� Strongly connected digraphs � be the partition of Q into n � m + 1 classes one of which is S and all others are � is a congruence of A . A =� is synchronizing and has S Return to our reasoning: let singletons. Then Clearly, the quotient as a unique sink. LATA 2008 – p.7/29
Strongly connected digraphs � be the partition of Q into n � m + 1 classes one of which is S and all others are � is a congruence of A . A =� is synchronizing and has S Return to our reasoning: let singletons. Then Clearly, the quotient as a unique sink. � S LATA 2008 – p.7/29
a b b a a; b a a a a a b b b b b b b b b b a a k � 1 t k � 1 � t � 1 k � t k ( k � 1) � 1 + 2 + � � � + ( k � 1) = 2 Automata with a Unique Sink k states has a unique k ( k � 1) � 2 If a synchronizing automata with sink, then it has a reset word of length . LATA 2008 – p.8/29
a a a b b b b b b k � 1 t k � 1 � t � 1 k � t k ( k � 1) � 1 + 2 + � � � + ( k � 1) = 2 Automata with a Unique Sink k states has a unique k ( k � 1) � 2 a If a synchronizing automata with b b a a; b sink, then it has a reset word of length . a a b b b b a 0 a LATA 2008 – p.8/29
a a a b b b b b b k � 1 t k � 1 � t � 1 k � t k ( k � 1) � 1 + 2 + � � � + ( k � 1) = 2 Automata with a Unique Sink k states has a unique k ( k � 1) � 2 a If a synchronizing automata with b b a a; b sink, then it has a reset word of length . a a b b b b a 0 a LATA 2008 – p.8/29
a a a b b b b b b k � 1 t k � 1 � t � 1 k � t k ( k � 1) � 1 + 2 + � � � + ( k � 1) = 2 Automata with a Unique Sink k states has a unique k ( k � 1) � 2 a If a synchronizing automata with b b a a; b sink, then it has a reset word of length . a a a b b b b a 0 a LATA 2008 – p.8/29
a a a b b b b b b k � 1 t k � 1 � t � 1 k � t k ( k � 1) � 1 + 2 + � � � + ( k � 1) = 2 Automata with a Unique Sink k states has a unique k ( k � 1) � 2 a If a synchronizing automata with b b a a; b sink, then it has a reset word of length . a a b b b b a 0 a LATA 2008 – p.8/29
a a a b b b b k � 1 t k � 1 � t � 1 k � t k ( k � 1) � 1 + 2 + � � � + ( k � 1) = 2 Automata with a Unique Sink k states has a unique k ( k � 1) � 2 a If a synchronizing automata with b b a a; b sink, then it has a reset word of length . a a a b b b b b b a 0 a LATA 2008 – p.8/29
a a a b b b b b b k � 1 t k � 1 � t � 1 k � t k ( k � 1) � 1 + 2 + � � � + ( k � 1) = 2 Automata with a Unique Sink k states has a unique k ( k � 1) � 2 a If a synchronizing automata with b b a a; b sink, then it has a reset word of length . a a b b b b a 0 a LATA 2008 – p.8/29
a b b k � 1 t k � 1 � t � 1 k � t k ( k � 1) � 1 + 2 + � � � + ( k � 1) = 2 Automata with a Unique Sink k states has a unique k ( k � 1) � 2 a If a synchronizing automata with b b a a; b sink, then it has a reset word of length . a a a a b b b b b b b b a 0 a LATA 2008 – p.8/29
a a a b b b b b b Automata with a Unique Sink k states has a unique k ( k � 1) � 2 a If a synchronizing automata with b b a a; b sink, then it has a reset word of length . a a b b b b a 0 a k � 1 steps and the t states still holds coins ( k � 1 � t � 1 ) is at most k � t . The total k ( k � 1) � 1 + 2 + � � � + ( k � 1) = The algorithm makes at most 2 length of the segment added in the step when length is . LATA 2008 – p.8/29
A =� u ( n � m +1)( n � m ) Q : u � S 2 S 2 v ( m � 1) S :v Q:uv � S :v uv A ( n � m + 1)( n � m ) 2 2 + ( m � 1) � ( n � 1) : 2 Strongly connected digraphs A =� is n � m + 1 Return to our reasoning: the quotient synchronizing with a unique sink and has states. LATA 2008 – p.9/29
Q : u � S S 2 v ( m � 1) S :v Q:uv � S :v uv A ( n � m + 1)( n � m ) 2 2 + ( m � 1) � ( n � 1) : 2 Strongly connected digraphs A =� is n � m + 1 A =� has a reset word u of length ( n � m +1)( n � m ) Return to our reasoning: the quotient 2 synchronizing with a unique sink and has states. Hence, . LATA 2008 – p.9/29
