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Floyd Warshall Algorithm for All-Pairs Shortest Path

Problem and examples

The Floyd Warshall Algorithm solves the problem of finding the shortest path between all pairs of vertices in a weighted, directed graph.

The graph is given as an n × n adjacency matrix matrix, where matrix[i][j] represents the weight of the edge from node i to node j. A value of -1 indicates that there is no direct edge between the two nodes.

Your task is to compute the shortest distances between every pair of nodes. If there is no path between two nodes, the corresponding distance should remain -1.

Input Matrix Output Matrix Description
[[0, 3, -1], [-1, 0, 1], [4, -1, 0]] [[0, 3, 4], [5, 0, 1], [4, 7, 0]] Shortest paths computed between all vertex pairs
[[0, 5, -1], [-1, 0, 2], [3, -1, 0]] [[0, 5, 7], [5, 0, 2], [3, 8, 0]] Path from 0 to 2 via 1 with weight 5+2=7
[[0, -1, -1], [-1, 0, -1], [-1, -1, 0]] [[0, -1, -1], [-1, 0, -1], [-1, -1, 0]] No edges means no shorter paths can be found

How it works

Understanding the Problem

The Floyd Warshall Algorithm is used to find the shortest distances between every pair of nodes in a weighted graph. This is also called the All-Pairs Shortest Path problem. The graph may be directed or undirected and can contain positive or negative edge weights—but no negative weight cycles.

You are given a matrix where matrix[i][j] represents the weight of the edge from node i to node j. If there is no direct edge between i and j, the value is -1, which we will interpret as "infinity" (no connection). Our goal is to fill this matrix with the shortest path values between all pairs of nodes.

Step-by-Step Solution with Example

Step 1: Convert -1 to Infinity

Since -1 means there's no direct path between two nodes, we replace it with a large number (e.g., 1e9 or Infinity) so that it doesn't interfere in the minimum comparisons during the algorithm.

Step 2: Initialize the Graph

Create a distance matrix from the input graph. The diagonal entries (i.e., distance from node to itself) should be 0, and all other entries should be as per the input (or converted to infinity if there’s no edge).

Step 3: Run Floyd Warshall Triple Loop

We now go through each node k as an intermediate node. For each pair of nodes (i, j), we check whether going from i → k → j is shorter than the direct path i → j. If it is, we update the distance.

The idea is to allow all nodes to be potential "intermediate stations" between any two other nodes.

Step 4: Convert Infinity Back to -1

Once the algorithm completes, any value in the distance matrix that is still Infinity means no path exists. So, we convert those back to -1 to match the input format.

Example:


// Input matrix
[
  [0,  3, -1],
  [-1, 0,  1],
  [4, -1, 0]
]

// After converting -1 to Infinity
[
  [0,  3,  INF],
  [INF, 0, 1],
  [4, INF, 0]
]

// Run Floyd Warshall updates:
→ After considering node 0:
  no changes

→ After considering node 1:
  matrix[0][2] = min(INF, 3 + 1) = 4
  matrix[2][1] = min(INF, 4 + 3) = 7

→ After considering node 2:
  matrix[1][0] = min(INF, 1 + 4) = 5

// Final result after converting back INF to -1
[
  [0,  3,  4],
  [5,  0,  1],
  [4,  7,  0]
]

Edge Cases

  • No path between nodes: Represented as -1 initially and should stay -1 in the final matrix.
  • Self loops: Distance from a node to itself is always 0.
  • Disconnected Graph: If a node is completely disconnected, all entries to and from that node (except to itself) remain -1.
  • Negative edge weights: Allowed, as long as there are no negative weight cycles.

Finally

The Floyd Warshall Algorithm is elegant and powerful for dense graphs and small-to-medium size node counts. It ensures that by the end of the triple-loop process, all indirect paths are checked, and the shortest ones are stored.

This algorithm runs in O(n³) time, so it is not ideal for very large graphs. But when you need shortest paths between every pair of vertices, this algorithm is often the go-to solution.

  1. Let matrix be the input adjacency matrix of size n × n.
  2. Initialize a distance matrix dist with the same values as matrix.
  3. Replace all -1 values (except diagonals) in dist with Infinity.
  4. For each node k from 0 to n-1:
    1. For each node i from 0 to n-1:
    2. For each node j from 0 to n-1:
    3. If dist[i][k] + dist[k][j] < dist[i][j], update dist[i][j].
  5. Convert all Infinity values in dist back to -1.

