Vanishing Signatures A Complete Dichotomy Rises from the Capture of - - PowerPoint PPT Presentation

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. .

A Complete Dichotomy Rises from the Capture of Vanishing Signatures

Heng Guo (joint work with Jin-Yi Cai and Tyson Williams)

University of Wisconsia-Madison

Palo Alto June 3rd 2013

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 1 / 29

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Counting Problems

Contents

. .

1 Counting Problems

. .

2 Dichotomy

. .

3 Vanishing Signatures

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 2 / 29

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Counting Problems

Counting problems

Computational Counting problems appear often in statistical physics, machine learning, quantum computation, information theory, and so on. e quantity to be computed is usually expressed as a sum of products. e expectation of any random variable; Approximate an integral by a weighted sum; Classical simulation of quantum circuits; Partition functions. Ising model, Potts model, Hard-core gas model, …

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 3 / 29

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Counting Problems

Counting problems

Computational Counting problems appear often in statistical physics, machine learning, quantum computation, information theory, and so on. e quantity to be computed is usually expressed as a sum of products. e expectation of any random variable; Approximate an integral by a weighted sum; Classical simulation of quantum circuits; Partition functions. Ising model, Potts model, Hard-core gas model, …

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 3 / 29

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Counting Problems

Counting problems

Computational Counting problems appear often in statistical physics, machine learning, quantum computation, information theory, and so on. e quantity to be computed is usually expressed as a sum of products. e expectation of any random variable; Approximate an integral by a weighted sum; Classical simulation of quantum circuits; Partition functions. Ising model, Potts model, Hard-core gas model, …

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 3 / 29

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Counting Problems

Partition functions

Let us take a closer look at the partition functions. Ising model (without an external eld): ∑

σ:V→{+,−}

βn(σ), where n(σ) is the number of (+, +) and (−, −) neighbours in the graph given σ. We can rewrite it in the following form:

V 0 1 i j E

fI i j where fI 0 0 fI 1 1 , fI 0 1 fI 1 0 1.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 4 / 29

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Counting Problems

Partition functions

Let us take a closer look at the partition functions. Ising model (without an external eld): ∑

σ:V→{+,−}

βn(σ), where n(σ) is the number of (+, +) and (−, −) neighbours in the graph given σ. We can rewrite it in the following form: ∑

σ:V→{0,1}

∏

(i,j)∈E

fI(σ(i), σ(j)), where fI(0, 0) = fI(1, 1) = β, fI(0, 1) = fI(1, 0) = 1.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 4 / 29

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Counting Problems

Partition functions

Hard-core gas model: ∑

V′⊆V

λ|V′|1{V′ is an independent set} We can rewrite it in the following form:

V 0 1 i j E

f i j

i V

g i where f 0 0 f 1 0 f 0 1 1, f 1 1 0, g 0 1 and g 1 .

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 5 / 29

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Counting Problems

Partition functions

Hard-core gas model: ∑

V′⊆V

λ|V′|1{V′ is an independent set} We can rewrite it in the following form: ∑

σ:V→{0,1}

∏

(i,j)∈E

f(σ(i), σ(j)) ∏

i∈V

g(σ(i)), where f(0, 0) = f(1, 0) = f(0, 1) = 1, f(1, 1) = 0, g(0) = 1 and g(1) = λ.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 5 / 29

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Counting Problems

Perfect matchings

#Perfect-Matching: ∑

E′⊆E

1{E′ is a perfect matching} We can rewrite it in the following form:

E 0 1 v V

f

E v

where

E v is the assignment

restricted to the set E v of incident edges of v, and f is the E-O function.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 6 / 29

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Counting Problems

Perfect matchings

#Perfect-Matching: ∑

E′⊆E

1{E′ is a perfect matching} We can rewrite it in the following form: ∑

σ:E→{0,1}

∏

v∈V

f(σ |E(v)), where σ |E(v) is the assignment σ restricted to the set E(v) of incident edges of v, and f is the E-O function.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 6 / 29

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Counting Problems

Common features

Instance is a graph. Vertices and edges are associated with some functions. Functions take assignments on adjacent edges/vertices as inputs. Quantity to compute is an exponential sum over all possible assignments.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 7 / 29

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Counting Problems

Common features

Instance is a graph. Vertices and edges are associated with some functions. Functions take assignments on adjacent edges/vertices as inputs. Quantity to compute is an exponential sum over all possible assignments.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 7 / 29

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Counting Problems

Common features

Instance is a graph. Vertices and edges are associated with some functions. Functions take assignments on adjacent edges/vertices as inputs. Quantity to compute is an exponential sum over all possible assignments.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 7 / 29

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Counting Problems

Frameworks

Counting problems are often parameterized by constraint functions. Frameworks specify where to put the functions and to sum over what assignments. . .

