Strong inapproximability of the shortest reset word - Google Sites

Strong inapproximability of the shortest reset word Paweł Gawrychowski1 1 University

Damian Straszak2

of Warsaw (supported by WCMCS) 2 EPFL

September 4, 2015

Paweł Gawrychowski, Damian Straszak

( University Shortest of Warsaw reset word (supported by WCMCS), EPFL) September 4, 2015

1 / 22

We consider deterministic finite automata with total transition functions: A = hQ, Σ, δi

states set

input alphabet

transition function, δ :Q×Σ→Q

Each element of Σ induces via δ a transformation of Q. We will view an automaton as a set of such transformations.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

2 / 22

We consider deterministic finite automata with total transition functions: A = hQ, Σ, δi

states set

input alphabet

transition function, δ :Q×Σ→Q

Each element of Σ induces via δ a transformation of Q. We will view an automaton as a set of such transformations.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

2 / 22

We consider the following question:

Reset words Is there a word w ∈ Σ∗ such that the state of A after reading w does not depend on the chosen starting state? w synchronizes (resets) A ⇔ |Qw| = 1

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

3 / 22

We consider the following question:

Reset words Is there a word w ∈ Σ∗ such that the state of A after reading w does not depend on the chosen starting state? w synchronizes (resets) A ⇔ |Qw| = 1

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

3 / 22

Example w = abba 0

a

b b a,b

2

1 a

{0, 1, 2}w = {2}, w resets A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

4 / 22

Example w = abba 0

a

b b a,b

2

1 a

{0, 1, 2}w = {2}, w resets A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

4 / 22

Example w = abba 0

a

b b a,b

2

1 a

{0, 1, 2}w = {2}, w resets A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

4 / 22

Example w = abba 0

a

b b a,b

2

1 a

{0, 1, 2}w = {2}, w resets A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

4 / 22

Example w = abba 0

a

b b a,b

2

1 a

{0, 1, 2}w = {2}, w resets A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

4 / 22

Example w = abba 0

a

b b a,b

2

1 a

{0, 1, 2}w = {2}, w resets A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

4 / 22

Example w = abba 0

a

b b a,b

2

1 a

{0, 1, 2}w = {2}, w resets A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

4 / 22

Example w = abba 0

a

b b a,b

2

1 a

{0, 1, 2}w = {2}, w resets A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

4 / 22

Example w = abba 0

a

b b a,b

2

1 a

{0, 1, 2}w = {2}, w resets A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

4 / 22

Natural questions: 1

how to check if an automaton admits a reset word?

2

what can be said about the shortest such word?

Observation To check if an automaton admits a reset word, it is enough to check whether for any pair of states q, q 0 ∈ Q there exists a word w such that |{q, q 0 }w| = 1.
Implies a polynomial time algorithm and an (n − 1) n2 ≈ 12 n3 bound on the length of the shortest synchronizing word (if one exists).

ˇ Conjecture (Cerný 1964) If an automaton is synchronizing then it admits a synchronizing word of length (n − 1)2 . However, the best bound known so far is 16 n3 . Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

5 / 22

Natural questions: 1

how to check if an automaton admits a reset word? easy

2

what can be said about the shortest such word?

Observation To check if an automaton admits a reset word, it is enough to check whether for any pair of states q, q 0 ∈ Q there exists a word w such that |{q, q 0 }w| = 1.
Implies a polynomial time algorithm and an (n − 1) n2 ≈ 12 n3 bound on the length of the shortest synchronizing word (if one exists).

ˇ Conjecture (Cerný 1964) If an automaton is synchronizing then it admits a synchronizing word of length (n − 1)2 . However, the best bound known so far is 16 n3 . Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

5 / 22

Natural questions: 1

how to check if an automaton admits a reset word? easy

2

what can be said about the shortest such word? not that easy!

Observation To check if an automaton admits a reset word, it is enough to check whether for any pair of states q, q 0 ∈ Q there exists a word w such that |{q, q 0 }w| = 1.
Implies a polynomial time algorithm and an (n − 1) n2 ≈ 12 n3 bound on the length of the shortest synchronizing word (if one exists).

