Hopfield网络解决TSP问题算法源代码

#include <stdio.h> #include <math.h> #include <stdlib.h> #include <time.h> #include <conio.h> #define N 7 /* number of cities */ seven() { float u0,t,tao; float v0,v1,v2,v3,v4,v5,v6,dv,delta,s,dist; float u[8][8],vb[9][9],v[9][9],du[8][8]; float A,B,C,D,J,K; int x,i,y,j,e; int n,column[9]; /* cycling times */ float d[8][8]={0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, /* distance */ 0.0,0.0,1.3,1.25,14.0,14.0,14.0,14.0, /* between */ 0.0,1.3,0.0,14.0,1.35,1.95,2.345,14.0,/* 7 cities */ 0.0,1.25,14.0,0.0,1.3,2.725,1.95,14.0, 0.0,14.0,1.35,1.3,0.0,1.425,1.325,14.0, 0.0,14.0,1.95,2.725,1.425,0.0,14.0,1.45, 0.0,14.0,2.345,1.95,1.325,14.0,0.0,1.375, 0.0,14.0,14.0,14.0,14.0,1.45,1.375,0.0}; u0=0.02; /* step 1:initialization */ A=3.3,B=3.3,C=1.5,D=1.8; /* step 1: */ J=30.0,K=30.0; /* step 1: */ t=0.005; tao=100.0; u[0][0]=0.5*u0*log(N-1); /* step 1: */ n=1; /* step 1: */ printf("u[0][0]=%8.6f\n",u[0][0]); /* */ srand(time(NULL)); /* second as seed */ for(x=1;x<8;x++) /* initial input */ { for(i=1;i<8;i++) /* values */ { /* of 49 cells */ delta=(rand()%2000)/1000.0-1.0; u[x][i]=u[0][0]+delta; printf("u[%d][%d]=%6.4f ",x,i,u[x][i]); } printf("\n"); } for(x=0;x<8;x++) /* step 02: output values */ for(i=0;i<8;i++) /* step 02: of cell [x][i]*/ vb[x][i]=0.0; /* step 02: at time t0 */ do { e=0; for(x=1;x<8;x++) /* step 2: */ for(i=1;i<8;i++) /* step 2: */ { v[x][i]=0.5*(1.0+tanh(u[x][i]/u0)); /* step 2:new output value */ dv=v[x][i]-vb[x][i]; /* step 2: variated value */ vb[x][i]=v[x][i]; if(fabs(dv)>0.001) /* step 3: one of criteria */ e=1; /* step 3: for end */ } for(x=1;x<8;x++) { v[x][0]=v[x][7]; v[x][8]=v[x][1]; } if(e==0) /* if all dv<0.001 */ break; /* stop */ else /* if any dv>0.001 */ n+=1; /* continue */ for(x=1;x<8;x++) /* step 3: */ for(i=1;i<8;i++) /* */ { v1=0.0; /* start point */ v2=0.0; v3=0.0; v4=0.0; v5=0.0; v6=0.0; v0=-u[x][i]/tao; /* step 3: ---- term 0 */ for(j=1;j<8;j++) /* step 3: total values of */ if(j!=i) /* step 3: cells in line x */ v1=v1+v[x][j]; v1=-A*v1; /* step 3: ---- term 1 */ for(y=1;y<8;y++) /* step 3: total values of */ if(y!=x) /* step 3: cells in column i */ v2=v2+v[y][i]; /* */ v2=-B*v2; /* step 3: ---- term 2 */ for(y=1;y<8;y++) /* step 3: total values */ for(j=1;j<8;j++) /* step 3: of all cells */ v3=v3+v[y][j]; /* */ v3=-C*(v3-N); /* step 3: ---- term 3 */ for(j=1;j<8;j++) /* step 3: possible distance */ v4=v4+d[x][j]*(v[j][i+1]+v[j][i-1]);/* step 3: between any 2 */ v4=-D*v4; /* step 3: ---- term 4 */ for(j=1;j<8;j++) /* two more restriction */ v5=v5+v[x][j]; /* conditions : v5 and v6 */ v5=-J*(v5-1.0); for(y=1;y<8;y++) v6=v6+v[y][i]; v6=-K*(v6-1.0); du[x][i]=v0+v1+v2+v3+v4+v5+v6; /* step 3: value of du/dt */ u[x][i]=u[x][i]+du[x][i]*t; /* step 4: new input */ } printf("\n"); }while(n<5000); /* the other criteria */ for(i=1;i<8;i++) { for(j=1;j<8;j++) printf("u%d,%d=%7.5f ",i,j,u[i][j]); /* result: input values */ printf("\n"); } printf("\n"); printf("times=%d\n",n); /* result: times */ for(x=1;x<8;x++) /* display the v matrix */ { /* in float number */ for(y=1;y<8;y++) printf("%7.4f ",v[x][y]); printf("\n"); } for(x=1;x<8;x++) /* dispaly the v matrix */ { /* in integer number */ for(y=1;y<8;y++) { v[x][y]=floor(v[x][y]+0.38); printf("%.0f ",v[x][y]); } printf("\n"); } getch(); } main() { seven(); }