# Tag Info

24

The most recent paper on this question seems to be: Noam Livne, A note on #P-completeness of NP-witnessing relations, Information Processing Letters, Volume 109, Issue 5, 15 February 2009, Pages 259–261 http://www.sciencedirect.com/science/article/pii/S0020019008003141 which gives some sufficient conditions. Interestingly the introduction states "To date,...

21

There are several algorithms that count the simple paths of length $k$ in $f(k)n^{k/2+O(1)}$ time, which is a whole lot better than brute force ($O(n^k)$ time). See e.g. Vassilevska and Williams, 2009.

20

This is still #P-complete . This problem is usually referred to as montone (#)SAT. Monotone #2-SAT is already #P-complete (this is equivalent to counting vertex covers of a graph).  Roth, Dan. "On the hardness of approximate reasoning." Artificial Intelligence 82.1-2 (1996): 273-302.

19

It's #P-complete (Valiant, 1979) so you're unlikely to do a whole lot better than brute force, if you want the exact answer. Approximations are discussed by Roberts and Kroese (2007). B. Roberts and D. P. Kroese, "Estimating the number of $s$--$t$ paths in a graph". Journal of Graph Algorithms and Applications, 11(1):195-214, 2007. L. G. Valiant, "The ...

19

The #P-completeness proof of counting simple s-t paths in both undirected and directed graphs can be found in: Leslie G. Valiant: The Complexity of Enumeration and Reliability Problems. SIAM J. Comput. 8(3): 410-421 (1979) From the paper: ... 4. Some #P-complete problems ... 14. S-T PATHS (i.e. SELF AVOIDING WALKS) (directed or undirected) Input: $G; s,t \... 17 One interesting example from number theory is expressing a positive integer as a sum of four squares. This can be done relatively easily in random polynomial time (see my 1986 article with Rabin at https://dx.doi.org/10.1002%2Fcpa.3160390713), and if I remember correctly, there is now even a deterministic polynomial-time solution. But counting the number ... 17 A very nice and simple example from Graph Theory is counting the number of Eularian circuits in an undirected graph. The decision version is easy (... and the Seven Bridges of Königsberg problem has no solution :-) The counting version is #P-hard: Graham R. Brightwell, Peter Winkler: Counting Eulerian Circuits is #P-Complete. ALENEX/ANALCO 2005: 259-262 16 This isn't my answer, but Terrence Tao gave a beautiful answer to this question on MathOverflow. Here are the first few lines of his answer. To read the complete answer, follow the link. There is a folklore observation that if one was able to quickly count the number of prime factors of an integer n, then one would likely be able to quickly factor n ... 15 #3-Regular Bipartite Planar Vertex Cover is #P-Complete As counting vertex covers is exactly the same as counting satisfying assignments of a monotone #2-SAT instance, the above result implies that it is #P-complete to count satisfying assignments of a #2-SAT instance which is monotone and 3-regular and bipartite and planar. This in turn means that, in ... 15 I think that whether p>0 can be decided in polynomial time. The problem in question can be easily cast as the edge-disjoint paths problem, where the underlying graph is a planar graph consisting of m+1 layers each of which contains n vertices, plus m degree-4 vertices to represent the possible adjacent swaps. Note that the planarity of this graph ... 14 Assuming$\mathsf{PH}$does not collapse and that Graph Isomorphism is not in$\mathsf{P}$, then$\# GI$(the counting version of graph isomorphism) satisfies your conditions. This is because$\# GI \equiv_m^p GI$. 14 If the function f is in #P, then given an input string x of some length N, the value f(x) is a nonnegative number bounded by$2^{poly(N)}$. (This follows from the definition, in terms of number of accepting paths of an NP verifier.) This means that many functions f lie outside of #P for uninteresting reasons---either because f is negative, or, in the case ... 14 This problem is Monotone-SAT. It is #P-Complete under Cook Reductions. It is one of those problems that are "easy to decide but hard to count." I recommend the following paper. Self-Reducibility of Hard Counting Problems with Decision Version in P 13 Something that has not been mentioned so far (as far as I can see) and that holds in the unrelativized world is the following: $$PH \subseteq PP \quad\mbox{ if }\quad QMA = PP.$$ This was observed by Vyalyi in this paper and comes from the strengthening of two theorems: Toda's theorem - Vyalyi shows that one query to a$\sharp P$oracle is enough for a "$...

13

The output number of paths may be $\Omega(2^N/n)$ (choose $s$ arbitrarily and then choose $t$ as the vertex that is the endpoint of the largest number of the $2^N$ walks from $s$) which requires $\Omega(N)$ bits to write down explicitly; this is exponential in the input size. On the other hand, the matrix powering approach has complexity polynomial in the ...