ˇ Conjecture (Cerný 1964) If an automaton is synchronizing then it admits a synchronizing word of length (n − 1)2 . However, the best bound known so far is 16 n3 . Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

5 / 22

Natural questions: 1

how to check if an automaton admits a reset word? easy

2

what can be said about the shortest such word? not that easy!

Observation To check if an automaton admits a reset word, it is enough to check whether for any pair of states q, q 0 ∈ Q there exists a word w such that |{q, q 0 }w| = 1.
Implies a polynomial time algorithm and an (n − 1) n2 ≈ 12 n3 bound on the length of the shortest synchronizing word (if one exists).

ˇ Conjecture (Cerný 1964) If an automaton is synchronizing then it admits a synchronizing word of length (n − 1)2 . However, the best bound known so far is 16 n3 . Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

5 / 22

Why (n − 1)2 ?

b b a

0

7

b

b

a 1

a

a

6

2

a

a 5

b

b

3 a

4

a

b

b Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

6 / 22

Why (n − 1)2 ?

b b a

0

7

b

b

a 1

a

a

6

2

a

a 5

b

b

3 a

4

a

b

b Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

6 / 22

Given a synchronizing automaton A = hQ, Σ, δi we want to find its shortest synchronizing word. INPUT: OUTPUT:

description of A and an integer k is there word of length at most k that synchronizes A?

This is NP-complete (Eppstein 1990). However...

...do we really need a shortest synchronizing word? Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

7 / 22

Given a synchronizing automaton A = hQ, Σ, δi we want to find its shortest synchronizing word. INPUT: OUTPUT:

description of A and an integer k is there word of length at most k that synchronizes A?

This is NP-complete (Eppstein 1990). However...

...do we really need a shortest synchronizing word? Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

7 / 22

Given a synchronizing automaton A = hQ, Σ, δi we want to find its shortest synchronizing word. INPUT: OUTPUT:

description of A and an integer k is there word of length at most k that synchronizes A?

This is NP-complete (Eppstein 1990). However...

...do we really need a shortest synchronizing word? Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

7 / 22

Given a synchronizing automaton A = hQ, Σ, δi we want to find its shortest synchronizing word. INPUT: OUTPUT:

description of A and an integer k is there word of length at most k that synchronizes A?

This is NP-complete (Eppstein 1990). However...

...do we really need a shortest synchronizing word? Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

7 / 22

Syn(A) is the length of a shortest word synchronizing A. We want to solve the following problem in polynomial time.

S YN A PPX(Σ, α) INPUT: OUTPUT:

a synchronizing n-state automaton A over an alphabet Σ word of length at most α · Syn(A) synchronizing A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

8 / 22

Syn(A) is the length of a shortest word synchronizing A. We want to solve the following problem in polynomial time.

S YN A PPX(Σ, α) INPUT: OUTPUT:

a synchronizing n-state automaton A over an alphabet Σ word of length at most α · Syn(A) synchronizing A.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

8 / 22

Find a reset word of length at most α · Syn(A) in polynomial time.

Simple observation By iteratively "merging" pairs of states we can guarantee α = n − 1.

Generalization By iteratively "merging" k -tuples of states we can guarantee α = d kn−1 −1 e.

Hmmmm. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

9 / 22

Find a reset word of length at most α · Syn(A) in polynomial time.

Simple observation By iteratively "merging" pairs of states we can guarantee α = n − 1.

Generalization By iteratively "merging" k -tuples of states we can guarantee α = d kn−1 −1 e.

Hmmmm. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

9 / 22

Find a reset word of length at most α · Syn(A) in polynomial time.

Simple observation By iteratively "merging" pairs of states we can guarantee α = n − 1.

Generalization By iteratively "merging" k -tuples of states we can guarantee α = d kn−1 −1 e.

Hmmmm. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

9 / 22

Find a reset word of length at most α · Syn(A) in polynomial time.

Simple observation By iteratively "merging" pairs of states we can guarantee α = n − 1.

Generalization By iteratively "merging" k -tuples of states we can guarantee α = d kn−1 −1 e.

Hmmmm. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

9 / 22

Theorem (this paper) For every constant ε > 0, S YN A PPX({0, 1}, n1−ε ) is not solvable in polynomial time, unless P = NP.

