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Synchronizing Automata III M. V. Volkov Ural State University, - - PowerPoint PPT Presentation

Synchronizing Automata III M. V. Volkov Ural State University, Ekaterinburg, Russia LATA 2008 p.1/29 A w 2 A Q j Q : w j = 1 Q : v = f ( q ; v ) j q 2 Q g w Synchronizing automata Recap A = h Q;


slide-1
SLIDE 1

Synchronizing Automata – III

  • M. V. Volkov

Ural State University, Ekaterinburg, Russia

LATA 2008 – p.1/29

slide-2
SLIDE 2

Synchronizing automata – Recap

We consider DFA:

A = hQ; ; Æ i. A w 2
  • A
Q jQ : w j = 1 Q : v = fÆ (q ; v ) j q 2 Qg w

LATA 2008 – p.2/29

slide-3
SLIDE 3

Synchronizing automata – Recap

We consider DFA:

A = hQ; ; Æ i.

The DFA

A is called synchronizing if there exists a

word

w 2
  • whose action resets
A , that is, leaves

the automaton in one particular state no matter which state in

Q it started at. jQ : w j = 1 Q : v = fÆ (q ; v ) j q 2 Qg w

LATA 2008 – p.2/29

slide-4
SLIDE 4

Synchronizing automata – Recap

We consider DFA:

A = hQ; ; Æ i.

The DFA

A is called synchronizing if there exists a

word

w 2
  • whose action resets
A , that is, leaves

the automaton in one particular state no matter which state in

Q it started at. jQ : w j =
  • 1. Here
Q : v = fÆ (q ; v ) j q 2 Qg. w

LATA 2008 – p.2/29

slide-5
SLIDE 5

Synchronizing automata – Recap

We consider DFA:

A = hQ; ; Æ i.

The DFA

A is called synchronizing if there exists a

word

w 2
  • whose action resets
A , that is, leaves

the automaton in one particular state no matter which state in

Q it started at. jQ : w j =
  • 1. Here
Q : v = fÆ (q ; v ) j q 2 Qg.

Any

w with this property is said to be a reset word for

the automaton.

LATA 2008 – p.2/29

slide-6
SLIDE 6

Synchronizing automata – Recap

1 2 3

a b b b b a a a abbbabbba 9 = (4
  • 1)
2 (n
  • 1)
2

LATA 2008 – p.3/29

slide-7
SLIDE 7

Synchronizing automata – Recap

1 2 3

a b b b b a a a

A reset word is

abbbabbba of length 9 = (4
  • 1)
2, thus the

automaton reaches the ˇ Cerný bound

(n
  • 1)
2.

LATA 2008 – p.3/29

slide-8
SLIDE 8

Synchronizing automata – Recap

1 2 3

a b b b b a a a

A reset word is

abbbabbba of length 9 = (4
  • 1)
2, thus the

automaton reaches the ˇ Cerný bound

(n
  • 1)
2.

Observe that this automaton is strongly connected (as a digraph). In fact, almost all our examples have been strongly connected.

LATA 2008 – p.3/29

slide-9
SLIDE 9

Strongly connected digraphs

Studying synchronizing automata, it is natural to restrict to the strongly connected case.

A = hQ; ; Æ i n S A m = jS j q 2 S w 2
  • Q:w
= fq g a 2
  • Q:w
a = fÆ (q ; a)g w a Æ (q ; a) 2 S Æ S
  • S
S S

LATA 2008 – p.4/29

slide-10
SLIDE 10

Strongly connected digraphs

Studying synchronizing automata, it is natural to restrict to the strongly connected case. For instance, it suffices to prove the ˇ Cerný conjecture for this case, the general case would be an easy consequence.

A = hQ; ; Æ i n S A m = jS j q 2 S w 2
  • Q:w
= fq g a 2
  • Q:w
a = fÆ (q ; a)g w a Æ (q ; a) 2 S Æ S
  • S
S S

LATA 2008 – p.4/29

slide-11
SLIDE 11

Strongly connected digraphs

Studying synchronizing automata, it is natural to restrict to the strongly connected case. For instance, it suffices to prove the ˇ Cerný conjecture for this case, the general case would be an easy consequence. Let

A = hQ; ; Æ i be a synchronizing automaton with n

states.

