The Complexity of Counting Problems Tyson Williams (University of - - PowerPoint PPT Presentation

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The Complexity of Counting Problems Tyson Williams (University of - - PowerPoint PPT Presentation

Nov 05, 2022 214 likes •1.35k views

The Complexity of Counting Problems Tyson Williams (University of Wisconsin-Madison) Joint with: Jin-Yi Cai and Heng Guo (University of Wisconsin-Madison) Michael Kowalczyk (Northern Michigan University) Tyson Williams (UW-M) Counting

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The Complexity of Counting Problems

Tyson Williams (University of Wisconsin-Madison) Joint with: Jin-Yi Cai and Heng Guo (University of Wisconsin-Madison) Michael Kowalczyk (Northern Michigan University)

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 1 / 32

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Complexity Theory Review

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 2 / 32

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In the beginning, there was SAT Problem: SAT Input: A Boolean formula (in conjunctive normal form). Output: “Yes” if there is a satisfying assignment “No” otherwise.

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 3 / 32

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In the beginning, there was SAT Problem: SAT Input: A Boolean formula (in conjunctive normal form). Output: “Yes” if there is a satisfying assignment “No” otherwise. Example input: (w ∨ x ∨ y ∨ z) ∧ (x ∨ y) ∧ (x ∨ y ∨ z) ∧ (w ∨ x ∨ y ∨ z)

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 3 / 32

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In the beginning, there was SAT Problem: SAT Input: A Boolean formula (in conjunctive normal form). Output: “Yes” if there is a satisfying assignment “No” otherwise. Example input: (w ∨ x ∨ y ∨ z) ∧ (x ∨ y) ∧ (x ∨ y ∨ z) ∧ (w ∨ x ∨ y ∨ z) Example output: Yes

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 3 / 32

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In the beginning, there was SAT Problem: SAT Input: A Boolean formula (in conjunctive normal form). Output: “Yes” if there is a satisfying assignment “No” otherwise. Example input: (w ∨ x ∨ y ∨ z) ∧ (x ∨ y) ∧ (x ∨ y ∨ z) ∧ (w ∨ x ∨ y ∨ z) Example output: Yes Satisfying assignment: w = x = True y = z = False

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 3 / 32

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In the beginning, there was SAT Problem: SAT Input: A Boolean formula (in conjunctive normal form). Output: “Yes” if there is a satisfying assignment “No” otherwise. Example input: (w ∨ x ∨ y ∨ z) ∧ (x ∨ y) ∧ (x ∨ y ∨ z) ∧ (w ∨ x ∨ y ∨ z) Example output: Yes Satisfying assignment: w = x = True y = z = False Theorem (Cook ’71, Levin ’73) SAT ∈ NP-Complete.

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 3 / 32

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And SAT begat 3SAT Problem: 3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. Output: “Yes” if there is a satisfying assignment “No” otherwise.

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 4 / 32

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And SAT begat 3SAT Problem: 3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. Output: “Yes” if there is a satisfying assignment “No” otherwise. Example input: (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ y) ∧ (w ∨ x ∨ y)

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 4 / 32

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And SAT begat 3SAT Problem: 3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. Output: “Yes” if there is a satisfying assignment “No” otherwise. Example input: (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ y) ∧ (w ∨ x ∨ y) Example output: No

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And SAT begat 3SAT Problem: 3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. Output: “Yes” if there is a satisfying assignment “No” otherwise. Example input: (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ y) ∧ (w ∨ x ∨ y) Example output: No Theorem 3SAT ∈ NP-Complete.

