Synchronizing automata: new techniques and results
Raphaël Jungers
UCLouvain
Jiao Tong Univ., Apr. 2015
Joint work with François Gonze
Synchronizing automata: new techniques and results Raphal Jungers - - PowerPoint PPT Presentation
Synchronizing automata: new techniques and results Raphal Jungers UCLouvain Jiao Tong Univ., Apr. 2015 Joint work with Franois Gonze Sy Sync nchroniz hronizing ing au automa omata ta Synchronizing word (or reset sequence)
Jiao Tong Univ., Apr. 2015
Joint work with François Gonze
Synchronizing word (or reset sequence) Cerny’s conjecture (1964): If a graph is synchronizing, then it admits a synchronizing sequence of length at most (n-1)2. Definition: A (complete deterministic) automaton is synchronizing if there is a sequence of colors such that all the paths compatible with this sequence end in the same node. Connected with the Road coloring conjecture
[2007 Trahtman] [1977 Adler et al.]
word (though I would love to!)
Develop new tools Proof of concept Mostly ideas, very few technical considerations
Černý’s conjecture Approach: the triple rendezvous time A tool: the synchronizing probability function Counter examples
approaches
a related quantity)
approaches
a related quantity)
Theorem [1990 Eppstein]: Synchronizing graphs are Recognizable in polynomial time. [Cerny, 1960’s] 1 2 3 2-3 1-3 1-2 1-1 3-3 2-2 Length (n-1)²
2-n
n³/2-3/2 n²+n+1
n(n-1)²/2
7/27 n³ - 17/18 n² + 17/6 n – 3
– The best so far!
Cerny’s conjecture (1964): If a graph is synchronizing, then it admits a synchronizing sequence of length at most (n-1)2 Known upper bounds on the shortest synchronizing word:
n
[Gonze, Trahtman, J. 2015]
– NP-hard [1990 Eppstein] – Apx-hard [2010 Berlinkov]
strongly transitive
Cerny’s conjecture (1964): If a graph is synchronizing, then it admits a synchronizing sequence of length at most (n-1)2.
circular of prime size
approaches
a related quantity)
Theorem [1990 Eppstein]: Synchronizing graphs are Recognizable in polynomial time. 1 2 3 2-3 1-3 1-2 1-1 3-3 2-2
Eppstein’s square graph gives a poor strategy to find a short synchronizing word
the « mouse »
mouse must pick up a node where to catch him
a particular sequence of colors of length t
1 2 3 4 5 6 and the « cat »
the « mouse »
1 2 3 4 5 6 and the « cat »
ensure to get caught, whatever strategy (of length t) the mouse chooses
(i.e. a probability function on the nodes)
1 2
ensure to get caught, whatever strategy (of length t) the mouse chooses
(i.e. a probability function on the nodes)
1 2
ensure to get caught, whatever strategy (of length t) the mouse chooses
k(0)=1/2 k(1)=1
(i.e. a probability function on the nodes)
1 2
ensure to get caught, whatever strategy (of length t) the mouse chooses
p(1)=1/2 p(2)=1/2
ensure to get caught, whatever strategy (of length t) the mouse chooses
(i.e. a probability function on the nodes)
1 2 k(0)=1/2 k(1)=1
ensure to get caught, whatever strategy (of length t) the mouse chooses
(i.e. a probability function on the nodes)
1 2 k(0)=1/2 k(1)=1
probabilistic as well
length t if and only if k(t)=1
k((n-1)²)=1
1 2 1/3 2/3
ensure to get caught, whatever strategy (of length t) the mouse chooses
ensure to get caught, whatever strategy (of length t) the mouse chooses
practice
shorter products
at most n different columns (n is the number of nodes)
ensure to get caught, whatever strategy (of length t) the mouse chooses
practice
shorter products
at most n different rows (n is the number of nodes)
Proof: if not, then P’ = P’
decrease anymore Theorem: If k(t)<1, then k(t+(n-1))>k(t) Proof: suppose k(t)=k(t+1)
t t+1 t t+1 t t+2 t+1 t t+1 t
is always higher (or equal) than Cerny’s automaton
t=1+ (n+1) i
approaches
– a new upper bound (on a related quantity) – a counterexample
1 2 3 2-3 1-3 1-2 1-1 3-3 2-2
35
First observation: we can represent A(t) on a graph!
5 1 2 3 4 5 6 Example for t=1:
First observation: we can represent A(t) on a graph!
5
First observation: we can represent A(t) on a graph!
5
Lemma: There is always an
with a disconnected union of singletons, pairs, and odd cycles
5
First observation: we can represent A(t) on a graph!
5
Lemma: There is always an
with a disconnected union of singletons, pairs, and odd cycles
5
First observation: we can represent A(t) on a graph!
5
Lemma: There is always an
with a disconnected union of singletons, pairs, and odd cycles
5
First observation: we can represent A(t) on a graph!
5
Lemma: There is always an
with a disconnected union of singletons, pairs, and odd cycles
5
First observation: we can represent A(t) on a graph!
5
Lemma: There is always an
with a disconnected union of singletons, pairs, and odd cycles
5
n1 « singleton » (n1=1 here) A pair An odd cycle
First observation: we can represent A(t) on a graph!
5
Lemma: There is always an
with a disconnected union of singletons, pairs, and odd cycles
5
Lemma: The SPF is equal to 2/(n+n1) (when the optimal decomposition is known). In this case the dimension of the optimal primal solutions P_t is the number of pairs.
n1 « singleton » (n1=1 here) A pair An odd cycle
5 n1 « singleton » (n1=1 here)
If n-2 singletons, If n-3 singletons, If n-4 singletons,
5 n1 « singleton » (n1=1 here)
First observation: we can represent A(t) on a graph!
5
Lemma: There is always an
with a disconnected union of singletons, pairs, and odd cycles
5
Lemma: The SPF is equal to 2/(n+n1) (when the optimal decomposition is known). In this case the dimension of the optimal primal solutions P_t is the number of pairs.
n1 « singleton » (n1=1 here)
First observation: we can represent A(t) on a graph!
5
Lemma: There is always an
with a disconnected union of singletons, pairs, and odd cycles
5
Lemma: The SPF is equal to 2/(n+n1) (when the optimal decomposition is known). In this case the dimension of the optimal primal solutions P_t is the number of pairs.
n1 « singleton » (n1=1 here)
Lemma: the dimension
Solutions has to decrease If k(t) remains constant
48
If n-2 singletons, at most 1 pair If n-3 singletons, at most 1 pair
If n-4 singletons, at most 2 pairs 1 step 1 step 2 steps
49
If n-2 singletons, at most 1 pair If n-3 singletons, at most 1 pair
If n-4 singletons, at most 2 pairs 1 step 1 step 2 steps
approaches
– a new upper bound (on a related quantity) – a counterexample
is always higher (or equal) than Cerny’s automaton
t=1+ (n+1) i
is always higher (or equal) than Cerny’s automaton
t=1+ (n+1) i
Automaton with 9 states, k(11)=2/9 and T3=12 = n+3 Contradicts both conjectures
Counterexample in black Černý’s automaton with 9 states in dashed
Extension of the family to 11 and 13 states It can be extended to any odd number
approaches
a related quantity)
– What with other automata: non-synchronizing automata, Non-deterministic... – Particular cases – Improve the bound on T3 O(n)? – Use our concepts to generate slowly synchronizing automata
The connection seems to bear some sense and suggests new questions.
n+3 < B <n²/6.4
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