A Constructive Algorithm to Prove P=NP Duan Wen-Qi ([email protected]) College of Economics and Management, Zhejiang Normal University, Jinhua, 321004, China Abstract: After reducing the undirected Hamiltonian cycle problem into the TSP problem with cost 0 or 1, we developed an effective algorithm to compute the optimal tour of the transformed TSP. Our algorithm is described as a growth process: initially, constructing 4-vertexes optimal tour; next, one new vertex being added into the optimal tour in such a way to obtain the new optimal tour; then, repeating the previous step until all vertexes are included into the optimal tour. This paper has shown that our constructive algorithm can solve the undirected Hamiltonian cycle problem in polynomial time. According to Cook-Levin theorem, we argue that we have provided a constructive proof of P=NP. This paper, taking Hamiltonian cycle as our object, be the total cost of the optimal tour, which corresponds wishes to develop a constructive algorithm to prove to the m -TSP composed of v1 , v2 , …, and v m . * P=NP, which is one of the seven Millennium Prize When C m 1 ,if any c(vi , v j ) 1 and the edge Problems selected by the Clay Mathematics Institute, (,)vi v j can appear in any optimal tour (whatever one and is also a major unsolved problem in computer or multiples optimal tours exist) of the m -TSP, we science. NP represents the class of questions for which call this kind of edges as optimizing edges (denoted as ** there is no known way to find an answer quickly, but (,)vi v j ) because of their key role playing in our ** an answer can be verified in polynomial time. For the algorithm. Let H m (vi , v j ) |i , j [1, m ] , which hardest NP problems (i.e., NP-complete problems), contains all optimizing edges. Evidently, whether an given an efficient algorithm for any one of them, we edge is optimizing edge depends on the specific [1-4] can find an efficient algorithm for all of them . m -TSP. H m will change if the value of m is In terms of graph theory, a Hamiltonian cycle is a changed. For each optimizing edge, we construct an cycle in a graph that visits each vertex exactly once. optimal tour which must contain that edge. Therefore, For a given graph (whether directed or undirected), we will construct KHm m optimal tours for the i determining the existence of such cycles is the m -TSP. Let m ( i 1,2, , K m ) be the ith Hamiltonian cycle problem, which is NP-complete optimal tour which contains the ith optimizing edge [3] i problem . According to the Cook-Levin theorem, if in H m , and m m |i 1,2, , K m be the set * we solve the undirected Hamiltonian cycle problem in of K m optimal tours. When C m =0, it will happen polynomial time, we also provides a constructive proof H m , K m =0, and m . In order to keep * of P=NP. Following is our constructive algorithm. We consistence, for those m -TSP problems with C m 0 , first reduce the undirected Hamiltonian cycle problem we define H m as the set of (1) edges must be of cost into a special TSP problem. Let GVE (,) be an 0 and appear in any optimal tour whose total cost is 0, instance of the undirected Hamiltonian cycle problem. and (2) edges must be of cost 1 and appear in any tour Consider a complete graph GVE (,) where whose total cost is 1. Considering that edges in H m E(,)| u v u , v in Vand u v . Assign a cost to will play the same role as edges in H m , we also call each edge in E as follows: (1) c( u , v ) 0 if them as optimizing edges, call tours containing (,)u v in E ; (2) c( u , v ) 1 if (,)u v not in E . optimizing edge with cost 0 as optimal tours and tours Evidently, if the total cost of the optimal tour of G is containing one optimizing edge with cost 1 as “optimal 0, then graph G must have at least one Hamiltonian tours”. We define KHm m as the number of edges i , ( ) cycle. Otherwise, there is no Hamiltonian cycle in in H m m i 1,2, , K