S A m = jS j q 2 S w 2
  • Q:w
= fq g a 2
  • Q:w
a = fÆ (q ; a)g w a Æ (q ; a) 2 S Æ S
  • S
S S

LATA 2008 – p.4/29

slide-12
SLIDE 12

Strongly connected digraphs

Studying synchronizing automata, it is natural to restrict to the strongly connected case. For instance, it suffices to prove the ˇ Cerný conjecture for this case, the general case would be an easy consequence. Let

A = hQ; ; Æ i be a synchronizing automaton with n
  • states. Consider the set
S of all states to which A can

be synchronized and let

m = jS
  • j. If
q 2 S, then there

exists a reset word

w 2
  • such that
Q:w = fq
  • g. For

each

a 2 , we have Q:w a = fÆ (q ; a)g whence w a also

is a reset word and

Æ (q ; a) 2
  • S. Thus, restricting the

function

Æ to S
  • , we get a subautomaton
S with the

state set

S. S

LATA 2008 – p.4/29

slide-13
SLIDE 13

Strongly connected digraphs

Studying synchronizing automata, it is natural to restrict to the strongly connected case. For instance, it suffices to prove the ˇ Cerný conjecture for this case, the general case would be an easy consequence. Let

A = hQ; ; Æ i be a synchronizing automaton with n
  • states. Consider the set
S of all states to which A can

be synchronized and let

m = jS
  • j. If
q 2 S, then there

exists a reset word

w 2
  • such that
Q:w = fq
  • g. For

each

a 2 , we have Q:w a = fÆ (q ; a)g whence w a also

is a reset word and

Æ (q ; a) 2
  • S. Thus, restricting the

function

Æ to S
  • , we get a subautomaton
S with the

state set

  • S. Obviously,
S is synchronizing and

strongly connected.

LATA 2008 – p.4/29

slide-14
SLIDE 14

Strongly connected digraphs

If the ˇ Cerný conjecture holds true for strongly connected synchronizing automata,

S has a reset

word

v of length (m
  • 1)
2.
  • Q
n
  • m
+ 1 S
  • A

LATA 2008 – p.5/29

slide-15
SLIDE 15

Strongly connected digraphs

If the ˇ Cerný conjecture holds true for strongly connected synchronizing automata,

S has a reset

word

v of length (m
  • 1)
2.

Now consider the partition

  • f
Q into n
  • m
+ 1 classes
  • ne of which is
S and all others are singletons. Then
  • is a congruence of the automaton
A .

LATA 2008 – p.5/29

slide-16
SLIDE 16

Strongly connected digraphs

If the ˇ Cerný conjecture holds true for strongly connected synchronizing automata,

S has a reset

word

v of length (m
  • 1)
2.

Now consider the partition

  • f
Q into n
  • m
+ 1 classes
  • ne of which is
S and all others are singletons. Then
  • is a congruence of the automaton
A .

We recall the notion of a congruence and the related notion of the quotient automaton w.r.t. a congruence in the next slide. They will be essentially used in this lecture!

LATA 2008 – p.5/29

slide-17
SLIDE 17

Congruences and Quotient Automata

An equivalence

  • n the state set
Q of a DFA A = hQ; ; Æ i is called a congruence if (p; q ) 2
  • implies
  • Æ
(p; a); Æ (q ; a)
  • 2
for all p; q 2 Q and all a 2 .
  • [q
  • q
A = hQ= ; ; Æ
  • i
Q= = f[q ℄
  • j
q 2 Qg Æ
  • Æ
  • ([q
  • ;
a) = [Æ (q ; a)℄
  • a
a a a b b b b
  • b
a a b

LATA 2008 – p.6/29

slide-18
SLIDE 18

Congruences and Quotient Automata

An equivalence

  • n the state set
Q of a DFA A = hQ; ; Æ i is called a congruence if (p; q ) 2
  • implies
  • Æ
(p; a); Æ (q ; a)
  • 2
for all p; q 2 Q and all a 2 . For being a congruence, [q ℄ is the
  • class

containing the state

q. A = hQ= ; ; Æ
  • i
Q= = f[q ℄
  • j
q 2 Qg Æ
  • Æ
  • ([q
  • ;
a) = [Æ (q ; a)℄
  • a
a a a b b b b
  • b
a a b

LATA 2008 – p.6/29

slide-19
SLIDE 19

Congruences and Quotient Automata

An equivalence

  • n the state set
Q of a DFA A = hQ; ; Æ i is called a congruence if (p; q ) 2
  • implies
  • Æ
(p; a); Æ (q ; a)
  • 2
for all p; q 2 Q and all a 2 . For being a congruence, [q ℄ is the
  • class

containing the state

q. A = hQ= ; ; Æ
  • i
Q= = f[q ℄
  • j
q 2 Qg Æ
  • Æ
  • ([q
  • ;
a) = [Æ (q ; a)℄
  • 1

3 2 4

a a a a b b b b
  • b
a a b

LATA 2008 – p.6/29

slide-20
SLIDE 20

Congruences and Quotient Automata

An equivalence

  • n the state set
Q of a DFA A = hQ; ; Æ i is called a congruence if (p; q ) 2
  • implies
  • Æ
(p; a); Æ (q ; a)
  • 2
for all p; q 2 Q and all a 2 . For being a congruence, [q ℄ is the
  • class

containing the state

q. A = hQ= ; ; Æ
  • i
Q= = f[q ℄
  • j
q 2 Qg Æ
  • Æ
  • ([q
  • ;
a) = [Æ (q ; a)℄
  • 1

3 2 4

a a a a b b b b
  • b
a a b

LATA 2008 – p.6/29

slide-21
SLIDE 21

Congruences and Quotient Automata

An equivalence

  • n the state set
Q of a DFA A = hQ; ; Æ i is called a congruence if (p; q ) 2
  • implies
  • Æ
(p; a); Æ (q ; a)
  • 2
for all p; q 2 Q and all a 2 . For being a congruence, [q ℄ is the
  • class

containing the state

q.

