From Holant to #CSP and Back: Dichotomy for Holant^c Problems

Abstract

We explore the intricate interdependent relationship among counting problems, considered from three frameworks for such problems: Holant Problems, counting CSP and weighted H-colorings. We consider these problems for general complex valued functions that take boolean inputs. We show that results from one framework can be used to derive results in another, and this happens in both directions. Holographic reductions discover an underlying unity, which is only revealed when these counting problems are investigated in the complex domain ℂ. We prove three complexity dichotomy theorems, leading to a general theorem for Holant^c problems. This is the natural class of Holant problems where one can assign constants 0 or 1. More specifically, given any signature grid on G=(V,E) over a set of symmetric functions, we completely classify the complexity to be in P or #P-hard, according to , of

$$\sum_{\sigma: E \rightarrow \{0,1\}}\prod_{v\in V} f_v(\sigma \vert _{E(v)}),$$

where (0, 1 are the unary constant 0, 1 functions). Not only is holographic reduction the main tool, but also the final dichotomy can be only naturally stated in the language of holographic transformations. The proof goes through another dichotomy theorem on Boolean complex weighted #CSP.

Appendices

Appendix A: Some Known Dichotomy Results

In this section, we review three dichotomy theorems from [9].

Theorem 7

([9])

Let be a set of symmetric non-degenerate signatures over ℂ. Then is computable in polynomial time in the following three cases. In all other cases, is #P-hard.

• 1. Every signature in is of arity no more than two;

• 2. There exist two constants a and b (not both zero, depending only on ), such that for every signature one of the two conditions is satisfied: (1) for every k=0,1,…,n−2, we have ax _k +bx _k+1−ax _k+2=0; (2) n=2 and the signature [x ₀,x ₁,x ₂] is of form [2aλ,bλ,−2aλ].

• 3. For every signature , one of the two conditions is satisfied: (1) For every k=0,1,…,n−2, we have x _k +x _k+2=0; (2) n=2 and the signature [x ₀,x ₁,x ₂] is of form [λ,0,λ].

Theorem 8

[9] Let be a set of real symmetric signatures, and let and be three families of signatures defined as

Then is computable in polynomial time if (1) After removing unary signatures from , it falls in one of the three cases of Theorem 7 (this implies is computable in polynomial time) or (2) (Without removing any unary signature) . Otherwise, is #P-hard.

Definition 5

A k-ary function f(x ₁,…,x _k ) is affine if it has the form

$$\chi_{AX=0} \cdot \sqrt{-1}^{\sum_{j=1}^{n}\langle \alpha_j,X\rangle}$$

where X=(x ₁,x ₂,…,x _k ,1), A is matrix over $\mathbb{F}_{2}$, α _j is a vector over $\mathbb{F}_{2}$, and χ is a 0–1 indicator function such that χ _AX=0 is 1 iff AX=0. Note that the inner product 〈α _j ,X〉 is calculated over $\mathbb{F}_{2}$, while the summation $\sum_{j=1}^{n}$ on the exponent of $i = \sqrt{-1}$ is evaluated as a sum mod 4 of 0–1 terms. We use to denote the set of all affine functions.

We use to denote the set of functions which can be expressed as a product of unary functions, binary equality functions ([1,0,1]) and binary disequality functions ([0,1,0]).

Theorem 9

[9]

Suppose is a class of functions mapping Boolean inputs to complex numbers. If or , then #CSP() is computable in polynomial time. Otherwise, #CSP() is #P-hard.

As we mentioned in [9], the class is a natural generalization of the symmetric signatures family . It is easy to show that the set of symmetric signatures in is exactly .

Appendix B: Some Useful Reductions

In this section, we list some useful simple reductions: reduction between Holant and #CSP, reduction between bipartite and non-bipartite settings, and holographic reduction.

Proposition 1

.

This says that #CSP is the same as Holant problems with Equality functions given for free.

Proposition 2

.

That is, we can transform every edge to a path of length 2 with the new vertex given (=₂)=[1,0,1].

Proposition 3

.

Binary Equality functions on both sides allow the transfer of signatures.

Proposition 4

For any T∈GL ₂(ℂ), .

This is a restatement of Valiant’s Holant Theorem.

Proposition 5

Let T be an orthogonal transformation (TT ^T=I). Then .

This follows from the invariance of (=₂)=[1,0,1] under an orthogonal transformation, and Propositions 2 and 4.

Appendix C: List of Matrices in

In this section, we explicitly list all the matrices in the family

which is defined and used in the statement of Theorem 6. The following condition is given in Theorem 6:

(2)

Together with the condition in Theorem 7, this gives an effective tractability condition which is both necessary and sufficient for Holant^c problems, by Theorem 6.

