# Problems Between P and NPC

Factoring and graph isomorphism are problems in NP that are not known to be in P nor to be NP-Complete. What are some other (sufficiently different) natural problems that share this property? Artificial examples coming directly from the proof of Ladner's theorem do not count.

Are any of these example provably NP-intermediate, assuming only some "reasonable" hypothesis?

• There is a similar question asked here that may be useful: cstheory.stackexchange.com/questions/52/… Commented Aug 17, 2010 at 2:13
• Related question at MO, with several pointers specifically for problems in NP and co-NP but not known to be in P: mathoverflow.net/questions/31821/… Commented Aug 17, 2010 at 2:29
• There are several complexity classes between P and NP-complete which are currently regarded as interesting: PPAD, problems that are UGC-equivalent, NP $\cap$ co-NP, BPP,.... If you are asking for a big list, could you make this a community wiki please? Commented Aug 17, 2010 at 2:36
• Thank you. I am aware of Ladner's Theorem. I guess I was asking for "natural problems." I guess PPAD has Nash Equilibria, so that counts... Commented Aug 17, 2010 at 3:05

Here's a collection of some of the responses of problems between P and NPC:

• Yes, this procedure works, as the "official" answer. Commented Sep 9, 2010 at 0:06
• It would be great to be able to add an answer to one's watchlist. This would definitely be on mine. Commented Sep 27, 2010 at 21:33
• I'm removing Planar MAX 2-SAT from the list, it was shown to be NP-complete by Guibas et al. in "Approximating polygons and subdivisions with minimum link paths" (springerlink.com/content/y234m35416w043v1) Commented Feb 24, 2011 at 21:50
• Are any of these examples provably NP-intermediate, assuming only some "reasonable" hypothesis (i.e., a hypothesis less trivial than "this problem is NP-intermediate")? If so, it would be interesting to mention that in this list. Commented Jun 25, 2011 at 17:57
• @Timothy Chow: The example above assuming $NEXP \neq EXP$ is provably intermediate, that is, assuming $NEXP \neq EXP$, the padded version of a $NEXP$-complete problem is provably neither $NP$-complete by Mahaney nor in $P$, as that would contradict $NEXP \neq EXP$. Commented Aug 9, 2011 at 20:44

The sums of square roots problem: Given two sequences $$a_1, a_2, \dots, a_n$$ and $$b_1, b_2, \dots, b_n$$ of positive integers, is $$A := \sum_i \sqrt{a_i}$$ less than, equal to, or greater than $$B := \sum_i \sqrt{b_i}$$?

• The problem has a trivial $$O(n)$$-time algorithm on the real RAM—Just compute the sums and compare them!—but this does not imply membership in P.

• There is an obvious finite-precision algorithm, but it is not known whether a polynomial number of bits of precision is sufficient for correctness. (See https://topp.openproblem.net/p33 for details.)

• The Pythogorean theorem implies that the length of any polygonal curve whose vertices and integer endpoints is a sum of square roots of integers. Thus, the sum-of-roots problem is inherent in several planar computational geometry problems, including Euclidean minimum spanning trees, Euclidean shortest paths, minimum-weight triangulations, and the Euclidean traveling salesman problem. (The Euclidean MST problem can be solved in polynomial time without resolving the sum-of-roots problem, thanks to the underlying matroid structure and the fact that the EMST is a subgraph of the Delaunay triangulation.)

• There is a polynomial-time randomized algorithm, due to Johannes Blömer, to decide whether the two sums are equal. However, if the answer is no, Blömer's algorithm does not determine which sum is larger.

• The decision version of this problem (Is $$A > B$$?) is not even known to be in NP. However, Blömer's algorithm implies that if the decision problem is in NP, then it is also in co-NP. Thus, the problem is unlikely to be NP-complete.