The quotient

A = is the DFA hQ= ; ; Æ
  • i where
Q= = f[q ℄
  • j
q 2 Qg and the function Æ is defined

by the rule

Æ
  • ([q
  • ;
a) = [Æ (q ; a)℄ .

1 3 2 4

a a a a b b b b
  • b
a a b

LATA 2008 – p.6/29

slide-22
SLIDE 22

Congruences and Quotient Automata

An equivalence

  • n the state set
Q of a DFA A = hQ; ; Æ i is called a congruence if (p; q ) 2
  • implies
  • Æ
(p; a); Æ (q ; a)
  • 2
for all p; q 2 Q and all a 2 . For being a congruence, [q ℄ is the
  • class

containing the state

q.

The quotient

A = is the DFA hQ= ; ; Æ
  • i where
Q= = f[q ℄
  • j
q 2 Qg and the function Æ is defined

by the rule

Æ
  • ([q
  • ;
a) = [Æ (q ; a)℄ .

1 3 2 4

a a a a b b b b
  • 1,2

3,4

b a a b

LATA 2008 – p.6/29

slide-23
SLIDE 23

Strongly connected digraphs

Return to our reasoning: let

be the partition of Q into n
  • m
+ 1 classes one of which is S and all others are
  • singletons. Then
is a congruence of A . A = S
  • LATA 2008 – p.7/29
slide-24
SLIDE 24

Strongly connected digraphs

Return to our reasoning: let

be the partition of Q into n
  • m
+ 1 classes one of which is S and all others are
  • singletons. Then
is a congruence of A .

Clearly, the quotient

A = is synchronizing and has S

as a unique sink.

  • LATA 2008 – p.7/29
slide-25
SLIDE 25

Strongly connected digraphs

Return to our reasoning: let

be the partition of Q into n
  • m
+ 1 classes one of which is S and all others are
  • singletons. Then
is a congruence of A .

Clearly, the quotient

A = is synchronizing and has S

as a unique sink. S

  • LATA 2008 – p.7/29
slide-26
SLIDE 26

Automata with a Unique Sink

If a synchronizing automata with

k states has a unique

sink, then it has a reset word of length

  • k
(k 1) 2

.

a; b a a a b b b b a a b b a a b b a b b a b b k
  • 1
t k
  • 1
  • t
  • 1
k
  • t
  • 1
+ 2 +
  • +
(k
  • 1)
= k (k 1) 2

LATA 2008 – p.8/29

slide-27
SLIDE 27

Automata with a Unique Sink

If a synchronizing automata with

k states has a unique

sink, then it has a reset word of length

  • k
(k 1) 2

.

a; b a a a b b b b a a b b a a b b a b b a b b k
  • 1
t k
  • 1
  • t
  • 1
k
  • t
  • 1
+ 2 +
  • +
(k
  • 1)
= k (k 1) 2

LATA 2008 – p.8/29

slide-28
SLIDE 28

Automata with a Unique Sink

If a synchronizing automata with

k states has a unique

sink, then it has a reset word of length

  • k
(k 1) 2

.

a; b a a a b b b b a a b b a a b b a b b a b b k
  • 1
t k
  • 1
  • t
  • 1
k
  • t
  • 1
+ 2 +
  • +
(k
  • 1)
= k (k 1) 2

LATA 2008 – p.8/29

slide-29
SLIDE 29

Automata with a Unique Sink

If a synchronizing automata with

k states has a unique

sink, then it has a reset word of length

  • k
(k 1) 2

.

a; b a a a b b b b a a b b a a a b b a b b a b b k
  • 1
t k
  • 1
  • t
  • 1
k
  • t
  • 1
+ 2 +
  • +
(k
  • 1)
= k (k 1) 2

LATA 2008 – p.8/29

slide-30
SLIDE 30

Automata with a Unique Sink

If a synchronizing automata with

k states has a unique

sink, then it has a reset word of length

  • k
(k 1) 2

.

a; b a a a b b b b a a b b a a b b a b b a b b k
  • 1
t k
  • 1
  • t
  • 1
k
  • t
  • 1
+ 2 +
  • +
(k
  • 1)
= k (k 1) 2