As noted before, the set of symmetric signatures in is exactly . We note that [1,0,1]T ^⊗2, [1,0]T and [0,1]T are all symmetric, as is the requirement in Theorem 6. Thus we can replace by in the expression above.

It is obvious that the family is closed under a scalar multiplication. Thus we list them up to a scalar multiple.

After a scalar multiple, symmetric binary signatures in are precisely

$$[1, 0, \pm 1],\quad [1, 0, \pm i],\quad [1, \pm 1, -1],\quad [1, \pm i, 1], \quad [0,1,0].$$

Also the unary signatures in , up to a scalar multiple, are

$$[1, \pm 1],\quad [1,\pm i],\quad [1,0],\quad [0,1].$$

Before we enumerate all the possibilities, we make two observations to simplify this process:

• 1. If we exchange the two columns of T, the signature [1,0,1]T ^⊗2 becomes its reversal, and the two numbers in [1,0]T are interchanged. Similarly the two numbers in [0,1]T are interchanged as well. The effect of exchanging the two columns of T is the same as replacing T by . On the other hand, the holographic transformation amounts exchanging input values 0 and 1 for functions in , and this operation keeps invariant.

• 2. If we multiply on the right side of T, then [y ₀,y ₁,y ₂]≜[1,0,1]T ^⊗2 becomes [y ₀,−y ₁,y ₂]. This operation also preserves both the set of binary and the set of unary signatures, respectively, listed for , up to a scalar factor. On the other hand, the effect of the holographic transformation is to transform an original function f to $f \cdot (-1)^{\sum_{i} x_{i}}$, and thus it is in iff the original .

  Similarly, we can multiply on the right side of T. Of course the invariance under implies that of .

What has been shown is that Condition (2) is invariant under the right action on by the group generated by and . By , we may consider only those T’s such that [1,0,1]T ^⊗2∈{[1,0,1],[1,0,i],[1,i,1],[0,1,0]}, up to a scalar factor. If we further normalize by the reversal action we may consider only those T’s such that [1,0]T∈{[1,±1],[1,i],[1,0]}, up to a scalar factor. However this reversal action is only partially closed for {[1,0,1],[1,0,i],[1,i,1],[0,1,0]}, with the exception [1,0,i] which is changed to [1,0,−i]. Thus we may have two extra cases to consider: [1,0,1]T ^⊗2=[1,0,i] and, [1,0]T=[0,1] or [1,−i]. But for these two cases if we apply first , followed by , we obtain [1,0,1]T ^⊗2=[1,0,i] and, [1,0]T=[1,0] or [1,1] respectively. Hence we can eliminate these two cases.

To summarize, to enumerate all T satisfying Condition (2) we only need to consider

up to a scalar factor. In the following, we denote by $\alpha=(1+i)/\sqrt{2}=\sqrt{i}$.

If [1,0,1]T ^⊗2=γ[1,0,1], [1,0]T=λ[1,1], we have

$$T=\left[\begin{array}{c@{\quad }c}1 & 1 \\1 & -1\end{array}\right],\left[\begin{array}{c@{\quad }c}1 & 1 \\-1 & 1\end{array}\right].$$

If [1,0,1]T ^⊗2=γ[1,0,1], [1,0]T=λ[1,−1], we have

$$T=\left[\begin{array}{c@{\quad }c}1 & -1 \\1 & 1\end{array}\right],\left[\begin{array}{c@{\quad }c}1 & -1 \\-1 & -1\end{array}\right].$$

For [1,0,1]T ^⊗2=γ[1,0,1], [1,0]T=λ[1,i], there is no solution.

If [1,0,1]T ^⊗2=γ[1,0,1], [1,0]T=λ[1,0], we have

$$T=\left[\begin{array}{c@{\quad }c}1 & 0 \\0 & \pm 1\end{array}\right].$$

If [1,0,1]T ^⊗2=γ[1,0,i], [1,0]T=λ[1,1], we have

$$T=\left[\begin{array}{c@{\quad }c}1 & 1 \\\alpha^3 & \alpha \end{array}\right],\left[\begin{array}{c@{\quad }c}1 & 1 \\-\alpha^3 & -\alpha \end{array}\right].$$

If [1,0,1]T ^⊗2=γ[1,0,i], [1,0]T=λ[1,−1], we have

$$T=\left[\begin{array}{c@{\quad }c}1 & -1 \\\alpha^3 & -\alpha \end{array}\right],\left[\begin{array}{c@{\quad }c}1 & -1 \\-\alpha^3 & \alpha \end{array}\right].$$