• Well, if we take just 1000 random integers, not too large, then there are about $2^{999}$ ways to divide them into two sets, so I would expect that two of these sums are within 900 or more bits within each other (and within half of the total sum). On the other hand, finding the "worst" two sequences to compare out of these $2^{999}$ possibilities is also very, very hard. Commented Sep 11, 2016 at 19:30
• The question asks for a problem in NP, but this answer says the problem is not known to be in NP? Commented Jun 10 at 19:16

My favorite problem in this class (I'll phrase it as a functional problem, but it's easy to turn into a decision problem in the standard way): compute the rotation distance between two binary trees (equivalently, the flip distance between two triangulations of a convex polygon).

• That's a neat problem: I didn't realize it was in limbo. Commented Aug 17, 2010 at 3:29
• Yeah I didn't know about it either! For all these problems/answers, I wonder if they are in Limbo because we think they really are or if they're more like PRIMES... Commented Aug 17, 2010 at 5:21
• A very pretty and important but older reference is Thurston, Sleator, and Tarjan, "Rotation distance, triangulations, and hyperbolic geometry", STOC'86 and JAMS'88. For a recent reference that explicitly mentions the complexity of the problem as still being open, see Lucas, "An improved kernel size for rotation distance in binary trees", IPL 2010, dx.doi.org/10.1016/j.ipl.2010.04.022 Commented Aug 17, 2010 at 6:01
• Interesting. Exploring the rotation space is also an active area of research it seems. "The rotation graph of k-ary trees is Hamiltonian", IPL 2008, dx.doi.org/10.1016/j.ipl.2008.09.013 Commented Nov 2, 2010 at 22:39
• See also "A Note on the Flip Distance Problem for Edge-Labeled Triangulations" (Arxiv, 2018). Commented Aug 22, 2018 at 0:33

A problem that is mentioned neither in this list or the MO list is the turnpike problem. Given a multiset of n(n-1)/2 numbers, each number representing the distance between two points on the line, reconstruct the positions of the original points.

Note that what makes this nontrivial is that for a given number d in the multiset, you don't know which pair of points is d units apart.

While it is known that for any given instance there are only a polynomial number of solutions, it is not known how to find one !

• Thanks - this is a good one! Reminds me of some other "localization" problems. Is it actually thought to be not in p? Commented Aug 17, 2010 at 5:18
• I am not aware that turnpike is directly linked to known problems in complexity. However, there's a "wrong direction" relation to factoring, in that the turnpike problem cam be phrased as a factoring problem on an appropriately chosen polynomial. Commented Aug 17, 2010 at 5:22
• Are there known unlikely consequences of this problem being NP-complete, as there are for Graph Isomorphism (PH collapsing)? Commented Aug 17, 2010 at 5:37
• not that I'm aware of. it hasn't been studied that much, which is a pity, because it's so natural. Commented Aug 17, 2010 at 5:46
• You encounter a similar problem in bioinformatics: Given a set of potentially/hopefully overlapping, randomly created substrings of a string much longer than the individual pieces; compute the original string. (gene sequencing) Commented Oct 26, 2010 at 22:02

Here is a list of problems that may or may not qualify as "sufficiently" different. By the same proof as for Graph Isomorphism, if any of them is NP-complete, then the Polynomial Hierarchy collapses to the second level. I do not think there is any broad consensus as to which of these "ought" to be in P.