LATA 2008 – p.8/29

slide-31
SLIDE 31

Automata with a Unique Sink

If a synchronizing automata with

k states has a unique

sink, then it has a reset word of length

  • k
(k 1) 2

.

a; b a a a b b b b a a b b a a a b b a b b a b b k
  • 1
t k
  • 1
  • t
  • 1
k
  • t
  • 1
+ 2 +
  • +
(k
  • 1)
= k (k 1) 2

LATA 2008 – p.8/29

slide-32
SLIDE 32

Automata with a Unique Sink

If a synchronizing automata with

k states has a unique

sink, then it has a reset word of length

  • k
(k 1) 2

.

a; b a a a b b b b a a b b a a b b a b b a b b k
  • 1
t k
  • 1
  • t
  • 1
k
  • t
  • 1
+ 2 +
  • +
(k
  • 1)
= k (k 1) 2

LATA 2008 – p.8/29

slide-33
SLIDE 33

Automata with a Unique Sink

If a synchronizing automata with

k states has a unique

sink, then it has a reset word of length

  • k
(k 1) 2

.

a; b a a a b b b b a a b b a a b b a b b a b b k
  • 1
t k
  • 1
  • t
  • 1
k
  • t
  • 1
+ 2 +
  • +
(k
  • 1)
= k (k 1) 2

LATA 2008 – p.8/29

slide-34
SLIDE 34

Automata with a Unique Sink

If a synchronizing automata with

k states has a unique

sink, then it has a reset word of length

  • k
(k 1) 2

.

a; b a a a b b b b a a b b a a b b a b b a b b

The algorithm makes at most

k
  • 1 steps and the

length of the segment added in the step when

t states

still holds coins ( k

  • 1
  • t
  • 1) is at most
k
  • t. The total

length is

  • 1
+ 2 +
  • +
(k
  • 1)
= k (k 1) 2

.

LATA 2008 – p.8/29

slide-35
SLIDE 35

Strongly connected digraphs

Return to our reasoning: the quotient

A = is

synchronizing with a unique sink and has

n
  • m
+ 1

states.

A = u (nm+1)(nm) 2 Q : u
  • S
S v (m
  • 1)
2 S :v Q:uv
  • S
:v uv A (n
  • m
+ 1)(n
  • m)
2 + (m
  • 1)
2
  • (n
  • 1)
2 :

LATA 2008 – p.9/29

slide-36
SLIDE 36

Strongly connected digraphs

Return to our reasoning: the quotient

A = is

synchronizing with a unique sink and has

n
  • m
+ 1
  • states. Hence,
A = has a reset word u of length (nm+1)(nm) 2

.

Q : u
  • S
S v (m
  • 1)
2 S :v Q:uv
  • S
:v uv A (n
  • m
+ 1)(n
  • m)
2 + (m
  • 1)
2
  • (n
  • 1)
2 :

LATA 2008 – p.9/29

slide-37
SLIDE 37

Strongly connected digraphs

Return to our reasoning: the quotient

A = is

synchronizing with a unique sink and has

n
  • m
+ 1
  • states. Hence,
A = has a reset word u of length (nm+1)(nm) 2

. Then

Q : u
  • S.
S v (m
  • 1)
2 S :v Q:uv
  • S
:v uv A (n
  • m
+ 1)(n
  • m)
2 + (m
  • 1)
2
  • (n
  • 1)
2 :

LATA 2008 – p.9/29

slide-38
SLIDE 38

Strongly connected digraphs

Return to our reasoning: the quotient

A = is

synchronizing with a unique sink and has

n
  • m
+ 1
  • states. Hence,
A = has a reset word u of length (nm+1)(nm) 2

. Then

Q : u
  • S.

Recall that we have assumed that the automaton

S

has a reset word

v of length (m
  • 1)
2. S :v Q:uv
  • S
:v uv A (n
  • m
+ 1)(n
  • m)
2 + (m
  • 1)
2
  • (n
  • 1)
2 :

LATA 2008 – p.9/29

slide-39
SLIDE 39

Strongly connected digraphs

Return to our reasoning: the quotient

A = is

synchronizing with a unique sink and has

n
  • m
+ 1
  • states. Hence,
A = has a reset word u of length (nm+1)(nm) 2

. Then

Q : u
  • S.

Recall that we have assumed that the automaton

S

has a reset word

v of length (m
  • 1)
  • 2. Then
S :v is a

singleton, whence also

Q:uv
  • S
:v is a singleton. uv A (n
  • m
+ 1)(n
  • m)
2 + (m
  • 1)
2
  • (n
  • 1)
2 :

LATA 2008 – p.9/29

slide-40
SLIDE 40

Strongly connected digraphs

Return to our reasoning: the quotient

A = is

synchronizing with a unique sink and has

n
  • m
+ 1
  • states. Hence,
A = has a reset word u of length (nm+1)(nm) 2

. Then

Q : u
  • S.

