 q<3
(n-1)2 9 16 25 36 49 64 81
100
Max of minimal length
q=2 8
q=3 8
q=4 8
q=2 15
q=3 15
q=4 15 23 22 32 31 30 44 42 58 74 92
The growing gap between (n-1)2 and Max of 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, of size 4, one of size 5 and one of size 6.

21. THANKS !