• Graph Automorphism (determine if a graph has a nontrivial automorphism). Reduces to Graph Isomorphism, but not known (not thought?) to be GI-hard.
• Group Isomorphism and Automorphism (where the groups are given by their multiplication tables). Again, reduces to Graph Isomorphism, but not thought to be GI-hard.
• Ring Isomorphism and Automorphism. In a sense, this is the grand-daddy of all the above problems, since integer factoring is equivalent to finding a nontrivial automorphism of a ring, and Graph Isomorphism reduces to Ring Isomorphism. See Neeraj Kayal, Nitin Saxena. Complexity of Ring Morphism Problems. Computational Complexity 15(4): 342-390 (2006). (Interestingly, determining if a ring has a nontrivial automorphism is in $P$.)
• This post by Bill Gasarch contains some other problems with the taste of Ramsey theory that look like they might be intermediate.
• By Mahaney's Theorem, no sparse set can be NP-complete. But we also know that there are sparse sets in $NP$ - $P$ iff $NEXP$ is not equal to $EXP$. So assuming $NEXP \neq EXP$, the padded version of any $NEXP$-complete problem is of intermediate complexity. (Such a set cannot be in $P$ unless $NEXP = EXP$, contradicting our assumption.) There are plenty of natural $NEXP$-complete problems.
• I like the last example. Do you have any references about it? Commented Aug 17, 2010 at 3:16
• S. R. Mahaney. Sparse complete sets for NP: Solution of a conjecture by Berman and Hartmanis. Journal of Computer and System Sciences 25:130-143. 1982. dx.doi.org/10.1016/0022-0000(82)90002-2 Sparse sets in NP - P iff NEXP neq EXP: J. Hartmanis, N. Immerman, V. Sewelson, Sparse sets in NP-P: EXPTIME versus NEXPTIME, Information and Control, Volume 65, Issues 2-3, May-June 1985, Pages 158-181. dx.doi.org/10.1016/S0019-9958(85)80004-8 Commented Aug 17, 2010 at 4:02
• This is a nice list, though the first three are quite similar :) I like the last example as well. Commented Aug 17, 2010 at 19:46

The Minimum Circuit Size Problem (MCSP) is my favorite "natural" problem in NP that's not known to be NP-complete: Given the truth-table (of size n=2^m) of an m-variate Boolean function f, and given a number s, does f have a circuit of size s? If MCSP is easy, then there is no cryptographically-secure one-way function. This problem and its variants provided much of the motivation for the study of "brute-force" algorithms in Russia, leading to Levin's work on NP-completeness. This problem can also be viewed in terms of resource-bounded Kolmogorov complexity: asking if a string can be recovered quickly from a short description. This version of the problem was studied by Ko; the name MCSP was used first by Cai and Kabanets, as far as I know. More references can be found in some papers of mine: http://ftp.cs.rutgers.edu/pub/allender/KT.pdf http://ftp.cs.rutgers.edu/pub/allender/pervasive.reach.pdf

Monotone self-duality

For any boolean function $$f=f(x_1, x_2, ..., x_n)$$, it's dual is $$f^d=\bar{f}(\bar{x_1}, \bar{x_2}, ..., \bar{x_n})$$. Given $$f(x_1, x_2, ..., x_n)$$ represented by a CNF formula, we have to decide whether $$f=f^d$$.

This problem is in co-NP[$$\log^2 n$$], i.e., it is decidable with $$O(\log^2n/ \log\log n)$$ nondeterministic steps. Thus, it has a quasi-polynomial time algorithm ($$O(n^{\log n/\log \log n})$$ time), and hence is unlikely to be co-NP-hard.

It is still open whether this problem is in P or not. More details can be found in the 2008 paper "Computational aspects of monotone dualization: A brief survey" by Thomas Eiter, Kazuhisa Makino, and Georg Gottlob.

Knot triviality: Given a closed polygonal chain in 3-space, is it unknotted (ie, ambient-isotopic to a flat circle)?

This is known to be in NP by deepish results in normal surface theory, but no poly-time algorithm or NP-hardness proof is known.

• It might be worth mentioning that, as with many potentially NP-intermediate problems, a slight variant is known to be NP-complete. Namely, 3-manifold knot genus is NP-complete: given a closed polygonal chain in a triangulated 3-manifold and an integer g, is the knot the boundary of a surface of genus at most g? (Being the unknot is equivalent to genus 0.) doi.acm.org.proxy.uchicago.edu/10.1145/509907.510016 Commented Oct 25, 2010 at 2:45
• It's also contained in co-AM (Hara, Tani, Yamamoto), so not NPC unless the polynomial hierarchy collapses. Commented Oct 29, 2010 at 4:13
• Actually, that's still open. Tasos Sidiropoulos found a bug in the Hara-Tani-Yamamoto proof. Commented Oct 29, 2010 at 5:55
• Since the time this answer was first posted, Kuperberg placed it in $\mathsf{coNP}$ conditional on the Generalized Riemann Hypothesis, and Lackenby placed it unconditonally in $\mathsf{coNP}$. Commented Dec 24, 2017 at 19:11

It is not known if it is possible to decide in polynomial time if player 1 has a winning strategy in a parity game (from a given starting position). The problem is, however, contained in NP and co-NP and even in UP and co-UP.