Recall that we have assumed that the automaton

S

has a reset word

v of length (m
  • 1)
  • 2. Then
S :v is a

singleton, whence also

Q:uv
  • S
:v is a singleton.

Thus,

uv is reset word for A , and the length of this

word does not exceed

(n
  • m
+ 1)(n
  • m)
2 + (m
  • 1)
2
  • (n
  • 1)
2 :

LATA 2008 – p.9/29

slide-41
SLIDE 41

Strongly connected digraphs

Thus, we assume that our synchronizing automata are strongly connected as digraphs.

q w q w q q ab 3 ab 3 ab 3 ab 3 ab 3 a ab 3 ab 3 ab ab 3 ab 3 ab 2 a b b b b a a a

LATA 2008 – p.10/29

slide-42
SLIDE 42

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.

q w q w q q ab 3 ab 3 ab 3 ab 3 ab 3 a ab 3 ab 3 ab ab 3 ab 3 ab 2 a b b b b a a a

LATA 2008 – p.10/29

slide-43
SLIDE 43

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. That is, to every state
q of the automaton one

can assign an instruction (a reset word)

w q such that

following

w q one will surely arrive at q from any initial

state.

ab 3 ab 3 ab 3 ab 3 ab 3 a ab 3 ab 3 ab ab 3 ab 3 ab 2 a b b b b a a a

LATA 2008 – p.10/29

slide-44
SLIDE 44

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. That is, to every state
q of the automaton one

can assign an instruction (a reset word)

w q such that

following

w q one will surely arrive at q from any initial

state.

ab 3 ab 3 ab 3 ab 3 ab 3 a

1

ab 3 ab 3 ab

2

ab 3 ab 3 ab 2

3

a b b b b a a a

LATA 2008 – p.10/29

slide-45
SLIDE 45

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

slide-46
SLIDE 46

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

slide-47
SLIDE 47

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

slide-48
SLIDE 48

Road Coloring

LATA 2008 – p.12/29

slide-49
SLIDE 49

Road Coloring

For the green node: blue-blue-red-blue-blue-red-blue-blue-red.

LATA 2008 – p.12/29

slide-50
SLIDE 50

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

slide-51
SLIDE 51

Road Coloring

Now suppose that we have a transport network, that is, a strongly connected digraph.

LATA 2008 – p.13/29

slide-52
SLIDE 52

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

slide-53
SLIDE 53

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

slide-54
SLIDE 54

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

slide-55
SLIDE 55

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

  • ut-degree condition.

LATA 2008 – p.13/29

slide-56
SLIDE 56

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.

  • =
(V ; E ) k > 1 v 2 V i = 0; 1; : : : ; k
  • 1
V i = fv 2 V j 9 v v i (mo d k )g: V = k 1 S i=0 V i V i \ V j = ? i 6= j

LATA 2008 – p.14/29

slide-57
SLIDE 57

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. To see why primitivity is necessary, suppose that

  • =
(V ; E ) is a strongly connected digraph and k > 1 is

a common divisor of lengths of its cycles.

v 2 V i = 0; 1; : : : ; k
  • 1
V i = fv 2 V j 9 v v i (mo d k )g: V = k 1 S i=0 V i V i \ V j = ? i 6= j

LATA 2008 – p.14/29

slide-58
SLIDE 58

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. To see why primitivity is necessary, suppose that

  • =
(V ; E ) is a strongly connected digraph and k > 1 is

a common divisor of lengths of its cycles. Take a vertex

v 2 V and, for i = 0; 1; : : : ; k
  • 1, let
V i = fv 2 V j 9 path from v 0 to v of length i (mo d k )g: V = k 1 S i=0 V i V i \ V j = ? i 6= j

LATA 2008 – p.14/29

slide-59
SLIDE 59

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. To see why primitivity is necessary, suppose that

  • =
(V ; E ) is a strongly connected digraph and k > 1 is

a common divisor of lengths of its cycles. Take a vertex

v 2 V and, for i = 0; 1; : : : ; k
  • 1, let
V i = fv 2 V j 9 path from v 0 to v of length i (mo d k )g:

Clearly,

V = k 1 S i=0 V i. V i \ V j = ? i 6= j

LATA 2008 – p.14/29

slide-60
SLIDE 60

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. To see why primitivity is necessary, suppose that

  • =
(V ; E ) is a strongly connected digraph and k > 1 is

a common divisor of lengths of its cycles. Take a vertex

v 2 V and, for i = 0; 1; : : : ; k
  • 1, let
V i = fv 2 V j 9 path from v 0 to v of length i (mo d k )g:

Clearly,

V = k 1 S i=0 V
  • i. We claim that
V i \ V j = ? if i 6= j.