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And 3SAT begat 1-in-3SAT and NAE-3SAT Problem: 1-in-3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. Problem: NAE-3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ z) ∧ (x ∨ y ∨ z)

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 5 / 32

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And 3SAT begat 1-in-3SAT and NAE-3SAT Problem: 1-in-3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. Output: “Yes” if each clause has exactly 1 true literal “No” otherwise. Problem: NAE-3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ z) ∧ (x ∨ y ∨ z)

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 5 / 32

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And 3SAT begat 1-in-3SAT and NAE-3SAT Problem: 1-in-3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. Output: “Yes” if each clause has exactly 1 true literal “No” otherwise. Problem: NAE-3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. Output: “Yes” if the literals in each clause are not all equal “No” otherwise. (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ z) ∧ (x ∨ y ∨ z)

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 5 / 32

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And 3SAT begat 1-in-3SAT and NAE-3SAT Problem: 1-in-3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. Output: “Yes” if each clause has exactly 1 true literal “No” otherwise. Problem: NAE-3SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has exactly 3 literals. Output: “Yes” if the literals in each clause are not all equal “No” otherwise. (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) Theorem 1-in-3SAT, NAE-3SAT ∈ NP-Complete.

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Then came the monotone versions Definition A Boolean formula is monotone if the formula contains no negations.

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Then came the monotone versions Definition A Boolean formula is monotone if the formula contains no negations. What is the complexity of... Mon-3SAT? Mon-1-in-3SAT? Mon-NAE-3SAT?

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Then came the monotone versions Definition A Boolean formula is monotone if the formula contains no negations. What is the complexity of... Mon-3SAT? Mon-1-in-3SAT ∈ NP-Complete Mon-NAE-3SAT ∈ NP-Complete

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Then came the monotone versions Definition A Boolean formula is monotone if the formula contains no negations. What is the complexity of... Mon-3SAT ∈ P Mon-1-in-3SAT ∈ NP-Complete Mon-NAE-3SAT ∈ NP-Complete

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What else is tractable? Problem: Horn-SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has at most 1 positive literal. Output: “Yes” if there is a satisfying assignment “No” otherwise.

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What else is tractable? Problem: Horn-SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has at most 1 positive literal. Output: “Yes” if there is a satisfying assignment “No” otherwise. Example input: (w ∨ y ∨ z) ∧ (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ y)

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What else is tractable? Problem: Horn-SAT Input: A Boolean formula (in conjunctive normal form) such that each clause has at most 1 positive literal. Output: “Yes” if there is a satisfying assignment “No” otherwise. Example input: (w ∨ y ∨ z) ∧ (w ∨ x ∨ z) ∧ (x ∨ y ∨ z) ∧ (w ∨ y ∨ z) ∧ (w ∨ x ∨ y) Theorem Horn-SAT ∈ P.

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Constraint Satisfaction Problems These types of problems called Constraint Satisfaction Problems (CSPs).

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Constraint Satisfaction Problems These types of problems called Constraint Satisfaction Problems (CSPs). Definition The CSP defined by a set of constraints F is denoted by CSP(F).

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Constraint Satisfaction Problems These types of problems called Constraint Satisfaction Problems (CSPs). Definition The CSP defined by a set of constraints F is denoted by CSP(F). Examples: SAT has F = {ORk | k ∈ N} ∪ {NOT2} 3SAT has F = {OR3, NOT2} 1-in-3SAT has F = {EXACTLY-ONE3, NOT2} NAE-3SAT has F = {NOT-All-EQUAL3, NOT2} Mon-3SAT has F = {OR3} Mon-1-in-3SAT has F = {EXACTLY-ONE3} Mon-NAE-3SAT has F = {NOT-All-EQUAL3}

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One theorem to rule them all

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One theorem to rule them all Theorem (Schaefer’s dichotomy theorem ’78) For any set of constraint functions F, the problem CSP(F) is NP-Complete unless one of the following conditions holds, in which case the problem is in P:

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One theorem to rule them all Theorem (Schaefer’s dichotomy theorem ’78) For any set of constraint functions F, the problem CSP(F) is NP-Complete unless one of the following conditions holds, in which case the problem is in P: All constraints in F...

1

that are not constantly false are true when all its arguments are true;

2

that are not constantly false are true when all its arguments are false;

3

are equivalent to a conjunction of binary clauses;

4

are equivalent to a conjunction of Horn clauses;

5

are equivalent to a conjunction of dual-Horn clauses;

6

are equivalent to a conjunction of affine formula.

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One theorem to rule them all Theorem (Schaefer’s dichotomy theorem ’78) For any set of constraint functions F, the problem CSP(F) is NP-Complete unless one of the following conditions holds, in which case the problem is in P: All constraints in F...