S 2 v ( m � 1) S :v Q:uv � S :v uv A ( n � m + 1)( n � m ) 2 2 + ( m � 1) � ( n � 1) : 2 Strongly connected digraphs A =� is n � m + 1 A =� has a reset word u of length ( n � m +1)( n � m ) Q : u � S . Return to our reasoning: the quotient 2 synchronizing with a unique sink and has states. Hence, . Then LATA 2008 – p.9/29
S :v Q:uv � S :v uv A ( n � m + 1)( n � m ) 2 2 + ( m � 1) � ( n � 1) : 2 Strongly connected digraphs A =� is n � m + 1 A =� has a reset word u of length ( n � m +1)( n � m ) Q : u � S . Return to our reasoning: the quotient 2 synchronizing with a unique sink and has S 2 . v of length ( m � 1) states. Hence, . Then Recall that we have assumed that the automaton has a reset word LATA 2008 – p.9/29
uv A ( n � m + 1)( n � m ) 2 2 + ( m � 1) � ( n � 1) : 2 Strongly connected digraphs A =� is n � m + 1 A =� has a reset word u of length ( n � m +1)( n � m ) Q : u � S . Return to our reasoning: the quotient 2 synchronizing with a unique sink and has S 2 . Then v of length ( m � 1) S :v is a states. Hence, Q:uv � S :v is a singleton. . Then Recall that we have assumed that the automaton has a reset word singleton, whence also LATA 2008 – p.9/29
Strongly connected digraphs A =� is n � m + 1 A =� has a reset word u of length ( n � m +1)( n � m ) Q : u � S . Return to our reasoning: the quotient 2 synchronizing with a unique sink and has S 2 . Then v of length ( m � 1) S :v is a states. Hence, Q:uv � S :v is a singleton. . Then uv is reset word for A , and the length of this Recall that we have assumed that the automaton has a reset word singleton, whence also ( n � m + 1)( n � m ) 2 2 + ( m � 1) � ( n � 1) : Thus, 2 word does not exceed LATA 2008 – p.9/29
q w q w q q a a 3 3 3 ab ab ab 3 3 b ab ab a b b 3 3 2 3 3 ab ab ab ab ab ab b a a Strongly connected digraphs Thus, we assume that our synchronizing automata are strongly connected as digraphs. LATA 2008 – p.10/29
q w q w q q a a 3 3 3 ab ab ab 3 3 b ab ab a b b 3 3 2 3 3 ab ab ab ab ab ab b a a Strongly connected digraphs Thus, we assume that our synchronizing automata are strongly connected as digraphs. Observe that such an automaton can be reset to any state. LATA 2008 – p.10/29
a a 3 3 3 ab ab ab 3 3 b ab ab a b b 3 3 2 3 3 ab ab ab ab ab ab b a a Strongly connected digraphs q of the automaton one Thus, we assume that our synchronizing automata are w q such that strongly connected as digraphs. w q from any initial q one will surely arrive at Observe that such an automaton can be reset to any state. That is, to every state can assign an instruction (a reset word) following state. LATA 2008 – p.10/29
Strongly connected digraphs q of the automaton one Thus, we assume that our synchronizing automata are w q such that strongly connected as digraphs. w q from any initial q one will surely arrive at Observe that such an automaton can be reset to any state. That is, to every state a can assign an instruction (a reset word) a 3 3 3 ab ab ab 3 3 following b ab ab a state. b b 3 3 2 3 3 ab ab ab ab ab ab 0 1 b a a 3 2 LATA 2008 – p.10/29
Road Coloring Now think of the automaton as of a scheme of a transport network in which arrows correspond to roads and labels are treated as colors of the roads. LATA 2008 – p.11/29
Road Coloring Now think of the automaton as of a scheme of a transport network in which arrows correspond to roads and labels are treated as colors of the roads. LATA 2008 – p.11/29
Road Coloring Now think of the automaton as of a scheme of a transport network in which arrows correspond to roads and labels are treated as colors of the roads. Then for each node there is a sequence of colors that brings one to the chosen node from anywhere. LATA 2008 – p.11/29
Road Coloring LATA 2008 – p.12/29
