Approximate solution for Travelling Salesman Problem using MST

Greedy Approximate Algorithm for K Centers Problem

Last Updated : 14 Mar, 2024

Given n cities and distances between every pair of cities, select k cities to place warehouses (or ATMs or Cloud Server) such that the maximum distance of a city to a warehouse (or ATM or Cloud Server) is minimized.

For example consider the following four cities, 0, 1, 2, and 3, and the distances between them, how to place 2 ATMs among these 4 cities so that the maximum distance of a city to an ATM is minimized.

kcenters1

There is no polynomial-time solution available for this problem as the problem is a known NP-Hard problem. There is a polynomial-time Greedy approximate algorithm, the greedy algorithm provides a solution that is never worse than twice the optimal solution. The greedy solution works only if the distances between cities follow Triangular Inequality (The distance between two points is always smaller than the sum of distances through a third point).

The 2-Approximate Greedy Algorithm:

Choose the first center arbitrarily.

Choose remaining k-1 centers using the following criteria.

Let c1, c2, c3, … ci be the already chosen centers. Choose

(i+1)’th center by picking the city which is farthest from already

selected centers, i.e, the point p which has following value as maximum

Min[dist(p, c1), dist(p, c2), dist(p, c3), …. dist(p, ci)]

greedyAlgo

Example (k = 3 in the above-shown Graph):

Let the first arbitrarily picked vertex be 0.

The next vertex is 1 because 1 is the farthest vertex from 0.

Remaining cities are 2 and 3. Calculate their distances from already selected centers (0 and 1). The greedy algorithm basically calculates the following values.

Minimum of all distanced from 2 to already considered centers

Min[dist(2, 0), dist(2, 1)] = Min[7, 8] = 7

Minimum of all distanced from 3 to already considered centers

Min[dist(3, 0), dist(3, 1)] = Min[6, 5] = 5

After computing the above values, city 2 is picked as the value corresponding to 2 is maximum.

Note that the greedy algorithm doesn’t give the best solution for k = 2 as this is just an approximate algorithm with a bound as twice optimal.

Proof that the above greedy algorithm is 2 approximate.

Let OPT be the maximum distance of a city from a center in the Optimal solution. We need to show that the maximum distance obtained from the Greedy algorithm is 2*OPT.
The proof can be done using contradiction.

Assume that the distance from the furthest point to all centers is > 2·OPT.

This means that distances between all centers are also > 2·OPT.

We have k + 1 points with distances > 2·OPT between every pair.

Each point has a center of the optimal solution with distance <= OPT to it.

There exists a pair of points with the same center X in the optimal solution (pigeonhole principle: k optimal centers, k+1 points)
The distance between them is at most 2·OPT (triangle inequality) which is a contradiction.

Implementation:

C++

// C++ program for the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
int maxindex(int* dist, int n) 
{ 
    int mi = 0; 
    for (int i = 0; i < n; i++) { 
        if (dist[i] > dist[mi]) 
            mi = i; 
    } 
    return mi; 
} 
  
void selectKcities(int n, int weights[4][4], int k) 
{ 
    int* dist = new int[n]; 
    vector<int> centers; 
    for (int i = 0; i < n; i++) { 
        dist[i] = INT_MAX; 
    } 
  
    // index of city having the 
    // maximum distance to it's 
    // closest center 
    int max = 0; 
    for (int i = 0; i < k; i++) { 
        centers.push_back(max); 
        for (int j = 0; j < n; j++) { 
  
            // updating the distance 
            // of the cities to their 
            // closest centers 
            dist[j] = min(dist[j], weights[max][j]); 
        } 
  
        // updating the index of the 
        // city with the maximum 
        // distance to it's closest center 
        max = maxindex(dist, n); 
    } 
  
    // Printing the maximum distance 
    // of a city to a center 
    // that is our answer 
    cout << endl << dist[max] << endl; 
  