1

that are not constantly false are true when all its arguments are true;

2

that are not constantly false are true when all its arguments are false;

3

are equivalent to a conjunction of binary clauses;

4

are equivalent to a conjunction of Horn clauses;

5

are equivalent to a conjunction of dual-Horn clauses;

6

are equivalent to a conjunction of affine formula.

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One theorem to rule them all Theorem (Schaefer’s dichotomy theorem ’78) For any set of constraint functions F, the problem CSP(F) is NP-Complete unless one of the following conditions holds, in which case the problem is in P: All constraints in F...

1

that are not constantly false are true when all its arguments are true;

2

that are not constantly false are true when all its arguments are false;

3

are equivalent to a conjunction of binary clauses;

4

are equivalent to a conjunction of Horn clauses;

5

are equivalent to a conjunction of dual-Horn clauses;

6

are equivalent to a conjunction of affine formula.

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Three takeaways from Schaefer’s dichotomy theorem

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Three takeaways from Schaefer’s dichotomy theorem Theorist: describes the line between easy problems and hard problems

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Three takeaways from Schaefer’s dichotomy theorem Theorist: describes the line between easy problems and hard problems Practitioner (i.e. employee of Google): a complexity dictionary

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Three takeaways from Schaefer’s dichotomy theorem Theorist: describes the line between easy problems and hard problems Practitioner (i.e. employee of Google): a complexity dictionary Observation: no problems of intermediate complexity

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Three takeaways from Schaefer’s dichotomy theorem Theorist: describes the line between easy problems and hard problems Practitioner (i.e. employee of Google): a complexity dictionary Observation: no problems of intermediate complexity Theorem (Ladner’s theorem ’75) If P = NP, then there exists problems in NP of intermediate complexity.

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http://commons.wikimedia.org/wiki/File:P_np_np-complete_np-hard.svg Tyson Williams (UW-M) Counting Complexity Google Madison 2013 11 / 32

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A Motivation for Counting

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Independent Set Want large independent set.

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Independent Set Want large independent set.

http://commons.wikimedia.org/wiki/File:Independent_set_graph.svg Tyson Williams (UW-M) Counting Complexity Google Madison 2013 13 / 32

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The Facts Problem: IndependentSet Input: A graph G and k ∈ N. Output: “Yes” if G contains an independent set of size at least k “No” otherwise.

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The Facts Problem: IndependentSet Input: A graph G and k ∈ N. Output: “Yes” if G contains an independent set of size at least k “No” otherwise. IndependentSet ∈ NP-Complete

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The Facts Problem: IndependentSet Input: A graph G and k ∈ N. Output: “Yes” if G contains an independent set of size at least k “No” otherwise. IndependentSet ∈ NP-Complete Next question: how close to optimal can we get?

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A Different Approach Let I(G) be the set of independent sets in G.

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A Different Approach Let I(G) be the set of independent sets in G. Want to randomly sample I from I(G) such that Pr(I) ∝ w(I).

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A Different Approach Let I(G) be the set of independent sets in G. Want to randomly sample I from I(G) such that Pr(I) ∝ w(I). Then must have Pr(I) = w(I) Z(G), where Z(G) =

  • I∈I(G)

w(I).

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A Different Approach Let I(G) be the set of independent sets in G. Want to randomly sample I from I(G) such that Pr(I) ∝ w(I). Then must have Pr(I) = w(I) Z(G), where Z(G) =

  • I∈I(G)

w(I). Know as the partition function in statistical physics.

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Weight functions Pr(I) = w(I) Z(G) where Z(G) =

  • I∈I(G)

w(I).

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Weight functions Pr(I) = w(I) Z(G) where Z(G) =

  • I∈I(G)

w(I). If w(I) = 1, then Z(G) = |I(G)| is the number of independent sets.

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Weight functions Pr(I) = w(I) Z(G) where Z(G) =

  • I∈I(G)

w(I). If w(I) = 1, then Z(G) = |I(G)| is the number of independent sets. Statistical physicists considered w(I) = λ|I| from some nonnegative λ ∈ R.