Previous work It was only known that S YN A PPX({0, 1}, log n) is not solvable in polynomial time, unless P = NP (Gerbush and Heeringa 2011 + Berlinkov 2013, reduction from set cover) and it was conjectured that O(log n) can be actually achieved. Essentially tight due to the d kn−1 −1 e algorithm, for some definition of essentially.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

10 / 22

Theorem (this paper) For every constant ε > 0, S YN A PPX({0, 1}, n1−ε ) is not solvable in polynomial time, unless P = NP.

Previous work It was only known that S YN A PPX({0, 1}, log n) is not solvable in polynomial time, unless P = NP (Gerbush and Heeringa 2011 + Berlinkov 2013, reduction from set cover) and it was conjectured that O(log n) can be actually achieved. Essentially tight due to the d kn−1 −1 e algorithm, for some definition of essentially.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

10 / 22

Theorem (this paper) For every constant ε > 0, S YN A PPX({0, 1}, n1−ε ) is not solvable in polynomial time, unless P = NP.

Previous work It was only known that S YN A PPX({0, 1}, log n) is not solvable in polynomial time, unless P = NP (Gerbush and Heeringa 2011 + Berlinkov 2013, reduction from set cover) and it was conjectured that O(log n) can be actually achieved. Essentially tight due to the d kn−1 −1 e algorithm, for some definition of essentially.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

10 / 22

Theorem (this talk) For some constant ε > 0, S YN A PPX({0, 1, 2}, nε ) is not solvable in polynomial time, unless P = NP.

Previous work It was only known that S YN A PPX({0, 1}, log n) is not solvable in polynomial time, unless P = NP (Gerbush and Heeringa 2011 + Berlinkov 2013, reduction from set cover) and it was conjectured that O(log n) can be actually achieved. Essentially tight due to the d kn−1 −1 e algorithm, for some definition of essentially.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

10 / 22

Let us start with a (much) weaker version.

Theorem For every constant ε > 0, S YN A PPX({0, 1, 2}, 2 − ε) is not solvable in polynomial time, unless P = NP.

Idea We will show that S YN A PPX({0, 1, 2}, 2 − ε) can be used to solve 3-SAT by the following reduction: given an N-variable 3-CNF formula φ consisting of M clauses we can build in polynomial time a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≈ N,

2

otherwise Syn(Aφ ) ≥ 2N.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

11 / 22

Let us start with a (much) weaker version.

Theorem For every constant ε > 0, S YN A PPX({0, 1, 2}, 2 − ε) is not solvable in polynomial time, unless P = NP.

Idea We will show that S YN A PPX({0, 1, 2}, 2 − ε) can be used to solve 3-SAT by the following reduction: given an N-variable 3-CNF formula φ consisting of M clauses we can build in polynomial time a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≈ N,

2

otherwise Syn(Aφ ) ≥ 2N.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

11 / 22

Let us start with a (much) weaker version.

Theorem For every constant ε > 0, S YN A PPX({0, 1, 2}, 2 − ε) is not solvable in polynomial time, unless P = NP.

Idea We will show that S YN A PPX({0, 1, 2}, 2 − ε) can be used to solve 3-SAT by the following reduction: given an N-variable 3-CNF formula φ consisting of M clauses we can build in polynomial time a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≈ N,

2

otherwise Syn(Aφ ) ≥ 2N.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

11 / 22

Let us start with a (much) weaker version.

Theorem For every constant ε > 0, S YN A PPX({0, 1, 2}, 2 − ε) is not solvable in polynomial time, unless P = NP.

Idea We will show that S YN A PPX({0, 1, 2}, 2 − ε) can be used to solve 3-SAT by the following reduction: given an N-variable 3-CNF formula φ consisting of M clauses we can build in polynomial time a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≈ N,

2

otherwise Syn(Aφ ) ≥ 2N.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

11 / 22

Aφ consists of M gadgets corresponding to the clauses. They all share the same sink state s.

Gadget for a clause x3 ∨ x5 and N = 6 0, 1 0

q30

0, 1

q40

0

q0ε

0, 1

q1ε

0, 1

q500

0, 1

q501

0, 1

q510

0, 1

q511

0, 1

1

q2ε 1

0

q31

0, 1

q41

1

q600 q601

0, 1 0, 1

q610

s

0, 1

q611

Additionally, every non-sink node is connected to the root of the gadget by a transition labeled with 2. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

12 / 22

Aφ consists of M gadgets corresponding to the clauses. They all share the same sink state s.