    // Printing the cities that 
    // were chosen to be made 
    // centers 
    for (int i = 0; i < centers.size(); i++) { 
        cout << centers[i] << " "; 
    } 
    cout << endl; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 4; 
    int weights[4][4] = { { 0, 4, 8, 5 }, 
                          { 4, 0, 10, 7 }, 
                          { 8, 10, 0, 9 }, 
                          { 5, 7, 9, 0 } }; 
    int k = 2; 
  
    // Function Call 
    selectKcities(n, weights, k); 
} 
// Contributed by Balu Nagar

Java

// Java program for the above approach 
import java.util.*; 
  
class GFG{ 
  
static int maxindex(int[] dist, int n) 
{ 
    int mi = 0; 
    for(int i = 0; i < n; i++) 
    { 
        if (dist[i] > dist[mi]) 
            mi = i; 
    } 
    return mi; 
} 
  
static void selectKcities(int n, int weights[][],  
                          int k) 
{ 
    int[] dist = new int[n]; 
    ArrayList<Integer> centers = new ArrayList<>(); 
    for(int i = 0; i < n; i++)  
    { 
        dist[i] = Integer.MAX_VALUE; 
    } 
  
    // Index of city having the 
    // maximum distance to it's 
    // closest center 
    int max = 0; 
    for(int i = 0; i < k; i++) 
    { 
        centers.add(max); 
        for(int j = 0; j < n; j++) 
        { 
              
            // Updating the distance 
            // of the cities to their 
            // closest centers 
            dist[j] = Math.min(dist[j],  
                               weights[max][j]); 
        } 
  
        // Updating the index of the 
        // city with the maximum 
        // distance to it's closest center 
        max = maxindex(dist, n); 
    } 
  
    // Printing the maximum distance 
    // of a city to a center 
    // that is our answer 
    System.out.println(dist[max]); 
  
    // Printing the cities that 
    // were chosen to be made 
    // centers 
    for(int i = 0; i < centers.size(); i++)  
    { 
        System.out.print(centers.get(i) + " "); 
    } 
    System.out.print("\n"); 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int n = 4; 
    int[][] weights = new int[][]{ { 0, 4, 8, 5 },  
                                   { 4, 0, 10, 7 },  
                                   { 8, 10, 0, 9 },  
                                   { 5, 7, 9, 0 } }; 
    int k = 2; 
  
    // Function Call 
    selectKcities(n, weights, k); 
} 
} 
  
// This code is contributed by nspatilme

Python3

# Python3 program for the above approach 
def maxindex(dist, n): 
    mi = 0
    for i in range(n): 
        if (dist[i] > dist[mi]): 
            mi = i 
    return mi 
  
def selectKcities(n, weights, k): 
    dist = [0]*n 
    centers = [] 
  
    for i in range(n): 
        dist[i] = 10**9
          
    # index of city having the 
    # maximum distance to it's 
    # closest center 
    max = 0
    for i in range(k): 
        centers.append(max) 
        for j in range(n): 
  
            # updating the distance 
            # of the cities to their 
            # closest centers 
            dist[j] = min(dist[j], weights[max][j]) 
  
        # updating the index of the 
        # city with the maximum 
        # distance to it's closest center 
        max = maxindex(dist, n) 
  
    # Printing the maximum distance 
    # of a city to a center 
    # that is our answer 
    # print() 
    print(dist[max]) 
  
    # Printing the cities that 
    # were chosen to be made 
    # centers 
    for i in centers: 
        print(i, end = " ") 
  
# Driver Code 
if __name__ == '__main__': 
    n = 4
    weights = [ [ 0, 4, 8, 5 ], 
              [ 4, 0, 10, 7 ], 
              [ 8, 10, 0, 9 ], 
              [ 5, 7, 9, 0 ] ] 
    k = 2
  