• Can you give a reference? Sounds interesting. Commented Aug 17, 2010 at 15:11
• M. Jurdzinski. Deciding the Winner in Parity Games is in UP \cap co-Up. Information Processing Letters 68(3):119-124. 1998. Should at least be a good starting point. Commented Aug 17, 2010 at 16:43
• The recent paper "A Pumping Algorithm for Ergodic Stochastic Mean Payoff Games with Perfect Information" also shows that even a generalization of the parity game can be solved in pseudo-polynomial time. In particular, they show that a game called BWR game has a pseudo-polynomial time algorithm when there is a constant number of "random nodes". The parity game is the case where there is no random nodes.
– Danu
Commented Dec 23, 2010 at 0:52
• It was shown recently that parity games can be solved in quasipolynomial time, see here for example. Commented Mar 12, 2017 at 21:28

You get a very long list of problems if willing to accept approximation problems, such as approximating Max-Cut to within a factor 0.878. We don't know if it's NP-hard or in P (only know NP-hardness assuming the Uniuqe Games Conjecture).

• Yes, that was a silly comment that I started to delete as soon as it was posted. Thank you. :) Commented Aug 17, 2010 at 2:44
• Thanks! But I guess I wasn't thinking as much about approximation problems, but more of natural problems. Commented Aug 17, 2010 at 3:07
• Arguably, these are natural problems since they correspond to what is achievable by a natural set of techniques, in this case, semidefinite programming. Commented Aug 17, 2010 at 3:15
• I guess "natural" is a vague criterion... Commented Aug 17, 2010 at 5:23

In a monotone CNF formula every clause contains only positive literals or only negative literals. In an intersecting monotone CNF formula every positive clause has some variable in common with every negative clause.

The decision problem

INTERSECTING MONOTONE SAT
Input: intersecting monotone CNF formula $f$
Question: is $f$ satisfiable?

has an $n^{o(\log\ n)}$ algorithm dating back to 1996, but is not known to be in P. (Of course, it might turn out to be in P, but that would be a major result.)

• Thomas Eiter and Georg Gottlob, Hypergraph Transversal Computation and Related Problems in Logic and AI, JELIA 2002. doi: 10.1007/3-540-45757-7_53

The Pigeonhole Version of Subset Sum (or Subset Sum Equality).

Given:

$$a_k \in \mathbb{Z}_{>0}$$ $$\sum_{k=0}^{n-1} a_k < 2^n - 1$$

By the pigeonhole principle, there must exist two disjoint subsets, $S_1, S_2 \subseteq \{ 1, \dots, n \}$ such that:

$$\sum_{j \in S_1} a_j = \sum_{k \in S_2} a_k$$

The pigeonhole subset sum problem asks for such a solution. Originally stated in "Efficient approximation algorithms for the SUBSET-SUMS EQUALITY problem" by Bazgan, Santha and Tuza.

Is a given triangulated 3-manifold a 3-sphere? From Joe O'Rourke.

There are a lot of problems related to finding hidden subgroups. You mentioned factoring, but there is also the discrete log problem as well as others related to elliptic curves, etc.

Here's a problem in computational social choice which is not known to be in P, and may or may not be NP-complete.

Agenda control for balanced single-elimination tournaments:

Given: tournament graph $T$ on $n=2^k$ nodes, node $a$

Question: does there exist a permutation of the nodes (a bracket) so that a is the winner of the induced single-elimination tournament?