Gadget for a clause x3 ∨ x5 and N = 6 0, 1 0

q30

0, 1

q40

0

q0ε

0, 1

q1ε

0, 1

q500

0, 1

q501

0, 1

q510

0, 1

q511

0, 1

1

q2ε 1

0

q31

0, 1

q41

1

q600 q601

0, 1 0, 1

q610

s

0, 1

q611

Additionally, every non-sink node is connected to the root of the gadget by a transition labeled with 2. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

12 / 22

Aφ consists of M gadgets corresponding to the clauses. They all share the same sink state s.

Gadget for a clause x3 ∨ x5 and N = 6 0, 1 0

q30

0, 1

q40

0

q0ε

0, 1

q1ε

0, 1

q500

0, 1

q501

0, 1

q510

0, 1

q511

0, 1

1

q2ε 1

0

q31

0, 1

q41

1

q600 q601

0, 1 0, 1

q610

s

0, 1

q611

Additionally, every non-sink node is connected to the root of the gadget by a transition labeled with 2. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

12 / 22

Now it is not difficult to see that synchronizing words correspond to truth assignments: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥ 2N + 2.

Hence (2 − ε)-approximation allows us to distinguish between these two cases in polynomial time.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

13 / 22

Now it is not difficult to see that synchronizing words correspond to truth assignments: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥ 2N + 2.

Hence (2 − ε)-approximation allows us to distinguish between these two cases in polynomial time.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

13 / 22

Now it is not difficult to see that synchronizing words correspond to truth assignments: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥ 2N + 2.

Hence (2 − ε)-approximation allows us to distinguish between these two cases in polynomial time.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

13 / 22

Now it is not difficult to see that synchronizing words correspond to truth assignments: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥ 2N + 2.

Hence (2 − ε)-approximation allows us to distinguish between these two cases in polynomial time.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

13 / 22

Now we will try to prove a stronger statement.

Theorem For some constant ε > 0, S YN A PPX({0, 1, 2}, nε ) is not solvable in polynomial time, unless P = NP. Working directly with 3-SAT doesn’t really work. We need a problem with larger "gap" between yes-instances and no-instances.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

14 / 22

Now we will try to prove a stronger statement.

Theorem For some constant ε > 0, S YN A PPX({0, 1, 2}, nε ) is not solvable in polynomial time, unless P = NP. Working directly with 3-SAT doesn’t really work. We need a problem with larger "gap" between yes-instances and no-instances.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

14 / 22

Now we will try to prove a stronger statement.

Theorem For some constant ε > 0, S YN A PPX({0, 1, 2}, nε ) is not solvable in polynomial time, unless P = NP. Working directly with 3-SAT doesn’t really work. We need a problem with larger "gap" between yes-instances and no-instances.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

14 / 22

PCP theorems

A polynomial-time probabilistic machine V is called a (p(n), r (n), q(n))-PCP verifier for a language L ⊆ {0, 1}∗ if: for an input x of length n, given random access to a “proof” π ∈ {0, 1}∗ , V uses at most r (n) random bits, accesses at most q(n) locations of π, and outputs 0 or 1 if x ∈ L then there is a proof π, such that Pr[V (x, π) = 1] = 1, if x ∈ / L then for every proof π, Pr[V (x, π) = 1] ≤ p(n). We consider only nonadaptive verifiers, meaning that the accessed locations depend only on the input and the random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

15 / 22

PCP theorems

A polynomial-time probabilistic machine V is called a (p(n), r (n), q(n))-PCP verifier for a language L ⊆ {0, 1}∗ if: for an input x of length n, given random access to a “proof” π ∈ {0, 1}∗ , V uses at most r (n) random bits, accesses at most q(n) locations of π, and outputs 0 or 1 if x ∈ L then there is a proof π, such that Pr[V (x, π) = 1] = 1, if x ∈ / L then for every proof π, Pr[V (x, π) = 1] ≤ p(n). We consider only nonadaptive verifiers, meaning that the accessed locations depend only on the input and the random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