    # Function Call 
    selectKcities(n, weights, k) 
  
# This code is contributed by mohit kumar 29.

C#

using System; 
using System.Collections.Generic; 
  
public class GFG{ 
      
    static int maxindex(int[] dist, int n) 
    { 
        int mi = 0; 
        for(int i = 0; i < n; i++) 
        { 
            if (dist[i] > dist[mi]) 
                mi = i; 
        } 
        return mi; 
    } 
       
    static void selectKcities(int n, int[,] weights, 
                              int k) 
    { 
        int[] dist = new int[n]; 
        List<int> centers = new List<int>(); 
        for(int i = 0; i < n; i++) 
        { 
            dist[i] = Int32.MaxValue; 
        } 
       
        // Index of city having the 
        // maximum distance to it's 
        // closest center 
        int max = 0; 
        for(int i = 0; i < k; i++) 
        { 
            centers.Add(max); 
            for(int j = 0; j < n; j++) 
            { 
                   
                // Updating the distance 
                // of the cities to their 
                // closest centers 
                dist[j] = Math.Min(dist[j], 
                                   weights[max,j]); 
            } 
       
            // Updating the index of the 
            // city with the maximum 
            // distance to it's closest center 
            max = maxindex(dist, n); 
        } 
       
        // Printing the maximum distance 
        // of a city to a center 
        // that is our answer 
        Console.WriteLine(dist[max]); 
       
        // Printing the cities that 
        // were chosen to be made 
        // centers 
        for(int i = 0; i < centers.Count; i++) 
        { 
            Console.Write(centers[i] + " "); 
        } 
        Console.Write("\n"); 
    } 
       
    // Driver Code 
    static public void Main (){ 
          
        int n = 4; 
    int[,] weights = new int[,]{ { 0, 4, 8, 5 }, 
                                   { 4, 0, 10, 7 }, 
                                   { 8, 10, 0, 9 }, 
                                   { 5, 7, 9, 0 } }; 
    int k = 2; 
   
    // Function Call 
    selectKcities(n, weights, k); 
          
    } 
} 
  
// This code is contributed by avanitrachhadiya2155.

Javascript

<script> 
// Javascript program for the above approach 
      
    function maxindex(dist,n) 
    { 
        let mi = 0; 
    for(let i = 0; i < n; i++) 
    { 
        if (dist[i] > dist[mi]) 
            mi = i; 
    } 
    return mi; 
    } 
      
    function selectKcities(n,weights,k) 
    { 
        let dist = new Array(n); 
    let centers = []; 
    for(let i = 0; i < n; i++) 
    { 
        dist[i] = Number.MAX_VALUE; 
    } 
   
    // Index of city having the 
    // maximum distance to it's 
    // closest center 
    let max = 0; 
    for(let i = 0; i < k; i++) 
    { 
        centers.push(max); 
        for(let j = 0; j < n; j++) 
        { 
               
            // Updating the distance 
            // of the cities to their 
            // closest centers 
            dist[j] = Math.min(dist[j], 
                               weights[max][j]); 
        } 
   
        // Updating the index of the 
        // city with the maximum 
        // distance to it's closest center 
        max = maxindex(dist, n); 
    } 
   
    // Printing the maximum distance 
    // of a city to a center 
    // that is our answer 
    document.write(dist[max]+"<br>"); 
   
    // Printing the cities that 
    // were chosen to be made 
    // centers 
    for(let i = 0; i < centers.length; i++) 
    { 
        document.write(centers[i] + " "); 
    } 
    document.write("<br>"); 
    } 
      
    // Driver Code 
    let n = 4; 
    let weights = [ [ 0, 4, 8, 5 ], 
              [ 4, 0, 10, 7 ], 
              [ 8, 10, 0, 9 ], 
              [ 5, 7, 9, 0 ] ] 
    let k = 2 
    selectKcities(n, weights, k) 
      
    // This code is contributed by unknown2108 
</script>

Output

5
0 2

Time Complexity: O(n*k), as we are using nested loops to traverse n*k times.
Auxiliary Space: O(n+k), as we are using extra space for the array center.

Approximate solution for Travelling Salesman Problem using MST

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Greedy Approximate Algorithm for K Centers Problem

C++

Java

Python3

C#

Javascript

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