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Travelling Salesman Problem
One sales-person is in a city, he has to visit all other cities those are listed, the cost of traveling from one city to another city is also provided. Find the route where the cost is minimum to visit all of the cities once and return back to his starting city.
The graph must be complete for this case, so the sales-person can go from any city to any city directly.
Here we have to find minimum weighted Hamiltonian Cycle.
Input and Output
Input: Cost matrix of the matrix. 0 20 42 25 30 20 0 30 34 15 42 30 0 10 10 25 34 10 0 25 30 15 10 25 0 Output: Distance of Travelling Salesman: 80
Algorithm
travellingSalesman (mask, pos)
There is a table dp, and VISIT_ALL value to mark all nodes are visited
Input − mask value for masking some cities, position.
Output minus; Find the shortest route to visit all the cities.
Begin if mask = VISIT_ALL, then //when all cities are visited return cost[pos, 0] if dp[mask, pos] ≠ -1, then return dp[mask, pos] finalCost := ∞ for all cities i, do tempMask := (shift 1 left side i times) if mask AND tempMask = 0, then tempCpst := cost[pos, i] + travellingSalesman(mask OR tempMask, i) finalCost := minimum of finalCost and tempCost done dp[mask, pos] = finalCost return finalCost End
Example
#include<iostream>
#define CITY 5
#define INF 9999
using namespace std;
int cost[CITY][CITY] = {
{0, 20, 42, 25, 30},
{20, 0, 30, 34, 15},
{42, 30, 0, 10, 10},
{25, 34, 10, 0, 25},
{30, 15, 10, 25, 0}
};
int VISIT_ALL = (1 << CITY) - 1;
int dp[16][4]; //make array of size (2^n, n)
int travellingSalesman(int mask, int pos) {
if(mask == VISIT_ALL) //when all cities are marked as visited
return cost[pos][0]; //from current city to origin
if(dp[mask][pos] != -1) //when it is considered
return dp[mask][pos];
int finalCost = INF;
for(int i = 0; i<CITY; i++) {
if((mask & (1 << i)) == 0) { //if the ith bit of the result is 0, then it is unvisited
int tempCost = cost[pos][i] + travellingSalesman(mask | (1 << i), i); //as ith city is visited
finalCost = min(finalCost, tempCost);
}
}
return dp[mask][pos] = finalCost;
}
int main() {
int row = (1 << CITY), col = CITY;
for(int i = 0; i<row; i++)
for(int j = 0; j<col; j++)
dp[i][j] = -1; //initialize dp array to -1
cout << "Distance of Travelling Salesman: ";
cout <<travellingSalesman(1, 0); //initially mask is 0001, as 0th city already visited
}Output
Distance of Travelling Salesman: 80
Updated on: 17-Jun-2020
1K+ Views
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