Road Coloring For the green node: blue-blue-red-blue-blue-red-blue-blue-red. LATA 2008 – p.12/29
Road Coloring For the green node: blue-blue-red-blue-blue-red-blue-blue-red. For the yellow node: blue-red-red-blue-red-red-blue-red-red. LATA 2008 – p.12/29
Road Coloring Now suppose that we have a transport network, that is, a strongly connected digraph. LATA 2008 – p.13/29
Road Coloring Now suppose that we have a transport network, that is, a strongly connected digraph. We aim to help people to orientate in it, and as we have seen, a neat solution may consist in coloring the roads such that our digraph becomes a synchronizing automaton. LATA 2008 – p.13/29
Road Coloring Now suppose that we have a transport network, that is, a strongly connected digraph. We aim to help people to orientate in it, and as we have seen, a neat solution may consist in coloring the roads such that our digraph becomes a synchronizing automaton. When is such a coloring possible? LATA 2008 – p.13/29
Road Coloring Now suppose that we have a transport network, that is, a strongly connected digraph. We aim to help people to orientate in it, and as we have seen, a neat solution may consist in coloring the roads such that our digraph becomes a synchronizing automaton. When is such a coloring possible? In other words: which strongly connected digraphs may appear as underlying digraphs of synchronizing automata? LATA 2008 – p.13/29
Road Coloring Now suppose that we have a transport network, that is, a strongly connected digraph. We aim to help people to orientate in it, and as we have seen, a neat solution may consist in coloring the roads such that our digraph becomes a synchronizing automaton. When is such a coloring possible? In other words: which strongly connected digraphs may appear as underlying digraphs of synchronizing automata? An obvious necessary condition: all vertices should have the same out-degree . In what follows we refer to this as to the constant out-degree condition. LATA 2008 – p.13/29
� = ( V ; E ) k > 1 v 2 V i = 0 ; 1 ; : : : ; k � 1 0 V = f v 2 V j 9 v v i (mo d k ) g : i 0 k � 1 S V = V V \ V = ? i 6 = j i i j i =0 Primitivity A less obvious necessary condition is called aperiodicity or primitivity: the g.c.d. of lengths of all cycles should be equal to 1 . LATA 2008 – p.14/29
v 2 V i = 0 ; 1 ; : : : ; k � 1 0 V = f v 2 V j 9 v v i (mo d k ) g : i 0 k � 1 S V = V V \ V = ? i 6 = j i i j i =0 Primitivity A less obvious necessary condition is called aperiodicity or primitivity: � = ( V ; E ) is a strongly connected digraph and k > 1 is the g.c.d. of lengths of all cycles should be equal to 1 . To see why primitivity is necessary, suppose that a common divisor of lengths of its cycles. LATA 2008 – p.14/29
k � 1 S V = V V \ V = ? i 6 = j i i j i =0 Primitivity A less obvious necessary condition is called aperiodicity or primitivity: � = ( V ; E ) is a strongly connected digraph and k > 1 is the g.c.d. of lengths of all cycles should be equal to 1 . v 2 V and, for i = 0 ; 1 ; : : : ; k � 1 , let 0 To see why primitivity is necessary, suppose that V = f v 2 V j 9 path from v v of length i (mo d k ) g : i 0 to a common divisor of lengths of its cycles. Take a vertex LATA 2008 – p.14/29
V \ V = ? i 6 = j i j Primitivity A less obvious necessary condition is called aperiodicity or primitivity: � = ( V ; E ) is a strongly connected digraph and k > 1 is the g.c.d. of lengths of all cycles should be equal to 1 . v 2 V and, for i = 0 ; 1 ; : : : ; k � 1 , let 0 To see why primitivity is necessary, suppose that V = f v 2 V j 9 path from v v of length i (mo d k ) g : i 0 to a common divisor of lengths of its cycles. Take a vertex k � 1 S V = V i . i =0 Clearly, LATA 2008 – p.14/29