If [1,0,1]T ^⊗2=γ[1,0,i], [1,0]T=λ[1,i], we have

$$T=\left[\begin{array}{c@{\quad }c}1 & i \\\alpha & -\alpha \end{array}\right],\left[\begin{array}{c@{\quad }c}1 & i \\-\alpha & \alpha \end{array}\right].$$

If [1,0,1]T ^⊗2=γ[1,0,i], [1,0]T=λ[1,0], we have

$$T=\left[\begin{array}{c@{\quad}c}1 & 0 \\0 & \alpha \end{array}\right],\left[\begin{array}{c@{\quad}c}1 & 0 \\0 & -\alpha \end{array}\right].$$

If [1,0,1]T ^⊗2=γ[1,i,1], [1,0]T=λ[1,1], we have

$$T=\left[\begin{array}{c@{\quad}c}1 & 1 \\-\alpha^3 & \alpha^3\end{array}\right],\left[\begin{array}{c@{\quad}c}1 & 1 \\\alpha^3 & -\alpha^3\end{array}\right].$$

If [1,0,1]T ^⊗2=γ[1,i,1], [1,0]T=λ[1,−1], we have

$$T=\left[\begin{array}{c@{\quad}c}1 & -1 \\\alpha & \alpha \end{array}\right],\left[\begin{array}{c@{\quad}c}1 & -1 \\-\alpha & -\alpha \end{array}\right].$$

If [1,0,1]T ^⊗2=γ[1,i,1], [1,0]T=λ[1,i], we have

$$T=\left[\begin{array}{c@{\quad}c}1 & i \\0 & \sqrt{2}\end{array}\right],\left[\begin{array}{c@{\quad}c}1 & i \\0 & -\sqrt{2}\end{array}\right].$$

If [1,0,1]T ^⊗2=γ[1,i,1], [1,0]T=λ[1,0], we have

$$T=\left[\begin{array}{c@{\quad}c}\sqrt{2} & 0 \\i & 1\end{array}\right],\left[\begin{array}{c@{\quad}c}\sqrt{2} & 0 \\-i & 1\end{array}\right].$$

If [1,0,1]T ^⊗2=γ[0,1,0], [1,0]T=λ[1,1], we have

$$T=\left[\begin{array}{c@{\quad}c}1 & 1 \\i & -i\end{array}\right],\left[\begin{array}{c@{\quad}c}1 & 1 \\-i & i\end{array}\right].$$

If [1,0,1]T ^⊗2=γ[0,1,0], [1,0]T=λ[1,−1], we have

$$T=\left[\begin{array}{c@{\quad}c}1 & -1 \\-i & -i\end{array}\right],\left[\begin{array}{c@{\quad}c}1 & -1 \\i & i\end{array}\right].$$

If [1,0,1]T ^⊗2=γ[0,1,0], [1,0]T=λ[1,i], we have

$$T=\left[\begin{array}{c@{\quad}c}1 & i \\-i & -1\end{array}\right],\left[\begin{array}{c@{\quad}c}1 & i \\i & 1\end{array}\right].$$

For [1,0,1]T ^⊗2=γ[0,1,0], [1,0]T=λ[1,0], there is no solution.

Appendix D: An Orthogonal Transformation

In this section we give the detail of an orthogonal holographic transformation used in the proof of Lemma 1.

We are given x _k =Akα ^k−1+Bα ^k, where $A \not = 0$, and α≠±i. Let

$$S = \left[\begin{array}{c@{\quad }c}1 & \frac{B-1}{3} \\ \noalign {\vspace {3pt}}\alpha & A+ \frac{B-1}{3} \alpha \end{array}\right],$$

then the signature [x ₀,x ₁,x ₂,x ₃] can be expressed as

$$(x_0, x_1, x_1, x_2, x_1, x_2, x_2, x_3)^\mathtt{T}= S^{\otimes 3} (1, 1, 1, 0, 1, 0, 0, 0)^\mathtt{T}.$$

This identity can be verified by observing that

and we apply S ^⊗3 using properties of tensor product, , etc.

Let , then T=T ^T=T ⁻¹∈O ₂(ℂ) is orthogonal, and is upper triangular, where $u = \sqrt{1 + \alpha^{2}}$. As $\det R = \det T \det S = (-1) A \not = 0$, we have $uv \not = 0$. It follows that

This can be written as a symmetric signature form [u ³+3u ² w,u ² v,0,0]. Note that the entry $u^{2}v \not = 0$, which we can normalize to 1, after a scalar multiplication. This gives us the form [z,1,0,0] for some z∈ℂ.