Given a permutation $P_k$ on $2^k$ nodes of $V$ and a tournament graph $T$ on $V$, one can obtain a permutation $P_{k-1}$ on $2^{k-1}$ nodes as follows. For every $i>0$, consider $P_k[2i-1]$ and $P_k[2i]$ and the arc $e$ between them in $T$; let $P_{k-1}[i]=P_k[2i-1]$ if $e=(P_k[2i-1],P_k[2i])$ and $P_{k-1}[i]=P_k[2i]$ otherwise. That is, we match up pairs of nodes according to $P_k$ and use $T$ to decide which nodes (winners) move on to the next round $P_{k-1}$. Hence given a permutation on $2^k$ one can actually define $k$ rounds $P_{k-1},\ldots,P_0$ inductively as above, until the last permutation contains only one node. This defines a (balanced) single-elimination tournament on $2^k$ nodes. The node which remains after all the rounds is the winner of the tournament.

Agenda control for balanced single-elimination tournaments (graph formulation):

Given: tournament graph $T$ on $n=2^k$ nodes, node $a$

Question: does $T$ contain a (spanning) binomial arborescence on $2^k$ nodes rooted at $a$?

A binomial arborescence on $2^k$ nodes rooted at a node $x$ is defined recursively as $a$ binomial arborescence on $2^{k-1}$ nodes rooted at $x$ and a binomial arborescence on $2^{k-1}$ nodes rooted at a different node $y$ and an arc from $x$ to $y$. (If $k=0$, a binomial arborescence is just the root.) The spanning binomial arborescences in a tournament graph capture exactly the single-elimination tournaments which can be played, given the match outcome information in the tournament graph.

Some references:

1. Jérôme Lang, Maria Silvia Pini, Francesca Rossi, Kristen Brent Venable, Toby Walsh: Winner Determination in Sequential Majority Voting. IJCAI 2007: 1372-1377.
2. N. Hazon, P. E. Dunne, S. Kraus, and , M. Wooldridge. How to Rig Elections and Competitions. COMSOC 2008.
3. Thuc Vu, Alon Altman, Yoav Shoham. On the complexity of schedule control problems for knockout tournaments. AAMAS (1) 2009: 225-232.
4. V. Vassilevska Williams. Fixing a tournament. AAAI 2010.

Take a look at the class TFNP. It has a lot of search problems with an intermediate status.

• I should point out that it is the class of related function problems in $NP\cap coNP$. Commented Aug 17, 2010 at 3:08

The induced subgraph isomorphism problem has NP-incomplete "left-hand side restrictions" assuming that P is not equal to NP. See Y. Chen, M. Thurley, M. Weyer: Understanding the Complexity of Induced Subgraph Isomorphisms, ICALP 2008.

• Although this is an interesting result, if you check the paper it even says that the proof of intermediate complexity is essentially the same as Ladner's Theorem, except you do the diagonalization in the choice of the LHS restriction. So I don't know if this counts as a "natural" problem, rather than just a different encoding of Ladner's Theorem. Commented Aug 19, 2010 at 2:50
• Note also that these are source-and-target restrictions. The target (right hand side) has to be of special form, to enforce injectivity. Commented Sep 22, 2010 at 14:59

The complexity of planar Minimum Bisection problem is an intriguing open problem which is not known to be $NP$-hard. In contrast, the planar Maximum Bisection problem is $NP$-hard.

Minimum Bisection problem: Find a partition of the set of nodes into two equal size parts such that the number of crossing edges is minimized.

Karpinski, Approximability of the Minimum Bisection Problem: An Algorithmic Challenge

• do you have a reference to the problem definition? Commented Dec 23, 2010 at 17:14
• Reference is added. Commented Dec 23, 2010 at 19:19

The cutting stock problem with a constant number of object lengths. See this discussion for more information.

The gap version of the closest vector in lattice problem is the following: Given a basis for an $n$-dimensional lattice and a vector $v$, distinguish between the two cases where there is a lattice vector at distance at most one $1$ from $v$ or when every lattice vector is $\beta$-far from $v$, for some fixed gap parameter $\beta > 1$.

When $\beta = \sqrt{n}$, the problem is in $\mathsf{NP} \cap \mathsf{coNP}$ and thus unlikely to be $\mathsf{NP}$-complete (and is conjectured to be outside $\mathsf{P}$). This case is of central attention for lattice-based cryptography and the related dihedral hidden subgroup problem in quentum computing. When $\beta$ is much smaller, say $\beta = n^{o(1/\log \log n)}$, the problem becomes $\mathsf{NP}$-hard.