Implementation and efficiency

C implementation

#include <stdio.h>
#define INF 1000000

void floydWarshall(int matrix[][3], int n) {
    int dist[3][3];

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] == -1 && i != j)
                dist[i][j] = INF;
            else
                dist[i][j] = matrix[i][j];
        }
    }

    for (int k = 0; k < n; k++) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (dist[i][k] + dist[k][j] < dist[i][j])
                    dist[i][j] = dist[i][k] + dist[k][j];
            }
        }
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (dist[i][j] == INF)
                printf("-1 ");
            else
                printf("%d ", dist[i][j]);
        }
        printf("\n");
    }
}

int main() {
    int matrix[3][3] = {
        {0, 3, -1},
        {-1, 0, 1},
        {4, -1, 0}
    };

    floydWarshall(matrix, 3);
    return 0;
}

C++ implementation

#include <iostream>
#include <vector>
using namespace std;

const int INF = 1e9;

void floydWarshall(vector<vector<int>>& matrix) {
    int n = matrix.size();
    vector<vector<int>> dist = matrix;

    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
            if (dist[i][j] == -1 && i != j)
                dist[i][j] = INF;

    for (int k = 0; k < n; ++k)
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < n; ++j)
                if (dist[i][k] < INF && dist[k][j] < INF)
                    dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (dist[i][j] == INF) cout << "-1 ";
            else cout << dist[i][j] << " ";
        }
        cout << endl;
    }
}

int main() {
    vector<vector<int>> matrix = {
        {0, 3, -1},
        {-1, 0, 1},
        {4, -1, 0}
    };
    floydWarshall(matrix);
    return 0;
}

Python implementation

def floyd_warshall(matrix):
    n = len(matrix)
    dist = [
        [float('inf') if matrix[i][j] == -1 and i != j else matrix[i][j] for j in range(n)]
        for i in range(n)
    ]

    for k in range(n):
        for i in range(n):
            for j in range(n):
                if dist[i][k] != float('inf') and dist[k][j] != float('inf'):
                    dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])

    for i in range(n):
        for j in range(n):
            if dist[i][j] == float('inf'):
                dist[i][j] = -1

    for row in dist:
        print(row)

matrix = [
    [0, 3, -1],
    [-1, 0, 1],
    [4, -1, 0]
]
floyd_warshall(matrix)

Java implementation

public class FloydWarshall {
    static final int INF = 1000000;

    public static void main(String[] args) {
        int[][] matrix = {
            {0, 3, -1},
            {-1, 0, 1},
            {4, -1, 0}
        };
        int n = matrix.length;

        int[][] dist = new int[n][n];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == -1 && i != j)
                    dist[i][j] = INF;
                else
                    dist[i][j] = matrix[i][j];
            }
        }

        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (dist[i][k] + dist[k][j] < dist[i][j])
                        dist[i][j] = dist[i][k] + dist[k][j];
                }
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (dist[i][j] == INF)
                    System.out.print("-1 ");
                else
                    System.out.print(dist[i][j] + " ");
            }
            System.out.println();
        }
    }
}

JavaScript implementation

function floydWarshall(matrix) {
  const n = matrix.length;
  const dist = matrix.map((row, i) =>
    row.map((val, j) => (val === -1 && i !== j ? Infinity : val))
  );

  for (let k = 0; k < n; k++) {
    for (let i = 0; i < n; i++) {
      for (let j = 0; j < n; j++) {
        if (dist[i][k] !== Infinity && dist[k][j] !== Infinity) {
          dist[i][j] = Math.min(dist[i][j], dist[i][k] + dist[k][j]);
        }
      }
    }
  }

  for (let i = 0; i < n; i++) {
    for (let j = 0; j < n; j++) {
      if (dist[i][j] === Infinity) dist[i][j] = -1;
    }
  }

  console.log("Shortest distances:");
  dist.forEach(row => console.log(row));
}

const matrix = [
  [0, 3, -1],
  [-1, 0, 1],
  [4, -1, 0]
];

floydWarshall(matrix);

Go implementation

package main

import (
	"fmt"
	"math"
)

func floydWarshall(matrix [][]int) {
	n := len(matrix)
	dist := make([][]int, n)
	for i := range dist {
		dist[i] = make([]int, n)
		for j := range dist[i] {
			if matrix[i][j] == -1 && i != j {
				dist[i][j] = math.MaxInt32
			} else {
				dist[i][j] = matrix[i][j]
			}
		}
	}

	for k := 0; k < n; k++ {
		for i := 0; i < n; i++ {
			for j := 0; j < n; j++ {
				if dist[i][k] != math.MaxInt32 && dist[k][j] != math.MaxInt32 && dist[i][k]+dist[k][j] < dist[i][j] {
					dist[i][j] = dist[i][k] + dist[k][j]
				}
			}
		}
	}

