Two Approaches for Hamiltonian Circuit Problem using Satisfiability

Transcription

1 Two Approaches for Hamiltonian Circuit Problem using Satisfiability Jaya Thomas 1, and Narendra S. Chaudhari 2 1,2 Department of Computer science & Engineering, Indian Institute of Technology, M-Block I.E.T D.A.V.V Campus, Khandwa Road, Indore , India. {jayat@iiti.ac.in, nsc183@gmail.com, narendra@iiti.ac.in} Abstract. A Hamiltonian circuit is a cycle in a graph which visits each vertex exactly once and also returns to the starting vertex. Determining whether such cycles exist in graphs is the Hamiltonian Circuit problem. In this paper, we propose two different approaches to solve the problem. Both our approaches proceed, by reducing a given graph to its equivalent 3-SAT representation. Further, we explore the generated clauses to determine a satisfying truth value assignment. If the values assigned evaluate the clauses to true it indicates the presence of Hamiltonian circuit otherwise not. The two approaches differ in the number of constraints used for reduction. The approach discussed, is better than the already existing backtracking and heuristic based approaches. Keywords: Hamiltonian Circuit, Satisfiability, 3SAT, constraints, NP- Complete. 1 Introduction A Hamiltonian circuit in a connected graph is defined as a closed walk that traverses every vertex of G exactly once, except of course the starting vertex. The objective is to find a path of consecutive vertices along the edges, visiting every vertex exactly once and returning to the original vertex to complete a circuit. The general problem of trying to find such Hamiltonian Circuits in arbitrary graphs turned out to be very difficult to solve. It was shown by karp[1] that the problem belong to the class of NP-Complete problem[2][3], which are believed to be computationally intractable. The Hamiltonian circuit problem may be considered an important combinatorial optimization problem, with application in many real world situations such as routing and ordering problem[4][5]. The problem belongs to a recently studied class of problems defined on colored graphs having several applications in telecommunication networks, electric networks, multimodal transportation networks, among others, where one aims to ensure connectivity or other properties by means of a limited number of different connections. This problem is studied in different domains like determination for existence of Hamiltonian paths on random geometric networks with random faults, which is an

2 important issue, because if a network has this property it is possible to build a path to efficiently perform distributed computations based on end-to-end communication protocols, which allow distributed algorithms to treat an unreliable network as a reliable channel[6]. It is also used to embed a ring in a hypercube, which is to find a Hamiltonian cycle through every node of the hypercube[7]. In this paper we propose two different approaches to solve the problem using Boolean satisfiability concept. The methodology is firstly to determine the various constraints to be satisfied for the existence of Hamiltonian circuit in the given graph. These constraints are then mapped to equivalent Boolean clauses in conjunctive normal form(cnf) and can be solved using the SAT solvers. The satisfiability or unsatisfiability of the generated clause indicates the presence or absence respectively of the Hamiltonian circuit in the given graph. The organization of the paper is, in section 2, we will briefly discuss about the background of the problem. Section 3 gives the detailed description of the proposed methodology for both the approaches. Section 4, we give the algorithm used for implementation of the source code to generated the equivalent CNF. In section 5, we give the concluding remarks. 2 Background The problem was invented by Sir William Rowan Hamilton in 1859 as a game which required finding a route along the edges of a regular dodecahedron that would pass once and only once through every point. The problem was worked on and since 1936 some progress was made. G.A. Dirac, 1952 stated that If G is a simple graph with n( 3) vertices, and if the degree of each is at least 1/2n, then G is Hamiltonian. O.Ore, 1960 [8] gave the approach that If G is a simple graph with n( 3) vertices, and if the sum of the degrees of each pair of non-adjacent vertices is at least n, then G is Hamiltonian. Bondy and Chvatal[9], 1976 For G to be Hamiltonian, it is necessary and sufficient that [G] n be Hamiltonian. ([G] n is gotten from G by adding edges joining non-adjacent vertices whose sum of degrees is equal to, or greater than n). Fraudee, Dould, Jacobsen, Schelp (1989) If G is a 2-connected graph such that for every pair of nonadjacent nodes u and v, d(u)+d(v) (2n-1)/3 then G is Hamiltonian. There are two basic approaches for solving the problem one is the heuristic[12] and other backtracking[13]. In the backtracking approach is to search for all the potential solutions. These approaches employ pruning of some kind to restrict the amount of researching. The main advantage is it finds all solution, and can decide Hamilton Circuit exists or not. The demerit of the approach is in worst case, it needs exponential time. The other approach is heuristic with the advantage that it is fast, linear or low-order polynomial time, the demerit being may be cannot find the Hamilton Circuit. It is well known (e.g., see Robinson and Wormald [10]) that almost all regular graphs are Hamiltonian. It is also well known (e.g., see Woodall [11]) that all sufficiently dense graphs are Hamiltonian, and so in the thesis, author primarily considers sparse graphs. For Hamiltonian graphs, a constructive solution to HCP is to explicitly