Primitivity A less obvious necessary condition is called aperiodicity or primitivity: � = ( V ; E ) is a strongly connected digraph and k > 1 is the g.c.d. of lengths of all cycles should be equal to 1 . v 2 V and, for i = 0 ; 1 ; : : : ; k � 1 , let 0 To see why primitivity is necessary, suppose that V = f v 2 V j 9 path from v v of length i (mo d k ) g : i 0 to a common divisor of lengths of its cycles. Take a vertex k � 1 S V = V V \ V = ? if i 6 = j . i . We claim that i j i =0 Clearly, LATA 2008 – p.14/29
v v 0 v v n 0 ` + n m + n Primitivity v 2 V \ V i 6 = j . This means that in � there i j where v v : of length ` � i (mo d k ) and 0 to m � j (mo d k ) . Let are two paths from of length LATA 2008 – p.15/29
v v n 0 ` + n m + n Primitivity v 2 V \ V i 6 = j . This means that in � there i j where v v : of length ` � i (mo d k ) and 0 to m � j (mo d k ) . Let are two paths from of length v v 0 LATA 2008 – p.15/29
Primitivity v 2 V \ V i 6 = j . This means that in � there i j where v v : of length ` � i (mo d k ) and 0 to m � j (mo d k ) . Let are two paths from of length v v 0 . . . v to v n . 0 of length, say, ` + n and a cycle of length m + n . There is also a path Combining it with the two paths above we get a cycle of length LATA 2008 – p.15/29
V V ; V ; : : : ; V 0 1 k � 1 � V V i i +1 (mo d k ) � ` V V 0 1 V V ` (mo d k ) ` +1 (mo d k ) Primitivity k divides the length of any cycle in � , we have ` + n � i + n � 0 (mo d k ) and m + n � j + n � 0 (mo d k ) , i � j (mo d k ) , a contradiction. Since whence LATA 2008 – p.16/29
� ` V V 0 1 V V ` (mo d k ) ` +1 (mo d k ) Primitivity k divides the length of any cycle in � , we have ` + n � i + n � 0 (mo d k ) and m + n � j + n � 0 (mo d k ) , i � j (mo d k ) , a contradiction. Since V is a disjoint union of V ; V ; : : : ; V 0 1 k � 1 , and by the � leads from V V i to i +1 (mo d k ) . whence Thus, definition each arrow in LATA 2008 – p.16/29
Primitivity k divides the length of any cycle in � , we have ` + n � i + n � 0 (mo d k ) and m + n � j + n � 0 (mo d k ) , i � j (mo d k ) , a contradiction. Since V is a disjoint union of V ; V ; : : : ; V 0 1 k � 1 , and by the � leads from V V i to i +1 (mo d k ) . whence � definitely cannot be converted into a Thus, definition each arrow in ` originated V V 0 and 1 can terminate in the same vertex because Then V V ` (mo d k ) and in ` +1 (mo d k ) respectively. synchronizing automaton by any labelling of its arrows: for instance, no paths of the same length in they end in LATA 2008 – p.16/29
Road Coloring Conjecture The Road Coloring Conjecture claims that the two necessary conditions (constant out-degree and primitivity) are in fact sufficient. LATA 2008 – p.17/29
Road Coloring Conjecture The Road Coloring Conjecture claims that the two necessary conditions (constant out-degree and primitivity) are in fact sufficient. In other words: every strongly connected primitive digraph with constant out-degree admits a synchronizing coloring . LATA 2008 – p.17/29
Road Coloring Conjecture The Road Coloring Conjecture claims that the two necessary conditions (constant out-degree and primitivity) are in fact sufficient. In other words: every strongly connected primitive digraph with constant out-degree admits a synchronizing coloring . The Road Coloring Conjecture was explicitly stated by Adler, Goodwyn and Weiss in 1977 (Equivalence of topological Markov shifts, Israel J. Math., 27, 49–63). LATA 2008 – p.17/29
Road Coloring Conjecture The Road Coloring Conjecture claims that the two necessary conditions (constant out-degree and primitivity) are in fact sufficient. In other words: every strongly connected primitive digraph with constant out-degree admits a synchronizing coloring . The Road Coloring Conjecture was explicitly stated by Adler, Goodwyn and Weiss in 1977 (Equivalence of topological Markov shifts, Israel J. Math., 27, 49–63). In an implicit form it was present already in an earlier memoir by Adler and Weiss (Similarity of automorphisms of the torus, Memoirs Amer. Math. Soc., 98 (1970)) almost 40 years ago. LATA 2008 – p.17/29
