Travelling Salesman Problem - algorithm

Question

I am trying to develop a program in C++ from Travelling Salesman Problem Algorithm. I need a distance matrix and a cost matrix. After using all the formulas, i get a new resultant matrix. But I dont understand what that matrix shows.
Suppose the resultant matrix is:
1 2 3
4 5 6
7 8 9
Now I want to know what this matrix shows? Assume I have 3 cities to traverse.
Please tell me the flow. A sample program of this algorithm will be more favorable..
Thank you.
My Program is:
#include<iostream.h>
#include<conio.h>
#include <stdlib.h>
void main()
{
clrscr();
int a,b,c,d,ctr,j,Q=1,K=1 ;
float q0=0.7, p = 0.5 ;
int phe[3][3];
double dist[3][3] , mem[3][3],exp[3][3],eplt[3][3], rnd;
cout<<"enter the iterations, cities , ants ";
cin>>a>>b>>c;
for (int i=0;i<3;i++)
{
for (j=0;j<3;j++)
{
dist[i][j]=(double)rand()/(double)RAND_MAX;
if (i==j)
dist[i][j]=0;
}
}
for (i=0;i<3;i++)
{
for (j=0;j<3;j++)
{
cout<< dist[i][j]<<"\t";
}
cout<<"\n";
}
cout<<"pheromone matrix "<<endl;
for (i=0;i<3;i++)
{
for (j=0;j<3;j++)
{
if (i==j)
phe[i][j]=0;
else
phe[i][j]=1;
}
}
for ( i=0;i<3;i++)
{
for ( j=0;j<3;j++)
{
cout<< phe[i][j]<<"\t";
}
cout<<"\n";
}
cout<< "after iteration "<<endl;
for (i=0;i<3;i++)
{
ctr=0;
for (int k=0;k<3;k++)
{
// mem[i][k]=(rand()%b)+1;
// cout<<"memory"<<mem[i][k]<<"\n";
rnd= (double)rand()/(double)RAND_MAX;
cout<<"hhhhhhh"<<rnd;
if (rnd<=q0)
{
cout<<"Exploitation\n";
eplt[i][ctr] =(p*phe[i][k])+(Q/K);
}
else
{
cout<<"EXPLORATION\n";
eplt[i][ctr]= phe[i][k]/dist[i][k];
}
ctr++;
}
}
for (i=0;i<3;i++)
{
for (int k=0;k<3;k++)
{
cout <<eplt[i][k]<<"\t";
}
cout<<"\n";
}
getch();
}
OUTPUT:
enter the iterations, cities , ants 3
4
4
0 0.003967 0.335154
0.033265 0 0.2172
0.536973 0.195776 0
pheromone matrix
0 1 1
1 0 1
1 1 0
after iteration
hhhhhhh0.949919EXPLORATION
hhhhhhh0.356777EXPLOITATION
hhhhhhh0.356777EXPLOITATION
hhhhhhh0.356777EXPLOITATION
hhhhhhh0.356777EXPLOITATION
hhhhhhh0.356777EXPLOITATION
hhhhhhh0.949919EXPLORATION

Answer 1

First up, I'm guessing when you say My Program you mean The program in the paper since it is basically out of date C++. Standard library headers don't have .h appended, and conio.h is an MS-DOS header - most code that I've seen that uses that comes from Borland Turbo C++. Worth bearing in mind if you're going to try to compile that demo on a modern system.
Next up, what you're looking at is an adjacancy matrix. I don't believe that matrix is part of the output at all; I believe it is part of the model being used, for demonstration purposes. I believe, given you have a pheromone matrix, that what you're looking at here is Ant Colony Optimisation, a probabilistic method of solving the TSP and other problems that can be reduced to it.
From your output, it isn't clear where or how the result is being stored, and since this is homework, I am lazy and you're just asking for an outright answer, I'm not going to read that code. The premise of Ant Colony optimisation is that pheromone trails laid by ants, which walk the graph at random, decay over time (number of iterations). The longer it takes an ant to move along a particular vertex (distance), the more the laid pheromone decays. At this point, ants start to make decisions based on the strength of the laid pheromone along a path. So what happens is ants start to prefer certain routes over others, and continually re-inforce the pheromone along that path.
So, somewhere in there, there must be a matrix like the adjacancy matrix, storing the pheromone levels for each route. Combined with the length of the route, each iteration should detect a rate of decay.

Answer 2

Your input variables a, b, c are never used.
Your variable ctr is used in the exact same incremental way as the variable k of the same loop.
Your phenomone matrix indicates use of an ant colony optimization algorithm, why just not say it in your question ?
Such "iteration" should be, well, iterated, so probably the output you give us (which is not a normal output) is not the definitive solution, rather a provisory result of the algorithm.

Answer 3

In this post, implementation of simple solution is discussed.
Consider city 1 or 0 as the starting and ending point. Since route is
cyclic, we can consider any point as starting point.
Generate all (n-1)! permutations of cities.
Calculate cost of every permutation and keep track of minimum cost
permutation.
Return the permutation with minimum cost.
#include <bits/stdc++.h>
using namespace std;
int main(void){
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
int graph[n][n];
for(int i =0;i<n;i++){
for(int j =0;j<n;j++){
scanf("%d",&graph[i][j]);
}
}
vector<int> v;
int s = 0;
for(int i =0;i<n;i++){
if(i!=s){
v.push_back(i);
}
}
int ans = INT_MAX;
do{
int current_pathsum = 0;
int k = s;
for(int i = 0;i<v.size();i++){
current_pathsum += graph[k][v[i]];
k = v[i];
}
current_pathsum += graph[k][s];
ans = min(ans,current_pathsum);
}while(next_permutation(v.begin(),v.end()));
cout<<ans<<endl;
}
}