A graph $G = (V,E)$ is said to be labeled by $f$ if each vertex $v\in V$ is assigned a non-negative integer value $f(v)$ and each edge $e = uv\in E$ is assigned the value $|f(u) − f(v)|$. The labeling is graceful if $f : V\rightarrow \{0, 1, 2,\dots, |E|\}$ is an injection and if all edges of G have distinct labels from $\{1, 2,..., |E|\}$. A graph is said to be graceful if it admits a graceful labeling. The computational complexity of deciding whether a graph is graceful is not known.

1. J.A. Gallian. A dynamic survey of graph labeling. The Electronic Journal of Combinatorics, 2009.
2. D.S. Johnson. The NP-completeness column: An ongoing guide. J. Algorithms, 4(1):87–100, 1983.
3. D.S. Johnson. The NP-completeness column. ACM Transactions on Algorithms, 1(1):160–176, 2005.

The problem of finding Vapnik–Chervonenkis dimension is not known to be in $P$ neither known to be $NP$-complete. The problem can be solved by quasi-polynomial time algorithm ($O(n^{\log n})$). The problem can not be $NP$-complete unless $NP$ is contained in Quasi-polynomial time.

• I think it's worth mentioning the classes $\mathsf{LOGNP}$ and $\mathsf{NP}[\log^2 n]$ defined in the paper linked in the answer as well as the other natural problems in that paper that are shown to be in these classes, since any problem complete for these classes is likely to be of intermediate complexity. Commented Nov 21, 2013 at 17:58

In the linear divisibility problem, the input is two integers $a$ and $b$ and the task is to determine whether there exists an integer of the form $a \cdot x+1$ that divides $b$.

The linear divisibility problem is known to be $\gamma$-complete for NP but not known (AFAIK) to be NP-complete.

Garey and Johnson in their seminal "Computers and Intractability" say that (pp. 158-159):

A $\gamma$-reduction, in contrast to our other notions of reducibility, is nondeterministic in nature. Let us first introduce the notion of the relation $R_M$ computed by an NDTM program $M$:$$R_M=\{\langle x,y\rangle:there\ is\ a\ string\ z\ such\ that\ on\ input\ x\ and\ guess\ z\ M\ has\ output\ y\}$$ (where the definition of "output" is as in the computation of functions by DTMs).

We say that a language $L_1$ over alphabet $\Sigma_1$ is $\gamma$-reducible to a language $L_2$ over $\Sigma_2$ (written $L_1\propto_{\gamma}L_2$) if there is a polynomial time NDTM program $M$ such that for all $x\in\Sigma_1^*$ there is some $y\in\Sigma_2^*$ for which $\langle x,y\rangle\in R_M$ and such that for all $\langle x,y\rangle\in R_M$, $x\in L_1$ if and only if $y\in L_2$. In other words, there is at least one halting computation for $M$ on every input $x$ and, given an input $x$, all halting computations on $x$ yield outputs that are in $L_2$ if and only if $x\in L_1$.

• Can you define $\gamma$-completeness in your answer? Commented Aug 12, 2011 at 13:41

The problem of finding a minimum dominating set in tournament is not known to be in $P$ neither known to be $NP$-complete. The problem has quasi-polynomial time algorithm ($O(n^{\log n})$). If the problem has polynomial time algorithm then satisfiability can be solved in sub-exponential time.

The following problem is believed to be NP-Intermediate, i.e. it is in NP but neither in P nor NP-complete.

Exponentiating Polynomial Root Problem (EPRP)

Let $p(x)$ be a polynomial with $\deg(p) \geq 0$ with coefficients drawn from a finite field $GF(q)$ with $q$ a prime number, and $r$ a primitive root for that field. Determine the solutions of: $$p(x) = r^x$$ (or equivalently, the zeros of $p(x) - r^x$) where $r^x$ means exponentiating $r$.