1 Graph Homomorphisms

. .

2 Constraint Satisfaction Problems (#CSP)

. .

3 Holant Problems

e expressive power is increasing in order.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 8 / 29

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Counting Problems

Instance - signature grid

A signature grid Ω = (G, F, π) consists of a graph G = (V, E), where each vertex is labeled by a function fv ∈ F, and π : V → F is the labelling. . .

f1. f2

.

f1

.

f3

.

f1

.

f3

.

f4

.

f2

Figure: A signature grid

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 9 / 29

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Counting Problems

Holant problems

e Holant problem on instance Ω is to evaluate HolantΩ = ∑

σ:E→{0,1}

∏

v∈V

fv(σ |E(v)), a sum over all edge assignments σ : E → {0, 1}. Also known as: Read-Twice #CSP , Tensor Contraction …

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 10 / 29

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Counting Problems

Holant problems

e Holant problem on instance Ω is to evaluate HolantΩ = ∑

σ:E→{0,1}

∏

v∈V

fv(σ |E(v)), a sum over all edge assignments σ : E → {0, 1}. Also known as: Read-Twice #CSP , Tensor Contraction …

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 10 / 29

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Counting Problems

Holant problems

e Holant problem on instance Ω is to evaluate HolantΩ = ∑

σ:E→{0,1}

∏

v∈V

fv(σ |E(v)), a sum over all edge assignments σ : E → {0, 1}. Also known as: Read-Twice #CSP , Tensor Contraction …

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 10 / 29

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Counting Problems

Holant problems

e Holant problem on instance Ω is to evaluate HolantΩ = ∑

σ:E→{0,1}

∏

v∈V

fv(σ |E(v)), a sum over all edge assignments σ : E → {0, 1}. Also known as: Read-Twice #CSP , Tensor Contraction …

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 10 / 29

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Counting Problems

Symmetric functions

HolantΩ = ∑

σ

∏

v∈V

fv(σ |E(v)), When the function is E-O, the Holant counts perfect matchings in G. Such a function is symmetric. e output only depends on the Hamming weight of the input. List a symmetric function f by the Hamming weights: f0 f1 fn .

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 11 / 29

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Counting Problems

Symmetric functions

HolantΩ = ∑

σ

∏

v∈V

fv(σ |E(v)), When the function is E-O, the Holant counts perfect matchings in G. Such a function is symmetric. e output only depends on the Hamming weight of the input. List a symmetric function f by the Hamming weights: [f0, f1, . . . , fn].

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 11 / 29

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Counting Problems

Some examples

E-O: [0, 1, 0, . . . , 0]. e Holant counts perfect matchings. A-M-O: 1 1 0 0 . e Holant counts matchings. What about f 3 0 1 0 3 ?

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 12 / 29

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Counting Problems

Some examples

E-O: [0, 1, 0, . . . , 0]. e Holant counts perfect matchings. A-M-O: [1, 1, 0, . . . , 0]. e Holant counts matchings. What about f 3 0 1 0 3 ?

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 12 / 29

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Counting Problems

Some examples

E-O: [0, 1, 0, . . . , 0]. e Holant counts perfect matchings. A-M-O: [1, 1, 0, . . . , 0]. e Holant counts matchings. What about f = [3, 0, 1, 0, 3]?

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 12 / 29

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Counting Problems

Holant([3, 0, 1, 0, 3])

e input must be a 4-regular graph G. Consider the edge-vertex incidence graph of G. e problem becomes Holant 1 0 1 3 0 1 0 3 . We often use

2 to denote 1 0 1 .

. .

f. f

.

f

.

f

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f

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f

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f

.

f

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f. f

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f

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f

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f

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f

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f

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f

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 13 / 29

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Counting Problems

Holant([3, 0, 1, 0, 3])

e input must be a 4-regular graph G. Consider the edge-vertex incidence graph of G. e problem becomes Holant ([1, 0, 1] | [3, 0, 1, 0, 3]). We often use =2 to denote [1, 0, 1]. . .

f. f

.

f

.

f

.

f

.

f

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f

.

f

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f. f

.

f

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f

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f

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f

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f

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f

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 13 / 29

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Counting Problems

Holant([3, 0, 1, 0, 3])

e input must be a 4-regular graph G. Consider the edge-vertex incidence graph of G. e problem becomes Holant ([1, 0, 1] | [3, 0, 1, 0, 3]). We often use =2 to denote [1, 0, 1]. . .