15 / 22

PCP theorems

A polynomial-time probabilistic machine V is called a (p(n), r (n), q(n))-PCP verifier for a language L ⊆ {0, 1}∗ if: for an input x of length n, given random access to a “proof” π ∈ {0, 1}∗ , V uses at most r (n) random bits, accesses at most q(n) locations of π, and outputs 0 or 1 if x ∈ L then there is a proof π, such that Pr[V (x, π) = 1] = 1, if x ∈ / L then for every proof π, Pr[V (x, π) = 1] ≤ p(n). We consider only nonadaptive verifiers, meaning that the accessed locations depend only on the input and the random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

15 / 22

PCP theorems

A polynomial-time probabilistic machine V is called a (p(n), r (n), q(n))-PCP verifier for a language L ⊆ {0, 1}∗ if: for an input x of length n, given random access to a “proof” π ∈ {0, 1}∗ , V uses at most r (n) random bits, accesses at most q(n) locations of π, and outputs 0 or 1 if x ∈ L then there is a proof π, such that Pr[V (x, π) = 1] = 1, if x ∈ / L then for every proof π, Pr[V (x, π) = 1] ≤ p(n). We consider only nonadaptive verifiers, meaning that the accessed locations depend only on the input and the random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

15 / 22

PCP theorems

A polynomial-time probabilistic machine V is called a (p(n), r (n), q(n))-PCP verifier for a language L ⊆ {0, 1}∗ if: for an input x of length n, given random access to a “proof” π ∈ {0, 1}∗ , V uses at most r (n) random bits, accesses at most q(n) locations of π, and outputs 0 or 1 if x ∈ L then there is a proof π, such that Pr[V (x, π) = 1] = 1, if x ∈ / L then for every proof π, Pr[V (x, π) = 1] ≤ p(n). We consider only nonadaptive verifiers, meaning that the accessed locations depend only on the input and the random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

15 / 22

PCP theorems Arora et al. 1998 NP = PCP1/2 [O(log n), O(1)]. meaning that 3-SAT has a verifier which: 1

uses O(log n) random bits,

2

accesses O(1) locations of the proof,

3

accepts a wrong proof with probability ≤ 12 .

We need a stronger version with probability ≈ n1 . This can be achieved by (standard) amplification using a random walks on an expander. NP = PCP1/n [O(log n), O(log n)].

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

16 / 22

PCP theorems Arora et al. 1998 NP = PCP1/2 [O(log n), O(1)]. meaning that 3-SAT has a verifier which: 1

uses O(log n) random bits,

2

accesses O(1) locations of the proof,

3

accepts a wrong proof with probability ≤ 12 .

We need a stronger version with probability ≈ n1 . This can be achieved by (standard) amplification using a random walks on an expander. NP = PCP1/n [O(log n), O(log n)].

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

16 / 22

PCP theorems Arora et al. 1998 NP = PCP1/2 [O(log n), O(1)]. meaning that 3-SAT has a verifier which: 1

uses O(log n) random bits,

2

accesses O(1) locations of the proof,

3

accepts a wrong proof with probability ≤ 12 .

We need a stronger version with probability ≈ n1 . This can be achieved by (standard) amplification using a random walks on an expander. NP = PCP1/n [O(log n), O(log n)].

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

16 / 22

PCP theorems Arora et al. 1998 NP = PCP1/2 [O(log n), O(1)]. meaning that 3-SAT has a verifier which: 1

uses O(log n) random bits,

2

accesses O(1) locations of the proof,

3

accepts a wrong proof with probability ≤ 12 .

We need a stronger version with probability ≈ n1 . This can be achieved by (standard) amplification using a random walks on an expander. NP = PCP1/n [O(log n), O(log n)].

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

16 / 22

PCP theorems Arora et al. 1998 NP = PCP1/2 [O(log n), O(1)]. meaning that 3-SAT has a verifier which: 1

uses O(log n) random bits,

2

accesses O(1) locations of the proof,

3

accepts a wrong proof with probability ≤ 12 .

We need a stronger version with probability ≈ n1 . This can be achieved by (standard) amplification using a random walks on an expander. NP = PCP1/n [O(log n), O(log n)].

