Phys Rev. Repeat this process more than N times, and a simple counting argument implies that we must have visited the same node more than once, at least once. The NP-hard [18] knapsack problem asks us to maximize subject to the constraint that ≤ W. It has a huge variety of applications, particularly in economics and finance [50]. (2002) 2:181. = (2011) 53:217. doi: 10.1137/090771806, 64. 0 0 1 We now turn to NP problems whose formulations as Ising models are more subtle, due to the fact that constraints involve inequalities as opposed to equalities. Bravyi S, DiVincenzo DP, Oliveira RI, Terhal BM. and j Furthermore, it is not true that a directed acyclic graph is acyclic if the orientation of edges is ignored. Science (2002) 295:2427. doi: 10.1126/science.1068774, 11. However, to account for the fact that we may remove vertices, we will allow for yv = 1 if v is part of the feedback vertex set, and 0 otherwise. Since the graph is made up of m connected triangles, the only way to color m vertices if each vertex is in a distinct triangle, so there must be an element of each clause Ci which is true. Barahona F. On the computational complexity of Ising spin glass models. are chosen such that max(Mα) is minimized? Computers and Intractability: A Guide to the Theory of NP-Completeness. i Unfortunately, this technique is not efficient for AQO; the Hamiltonian contains N-body terms. The decision problem (does a path of total weight ≤ W exist?) The constraint that every edge has at least colored vertex is encoded in HA: Then, we want to minimize the number of colored vertices with HB: Choose B < A, as if we uncolor any vertex that ruins the solution, at least one edge will no longer connect to a colored vertex. Of course, in this special case10, one can prove that there is always a coloring for n ≥ 4 [52, 53]. To do this, we simply let. For simplicity, let us assume that the constraint Equation (21) can be satisfied for some choice of x. Proc Natl Acad Sci USA. x, for some vector c, given a constraint. j J 36. ^The ground state has H = 0 so long as the edge set is non-empty: any connected pair of edges is a clique of size 2. − We stress that this definition is only valid for states with HA = 0, since in these states each vertex has either 0 or 1 colored edges. Rev Mod Phy. with S an m × N matrix and b a vector with m components. The traveling salesman problem for a graph G = (V, E), where each edge uv in the graph has a weight Wuv associated to it, is to find the Hamiltonian cycle such that the sum of the weights of each edge in the cycle is minimized. count the number of colored edges. These are, by far, the most popular class of problems discussed in the AQO literature. This model can be solved exactly for the critical temperatures and a glassy phase is observed to exist at low temperatures. We then simply add. Babbush R, Perdomo-Ortiz A, O'Gorman B, Macready W, Aspuru-Guzik A. Boros E, Hammer PL, Tavares G. Preprocessing of unconstrained quadratic binary optimization. (1977) 21:491. ln We then add. 6. Thus H = 0 if and only if the clique cover problem is solved by the given coloring. J Chichester, NY: Wiley (1998). J β The equilibrium solution of the model, after some initial attempts by Sherrington, Kirkpatrick and others, was found by Giorgio Parisi in 1979 with the replica method. Magnetization then decays slowly as it approaches zero (or some small fraction of the original value—this remains unknown). Another possibility is that these Hamiltonians have small spectral gaps, and that alternative choices have much larger spectral gaps—this is a question we have not addressed at all in this paper. It is also straightforward to extend this, and find the smallest exact cover (this makes the problem NP-hard). S J … , If the ground state of this Hamiltonian has H = 0, there is an isomorphism. Given an undirected graph G = (V, E), and a set of n colors, is it possible to color each vertex in the graph with a specific color, such that no edge connects two vertices of the same color? Illinois J Math. He would like to thank Robert Lucas for pointing out that a compendium of ways to map famous NP problems to Ising glasses was lacking, Jacob Biamonte for encouraging publication, and Vicky Choi, Jacob Sanders, Federico Spedalieri, John Tran, and especially the reviewers, for many helpful comments on AQO and computer science. To solve this by finding the ground state of an Ising model, we use the same Hamiltonian as for the minimal spanning tree, except we add binary variables yv for v ∉ U which determine whether or not a node v is included in the tree. J Phys. Let us use the binary variable xe to denote whether or not an edge is colored; thus, the number of spins is |E| = O(ΔN), the size of the edge set; as before, Δ represents the maximal degree. For a directed graph, the feedback edge set problem is to find the smallest set of edges F ⊂ E such that (V, E − F) is a directed acyclic graph. (1982) A15:3241. Johnson MW, Amin MHS, Gildert S, Lanting T, Hamze F, Dickson N, et al. Given an undirected graph G = (V, E), what is the smallest number of vertices that can be “colored” such that every edge is incident to a colored vertex? and a variance The constraints on C are as follows: for each edge in C, let us color the two vertices it connects: i.e., let D = ∪e∈C ∂e. Zhou HJ. {\displaystyle i} which is NP-complete. Without loss of generality, let us label the vertices 1, …, N, and take the edge set (uv) to be directed—i.e., the order uv matters. 2. Note that, from this point on in this paper, we have not found any of the Ising formulations of this paper in the literature. The knapsack problem is the following problem: we have a list of N objects, labeled by indices α, with the weight of each object given by wα, and its value given by cα, and we have a knapsack which can only carry weight W. If xα is a binary variable denoting whether (1) or not (0) object α is contained in the knapsack, the total weight in the knapsack is. In 2020, physics researchers at Radboud University and Uppsala University announced they had observed a behavior known as self-induced spin glass in the atomic structure of neodymium. The complexity of stoquastic local Hamiltonian problems. Mézard M, Parisi G, Virasoro M. Spin Glass Theory and Beyond. J Phys. Each ground state of this spin model is equivalent to a solution of the minimax problem. ] This problem was described in Choi [49] but for simplicity, we repeat it here. Science (1983) 220:671. doi: 10.1126/science.220.4598.671, 24. We can also read off the color of each node (in one such coloring scheme) by looking at which xs are 1. i The Hamiltonian for SK model is very similar to the EA model: where Berlin: Springer (2003). The NP-hard [18] Steiner tree problem is somewhat similar to the problem above: given our costs cuv, we want to find a minimal spanning tree for a subset U ⊂ V of the vertices (i.e., a tree such that the sum of cuvs along all included edges is minimal). Rønnow, M. Troyer We present several efficient implementations of the simulated annealing algorithm for Ising spin glasses on sparse graphs. … Therefore, we see that any two spins can be linked with a ferromagnetic or an antiferromagnetic bond and the distribution of these is given exactly as in the case of Edwards–Anderson model. Each job i has length Li. The number of spins is |V| (⌊|V| + 1⌋ + 4 + 2|E|)/2 + |E|. In addition to unusual experimental properties, spin glasses are the subject of extensive theoretical and computational investigations. [ N Here A > 0 is a positive energy, and ∂v corresponds to the subset of E of edges which connect to v. Thus the ground states consist of HA = 0; if HA > 0, it is because there is a vertex where two of its edges are colored.