m be the ith optimal graph G . Therefore, we will prove P=NP provided tour or “optimal tour” which contains the ith i that we can develop a deterministic algorithm to solve optimizing edge, and m m |i 1,2, , K m be the transformed TSP problem G in polynomial time. the set of tours which will be used in vertex growth. The number of vertexes in G is NV , and When the new vertex v m1 is added into the * vi (i 1,2, , N ) denotes the ith vertex. Let C m m -TSP, the total cost of the optimal tour of 1 * , ( ), * * m 1 -TSP depends on C m H m H m Dm1 Table 1, and obtaining one optimal tour m1 with m1 , ( ) * Mindi, j c ( v m 1 ,) v i c ( v m 1 ,)| v j i j [1,] m m m . The optimal tour m1 is constructed in m1 * * m1(vi , v j ) | d i, j =D m 1 ,i j [1, m ] , and such a way that the lowest total cost C m1 in table 1 s m1 * m1 (,)|vi v j d i, j =D m1 1,i j [1,] m . Table will be realized. Following we describe in detail how * * 1 shows in detail how C m+1 depends on the specific to construct m1 under different situations. Case * , * , ( ), , * value of C m Dm1 H m H m m1 and one: D m1 2 . In this case, we first (1) randomly s * * m1 . It is worth noting that C N is the total cost of choose one edge (if C m 0 ), or (2) choose one * * the optimal tour of G . If C N 0 , it implies that optimizing edge (if C m 1) from one of the optimal there is at least one Hamiltonian cycle in graph G ; tour(s) in m (or m ), then connect v m1 to the otherwise there is no Hamiltonian cycle in graph G . two vertices of the previous chosen edge and delete * * s * Evidently, C m1 , Dm1 , m1 , and m1 are that edge. Case two: Dm1 1. In this case, there * very easy to compute if we know C m and H m must exist one, but only one vertex vl which makes ( ). H m Therefore, we will solve the transformed TSP (vm1 , v l ) 0 We pick out vl and (1) find out the problem G in polynomial time provided that our optimal tour which contains an edge (vl , v i ) 1 from ( ) ( ) algorithm can construct H m H m and m m m if ()H m m1 or (2) pick out another correctly from m=4 to m= N in polynomial time, vertex v j provided that (,)vl v j appears in one * which will be proved in the latter. optimal tour if (H m m1 )= or C m 0 , then Table 1 The total cost of the optimal tour with m 1vertexes connect two edges (,)vm1 v l and (,)vm1 v i (or (,)vm1 v j ), and delete the edge (vl , v i (or v j )) * * * D m1 C m 0 C m 1 from the corresponding optimal tour. Case three: if if * * ()H m m1 ()H m m1 D m1 0 . In this case, if C m 0 and ()H m m1 * ** then C m1 0 then CCm1 m 1 , we first find out an edge (,)vi v j so as to , if ()H m m1 (vm1 , v i ) 0 and (vm1 , v j ) 0 from m then s and()H m m1 pick out that tour which contains (,)vi v j , add two 0 ** edges (,) and (,) into, and delete the if ()H m m1 then CCm1 m vm1 v i vm1 v j * * edge (,) from it. If 0 and () then C m1 1 if ()H m m1 vi v j C m H m m1 s = , we randomly pick out one optimal tour from and()H m m1 ** , and connect the new vertex m1 to the vertex then CCm1 m 1 m v vi which satisfies with (vm1 , v i ) 0 and vi ’s if ()H m m1 , then ** adjacent vertex v j then delete the edge (,)vi v j * CCm1 m 1 C m1 1 * from that optimal tour. If C m 1 and ()H m m1 if ()H m m1 ** , we first find out one edge (,)vi v j that satisfies then CCm1 m 1 * ** with (vm1 , v i ) 0 and (vm1 , v j ) 0 from m , 2 C m1 2 CCm1 m 1 then pick out the optimal tour which contains the edge After introducing our general idea, we now (,)vi v j , add the edges (,)vm1 v i and (,)vm1 v j present the detailed process to solve the TSP problem. into, and delete (,)vi v j from that optimal tour. Lastly, * Step one: starting from m=4 , picking out four considering situation ()H m m1 and C m 1 s s vertices and making sure that the total cost of the ( ()H m m1 or ()H m m1 ), it is very * * optimal tour of the 4-TSP is larger than 0, i.e. 1C 4 similar to the Case two with C m 1 if we replace the s * .