Road Coloring Conjecture The original motivation for the Road Coloring Conjecture comes from symbolic dynamics, see Marie-Pierre Béal and Dominiues Perrin’s chapter “Symbolic Dynamics and Finite Automata” in Handbook of Formal Languages, Vol.I. LATA 2008 – p.18/29
Road Coloring Conjecture The original motivation for the Road Coloring Conjecture comes from symbolic dynamics, see Marie-Pierre Béal and Dominiues Perrin’s chapter “Symbolic Dynamics and Finite Automata” in Handbook of Formal Languages, Vol.I. However the conjecture is natural also from the viewpoint of the “reverse engineering” of synchronizing automata as presented here. LATA 2008 – p.18/29
Road Coloring Conjecture The original motivation for the Road Coloring Conjecture comes from symbolic dynamics, see Marie-Pierre Béal and Dominiues Perrin’s chapter “Symbolic Dynamics and Finite Automata” in Handbook of Formal Languages, Vol.I. However the conjecture is natural also from the viewpoint of the “reverse engineering” of synchronizing automata as presented here. The Road Coloring Conjecture has attracted much attention. There were several interesting partial results, and finally the problem was solved (in the affirmative) in August 2007 by Avraham Trahtman. LATA 2008 – p.18/29
Road Coloring Conjecture The original motivation for the Road Coloring Conjecture comes from symbolic dynamics, see Marie-Pierre Béal and Dominiues Perrin’s chapter “Symbolic Dynamics and Finite Automata” in Handbook of Formal Languages, Vol.I. However the conjecture is natural also from the viewpoint of the “reverse engineering” of synchronizing automata as presented here. The Road Coloring Conjecture has attracted much attention. There were several interesting partial results, and finally the problem was solved (in the affirmative) in August 2007 by Avraham Trahtman. Trahtman’s solution got much publicity this year. LATA 2008 – p.18/29
A = h Q; � ; Æ i � Q 0 � � 0 q � q ( ) 8 u 2 � 9 v 2 � q : uv = q :uv : 0 � ( q ; q ) 0 q � q � A A Stability Trahtman’s proof heavily depends on a neat idea of stability which is due to Karel Culik II, Juhani Karhumäki and Jarkko Kari (A note on synchronized automata and Road Coloring Problem, Int. J. Found. Comput. Sci., 13 (2002) 459–471). LATA 2008 – p.19/29
0 � ( q ; q ) 0 q � q � A A Stability Trahtman’s proof heavily depends on a neat idea of A = h Q; � ; Æ i stability which is due to Karel Culik II, Juhani � on Q as follows: Karhumäki and Jarkko Kari (A note on synchronized automata and Road Coloring Problem, Int. J. Found. 0 � � 0 q � q ( ) 8 u 2 � 9 v 2 � q : uv = q :uv : Comput. Sci., 13 (2002) 459–471). Let be a DFA. We define the relation LATA 2008 – p.19/29
� A A Stability Trahtman’s proof heavily depends on a neat idea of A = h Q; � ; Æ i stability which is due to Karel Culik II, Juhani � on Q as follows: Karhumäki and Jarkko Kari (A note on synchronized automata and Road Coloring Problem, Int. J. Found. 0 � � 0 q � q ( ) 8 u 2 � 9 v 2 � q : uv = q :uv : Comput. Sci., 13 (2002) 459–471). Let 0 be a DFA. We define the relation � is called the stability relation and any pair ( q ; q ) 0 is called stable . q � q such that LATA 2008 – p.19/29
A Stability Trahtman’s proof heavily depends on a neat idea of A = h Q; � ; Æ i stability which is due to Karel Culik II, Juhani � on Q as follows: Karhumäki and Jarkko Kari (A note on synchronized automata and Road Coloring Problem, Int. J. Found. 0 � � 0 q � q ( ) 8 u 2 � 9 v 2 � q : uv = q :uv : Comput. Sci., 13 (2002) 459–471). Let 0 be a DFA. We define the relation � is called the stability relation and any pair ( q ; q ) 0 is called stable . It is immediate that q � q � is a congruence of the automaton A . such that LATA 2008 – p.19/29