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Weight functions Pr(I) = w(I) Z(G) where Z(G) =

  • I∈I(G)

w(I). If w(I) = 1, then Z(G) = |I(G)| is the number of independent sets. Statistical physicists considered w(I) = λ|I| from some nonnegative λ ∈ R. If λ = 1, then Z(G) = |I(G)| again.

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Weight functions Pr(I) = w(I) Z(G) where Z(G) =

  • I∈I(G)

w(I). If w(I) = 1, then Z(G) = |I(G)| is the number of independent sets. Statistical physicists considered w(I) = λ|I| from some nonnegative λ ∈ R. If λ = 1, then Z(G) = |I(G)| again. If λ = 0, then Z(G) = 1.

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Weight functions Pr(I) = w(I) Z(G) where Z(G) =

  • I∈I(G)

w(I). If w(I) = 1, then Z(G) = |I(G)| is the number of independent sets. Statistical physicists considered w(I) = λ|I| from some nonnegative λ ∈ R. If λ = 1, then Z(G) = |I(G)| again. If λ = 0, then Z(G) = 1. Theorem (Sly,Sun ’12) For d ≥ 3 and λ > λc(d) = (d−1)d−1

(d−2)d , unless NP = RP there is no

approximation algorithm for the partition function with w(I) = λ|I|

  • n d-regular graphs.

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Local Constraints

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#VertexCover Definition A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at least one vertex in the set.

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#VertexCover Definition A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at least one vertex in the set.

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#VertexCover Definition A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at least one vertex in the set.

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#VertexCover Definition A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at least one vertex in the set.

X

  • Tyson Williams (UW-M)

Counting Complexity Google Madison 2013 18 / 32

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Systematic Approach to #VertexCover G = (V , E)

  • (u,v)∈E

OR(σ(u), σ(v)) = 1 · 1 · 1 · 1 · 1 · 1 = 1

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Systematic Approach to #VertexCover G = (V , E)

  • (u,v)∈E

OR(σ(u), σ(v)) = 1 · 1 · 1 · 1 · 1 · 1 = 1

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Systematic Approach to #VertexCover G = (V , E) σ : V → {0, 1}

1 1 1

  • (u,v)∈E

OR(σ(u), σ(v)) = 1 · 1 · 1 · 1 · 1 · 1 = 1

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Systematic Approach to #VertexCover G = (V , E) σ : V → {0, 1}

1 1 1

OR OR OR OR OR OR

  • (u,v)∈E

OR(σ(u), σ(v)) = 1 · 1 · 1 · 1 · 1 · 1 = 1

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Systematic Approach to #VertexCover G = (V , E) σ : V → {0, 1}

1 1 1

OR OR OR OR OR OR

  • (u,v)∈E

OR(σ(u), σ(v)) = 1 · 1 · 1 · 1 · 1 · 1 = 1

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Systematic Approach to #VertexCover G = (V , E) σ : V → {0, 1}

1 1

OR OR OR OR OR OR

  • (u,v)∈E

OR(σ(u), σ(v)) = 1 · 1 · 0 · 1 · 1 · 1 = 0

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Systematic Approach to #VertexCover G = (V , E) σ : V → {0, 1} #VertexCover(G) =

  • σ:V →{0,1}
  • (u,v)∈E

OR(σ(u), σ(v))

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Generalize

  • σ:V →{0,1}
  • (u,v)∈E

OR (σ(u), σ(v))

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Generalize

  • σ:V →{0,1}
  • (u,v)∈E

OR (σ(u), σ(v)) Input Output p q OR(p, q) 1 1 1 1 1 1 1

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Generalize

  • σ:V →{0,1}
  • (u,v)∈E

f (σ(u), σ(v)) Input Output p q OR(p, q) 1 1 1 1 1 1 1 Input Output p q f (p, q) w 1 x 1 y 1 1 z where w, x, y, z ∈ C

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Generalize Partition Function: Z(·) Z(G) =

  • σ:V →{0,1}
  • (u,v)∈E

f (σ(u), σ(v)) Input Output p q OR(p, q) 1 1 1 1 1 1 1 Input Output p q f (p, q) w 1 x 1 y 1 1 z where w, x, y, z ∈ C

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A Counting Dichotomy Theorem (Cai, Kowalczyk, W ’12) Over 3-regular graphs G, the exact counting problem for any (binary) complex-weighted function f Z(G) =

  • σ:V →{0,1}
  • (u,v)∈E

f (σ(u), σ(v)) is either computable in polynomial time or #P-hard.