	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			if dist[i][j] == math.MaxInt32 {
				fmt.Print("-1 ")
			} else {
				fmt.Printf("%d ", dist[i][j])
			}
		}
		fmt.Println()
	}
}

func main() {
	matrix := [][]int{
		{0, 3, -1},
		{-1, 0, 1},
		{4, -1, 0},
	}
	floydWarshall(matrix)
}

Rust implementation

fn floyd_warshall(matrix: Vec<Vec<i32>>) {
    let n = matrix.len();
    let mut dist = matrix.clone();

    for i in 0..n {
        for j in 0..n {
            if dist[i][j] == -1 && i != j {
                dist[i][j] = i32::MAX / 2;
            }
        }
    }

    for k in 0..n {
        for i in 0..n {
            for j in 0..n {
                if dist[i][k] < i32::MAX / 2 && dist[k][j] < i32::MAX / 2 {
                    dist[i][j] = dist[i][j].min(dist[i][k] + dist[k][j]);
                }
            }
        }
    }

    for row in dist.iter() {
        for &val in row.iter() {
            if val >= i32::MAX / 2 {
                print!("-1 ");
            } else {
                print!("{} ", val);
            }
        }
        println!();
    }
}

fn main() {
    let matrix = vec![
        vec![0, 3, -1],
        vec![-1, 0, 1],
        vec![4, -1, 0],
    ];
    floyd_warshall(matrix);
}

Kotlin implementation

fun floydWarshall(matrix: Array<IntArray>) {
    val n = matrix.size
    val dist = Array(n) { i ->
        IntArray(n) { j ->
            if (matrix[i][j] == -1 && i != j) Int.MAX_VALUE / 2 else matrix[i][j]
        }
    }

    for (k in 0 until n)
        for (i in 0 until n)
            for (j in 0 until n)
                if (dist[i][k] < Int.MAX_VALUE / 2 && dist[k][j] < Int.MAX_VALUE / 2)
                    dist[i][j] = minOf(dist[i][j], dist[i][k] + dist[k][j])

    for (i in 0 until n) {
        for (j in 0 until n) {
            if (dist[i][j] >= Int.MAX_VALUE / 2) print("-1 ")
            else print("${dist[i][j]} ")
        }
        println()
    }
}

fun main() {
    val matrix = arrayOf(
        intArrayOf(0, 3, -1),
        intArrayOf(-1, 0, 1),
        intArrayOf(4, -1, 0)
    )
    floydWarshall(matrix)
}

Swift implementation

import Foundation

func floydWarshall(_ matrix: [[Int]]) {
    let n = matrix.count
    var dist = matrix

    for i in 0..<n {
        for j in 0..<n {
            if matrix[i][j] == -1 && i != j {
                dist[i][j] = Int.max / 2
            }
        }
    }

    for k in 0..<n {
        for i in 0..<n {
            for j in 0..<n {
                if dist[i][k] < Int.max / 2 && dist[k][j] < Int.max / 2 {
                    dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
                }
            }
        }
    }

    for i in 0..<n {
        for j in 0..<n {
            if dist[i][j] >= Int.max / 2 {
                print("-1", terminator: " ")
            } else {
                print(dist[i][j], terminator: " ")
            }
        }
        print()
    }
}

let matrix = [
    [0, 3, -1],
    [-1, 0, 1],
    [4, -1, 0]
]
floydWarshall(matrix)

TypeScript implementation

function floydWarshall(matrix: number[][]): number[][] {
  const n = matrix.length;
  const dist: number[][] = matrix.map((row, i) =>
    row.map((val, j) => (val === -1 && i !== j ? Infinity : val))
  );

  for (let k = 0; k < n; k++) {
    for (let i = 0; i < n; i++) {
      for (let j = 0; j < n; j++) {
        if (dist[i][k] !== Infinity && dist[k][j] !== Infinity) {
          dist[i][j] = Math.min(dist[i][j], dist[i][k] + dist[k][j]);
        }
      }
    }
  }

  for (let i = 0; i < n; i++) {
    for (let j = 0; j < n; j++) {
      if (dist[i][j] === Infinity) dist[i][j] = -1;
    }
  }

  return dist;
}

const matrix: number[][] = [
  [0, 3, -1],
  [-1, 0, 1],
  [4, -1, 0]
];

console.log("Shortest distances:", floydWarshall(matrix));