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

16 / 22

PCP theorems Arora et al. 1998 NP = PCP1/2 [O(log n), O(1)]. meaning that 3-SAT has a verifier which: 1

uses O(log n) random bits,

2

accesses O(1) locations of the proof,

3

accepts a wrong proof with probability ≤ 12 .

We need a stronger version with probability ≈ n1 . This can be achieved by (standard) amplification using a random walks on an expander. NP = PCP1/n [O(log n), O(log n)].

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

16 / 22

Constraint satisfaction problems

Now we are ready to define a family of problems with larger "gap" between yes-instances and no-instances.

qCSP A qCSP over N boolean variables is a collection of M boolean constraints φ(C1 , C2 , . . . , CM ). Every constraint Ci is just a function {0, 1}N → {0, 1} which "depends" on at most q variables. Val(φ) is the maximum fraction of constraints of φ which can be simultaneously satisfied by a single assignment.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

17 / 22

Constraint satisfaction problems

Now we are ready to define a family of problems with larger "gap" between yes-instances and no-instances.

qCSP A qCSP over N boolean variables is a collection of M boolean constraints φ(C1 , C2 , . . . , CM ). Every constraint Ci is just a function {0, 1}N → {0, 1} which "depends" on at most q variables. Val(φ) is the maximum fraction of constraints of φ which can be simultaneously satisfied by a single assignment.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

17 / 22

Constraint satisfaction problems

Now we are ready to define a family of problems with larger "gap" between yes-instances and no-instances.

qCSP A qCSP over N boolean variables is a collection of M boolean constraints φ(C1 , C2 , . . . , CM ). Every constraint Ci is just a function {0, 1}N → {0, 1} which "depends" on at most q variables. Val(φ) is the maximum fraction of constraints of φ which can be simultaneously satisfied by a single assignment.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

17 / 22

qCSPs from the PCP theorem Given a 3-CNF n-variable formula φ, we can construct in polynomial time a qCSP instance f (φ) with q = O(log n), such that: 1

if φ is satisfiable then Val(f (φ)) = 1,

2

otherwise Val(f (φ)) ≤ n1 .

Take the PCP verifier with r = O(log n), q = O(log n) and p(n) = n1 . The proof length is ` ≤ q · 2r . 1

We create one boolean variable for every location in the proof.

2

We create one constraint for every possible sequence of r random bits.

3

Constraint corresponding to a particular sequence of r random bits evaluates to 1 if any only if the verifier accepts the proof for that sequence of random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

18 / 22

qCSPs from the PCP theorem Given a 3-CNF n-variable formula φ, we can construct in polynomial time a qCSP instance f (φ) with q = O(log n), such that: 1

if φ is satisfiable then Val(f (φ)) = 1,

2

otherwise Val(f (φ)) ≤ n1 .

Take the PCP verifier with r = O(log n), q = O(log n) and p(n) = n1 . The proof length is ` ≤ q · 2r . 1

We create one boolean variable for every location in the proof.

2

We create one constraint for every possible sequence of r random bits.

3

Constraint corresponding to a particular sequence of r random bits evaluates to 1 if any only if the verifier accepts the proof for that sequence of random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

18 / 22

qCSPs from the PCP theorem Given a 3-CNF n-variable formula φ, we can construct in polynomial time a qCSP instance f (φ) with q = O(log n), such that: 1

if φ is satisfiable then Val(f (φ)) = 1,

2

otherwise Val(f (φ)) ≤ n1 .

Take the PCP verifier with r = O(log n), q = O(log n) and p(n) = n1 . The proof length is ` ≤ q · 2r . 1

We create one boolean variable for every location in the proof.

2

We create one constraint for every possible sequence of r random bits.

3

Constraint corresponding to a particular sequence of r random bits evaluates to 1 if any only if the verifier accepts the proof for that sequence of random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

18 / 22

qCSPs from the PCP theorem Given a 3-CNF n-variable formula φ, we can construct in polynomial time a qCSP instance f (φ) with q = O(log n), such that: 1

if φ is satisfiable then Val(f (φ)) = 1,

2

otherwise Val(f (φ)) ≤ n1 .

Take the PCP verifier with r = O(log n), q = O(log n) and p(n) = n1 . The proof length is ` ≤ q · 2r . 1

We create one boolean variable for every location in the proof.