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Nonlocal Examples Problem: HamiltonianCycle Input: A graph G. Output: “Yes” if G contains an Hamiltonian cycle “No” otherwise.

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Nonlocal Examples Problem: HamiltonianCycle Input: A graph G. Output: “Yes” if G contains an Hamiltonian cycle “No” otherwise. Problem: Connected Input: A graph G. Output: “Yes” if G is connected “No” otherwise.

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Nonlocal Examples Problem: HamiltonianCycle Input: A graph G. Output: “Yes” if G contains an Hamiltonian cycle “No” otherwise. Problem: Connected Input: A graph G. Output: “Yes” if G is connected “No” otherwise. Confessions of a theorists:

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Nonlocal Examples Problem: HamiltonianCycle Input: A graph G. Output: “Yes” if G contains an Hamiltonian cycle “No” otherwise. Problem: Connected Input: A graph G. Output: “Yes” if G is connected “No” otherwise. Confessions of a theorists: Some proofs of this depending on definition of “local”.

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Nonlocal Examples Problem: HamiltonianCycle Input: A graph G. Output: “Yes” if G contains an Hamiltonian cycle “No” otherwise. Problem: Connected Input: A graph G. Output: “Yes” if G is connected “No” otherwise. Confessions of a theorists: Some proofs of this depending on definition of “local”. Formally, just think of these as conjectures.

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Symmetric Function Definition A function is symmetric if invariant under any permutation of its inputs.

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Symmetric Function Definition A function is symmetric if invariant under any permutation of its inputs. Examples: OR2 = [0, 1, 1] AND3 = [0, 0, 0, 1] EVEN-PARITY4 = [1, 0, 1, 0, 1] MAJORITY5 = [0, 0, 0, 1, 1, 1] (=6) = EQUALITY6 = [1, 0, 0, 0, 0, 0, 1]

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Constraint Graph for #CSP(F) Instance F = {EVEN-PARITY3, MAJORITY3, OR3}

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Constraint Graph for #CSP(F) Instance F = {EVEN-PARITY3, MAJORITY3, OR3} EVEN-PARITY3(x, y, z) ∧ MAJORITY3(x, y, z) ∧ OR3(x, y, z)

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 24 / 32

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SLIDE 80

Constraint Graph for #CSP(F) Instance F = {EVEN-PARITY3, MAJORITY3, OR3} EVEN-PARITY3(x, y, z) ∧ MAJORITY3(x, y, z) ∧ OR3(x, y, z) x y z EVEN-PARITY3 MAJORITY3 OR3

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 24 / 32

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SLIDE 81

Constraint Graph for #CSP(F) Instance F = {EVEN-PARITY3, MAJORITY3, OR3} EVEN-PARITY3(x, y, z) ∧ MAJORITY3(x, y, z) ∧ OR3(x, y, z) x y z EVEN-PARITY3 MAJORITY3 OR3 NOT planar, so NOT an instance of Pl-#CSP({EVEN-PARITY3, MAJORITY3, OR3})

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SLIDE 82

Constraint Graph for #CSP(F) Instance F = {EVEN-PARITY3, MAJORITY3, OR3} EVEN-PARITY3(x, y, z) ∧ MAJORITY3(x, y, z) ∧ OR3(x, y, z) x y z EVEN-PARITY3 MAJORITY3 OR3 NOT planar, so NOT an instance of Pl-#CSP({EVEN-PARITY3, MAJORITY3, OR3})