LATA 2008 – p.14/29

slide-61
SLIDE 61

Primitivity

Let

v 2 V i \ V j where i 6=
  • j. This means that in
there

are two paths from

v 0 to v: of length `
  • i
(mo d k ) and
  • f length
m
  • j
(mo d k ). v v v v n ` + n m + n

LATA 2008 – p.15/29

slide-62
SLIDE 62

Primitivity

Let

v 2 V i \ V j where i 6=
  • j. This means that in
there

are two paths from

v 0 to v: of length `
  • i
(mo d k ) and
  • f length
m
  • j
(mo d k ). v v v v n ` + n m + n

LATA 2008 – p.15/29

slide-63
SLIDE 63

Primitivity

Let

v 2 V i \ V j where i 6=
  • j. This means that in
there

are two paths from

v 0 to v: of length `
  • i
(mo d k ) and
  • f length
m
  • j
(mo d k ). v v

. . . There is also a path

v to v 0 of length, say, n.

Combining it with the two paths above we get a cycle

  • f length
` + n and a cycle of length m + n.

LATA 2008 – p.15/29

slide-64
SLIDE 64

Primitivity

Since

k divides the length of any cycle in , we have ` + n
  • i
+ n
  • (mo
d k ) and m + n
  • j
+ n
  • (mo
d k ),

whence

i
  • j
(mo d k ), a contradiction. V V ; V 1 ; : : : ; V k 1
  • V
i V i+1 (mo d k )
  • `
V V 1 V ` (mo d k ) V `+1 (mo d k )

LATA 2008 – p.16/29

slide-65
SLIDE 65

Primitivity

Since

k divides the length of any cycle in , we have ` + n
  • i
+ n
  • (mo
d k ) and m + n
  • j
+ n
  • (mo
d k ),

whence

i
  • j
(mo d k ), a contradiction.

Thus,

V is a disjoint union of V ; V 1 ; : : : ; V k 1, and by the

definition each arrow in

leads from V i to V i+1 (mo d k ).
  • `
V V 1 V ` (mo d k ) V `+1 (mo d k )

LATA 2008 – p.16/29

slide-66
SLIDE 66

Primitivity

Since

k divides the length of any cycle in , we have ` + n
  • i
+ n
  • (mo
d k ) and m + n
  • j
+ n
  • (mo
d k ),

whence

i
  • j
(mo d k ), a contradiction.

Thus,

V is a disjoint union of V ; V 1 ; : : : ; V k 1, and by the

definition each arrow in

leads from V i to V i+1 (mo d k ).

Then

definitely cannot be converted into a

synchronizing automaton by any labelling of its arrows: for instance, no paths of the same length

` originated

in

V 0 and V 1 can terminate in the same vertex because

they end in

V ` (mo d k ) and in V `+1 (mo d k ) respectively.

LATA 2008 – p.16/29

slide-67
SLIDE 67

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

slide-68
SLIDE 68

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

  • ut-degree admits a synchronizing coloring.

LATA 2008 – p.17/29

slide-69
SLIDE 69

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

  • ut-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

slide-70
SLIDE 70

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

  • ut-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

slide-71
SLIDE 71

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

slide-72
SLIDE 72

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

slide-73
SLIDE 73

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

slide-74
SLIDE 74

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

slide-75
SLIDE 75

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).
A = hQ; ; Æ i
  • Q
q
  • q
( ) 8u 2
  • 9v
2
  • q
: uv = q :uv :
  • (q
; q ) q
  • q
  • A
A

LATA 2008 – p.19/29

slide-76
SLIDE 76

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). Let
A = hQ; ; Æ i

be a DFA. We define the relation

  • n
Q as follows: q
  • q
( ) 8u 2
  • 9v
2
  • q
: uv = q :uv :
  • (q
; q ) q
  • q
  • A
A

LATA 2008 – p.19/29

slide-77
SLIDE 77

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). Let
A = hQ; ; Æ i

be a DFA. We define the relation

  • n
Q as follows: q
  • q
( ) 8u 2
  • 9v
2
  • q
: uv = q :uv : is called the stability relation and any pair (q ; q )

such that

q
  • q
0 is called stable.
  • A
A

LATA 2008 – p.19/29

slide-78
SLIDE 78

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). Let
A = hQ; ; Æ i

be a DFA. We define the relation

  • n
Q as follows: q
  • q
( ) 8u 2
  • 9v
2
  • q
: uv = q :uv : is called the stability relation and any pair (q ; q )

such that

q
  • q
0 is called stable. It is immediate that is a congruence of the automaton A . A

LATA 2008 – p.19/29

slide-79
SLIDE 79

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). Let
A = hQ; ; Æ i

be a DFA. We define the relation

  • n
Q as follows: q
  • q
( ) 8u 2
  • 9v
2
  • q
: uv = q :uv : is called the stability relation and any pair (q ; q )

such that

q
  • q
0 is called stable. It is immediate that is a congruence of the automaton A . Also observe

that

A is synchronizing iff all pairs are stable.