Note that, when $\deg(p)=0$ (the polynomial is a constant), this problem reverts to the Discrete Logarithm Problem, which is believed to be NP-Intermediate as well.

For additional details see my question and related discussion.

I don't know whether the weighted hypergraph isomorphism problem proposed in the answer by Thinh D. Nguyen cannot be simply shown to be GI complete. However, there is a GI-hard problem closely related to GI, which has not been reduced to GI yet, namely the string isomorphism problem (also called color isomorphism problem). This is the problem actually shown to be in quasi-polynomial time by László Babai. It is of independent interest, since it is equivalent to a number of decision problems in (permutation) group theory:

• Indeed, weighted hypergraph iso is GI-complete. Commented Aug 16, 2020 at 16:34

A problem that is not known to be either in FP or to be NP-hard is the problem of finding a minimal Steiner tree when the Steiner vertices are promised to fall on two straight line segments intersecting at an angle of 120°. If the angle between the line segments is less than 120°, then the problem is NP-hard. It is conjectured that when the angle is greater than 120°, then the problem is in FP.

Hence the following decision problem currently appears to be of intermediate complexity:

Minimal 120°-Steiner tree
Input: a finite set of points in the plane, lying on two line segments intersecting at an angle of 120°, and a positive rational number $q$.
Question: is there a Steiner tree of total length at most $q$?

Of course, this may actually be in P or be NP-complete, but then it seems we would have an interesting dichotomy at 120° instead of an intermediate problem. (The conjecture may also be false.)

• J. H. Rubinstein, D. A. Thomas, N. C. Wormald, Steiner Trees for Terminals Constrained to Curves, SIAM J. Discrete Math. 10(1) 1–17, 1997. doi:10.1137/S0895480192241190

A GI-hard problem not known to be NP-complete could be WEIGHTED_HYPERGRAPH_ISOMORPHISM. You are given two hypergraphs $G_1$ and $G_2$ on $n$ vertices with weighted hyper-edges, decide if there is a vertex permutation $\pi$ turning $G_1$ into $G_2$. See also: GI-hard graph problem not known to be $NP$-complete

• This is GI-complete, so from the complexity point of view not significantly different from GI itself. Commented Aug 16, 2020 at 16:32

A handful of natural problems are (promise) $$\mathsf{BQP\:complete}$$, and thus are not likely to be $$\mathsf{NP\:complete}$$ under the reasonable hypothesis that $$\mathsf{NP}\not\subseteq\mathsf{BQP}$$.

Adjusting the hypotheses to remove negative signs might negate the strength provided by interference, landing the problem in $$\mathsf{NP}$$ and not empowering it to be $$\mathsf{BQP\:complete}$$.

For example estimating an entry $$(A^m)_{ij}$$ of the $$m^{th}$$ power of a large, sparse matrix $$A$$ with $$\mathcal{O}(\exp m)$$ entries $$A_{ij}\in\{-1,0,1\}$$ is $$\mathsf{BQP\: complete}$$, as shown here.

I believe that when the entries $$A_{ij}\in\{0,1\}$$, e.g. when $$A$$ is an adjacency matrix of a large graph and $$(A^m)_{ij}$$ counts the number of $$m$$-length walks from vertex $$i$$ to vertex $$j$$, the estimation is amenable to Stockmeyer approximation, and thus likely in $$\mathsf{AM}$$, but still not $$\mathsf{NP\:complete}$$.

• Is the 01-version of this problem even in NP? Commented Aug 16, 2020 at 2:03
• I don’t know! If Stockmeyer counting works then it is maybe in AM... which by a reasonable assumption is equal to NP. But I don’t know if there’s another path to NP. (The BQP algorithm has a lot of similarities to the HHL algorithm. It might even be in P if it could be dequantized as part of the program initiated by Tang, but on that I’m doubtful.) Commented Aug 16, 2020 at 2:17

The Coffman–Graham algorithm solves scheduling N jobs (with unit length and DAG dependency) on K=2 machines with minimal span in polynomial time. When K is not a constant, the problem is known to be NPC. However, when K is a constant like 3, the problem is not known to be in P or NPC.