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SLIDE 83

Constraint Graph for #CSP(F) Instance F = {EVEN-PARITY3, MAJORITY3, OR2} EVEN-PARITY3(x, y, z) ∧ MAJORITY3(x, y, z) ∧ OR2(x, y), z x y z EVEN-PARITY3 MAJORITY3 OR2

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SLIDE 84

Constraint Graph for #CSP(F) Instance F = {EVEN-PARITY3, MAJORITY3, OR2} EVEN-PARITY3(x, y, z) ∧ MAJORITY3(x, y, z) ∧ OR2(x, y), z x y z EVEN-PARITY3 MAJORITY3 OR2 x OR2 y EVEN-PARITY3 z MAJORITY3 VALID instance of Pl-#CSP({EVEN-PARITY3, MAJORITY3, OR2})

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SLIDE 85

More Counting Dichotomies Theorem (Cai, Lu, Xia ’09) Let F be any set of complex-valued constraints in Boolean variables. Then #CSP(F) is either #P-hard or computable in polynomial time.

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SLIDE 86

More Counting Dichotomies Theorem (Cai, Lu, Xia ’09) Let F be any set of complex-valued constraints in Boolean variables. Then #CSP(F) is either #P-hard or computable in polynomial time. Theorem (Cai, Xia ’12) Let F be any set of complex-valued constraints. Then #CSP(F) is either #P-hard or computable in polynomial time.

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SLIDE 87

More Counting Dichotomies Theorem (Cai, Lu, Xia ’09) Let F be any set of complex-valued constraints in Boolean variables. Then #CSP(F) is either #P-hard or computable in polynomial time. Theorem (Cai, Xia ’12) Let F be any set of complex-valued constraints. Then #CSP(F) is either #P-hard or computable in polynomial time. Theorem (Guo, W ’13) Let F be any set of symmetric, complex-valued constraints in Boolean variables. Then Pl-#CSP(F) is either #P-hard or computable in polynomial time.

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SLIDE 88

Definition of Holant Function Partition Function

f f f f f f

  • σ:V →{0,1}
  • (u,v)∈E

f (σ(u), σ(v))

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SLIDE 89

Definition of Holant Function Partition Function

Assignments to vertices Functions on edges

f f f f f f

  • σ:V →{0,1}
  • (u,v)∈E

f (σ(u), σ(v))

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SLIDE 90

Definition of Holant Function Partition Function

Assignments to vertices Functions on edges

f f f f f f

  • σ:V →{0,1}
  • (u,v)∈E

f (σ(u), σ(v)) Holant Function

Assignment to edges Functions on vertices

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SLIDE 91

Definition of Holant Function Partition Function

Assignments to vertices Functions on edges

f f f f f f

  • σ:V →{0,1}
  • (u,v)∈E

f (σ(u), σ(v)) Holant Function

Assignment to edges Functions on vertices

=3 =3 =3 =3 f f f f f f

  • σ:E→{0,1}
  • v∈V

gv

  • σ |E(v)
  • Tyson Williams (UW-M)

Counting Complexity Google Madison 2013 26 / 32

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SLIDE 92

Definition of Holant Function Holant({f } | {=3}) is a counting problem defined

  • ver (2,3)-regular bipartite

graphs. Holant Function

Assignment to edges Functions on vertices

=3 =3 =3 =3 f f f f f f

  • σ:E→{0,1}
  • v∈V

gv

  • σ |E(v)
  • Tyson Williams (UW-M)

Counting Complexity Google Madison 2013 26 / 32

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SLIDE 93

Definition of Holant Function Holant({f } | {=3}) is a counting problem defined

  • ver (2,3)-regular bipartite

graphs. Degree 2 vertices take f . Degree 3 vertices take =3. Holant Function

Assignment to edges Functions on vertices

=3 =3 =3 =3 f f f f f f

  • σ:E→{0,1}
  • v∈V

gv

  • σ |E(v)
  • Tyson Williams (UW-M)

Counting Complexity Google Madison 2013 26 / 32

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SLIDE 94

Example Holant Problems Holant({OR2} | {=3}) is #VertexCover on 3-regular graphs.

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SLIDE 95

Example Holant Problems Holant({OR2} | {=3}) is #VertexCover on 3-regular graphs. Holant({NAND2} | {=3}) is #IndependentSet on 3-regular graphs.