Stability Trahtman’s proof heavily depends on a neat idea of A = h Q; � ; Æ i stability which is due to Karel Culik II, Juhani � on Q as follows: Karhumäki and Jarkko Kari (A note on synchronized automata and Road Coloring Problem, Int. J. Found. 0 � � 0 q � q ( ) 8 u 2 � 9 v 2 � q : uv = q :uv : Comput. Sci., 13 (2002) 459–471). Let 0 be a DFA. We define the relation � is called the stability relation and any pair ( q ; q ) 0 is called stable . It is immediate that q � q � is a congruence of the automaton A . Also observe A is synchronizing iff all pairs are stable. such that that LATA 2008 – p.19/29
� A A = � Stability 0 0 . ( q ; q ) with q 6 = q We say that a coloring of a digraph with constant out-degree is stable if the resulting automaton contains at least one stable pair LATA 2008 – p.20/29
� A A = � Stability 0 0 .The ( q ; q ) with q 6 = q We say that a coloring of a digraph with constant out-degree is stable if the resulting automaton contains at least one stable pair crucial observation by Culik, Karhumäki and Kari is Proposition CKK. Suppose every strongly connected primitive digraph with constant out-degree and more than 1 vertex has a stable coloring. Then the Road Coloring Conjecture holds true. LATA 2008 – p.20/29
� A A = � Stability 0 0 .The ( q ; q ) with q 6 = q We say that a coloring of a digraph with constant out-degree is stable if the resulting automaton contains at least one stable pair crucial observation by Culik, Karhumäki and Kari is Proposition CKK. Suppose every strongly connected primitive digraph with constant out-degree and more than 1 vertex has a stable coloring. Then the Road Coloring Conjecture holds true. The proof is rather straightforward: one inducts on the number of vertices in the digraph. LATA 2008 – p.20/29
Stability 0 0 .The ( q ; q ) with q 6 = q We say that a coloring of a digraph with constant out-degree is stable if the resulting automaton contains at least one stable pair crucial observation by Culik, Karhumäki and Kari is Proposition CKK. Suppose every strongly connected primitive digraph with constant out-degree and more than 1 vertex has a stable coloring. Then the Road � admits a stable Coloring Conjecture holds true. A is the resulting automaton, then the A = � admits a synchronizing The proof is rather straightforward: one inducts on the number of vertices in the digraph. If coloring and quotient automaton recoloring by the induction assumption. LATA 2008 – p.20/29
Stability A = � to a � . Then it remains to lift the correct coloring of synchronizing coloring of LATA 2008 – p.21/29
Stability A = � to a � . Then it remains to lift the correct coloring of synchronizing coloring of 3 6 5 2 1 4 LATA 2008 – p.21/29
Stability A = � to a � . Then it remains to lift the correct coloring of synchronizing coloring of 3 6 5 2 1 4 LATA 2008 – p.21/29
Stability 123 456 LATA 2008 – p.22/29
Stability 123 456 123 456 LATA 2008 – p.22/29
Stability 123 456 123 456 3 6 5 2 1 4 LATA 2008 – p.22/29
Road Coloring Conjecture Trahtman has managed to prove exactly what was needed to use Proposition CKK: every strongly connected primitive digraph with constant out-degree and more than 1 vertex has a stable coloring. Thus, Road Coloring Conjecture holds true. LATA 2008 – p.23/29
Road Coloring Conjecture Trahtman has managed to prove exactly what was needed to use Proposition CKK: every strongly connected primitive digraph with constant out-degree and more than 1 vertex has a stable coloring. Thus, Road Coloring Conjecture holds true. The proof is not difficult but still a bit too technical for a presentation at the end of our conference. LATA 2008 – p.23/29