LATA 2008 – p.19/29

slide-80
SLIDE 80

Stability

We say that a coloring of a digraph with constant

  • ut-degree is stable if the resulting automaton

contains at least one stable pair

(q ; q ) with q 6= q 0.
  • A
A =
  • LATA 2008 – p.20/29
slide-81
SLIDE 81

Stability

We say that a coloring of a digraph with constant

  • ut-degree is stable if the resulting automaton

contains at least one stable pair

(q ; q ) with q 6= q 0.The

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.

  • A
A =
  • LATA 2008 – p.20/29
slide-82
SLIDE 82

Stability

We say that a coloring of a digraph with constant

  • ut-degree is stable if the resulting automaton

contains at least one stable pair

(q ; q ) with q 6= q 0.The

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.

  • A
A =
  • LATA 2008 – p.20/29
slide-83
SLIDE 83

Stability

We say that a coloring of a digraph with constant

  • ut-degree is stable if the resulting automaton

contains at least one stable pair

(q ; q ) with q 6= q 0.The

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. If

admits a stable

coloring and

A is the resulting automaton, then the

quotient automaton

A = admits a synchronizing

recoloring by the induction assumption.

LATA 2008 – p.20/29

slide-84
SLIDE 84

Stability

Then it remains to lift the correct coloring of

A = to a

synchronizing coloring of

.

LATA 2008 – p.21/29

slide-85
SLIDE 85

Stability

Then it remains to lift the correct coloring of

A = to a

synchronizing coloring of

.

1 2 3 4 5 6

LATA 2008 – p.21/29

slide-86
SLIDE 86

Stability

Then it remains to lift the correct coloring of

A = to a

synchronizing coloring of

.

1 2 3 4 5 6

LATA 2008 – p.21/29

slide-87
SLIDE 87

Stability

123 456

LATA 2008 – p.22/29

slide-88
SLIDE 88

Stability

123 456 123 456

LATA 2008 – p.22/29

slide-89
SLIDE 89

Stability

123 456 123 456 1 2 3 4 5 6

LATA 2008 – p.22/29

slide-90
SLIDE 90

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

slide-91
SLIDE 91

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

slide-92
SLIDE 92

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.

  • n

LATA 2008 – p.24/29

slide-93
SLIDE 93

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.

  • n

LATA 2008 – p.24/29

slide-94
SLIDE 94

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.

Synchronizing random automata: what is the

expectation of the minimum length of reset words for the random automaton with

n states?

LATA 2008 – p.24/29

slide-95
SLIDE 95

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.

Synchronizing random automata: what is the

expectation of the minimum length of reset words for the random automaton with

n states? What is the

probability distribution of this random variable?

LATA 2008 – p.24/29

slide-96
SLIDE 96

Summary of Open Problems

The ˇ

Cerný Conjecture: is it true that, for any synchronizing automaton with

n states, there exists a

reset word of length

(n
  • 1)
2?
  • A
= hQ; ; Æ i n k w 2
  • (n
  • k
) 2 jQ : w j = k
  • n
  • LATA 2008 – p.25/29
slide-97
SLIDE 97

Summary of Open Problems

The ˇ

Cerný Conjecture: is it true that, for any synchronizing automaton with

n states, there exists a

reset word of length

(n
  • 1)
2? The Rank Conjecture: is it true that, for any DFA A = hQ; ; Æ i with n states and rank k, there is a w 2
  • f length
(n
  • k
) 2 such that jQ : w j = k?
  • n
  • LATA 2008 – p.25/29
slide-98
SLIDE 98

Summary of Open Problems

The ˇ

Cerný Conjecture: is it true that, for any synchronizing automaton with

n states, there exists a

reset word of length

(n
  • 1)
2? The Rank Conjecture: is it true that, for any DFA A = hQ; ; Æ i with n states and rank k, there is a w 2
  • f length
(n
  • k
) 2 such that jQ : w j = k? The hybrid ˇ

Cerný/Road Coloring problem. Let

be

a strongly connected primitive digraph with constant

  • ut-degree and
n vertices. What is the minimum

length of reset words for synchronizing colorings of

?

