For the algorithmic results we will use a reduction of counting to sampling. This is a standard argument, which has been formalised in particular in [73, Theorem 6.4]. The important hypothesis is called self-reducibility and means that a problem can be expressed as a combination of not too many smaller subproblems. Rather than check this hypothesis (which can require a problem to be encoded in an unusual way), we will now describe the argument for the particular case of Holant problems, illustrating exactly what type of samples are needed.
The aim is to approximate Z0(ϕ) for a closed circuit ϕ with Z0(ϕ) > 0. We will
consider the unweighted case, by assuming that wtϕ takes values in {0,1}, so Z0(ϕ) =
|Ω0|. First, the set of variablesJ is ordered, say as1, . . . , n. For eachk= 1, . . . , nin turn,
the procedure takes a large number t of samplesz1, . . . ,zt sampled “almost uniformly” from the set Ek−1 = {x ∈Ω0 | (x1, . . . , xk−1) = (y1, . . . , yk−1)} (y1, . . . , yk−1 will have
been chosen in steps1, . . . , k−1). This is the important sampling step we need to be able to implement. Thenykis defined to be the most common value in the listzk1, . . . , ztk, and αk is defined to be the proportion of these values equal toyk. The output is1/α1. . . αn.
This approximation relies on the telescoping product
1 Z0(ϕ) = |E1| |E0| |E2| |E1| · · · |En| |En−1| ≈α1· · ·αn whereEn={y}.
The calculations in [73, Theorem 6.4] show that this approximation gives an FPRAS as long as for anyε >0we can take a samplez∈Eksatisfying(1 +ε)−1|Ek|−1≤Pr(x=
z)≤(1 +ε)|Ek|−1 for all x∈E
k, in time polynomial in the size of ϕand in log(1/ε).
The important requirement is that we can sample to within a small error from the uniform distribution on the sets Ek – holding some variables constant. This ability to hold variables constant is a type of self-reducibility. For an FPRAS, we also need to be able to test whether there are any satisfying assignments at all, that is, Z0(ϕ)>0.
Theorem 3.1. #ParityNAE has an FPRAS.
Proof. We are given a labelled graph, which is naturally a closed circuit ϕ using con- straints of the formEvenJ,OddJ, and NAEJ.
As described above, it suffices to show that we can test Z0(ϕ), and that we can
sample uniformly from the subsets ofΩ0 where some variables are kept constant. But we
can test whetherZ0(ϕ)>0in polynomial time by Cornuéjols’ algorithm for the general
factor problem [39]. And degree-1 parity relations can be used to fix variables to take a particular value, so it suffices to show how to sample uniformly fromΩ0 (without holding
We will use the near-assignments chain to sample from assignments of ϕ. Define F:{0,1}J → Q≥0 by F(x) = wtϕ(x), if P i∈Jxi is even 0 otherwise.
(In fact,wtϕ(x) is always0 or 1.)
By Lemma 3.17, all the constraints of ϕ are windable. By Lemma 3.16, wtϕ is windable. By Lemma 3.15, (wtϕ)⊕ is even-windable. But F is a pinning of (wtϕ)⊕ (setting the parity bit to zero). A pinning of an even-windable function is even-windable - this is immediate from the characterisation in terms of 2-decompositions given in Section 3.3.3.
We will use the notation π,Ω,Ω0 from Theorem 3.11, for the near-assignment chain
on the pair(F, Eϕ).
Recall from Lemma 3.19 that θ(NAEJ) ≤ 3 and θ(EvenJ) = θ(OddJ) = 0, and
that all these weight-functions are strictly terraced. LetR= 3|V|2|E|2; by Lemma 3.23
we have1/R≤Z0(ϕ)/Z2(ϕ)≤Z0(ϕ)/(Z0(ϕ) +Z2(ϕ)) =π(Ω0).
By Cornuéjols’ algorithm, mentioned above, we get an assignmentywithwtϕ(y)>0
and in particularπ(y)≥2−|E|. Let δ >0be an error parameter, which will be specified later. Applying Theorem 3.11, by simulating the near-assignments Markov chain of
(F, E) for t ≥ (2|E|)4R2(log2R
δ +|E|log 2) steps we can take a sample z from near-
assignments of ϕsuch that
1 2 X x∈Ω |Pr(x=z)−π(x)| ≤δ/2R Thus 1 2 X x∈Ω0 |Pr(x=z|z∈Ω0)−F(x)/Z0(ϕ)| ≤δ/2
We sample from Ω0 by rejection sampling: run the simulation at least 2Rlog2δ times
and return the first sample in Ω0. The probability that this fails is small (at most
(1−12π(Ω0))2Rlog
2
δ ≤δ/2). To get the condition that(1+ε)−1|Ω0|−1≤Pr(x=z)≤(1+ ε)|Ω0|−1 for all x∈Ω0, as required by [73, Theorem 6.4], we can takeδ=ε/2|E|+1.
The reduction of counting to sampling in [73, Theorem 6.4] is stated in the setting of unweighted counting and uniform sampling. But the results generalise to weighted sums and non-uniform distributions. Instead of the uniform distribution on the sets
Ek, we need to be able to sample to within a small error from the distribution on
Ek proportional to wtϕ. The FPRAS described above would output wtϕ(z)/α1. . . αn
instead of1/α1. . . αn. The sampling condition becomes (1 +ε)−1πk(x)≤Pr(x=z)≤ (1 +ε)πk(x) whereπk(x) = wtϕ(x)/Pz∈Ekwtϕ(z).
• F is closed under taking weight-functions of connected circuits
• F contains Evenk,Oddk, andNAEk for allk≥1
• for all finite subsetsF0⊂ F there is an FPRAS forHolant(F0)
Proof. The first statement is Lemma 3.16. The second statement is Lemma 3.17. For the third statement, given F0, let F00 =F0∪ {Even1,Odd1}. For the decision
problem we can use Feder’s algorithm for coindependent relations [52, Theorem 4]. We can use Even1 and Odd1 to fix the value a variable takes. Otherwise the argument
proceeds as in the previous proof except taking R to be |E|2|V|2maxF
∈F0θ(F), and taking δ slightly smaller, for example δ =ε/2|E|+1(M/m)|V| where M is the maximum value taken by any function in F and m is the minimum non-zero value taken by any function in F. We find that Holant(F00), and therefore Holant(F0), has an FPRAS.