Traveling Salesman Problem (TSP) Implementation
Last Updated :
26 Nov, 2024
Given a 2d matrix cost[][] of size n where cost[i][j] denotes the cost of moving from city i to city j. The task is to complete a tour from city 0 (0-based index) to all other cities such that we visit each city exactly once and then at the end come back to city 0 at minimum cost.
Note the difference between Hamiltonian Cycle and TSP. The Hamiltonian cycle problem is to find if there exists a tour that visits every city exactly once. Here we know that Hamiltonian Tour exists (because the graph is complete) and in fact, many such tours exist, the problem is to find a minimum weight Hamiltonian Cycle.
Examples:
Input: cost[][] = [[0, 111], [112, 0]]
Output: 223
Explanation: We can visit 0->1->0 and cost = 111 + 112 = 223.
Input: cost[][] = [[0, 1000, 5000], [5000, 0, 1000], [1000, 5000, 0]]
Output: 3000
Explanation: We can visit 0->1->2->0 and cost = 1000 + 1000 + 1000 = 3000.
Approach:
The given graph is a complete graph, meaning there is an edge between every pair of nodes. A naive approach to solve this problem is to generate all permutations of the nodes, and calculate the cost for each permutation, and select the minimum cost among them.
An important observation in the Traveling Salesman Problem (TSP) is that the choice of the starting node does not affect the solution. This is because the optimal path forms a cyclic tour. For example, if the optimal tour is a1→a2→a3→a4→a1, starting from any other node, such as a2, results in the equivalent tour a2→a3→a4→a1→a2 with same total cost.
To simplify the problem, we can fix one node (e.g., node 1) as the starting point and only consider permutations of the remaining nodes.
C++
// C++ program to find the shortest possible route
// that visits every city exactly once and returns to
// the starting point
#include <bits/stdc++.h>
using namespace std;
int tsp(vector<vector<int>> &cost) {
// Number of nodes
int numNodes = cost.size();
vector<int> nodes;
// Initialize the nodes excluding the fixed
// starting point (node 0)
for (int i = 1; i < numNodes; i++)
nodes.push_back(i);
int minCost = INT_MAX;
// Generate all permutations of the remaining nodes
do {
int currCost = 0;
// Start from node 0
int currNode = 0;
// Calculate the cost of the current permutation
for (int i = 0; i < nodes.size(); i++) {
currCost += cost[currNode][nodes[i]];
currNode = nodes[i];
}
// Add the cost to return to the starting node
currCost += cost[currNode][0];
// Update the minimum cost if the current cost
// is lower
minCost = min(minCost, currCost);
} while (next_permutation(nodes.begin(), nodes.end()));
return minCost;
}
int main() {
vector<vector<int>> cost = {{0, 10, 15, 20},
{10, 0, 35, 25},
{15, 35, 0, 30},
{20, 25, 30, 0}};
int res = tsp(cost);
cout << res << endl;
return 0;
}
// Java program to find the shortest possible route
// that visits every city exactly once and returns to
// the starting point
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
class GfG {
static int tsp(int[][] cost) {
// Number of nodes
int numNodes = cost.length;
List<Integer> nodes = new ArrayList<>();
// Initialize the nodes excluding the fixed starting
// point (node 0)
for (int i = 1; i < numNodes; i++) {
nodes.add(i);
}
int minCost = Integer.MAX_VALUE;
// Generate all permutations of the remaining nodes
do {
int currCost = 0;
// Start from node 0
int currNode = 0;
// Calculate the cost of the current permutation
for (int i = 0; i < nodes.size(); i++) {
currCost += cost[currNode][nodes.get(i)];
currNode = nodes.get(i);
}
// Add the cost to return to the starting node
currCost += cost[currNode][0];
// Update the minimum cost if the current cost
// is lower
minCost = Math.min(minCost, currCost);
} while (nextPermutation(nodes));
return minCost;
}
// function to generate the next lexicographical
// permutation
static boolean nextPermutation(List<Integer> nodes) {
int i = nodes.size() - 2;
// Find the first pair where nodes[i] < nodes[i + 1]
while (i >= 0 && nodes.get(i) >= nodes.get(i + 1)) {
i--;
}
// No next permutation exists
if (i < 0) {
return false;
}
// Find the smallest element larger than nodes[i] to
// the right
int j = nodes.size() - 1;
while (nodes.get(j) <= nodes.get(i)) {
j--;
}
// Swap nodes[i] and nodes[j]
Collections.swap(nodes, i, j);
// Reverse the sequence from i + 1 to the end
Collections.reverse(
nodes.subList(i + 1, nodes.size()));
return true;
}
public static void main(String[] args) {