� n Summary of Open Problems � The complexity of polynomial approximations for the problem of finding the minimum length of reset words for a given synchronizing automaton. LATA 2008 – p.24/29
� n Summary of Open Problems � The complexity of polynomial approximations for the problem of finding the minimum length of reset words for a given synchronizing automaton. Recall that no polynomial algorithm, even non-deterministic, can find the exact value, but all known results are consistent with the existence of very good polynomial approximation algorithms for the problem. LATA 2008 – p.24/29
Summary of Open Problems � The complexity of polynomial approximations for the problem of finding the minimum length of reset words for a given synchronizing automaton. Recall that no polynomial algorithm, even non-deterministic, can find the exact value, but all known results are consistent � Synchronizing random automata: what is the with the existence of very good polynomial approximation algorithms for the problem. n states? expectation of the minimum length of reset words for the random automaton with LATA 2008 – p.24/29
Summary of Open Problems � The complexity of polynomial approximations for the problem of finding the minimum length of reset words for a given synchronizing automaton. Recall that no polynomial algorithm, even non-deterministic, can find the exact value, but all known results are consistent � Synchronizing random automata: what is the with the existence of very good polynomial approximation algorithms for the problem. n states? What is the expectation of the minimum length of reset words for the random automaton with probability distribution of this random variable? LATA 2008 – p.24/29
� A = h Q; � ; Æ i n k � 2 w 2 � ( n � k ) j Q : w j = k � � n � Summary of Open Problems � The ˇ n states, there exists a 2 ? ( n � 1) Cerný Conjecture: is it true that, for any synchronizing automaton with reset word of length LATA 2008 – p.25/29
� � n � Summary of Open Problems � The ˇ n states, there exists a 2 ? ( n � 1) Cerný Conjecture: is it true that, for any � The Rank Conjecture: is it true that, for any DFA synchronizing automaton with A = h Q; � ; Æ i with n states and rank k , there is a � of length 2 such that reset word of length w 2 � ( n � k ) j Q : w j = k ? LATA 2008 – p.25/29
Summary of Open Problems � The ˇ n states, there exists a 2 ? ( n � 1) Cerný Conjecture: is it true that, for any � The Rank Conjecture: is it true that, for any DFA synchronizing automaton with A = h Q; � ; Æ i with n states and rank k , there is a � of length 2 such that reset word of length w 2 � ( n � k ) j Q : w j = k ? � The hybrid ˇ � be n vertices. What is the minimum � ? Cerný/Road Coloring problem. Let a strongly connected primitive digraph with constant out-degree and length of reset words for synchronizing colorings of LATA 2008 – p.25/29
Summary of Open Problems � The ˇ n states, there exists a 2 ? ( n � 1) Cerný Conjecture: is it true that, for any � The Rank Conjecture: is it true that, for any DFA synchronizing automaton with A = h Q; � ; Æ i with n states and rank k , there is a � of length 2 such that reset word of length w 2 � ( n � k ) j Q : w j = k ? � The hybrid ˇ � be n vertices. What is the minimum � ? Cerný/Road Coloring problem. Let a strongly connected primitive digraph with constant out-degree and length of reset words for synchronizing colorings of For instance, the ˇ Cerný automata admit synchronizing recolorings with pretty short reset words. LATA 2008 – p.25/29
b a n � 1 b � � � a a a Summary of Open Problems b b n � 1 b b a a 0 n � 2 1 2 LATA 2008 – p.26/29
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