3 find a Hamiltonian cycle. Many heuristics have been designed that attempt to find a single Hamiltonian cycle as quickly as possible, solving the HCP quickly in most cases. For non-hamiltonian graphs, however, no Hamiltonian cycles exist and the heuristics fail. 3 Proposed Methodology In this paper we present two different approaches for converting Hamiltonian circuit problem to 3SAT. The two approaches differ in the number of constraints used for conversion. In both the approaches a propositional value to the edge is assigned and clauses are written considering each vertex in the graph. Due to difference in selection criteria/ constraints for determining the existence of Hamiltonian cycle, the number and the nature of the clauses generated in both the cases are different. In the coming section we are going to detail both the approaches. 3.1 Generalized Constraints reduction approach In this approach, we proceed by assigning the Boolean variable for each edge based on the direction in case of directed graph. The assignment is shown below e i, j = 1 [if the edge exists from node i to node j] 0 [otherwise] Thus the number of literals that will be used to represent the edges will depend on the number of ones, present in the adjacency matrix of the graph say N x N, where N is the total number of vertex present in the given graph. We assume that the graph does not contain any self loop, thus values corresponding to it in the adjacency matrix are set to zero. The literals corresponding to the edges are used, to represent the constraints that exist on any edge to be part of the Hamiltonian Cycle, and then these constraints are further formulated into respective equivalent clauses. For any edge to be included in the Hamiltonian circuit path, we need to ensure that certain constraints are met. If we select an edge from the graph, than for the edge to be part of Hamilton cycle, it must fulfill the four governing conditions:- Firstly, there must not be any close loop or cycle to ensure this condition, if an edge exists from node i to node j i.e. e i, j is true than a reverse edge from j to i is not allowed. e i, j j i e, (1) Secondly, if any vertex is selected to be the source vertex for an edge then, it cannot be selected again to be the source vertex of any other edge. In other words, if an

4 edge exists from source node i to node j than edge to no other edge is allowed from same source node i. e, e i, k (2) i j Thirdly, if any vertex is selected to be the destination vertex for an edge then, it cannot be selected again to be the destination vertex of any other edge. In other words, if an edge exists from node i to destination node j than edge from no other node is allowed to same destination j. e i, j k j e, (3) Finally, the clause set would be completed by considering each vertex present in the given graph. For each of the vertex one and only one of the outgoing edges must be a part of the complete sets of Hamiltonian cycle. The constraint can be generalized as N i, j 1 i e (4) SAT Clause for generalized approach. The constraints mentioned above are crucial for determining the Hamiltonian cycle in any graph. Each constraints mentioned above can be formulated as SAT instance. Considering, the constraints (1)(2) and (3) which are of the form a b the equivalent CNF representation is a b. The constraint (4) is already in CNF Illustrative Example Fig. 1. Undirected Graph with 5 vertices Considering the above example, let us apply the constraints to be satisfied in case the Hamiltonian cycle exists in the given graph. Considering, the constraint reduction to SAT only for the first node 1, for the remaining node refer the Appendix1. The first