2

We create one constraint for every possible sequence of r random bits.

3

Constraint corresponding to a particular sequence of r random bits evaluates to 1 if any only if the verifier accepts the proof for that sequence of random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

18 / 22

qCSPs from the PCP theorem Given a 3-CNF n-variable formula φ, we can construct in polynomial time a qCSP instance f (φ) with q = O(log n), such that: 1

if φ is satisfiable then Val(f (φ)) = 1,

2

otherwise Val(f (φ)) ≤ n1 .

Take the PCP verifier with r = O(log n), q = O(log n) and p(n) = n1 . The proof length is ` ≤ q · 2r . 1

We create one boolean variable for every location in the proof.

2

We create one constraint for every possible sequence of r random bits.

3

Constraint corresponding to a particular sequence of r random bits evaluates to 1 if any only if the verifier accepts the proof for that sequence of random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

18 / 22

qCSPs from the PCP theorem Given a 3-CNF n-variable formula φ, we can construct in polynomial time a qCSP instance f (φ) with q = O(log n), such that: 1

if φ is satisfiable then Val(f (φ)) = 1,

2

otherwise Val(f (φ)) ≤ n1 .

Take the PCP verifier with r = O(log n), q = O(log n) and p(n) = n1 . The proof length is ` ≤ q · 2r . 1

We create one boolean variable for every location in the proof.

2

We create one constraint for every possible sequence of r random bits.

3

Constraint corresponding to a particular sequence of r random bits evaluates to 1 if any only if the verifier accepts the proof for that sequence of random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

18 / 22

qCSPs from the PCP theorem Given a 3-CNF n-variable formula φ, we can construct in polynomial time a qCSP instance f (φ) with q = O(log n), such that: 1

if φ is satisfiable then Val(f (φ)) = 1,

2

otherwise Val(f (φ)) ≤ n1 .

Take the PCP verifier with r = O(log n), q = O(log n) and p(n) = n1 . The proof length is ` ≤ q · 2r . 1

We create one boolean variable for every location in the proof.

2

We create one constraint for every possible sequence of r random bits.

3

Constraint corresponding to a particular sequence of r random bits evaluates to 1 if any only if the verifier accepts the proof for that sequence of random bits.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

18 / 22

Now we can prove our theorem.

Idea Similarly as before, given an N-variables qCSP φ consisting of M clauses and q = O(log N), we want to construct a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥

1 Val(φ) (N

+ 1).

Almost the same construction works: for every constraint, we build a separate tree gadget, which "accumulates" the relevant variables. All gadget share the same sink state s. The size of the automaton corresponding to an n variable 3-CNF formula φ is polynomial in n. Hence, for some constant ε > 0 solving S YN A PPX({0, 1, 2}, nε ) in polynomial time would allow us to solve 3-SAT in polynomial time. ε ≈ 0.0095 can be achieved by keeping track of all the constants. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

19 / 22

Now we can prove our theorem.

Idea Similarly as before, given an N-variables qCSP φ consisting of M clauses and q = O(log N), we want to construct a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥

1 Val(φ) (N

+ 1).

Almost the same construction works: for every constraint, we build a separate tree gadget, which "accumulates" the relevant variables. All gadget share the same sink state s. The size of the automaton corresponding to an n variable 3-CNF formula φ is polynomial in n. Hence, for some constant ε > 0 solving S YN A PPX({0, 1, 2}, nε ) in polynomial time would allow us to solve 3-SAT in polynomial time. ε ≈ 0.0095 can be achieved by keeping track of all the constants. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

19 / 22

Now we can prove our theorem.

Idea Similarly as before, given an N-variables qCSP φ consisting of M clauses and q = O(log N), we want to construct a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥

1 Val(φ) (N

+ 1).

Almost the same construction works: for every constraint, we build a separate tree gadget, which "accumulates" the relevant variables. All gadget share the same sink state s. The size of the automaton corresponding to an n variable 3-CNF formula φ is polynomial in n. Hence, for some constant ε > 0 solving S YN A PPX({0, 1, 2}, nε ) in polynomial time would allow us to solve 3-SAT in polynomial time. ε ≈ 0.0095 can be achieved by keeping track of all the constants. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

19 / 22

Now we can prove our theorem.