f. f

.

f

.

f

.

f

.

f

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f

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f

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f. f

.

f

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f

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f

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f

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f

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f

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 13 / 29

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Counting Problems

Holant([3, 0, 1, 0, 3])

e input must be a 4-regular graph G. Consider the edge-vertex incidence graph of G. e problem becomes Holant ([1, 0, 1] | [3, 0, 1, 0, 3]). We often use =2 to denote [1, 0, 1]. . .

f. f

.

f

.

f

.

f

.

f

.

f

.

f

→ . .

f. f

.

f

.

f

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f

.

f

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f

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f

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

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2

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 13 / 29

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Counting Problems

Holant([3, 0, 1, 0, 3])

e input must be a 4-regular graph G. Consider the edge-vertex incidence graph of G. e problem becomes Holant ([1, 0, 1] | [3, 0, 1, 0, 3]). We often use =2 to denote [1, 0, 1]. . .

f. f

.

f

.

f

.

f

.

f

.

f

.

f

→ . .

f. f

.

f

.

f

.

f

.

f

.

f

.

f

.

=2

.

=2

.

=2

.

=2

.

=2

.

=2

.

=2

.

=2

.

=2

.

=2

.

=2

.

=2

.

=2

.

=2

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 13 / 29

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Counting Problems

Holographic transformation

For a 2-by-2 nonsingular matrix T, two functions f and g of arities m and n respectively, Valiant's Holant theorem states Holant (f | g) = Holant ( fT⊗m | (T−1)⊗ng ) is is what we call a holographic transformation. Here f is treated as a row vector

• f length 2m and g as a column vector of length 2n.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 14 / 29

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Counting Problems

Holant([3, 0, 1, 0, 3])

We apply the transformation Z =

1 √ 2

[ 1 1

i −i

] to Holant (=2| [3, 0, 1, 0, 3]). (=2)Z⊗2 = [0, 1, 0], which we denote by ̸=2. (Z−1)⊗4([3, 0, 1, 0, 3]) = 2[0, 0, 1, 0, 0]. e constant does not affect the complexity. erefore, Holant (=2| [3, 0, 1, 0, 3]) = Holant (̸=2| 2[0, 0, 1, 0, 0]) .

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 15 / 29

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Counting Problems

Holant([3, 0, 1, 0, 3])

What is Holant (̸=2| [0, 0, 1, 0, 0])? On the edge side, ̸=2 suggests an orientation. On vertices, 0 0 1 0 0 requires the orientation to be Eulerian. Holant

2

0 0 1 0 0 is actually the #E-O problem on 4-regular graphs. So is Holant 3 0 1 0 3 ! Two integer weighted problems are equivalent via a complex holographic transformation.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 16 / 29

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Counting Problems

Holant([3, 0, 1, 0, 3])

What is Holant (̸=2| [0, 0, 1, 0, 0])? On the edge side, ̸=2 suggests an orientation. On vertices, [0, 0, 1, 0, 0] requires the orientation to be Eulerian. Holant

2

0 0 1 0 0 is actually the #E-O problem on 4-regular graphs. So is Holant 3 0 1 0 3 ! Two integer weighted problems are equivalent via a complex holographic transformation.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 16 / 29

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Counting Problems

Holant([3, 0, 1, 0, 3])

What is Holant (̸=2| [0, 0, 1, 0, 0])? On the edge side, ̸=2 suggests an orientation. On vertices, [0, 0, 1, 0, 0] requires the orientation to be Eulerian. Holant (̸=2| [0, 0, 1, 0, 0]) is actually the #E-O problem on 4-regular graphs. So is Holant([3, 0, 1, 0, 3])! Two integer weighted problems are equivalent via a complex holographic transformation.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 16 / 29

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Counting Problems

Holant([3, 0, 1, 0, 3])

What is Holant (̸=2| [0, 0, 1, 0, 0])? On the edge side, ̸=2 suggests an orientation. On vertices, [0, 0, 1, 0, 0] requires the orientation to be Eulerian. Holant (̸=2| [0, 0, 1, 0, 0]) is actually the #E-O problem on 4-regular graphs. So is Holant([3, 0, 1, 0, 3])! Two integer weighted problems are equivalent via a complex holographic transformation.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 16 / 29

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Dichotomy

Contents

. .

1 Counting Problems

. .

2 Dichotomy

. .

3 Vanishing Signatures

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 17 / 29

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Dichotomy

Known tractable cases

Our goal is to determine the complexity of Holant problems with complex weights. Real-weighted dichotomy [Huang and Lu '12]. Tractable cases:

Equivalent to a problem on graphs of bounded degree 2; Equivalent to a tractable #CSP problem (via a holographic transformation).