13

It is known that, for $G$ of bounded treewidth, the Tutte polynomial $T(G;x,y)$ can be evaluated at any $(x,y)$ using $O(n)$ arithmetic operations. If $G$ is connected, then $t(G)=T(G;1,1)$.

13

The problem of counting such "imperfect" matchings in bipartite graphs is #P-complete. This has been proved by Les Valiant himself, on page 415 of the paper Leslie G. Valiant The Complexity of Enumeration and Reliability Problems SIAM J. Comput., 8(3), 410–421

11

You have some fundamental misunderstanding of what a language $L$ being in $\mathsf{NP} \cap \mathsf{coNP}$ means. You'd need to show that there exist two machines: $M$ which is an $\mathsf{NP}$ machine, and $M'$, which is a $\mathsf{coNP}$ machine, such that $L$ is decided by $M$, and $L$ is also decided by $M'$. Each one should be able to solve the problem ...

11

One such algorithm for $\#3\operatorname{SAT}$ is due to Kutzkov.

11

I don't know where this was first proved, but since EdgeCover has an expression as a Boolean domain Holant problem, it is included in many Holant dichotomy theorems. EdgeCover is included in the dichotomy theorem in (1). Theorem 6.2 (in the journal version or Theorem 6.1 in the preprint) shows that EdgeCover is #P-hard over planar 3-regular graphs. To see ...

11

There has been a series of breakthroughs on Dichotomy Theorems by Professor Jin-Yi Cai https://rjlipton.wordpress.com/2012/01/20/is-it-time-to-declare-victory-in-counting-complexity/ Another dichotomy theorem https://rjlipton.wordpress.com/2014/06/23/counting-edge-colorings-is-hard/ https://rjlipton.wordpress.com/2011/02/14/classifying-papers-on-classifying-...

11

Well, at least $\#\mathsf{P}$-hard. Given a SAT formula, construct a graph with two vertices, $v_x$ and $v_x'$, for every possible assignment of variables $\vec{x}$. If $x$ is a satisfying assignment for the formula, draw an edge between $v_x$ and $v_x'$; these are the only edges. It is easy to construct the circuit for this graph from the SAT formula, and ...

10

This post discusses #P-completeness of #Monotone-2SAT under weakly parsimonious reductions. If you negate all literals in a monotone 2-CNF formula $\phi$, you obtain a Horn 2-CNF formula $\psi$ with the same number of satisfying assignments.

10

A particularly striking example of a phase transition is the maximum degree bound for Exactly-$k$-SAT (X$k$SAT), in which each clause contains exactly $k$ distinct literals. The problem flips from being trivially easy (always satisfiable) to being NP-complete by adding one to the associated parameter. Let $f(k)$ denote the largest number such that any X$k$...

10

We're interested in additive approximations to #3SAT. i.e. given a 3CNF $\phi$ on $n$ variables count the number of satisfying assignments (call this $a$) up to additive error $k$. Here are some basic results for this: Case 1: $k=2^{n-1}-\mathrm{poly}(n)$ Here there is a deterministic poly-time algorithm: Let $m=2^n-2k = \mathrm{poly}(n)$. Now evaluate $\... 10 Finding$k$-path (simple paths of length$k$) in a graph is in$FPT$and can be done in$O^*(2^k)$with a randomized algorithm or$O^*(2.62^k)$deterministically. This is while Counting$k$-paths is$\#W$-hard. A more interesting example (decision is even in$P$while counting is parameterized-hard) would be counting$k$-matchings in bipartite graph. ... 9 By the work of Klivans and van Melkebeek (which relativizes), if E = DTIME($2^{O(n)}$) does not have circuits with PP gates of size$2^{o(n)}$then PH is in PP. The contrapositive says that if PH is not in PP then E has subexponential-size circuits with PP gates. That is consistent with the fact that an oracle proof of PH not in PP gives a relativized lower ... 9 Let$f(G, k_1, k_2)$be the counting problem that you have defined. Then $$g(G) = \sum_{k_1 = 0}^{|V_G| / 2} f(G, k_1, 2 k_1)$$ counts the number of matchings in$G$, which only uses a linear number of oracle calls to your problem. Since counting matchings in 3-regular planar graphs is #P-hard (see ref below), your questions (1), (2), and (3) all give a #P-... 9 The ODD EVEN DELTA problem is #P-hard, even on 3-regular bipartite planar graphs. Let$\mathcal{C}$be the set of vertex covers of a general graph$G$. Then, assuming$G\$ has no isolated vertices, the following equation holds (refer to the above article for the proof): $$|\mathcal{C}| = 2^{|V|} - \sum_{k = 2}^{|V|} \Delta_k \cdot 2^{|V|-k}$$ Counting ...

Only top voted, non community-wiki answers of a minimum length are eligible