LATA 2008 – p.25/29

slide-99
SLIDE 99

Summary of Open Problems

The ˇ

Cerný Conjecture: is it true that, for any synchronizing automaton with

n states, there exists a

reset word of length

(n
  • 1)
2? The Rank Conjecture: is it true that, for any DFA A = hQ; ; Æ i with n states and rank k, there is a w 2
  • f length
(n
  • k
) 2 such that jQ : w j = k? The hybrid ˇ

Cerný/Road Coloring problem. Let

be

a strongly connected primitive digraph with constant

  • ut-degree and
n vertices. What is the minimum

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

slide-100
SLIDE 100

Summary of Open Problems

n2 n1

1 2

a a a b b a b b a a b b n1
  • LATA 2008 – p.26/29
slide-101
SLIDE 101

Summary of Open Problems

n2 n1

1 2

a a a b b a b b a a b

The recolored automaton is reset by the word

b n1.
  • LATA 2008 – p.26/29
slide-102
SLIDE 102

Summary of Open Problems

n2 n1

1 2

a a a b b a b b a a b

The recolored automaton is reset by the word

b n1. Careful Road Coloring Problem. From the viewpoint
  • f transportation network the constant out-degree

condition does not seem to be natural.

  • LATA 2008 – p.26/29
slide-103
SLIDE 103

Summary of Open Problems

n2 n1

1 2

a a a b b a b b a a b

The recolored automaton is reset by the word

b n1. Careful Road Coloring Problem. From the viewpoint
  • f transportation network the constant out-degree

condition does not seem to be natural. We rather want to find a synchronizing coloring for arbitrary strongly connected primitive digraph

, the number of

colors being the maximal out-degree of

.

LATA 2008 – p.26/29

slide-104
SLIDE 104

Summary of Open Problems

But in the absence of the constant out-degree condition, the resulting automaton

A = hQ; ; Æ i is
  • incomplete. We need a suitable modification of the

notion of a synchronizing automaton for this case.

w = a 1
  • a
` a 1 ; : : : ; a ` 2
  • A
Æ (q ; a 1 ) q 2 Q Æ (q ; a i ) 1 < i
  • `
q 2 Q : a 1
  • a
i1 jQ : w j = 1 w

LATA 2008 – p.27/29

slide-105
SLIDE 105

Summary of Open Problems

But in the absence of the constant out-degree condition, the resulting automaton

A = hQ; ; Æ i is
  • incomplete. We need a suitable modification of the

notion of a synchronizing automaton for this case. We say that

w = a 1
  • a
` with a 1 ; : : : ; a ` 2 is a

careful reset word for

A if
  • Æ
(q ; a 1 ) is defined for all q 2 Q,
  • Æ
(q ; a i ) with 1 < i
  • ` is defined for all
q 2 Q : a 1
  • a
i1,
  • jQ
: w j = 1. w

LATA 2008 – p.27/29

slide-106
SLIDE 106

Summary of Open Problems

But in the absence of the constant out-degree condition, the resulting automaton

A = hQ; ; Æ i is
  • incomplete. We need a suitable modification of the

notion of a synchronizing automaton for this case. We say that

w = a 1
  • a
` with a 1 ; : : : ; a ` 2 is a

careful reset word for

A if
  • Æ
(q ; a 1 ) is defined for all q 2 Q,
  • Æ
(q ; a i ) with 1 < i
  • ` is defined for all
q 2 Q : a 1
  • a
i1,
  • jQ
: w j = 1.

In transport network terms this means that following the instruction

w is always possible and brings one to

the node which is independent of the initial node.

LATA 2008 – p.27/29

slide-107
SLIDE 107

Summary of Open Problems

The automaton

A is then said to be carefully

synchronizing.

LATA 2008 – p.28/29

slide-108
SLIDE 108

Summary of Open Problems

The automaton

A is then said to be carefully

synchronizing. The Careful Road Coloring Problem asks under which conditions strongly connected digraphs admit carefully synchronizing colorings.

LATA 2008 – p.28/29

slide-109
SLIDE 109

Summary of Open Problems

The automaton

A is then said to be carefully

synchronizing. The Careful Road Coloring Problem asks under which conditions strongly connected digraphs admit carefully synchronizing colorings. Is it true that every primitive strongly connected digraph has such a coloring? (The Careful Road Coloring Conjecture)

LATA 2008 – p.28/29

slide-110
SLIDE 110

Thanks

T h a n k y

  • u

v e r y m u c h f

  • r

y

  • u

r k i n d at t e n t i

  • n

!

LATA 2008 – p.29/29

slide-111
SLIDE 111

Thanks

T h a n k y

  • u

v e r y m u c h f

  • r

y

  • u

r k i n d at t e n t i

  • n

! T h a n k y

  • u

v e r y m u c h f

  • r

y

  • u

r k i n d at t e n t i

  • n

!

LATA 2008 – p.29/29

slide-112
SLIDE 112

Thanks

T h a n k y

  • u

v e r y m u c h f

  • r

y

  • u

r k i n d at t e n t i

  • n

! T h a n k y

  • u

v e r y m u c h f

  • r

y

  • u

r k i n d at t e n t i

  • n

! T h a n k y

  • u

v e r y m u c h f

  • r

y

  • u

r k i n d at t e n t i

  • n

!

LATA 2008 – p.29/29