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 18 / 29

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Dichotomy

Known tractable cases

Our goal is to determine the complexity of Holant problems with complex weights. Real-weighted dichotomy [Huang and Lu '12]. Tractable cases:

Equivalent to a problem on graphs of bounded degree 2; Equivalent to a tractable #CSP problem (via a holographic transformation).

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 18 / 29

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Dichotomy

Known tractable cases

Our goal is to determine the complexity of Holant problems with complex weights. Real-weighted dichotomy [Huang and Lu '12]. Tractable cases:

Equivalent to a problem on graphs of bounded degree 2; Equivalent to a tractable #CSP problem (via a holographic transformation).

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 18 / 29

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Dichotomy

Our Contribution

. eorem . . Let F be a set of complex-weighted Boolean symmetric functions. en Holant(F) is either tractable or #P-hard. A new class of tractable functions: vanishing signatures. A clear characterization regarding cases that can be transformed into tractable #CSPs. Everything else is hard.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 19 / 29

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Dichotomy

Our Contribution

. eorem . . Let F be a set of complex-weighted Boolean symmetric functions. en Holant(F) is either tractable or #P-hard. A new class of tractable functions: vanishing signatures. A clear characterization regarding cases that can be transformed into tractable #CSPs. Everything else is hard.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 19 / 29

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Vanishing Signatures

Contents

. .

1 Counting Problems

. .

2 Dichotomy

. .

3 Vanishing Signatures

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 20 / 29

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Vanishing Signatures

Vanishing

A set of signatures F is called vanishing if the value HolantΩ(F) is zero for every signature grid Ω. e signature 1 i is vanishing. .

1 i

.

1 i

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 21 / 29

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Vanishing Signatures

Vanishing

A set of signatures F is called vanishing if the value HolantΩ(F) is zero for every signature grid Ω. e signature [1, i] is vanishing. . .

1 i

.

1 i

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 21 / 29

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Vanishing Signatures

Vanishing

A set of signatures F is called vanishing if the value HolantΩ(F) is zero for every signature grid Ω. e signature [1, i] is vanishing. . .

[1, i]

.

[1, i]

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 21 / 29

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Vanishing Signatures

Vanishing

A set of signatures F is called vanishing if the value HolantΩ(F) is zero for every signature grid Ω. e signature [1, i] is vanishing. . .

[1, i]

.

[1, i]

. 1 · 1

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 21 / 29

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Vanishing Signatures

Vanishing

A set of signatures F is called vanishing if the value HolantΩ(F) is zero for every signature grid Ω. e signature [1, i] is vanishing. . .

[1, i]

.

[1, i]

.

1

1 · 1 + i · i

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 21 / 29

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Vanishing Signatures

Vanishing

A set of signatures F is called vanishing if the value HolantΩ(F) is zero for every signature grid Ω. e signature [1, i] is vanishing. . .

[1, i]

.

[1, i]

1 · 1 + i · i = 0

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 21 / 29

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Vanishing Signatures

Vanishing

We can view several unary signatures as a new one, which we call degenerate. It is the tensor product of the unary signatures. For example, f = g ⊗ g ⊗ h. . .

g

.

g

.

h

.

f

. However, such signatures are not symmetric. We need to introduce an

• peration of symmetrization.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 22 / 29

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Vanishing Signatures

Vanishing

Any degenerate signature containing more than half [1, i]'s is vanishing. For example, f = [1, i] ⊗ [1, i] ⊗ [0, 1]. . .

[1, i]

.

[1, i]

.

[0, 1]

.

f

. However, such signatures are not symmetric. We need to introduce an

• peration of symmetrization.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 22 / 29

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Vanishing Signatures

Vanishing

Any degenerate signature containing more than half [1, i]'s is vanishing. For example, f = [1, i] ⊗ [1, i] ⊗ [0, 1]. . .

[1, i]

.

[1, i]

.