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SLIDE 96

Example Holant Problems Holant({OR2} | {=3}) is #VertexCover on 3-regular graphs. Holant({NAND2} | {=3}) is #IndependentSet on 3-regular graphs. Holant({=2} | {AT-MOST-ONE}) Holant(AT-MOST-ONE)

  • is #Matching.

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SLIDE 97

Example Holant Problems Holant({OR2} | {=3}) is #VertexCover on 3-regular graphs. Holant({NAND2} | {=3}) is #IndependentSet on 3-regular graphs. Holant({=2} | {AT-MOST-ONE}) Holant(AT-MOST-ONE)

  • is #Matching.

Holant({=2} | {EXACTLY-ONE}) Holant(EXACTLY-ONE)

  • is #PerfectMatching.

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SLIDE 98

Final Dichotomy Theorem (Cai, Guo, W ’13) Let F be any set of symmetric, complex-valued constraints in Boolean variables. Then Holant(F) is either #P-hard or computable in polynomial time.

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SLIDE 99

A Proof Technique: Polynomial Interpolation

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SLIDE 100

Polynomial Interpolation 2 (distinct) points defines a (unique) line

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SLIDE 101

Polynomial Interpolation 2 (distinct) points defines a (unique) line 3 (distinct) points defines a (unique) quadratic (actually: polynomial of degree at most 2)

Tyson Williams (UW-M) Counting Complexity Google Madison 2013 30 / 32

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SLIDE 102

Polynomial Interpolation 2 (distinct) points defines a (unique) line 3 (distinct) points defines a (unique) quadratic (actually: polynomial of degree at most 2) Lemma Given n + 1 distinct points (xi, yi), there is a unique polynomial p(·) of degree at most n such that p(xi) = yi.

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SLIDE 103

Polynomial Interpolation 2 (distinct) points defines a (unique) line 3 (distinct) points defines a (unique) quadratic (actually: polynomial of degree at most 2) Lemma Given n + 1 distinct points (xi, yi), there is a unique polynomial p(·) of degree at most n such that p(xi) = yi. Furthermore, the coefficients of p can be computed in polynomial time.

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SLIDE 104

#PerfectMatching ≤T #Matching [Valiant ’79]

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SLIDE 105

#PerfectMatching ≤T #Matching [Valiant ’79] Given a graph G with n vertices.

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SLIDE 106

#PerfectMatching ≤T #Matching [Valiant ’79] Given a graph G with n vertices. Assume we can compute the number of matchings in any graph.

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SLIDE 107

#PerfectMatching ≤T #Matching [Valiant ’79] Given a graph G with n vertices. Assume we can compute the number of matchings in any graph. Goal is to compute the number of perfect matchings in G.

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SLIDE 108

#PerfectMatching ≤T #Matching [Valiant ’79] Given a graph G with n vertices. Assume we can compute the number of matchings in any graph. Goal is to compute the number of perfect matchings in G.

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SLIDE 109

#PerfectMatching ≤T #Matching [Valiant ’79] Given a graph G with n vertices. Assume we can compute the number of matchings in any graph. Goal is to compute the number of perfect matchings in G. Let mk be the number of matchings that omit k vertices.

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SLIDE 110

#PerfectMatching ≤T #Matching [Valiant ’79] Given a graph G with n vertices. Assume we can compute the number of matchings in any graph. Goal is to compute the number of perfect matchings in G. Let mk be the number of matchings that omit k vertices. Let G ℓ be the graph G after adding, for each vertex v, ℓ vertices incident only to v.

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SLIDE 111

#PerfectMatching ≤T #Matching [Valiant ’79] Given a graph G with n vertices. Assume we can compute the number of matchings in any graph. Goal is to compute the number of perfect matchings in G. Let mk be the number of matchings that omit k vertices. Let G ℓ be the graph G after adding, for each vertex v, ℓ vertices incident only to v. #Matching(G ℓ) =

n

  • k=0

mk(ℓ + 1)k.

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SLIDE 112

Thank You

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