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