5 clause corresponds to the first constraint, the next to the second one and so on. However, the number varies depending on the number of connected edges. x x 2, 1 )( x 1, 2 x 1, 3 )( x1, 2 x 1, 5 )( x 1, 2 x 3, 2 )( x1, 2 x x 1, 5 ) (5) ( 1, 2 1,3 3.2 Specialized Constraints Reduction Approach The above discussed approach, and the current approach both are suitable for a both directed and undirected graph. Let there be a graph G=(V,E),where V is the set of N vertex and E is the set of M edges, V =N and E =M. In this case we do not consider edge label as vertex to vertex end, but here we assign each edge of the graph a Boolean literal value. Two main constraints are applied in this approach. In the first constraint, we focus on the fact that for the presence of Hamiltonian circuit in a graph, each vertex must have atleast two edges connected to it. So, we assign a Boolean literal to each edge present in the graph, and then we explore each vertex one by one, by generating the clauses that satisfy the above constraint. The clauses are generated in the following manner. If a vertex has only two edges connecting it, then the Boolean literals can be directly written in conjunction form. In case, there are more than two edges connecting a particular vertex then the clause will be generated by taking two literals at a time and negating the rest, all the possible cases generated would be combined and written in conjunction. Then, finally on combining all the cases, we have clauses in Disjunction Normal form. The form is converted into CNF, using the demorgan s law. The clauses we obtain are in SAT with k literal each( where k=2 N E, N E number of edges). This is further converted into 3-SAT, i.e. each clause with 3 literals each. The clause generated by the above approach, may become true also for the presence of disjoint cycle. Thus, another constraint need to be introduced which take care of such violating conditions. Second constraint basically, finds all the cycle that may exist in the given graph, but not containing all the vertices. In other words we find, incomplete cycle that exist in the given graph. To, generate the clauses for these cycle, we will consider the Boolean literals assigned to the edges which are not part of the cycle, and we will consider all possible combinations of these literals, in pair, again the clauses generate in this case would be again in DNF, which need to be converted back to CNF. Another, note worthy point is that, if we consider only the first constraint, for 3- SAT clause generation, then also, we can determine the existence of Hamiltonian circuit subjected to monitoring constraint. Suppose, given a Hamilton circuit with n vertices must have exactly n edges. Thus, for the Hamilton Circuit to exist, out of m literals, at most n must be TRUE and m-n must be FALSE. This we can conclude by the fact, that we are assign Boolean literals to the edge, so if the edge literal

6 evaluates to true means the presence of the edge in the Hamiltonian circuit. The above discussed approach is illustrated with the help of example, in the next section Illustrative Example. e 6 6 e 5 5 e 4 e 7 1 e 8 4 e e 3 Fig. 2. Edge labeled with Boolean literals Now, applying the constraints on the vertex in figure 2. The clauses generated are:- (e 1 e 2 )( e 1 e 2 e 8 e 1 e 2 e 8 e 1 e 2 e 8 ) ( e 2 e 3 e 7 e 2 e 3 e 7 e 2 e 3 e 7 ) (e 3 e 4 )( e 4 e 5 e 8 e 4 e 5 e 8 e 4 e 5 e 8 )( e 5 e 6 e 7 e 5 e 6 e 7 e 5 e 6 e 7 ) Now considering the next constraint of in complete cycle i.e. not containing all the nodes like , , , etc. The clause generated for is: (e 6 e 7 e 7 e 8 e 6 e 8 ) and similarly can be done for the remaining cycles. e 2 4 Implementation In this section, we illustrate the algorithm used for the implementation of the source code to generate 3-CNF clauses. Suppose, we are given a graph G(V,M), where number of Vertex V be n, and the number of edges M be m. 4.1 Algorithm 1. Take input in the form of adjacency matrix x[n][n] for the graph from the user. 2. for i= 1: n do 3. For each k nodes connected to node i 4. Select a node l such that l belongs to k 5. Using constraint (1) generate a clause for (i, l) pair. 6. For all edges connected to node i 7. Apply constraint (2)