int[][] cost = { { 0, 10, 15, 20 },
{ 10, 0, 35, 25 },
{ 15, 35, 0, 30 },
{ 20, 25, 30, 0 } };
int res = tsp(cost);
System.out.println(res);
}
}
# Python program to find the shortest possible route
# that visits every city exactly once and returns to
# the starting point
from itertools import permutations
def tsp(cost):
# Number of nodes
numNodes = len(cost)
nodes = list(range(1, numNodes))
minCost = float('inf')
# Generate all permutations of the
# remaining nodes
for perm in permutations(nodes):
currCost = 0
currNode = 0
# Calculate the cost of the current permutation
for node in perm:
currCost += cost[currNode][node]
currNode = node
# Add the cost to return to the starting node
currCost += cost[currNode][0]
# Update the minimum cost if the current cost
# is lower
minCost = min(minCost, currCost)
return minCost
if __name__ == "__main__":
cost = [
[0, 10, 15, 20],
[10, 0, 35, 25],
[15, 35, 0, 30],
[20, 25, 30, 0]
]
res = tsp(cost)
print(res)
// C# program to find the shortest possible route
// that visits every city exactly once and returns to
// the starting point
using System;
using System.Collections.Generic;
using System.Linq;
class GfG {
static int tsp(List<List<int> > cost) {
// Number of nodes
int numNodes = cost.Count;
List<int> nodes = new List<int>();
// Initialize the nodes excluding the fixed starting
// point (node 0)
for (int i = 1; i < numNodes; i++) {
nodes.Add(i);
}
int minCost = int.MaxValue;
// Generate all permutations of the remaining nodes
var permutations
= GetPermutations(nodes, nodes.Count);
foreach(var perm in permutations) {
int currCost = 0;
// Start from node 0
int currNode = 0;
// Calculate the cost of the current permutation
foreach(int node in perm) {
currCost += cost[currNode][node];
currNode = node;
}
// Add the cost to return to the starting node
currCost += cost[currNode][0];
// Update the minimum cost if the current cost
// is lower
minCost = Math.Min(minCost, currCost);
}
return minCost;
}
// Helper function to generate all permutations of a
// list
static IEnumerable<IEnumerable<T> >
GetPermutations<T>(List<T> list, int length) {
if (length == 1)
foreach(var item in list)
yield return new T[] { item };
else
for (int i = 0; i < list.Count; i++) {
var subList = new List<T>(list);
subList.RemoveAt(i);
foreach(var perm in GetPermutations(
subList, length - 1))
yield return new T[] { list[i] }
.Concat(perm);
}
}
static void Main() {
List<List<int> > cost = new List<List<int> >() {
new List<int>{ 0, 10, 15, 20 },
new List<int>{ 10, 0, 35, 25 },
new List<int>{ 15, 35, 0, 30 },
new List<int>
{
20, 25, 30, 0
}
};
int res = tsp(cost);
Console.WriteLine(res);
}
}
// JavaScript program to find the shortest possible route
// that visits every city exactly once and returns to
// the starting point
function tsp(cost) {
// Number of nodes
const numNodes = cost.length;
const nodes = [];
// Initialize the nodes excluding the fixed starting
// point (node 0)
for (let i = 1; i < numNodes; i++) {
nodes.push(i);
}
let minCost = Infinity;
// Generate all permutations of the remaining nodes
const permutations = getPermutations(nodes);
for (let perm of permutations) {
let currCost = 0;
let currNode = 0; // Start from node 0
// Calculate the cost of the current permutation
for (let i = 0; i < perm.length; i++) {
currCost += cost[currNode][perm[i]];
currNode = perm[i];
}
// Add the cost to return to the starting node
currCost += cost[currNode][0];
// Update the minimum cost if the current cost is
// lower
minCost = Math.min(minCost, currCost);
}
return minCost;
}
// Helper function to generate all permutations of an array
function getPermutations(array) {
const result = [];
if (array.length === 1) {
return [ array ];
}
for (let i = 0; i < array.length; i++) {
const current = array[i];
const remaining
= array.slice(0, i).concat(array.slice(i + 1));
const remainingPermutations
= getPermutations(remaining);
for (let perm of remainingPermutations) {
result.push([ current ].concat(perm));
}
}
return result;
}
const cost = [
[ 0, 10, 15, 20 ], [ 10, 0, 35, 25 ], [ 15, 35, 0, 30 ],
[ 20, 25, 30, 0 ]
];
const res = tsp(cost);
console.log(res);
Time complexity: O(n!) where n is the number of vertices in the graph. This is because the algorithm uses the next_permutation function which generates all the possible permutations of the vertex set.
Auxiliary Space: O(n) as we are using a vector to store all the vertices.
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