Idea Similarly as before, given an N-variables qCSP φ consisting of M clauses and q = O(log N), we want to construct a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥

1 Val(φ) (N

+ 1).

Almost the same construction works: for every constraint, we build a separate tree gadget, which "accumulates" the relevant variables. All gadget share the same sink state s. The size of the automaton corresponding to an n variable 3-CNF formula φ is polynomial in n. Hence, for some constant ε > 0 solving S YN A PPX({0, 1, 2}, nε ) in polynomial time would allow us to solve 3-SAT in polynomial time. ε ≈ 0.0095 can be achieved by keeping track of all the constants. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

19 / 22

Now we can prove our theorem.

Idea Similarly as before, given an N-variables qCSP φ consisting of M clauses and q = O(log N), we want to construct a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥

1 Val(φ) (N

+ 1).

Almost the same construction works: for every constraint, we build a separate tree gadget, which "accumulates" the relevant variables. All gadget share the same sink state s. The size of the automaton corresponding to an n variable 3-CNF formula φ is polynomial in n. Hence, for some constant ε > 0 solving S YN A PPX({0, 1, 2}, nε ) in polynomial time would allow us to solve 3-SAT in polynomial time. ε ≈ 0.0095 can be achieved by keeping track of all the constants. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

19 / 22

Now we can prove our theorem.

Idea Similarly as before, given an N-variables qCSP φ consisting of M clauses and q = O(log N), we want to construct a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥

1 Val(φ) (N

+ 1).

Almost the same construction works: for every constraint, we build a separate tree gadget, which "accumulates" the relevant variables. All gadget share the same sink state s. The size of the automaton corresponding to an n variable 3-CNF formula φ is polynomial in n. Hence, for some constant ε > 0 solving S YN A PPX({0, 1, 2}, nε ) in polynomial time would allow us to solve 3-SAT in polynomial time. ε ≈ 0.0095 can be achieved by keeping track of all the constants. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

19 / 22

Now we can prove our theorem.

Idea Similarly as before, given an N-variables qCSP φ consisting of M clauses and q = O(log N), we want to construct a synchronizing automaton Aφ such that: 1

if φ is satisfiable then Syn(Aφ ) ≤ N + 2,

2

otherwise Syn(Aφ ) ≥

1 Val(φ) (N

+ 1).

Almost the same construction works: for every constraint, we build a separate tree gadget, which "accumulates" the relevant variables. All gadget share the same sink state s. The size of the automaton corresponding to an n variable 3-CNF formula φ is polynomial in n. Hence, for some constant ε > 0 solving S YN A PPX({0, 1, 2}, nε ) in polynomial time would allow us to solve 3-SAT in polynomial time. ε ≈ 0.0095 can be achieved by keeping track of all the constants. Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

19 / 22

Theorem (this paper) For every constant ε > 0, S YN A PPX({0, 1}, n1−ε ) is not solvable in polynomial time, unless P = NP. Here we need the notion of free bit complexity and a stronger PCP theorem (Håstad 1999 + Zuckerman 2006). This is somehow more involved.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

20 / 22

Theorem (this paper) For every constant ε > 0, S YN A PPX({0, 1}, n1−ε ) is not solvable in polynomial time, unless P = NP. Here we need the notion of free bit complexity and a stronger PCP theorem (Håstad 1999 + Zuckerman 2006). This is somehow more involved.

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

20 / 22

Open problems: 1

design an o(n)-approximation algorithm. Clever o(n)-approximation algorithms do exist for the max clique problem (Feige 2004).

2

hardness for non-multiplicative approximation?

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

21 / 22

Open problems: 1

design an o(n)-approximation algorithm. Clever o(n)-approximation algorithms do exist for the max clique problem (Feige 2004).

2

hardness for non-multiplicative approximation?

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

21 / 22

Open problems: 1

design an o(n)-approximation algorithm. Clever o(n)-approximation algorithms do exist for the max clique problem (Feige 2004).

2

hardness for non-multiplicative approximation?

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

21 / 22

Questions?

Paweł Gawrychowski, Damian Straszak

Shortest reset word

September 4, 2015

22 / 22