7 8. For all edges connected to node l 9. Apply constraint (3) 10. For each edge connected to i use constraint (4) 11. End for 12. Convert from k-cnf to3-cnf. 13. End 5 Conclusion In this paper, we have illustrated the two different approaches using which a given graph can be tested to check the presence of Hamiltonian circuit. To, do so we have actually reduced the given graph into 3SAT instances using all the constraints mandatory for the existence of Hamiltonian circuit. The clauses generated would be satisfied if and only if the Hamiltonian circuit exists. The approach is more efficient than existing backtracking and other heuristic based approaches. References 1. Karp, R.M.: Reducibility Among Combinatorial Problems in Complexity of Computer computations, ed. By R.E. Miller and J.W. Thatcher, Plenum Press, ew York, Garey, M.R. and Johnson, D.S.: Computers and Intractability, A Guide to the theory of NP-completeness, N.H Freeman and Co., San Francisco, Stephen Cook: The P versus NP Problem, official Problem Description, Millenium Problems, Clay Mathematics Institute, J.A. Bondy and U.S.R. Murty : Graph Theory with Applications, North Holland, Tutte, W.T. : On Hamiltonian Circuits, J. London of Math. Soc., Vol. 21(1946),pp Josep Diaz, Jordi Petit And Maria Serna: Faulty Random Geometric Networks, Parallel Processing Letters, World Scientific Publishing Company. 7. Yuh-Rong Leu and Sy-Yen Kuo,: Distributed Fault-Tolerant Ring Embedding and Reconfiguration in Hypercubes, ieee transactions on computers, vol. 48, no. 1, january O.Ore,: Notes on Hamilton Circuit, Amer. Math. Monthly, 65(1960), E. F. Schmeichel and S. L. Hakimi,: Pancyclic graphs and a conjecture of Bondy and Chvátal, Journal of Combinatorial Theory, Series B, Volume 17, Issue 1,, Pages 22-34, August R. Robinson and N. Wormald.: Almost all regular graphs are Hamiltonian. Random Structures and Algorithms, 5(2):Pages , E. W.Weisstein. Horton Graph.: mathworld.wolfram.com/ HortonGraph.html, Math World,A Wolfram Web Resource. 12. Masanori Arita and Akira Suyama and Masami Hagiya :A Heuristic Approach for Hamiltonian Path Problem with Molecules, Proceedings of the Second Annual Conference (GP-97) Pages , Ray-Shang Lo,Gen-Huey Chen: Embedding Hamiltonian Paths in Faulty Arrangement Graphs with the Backtracking Method, IEEE Transactions on Parallel and Distributed Systems,Volume 12 Issue 2, February 2001

8 Appendix Node 1: Clauses generation as per 1,2,3 and 4 ( x1, 2 x 2, 1)( x 1, 2 x 1, 3)( x1, 2 x 1, 5 )( x 1, 2 x 3, 2)( x1, 2 x1,3 x 1, 5 ) Node 2: ( x2, 3 x 3, 2 )( x 2, 3 x 2, 1 )( x 1, 2 x 3, 2)( x2, 3 x 1, 3 )( x2, 3 x 4, 3 )( x 2, 3 x 5, 3)( x2, 1 x 2, 3) Node 3: ( x3, 4 x 4, 3 )( x 3, 4 x 3, 5)( x3, 4 x 3, 1 )( x 3, 4 x 3, 2 )( x 3, 4 x 5, 4 )( x3, 4 x 3, 5 x 3, 2 x 3, 1) Node 4: ( x4, 5 5, 4 )( x 4, 5 x 4, 3) ( x4, 5 x 1, 5 )( x4, 5 x 3,5 )( x4, 5 x 4, 3 ) Node 5: ( x5, 1 x 1, 5 )( x 5, 1 x 5, 3 )( x 5, 1 x 5, 4 )( x 5, 1 x 2, 1 )( x 5, 1 x 3, 1 )( x5, 1 x5, 3 x 5, 4 )