[0, 1]

.

f

. However, such signatures are not symmetric. We need to introduce an

• peration of symmetrization.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 22 / 29

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Vanishing Signatures

Sum of vanishing

A key observation is that the linear sum of a set of vanishing signatures is still vanishing. Assume f f1 f2 f3, and f1 f2 f3 is vanishing. Holant

v V

f

E v V 1 2 3

Notice that

v V f v E v

0 because f1 f2 f3 is vanishing.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 23 / 29

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Vanishing Signatures

Sum of vanishing

A key observation is that the linear sum of a set of vanishing signatures is still vanishing. Assume f = f1 + f2 + f3, and {f1, f2, f3} is vanishing. Holant

v V

f

E v V 1 2 3

Notice that

v V f v E v

0 because f1 f2 f3 is vanishing.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 23 / 29

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Vanishing Signatures

Sum of vanishing

A key observation is that the linear sum of a set of vanishing signatures is still vanishing. Assume f = f1 + f2 + f3, and {f1, f2, f3} is vanishing. HolantΩ = ∑

σ

∏

v∈V

f(σ |E(v))

V 1 2 3

Notice that

v V f v E v

0 because f1 f2 f3 is vanishing.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 23 / 29

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Vanishing Signatures

Sum of vanishing

A key observation is that the linear sum of a set of vanishing signatures is still vanishing. Assume f = f1 + f2 + f3, and {f1, f2, f3} is vanishing. HolantΩ = ∑

σ

∏

v∈V

f(σ |E(v)) = ∑

τ:V→1,2,3

∑

σ

∏

v∈V

fτ(v)(σ |E(v))

V 1 2 3

Notice that

v V f v E v

0 because f1 f2 f3 is vanishing.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 23 / 29

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Vanishing Signatures

Sum of vanishing

A key observation is that the linear sum of a set of vanishing signatures is still vanishing. Assume f = f1 + f2 + f3, and {f1, f2, f3} is vanishing. HolantΩ = ∑

σ

∏

v∈V

f(σ |E(v)) = ∑

τ:V→1,2,3

∑

σ

∏

v∈V

fτ(v)(σ |E(v))

V 1 2 3

Notice that ∑

σ

∏

v∈V fτ(v)(σ |E(v)) = 0 because {f1, f2, f3} is vanishing.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 23 / 29

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Vanishing Signatures

Sum of vanishing

A key observation is that the linear sum of a set of vanishing signatures is still vanishing. Assume f = f1 + f2 + f3, and {f1, f2, f3} is vanishing. HolantΩ = ∑

σ

∏

v∈V

f(σ |E(v)) = ∑

τ:V→1,2,3

∑

σ

∏

v∈V

fτ(v)(σ |E(v)) = ∑

τ:V→1,2,3

0 = 0 Notice that ∑

σ

∏

v∈V fτ(v)(σ |E(v)) = 0 because {f1, f2, f3} is vanishing.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 23 / 29

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Vanishing Signatures

Symmetrization

Let Sn be the symmetric group of degree n. en for positive integers t and n with t ≤ n and unary signatures v, v1, . . . , vn−t, we dene Symt

n(v; v1, . . . , vn−t) =

∑

π∈Sn n

⊗

k=1

uπ(k), where the ordered sequence (u1, u2, . . . , un) = (v, . . . , v

t copies

, v1, . . . , vn−t).

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 24 / 29

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Vanishing Signatures

Examples

For example,

Sym2

3([1, i]; [0, 1]) = 2[0, 1] ⊗ [1, i] ⊗ [1, i] + 2[1, i] ⊗ [0, 1] ⊗ [1, i] + 2[1, i] ⊗ [1, i] ⊗ [0, 1]

= 2[0, 1, 2i, −3].

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 25 / 29

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Vanishing Signatures

Vanishing degrees

. Denition . . A nonzero symmetric signature f of arity n has positive vanishing degree k ≥ 1, which is denoted by vd+(f) = k, if k ≤ n is the largest positive integer such that there exists n − k unary signatures v1, . . . , vn−k satisfying f = Symk

n([1, i]; v1, . . . , vn−k).

If f cannot be expressed as such a symmetrization form, we dene vd+(f) = 0. If f is the all zero signature, dene vd+(f) = n + 1. We dene vd−(f) similarly, using −i instead of i.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 26 / 29

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Vanishing Signatures

Characterization

. eorem . . A signature f is vanishing if and only if 2 vdσ(f) > arity(f) for σ = + or −. is result also generalizes to a set of signatures. vd+([0, 1, 2i, −3]) = 2, so [0, 1, 2i, −3] is vanishing. vd+([1, 0, 1]) = vd−([1, 0, 1]) = 1. It is not vanishing.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 27 / 29

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Vanishing Signatures

Related tractable cases

Vanishing signatures are by themselves tractable. Some unary and binary (non-vanishing) signatures can be combined with them and remain tractable. Technically there are two categories, but the basic idea is that for these problems a given instance is either vanishing or of bounded degree 2.

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 28 / 29

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Vanishing Signatures

ank you!

Heng Guo (CS, UW-Madison) Complex Holant STOC 2013 29 / 29