Floyd-Warshall algorithm

Given a weighted graph, return the shortest distances between every pair of nodes within the graph, where the graph can contain negative weights. It solves the all-pair shortest path problem.

Restrictions

• must be a weighted graph
  • weights can be negative
• must not contain a negative weight cycle
• able to be used to determine Transitive Closure

Abstract

Operates by repeatedly lowering the distance between two nodes by checking if the shortest path between and involving an intermediate node is shorter. A good explanation is contained in this YouTube video. The $k$-th instance of the outermost loop represents the algorithm has found shortest distances out of all possible paths involving $k-1$ nodes or less.

More details

The beauty behind this algorithm relies on its algorithmic technique, namely Dynamic Programming. We assume that all the nodes has an intermediate node, and we use every single node as an intermediate node one by one. When we consider a node, say $k$, we have considered nodes from 0 to $k-1$ already. Hence, we use the shortest path built by the previous nodes to build shorter paths with $k$ included.

The loop within the algorithm can be grossly simplified: if you can get from $i$ to $k$, then from $k$ to $j$ faster than from $i$ to $j$ through any path you’ve found so far, then the path from $i$ to $j$ through $k$ (the intermediate) becomes the new shortest path.

Efficiency

Time complexity: $O(n^3)$

Implementation

Python 3 (NetworkX) - Alex's Method

import networkx as nx
 
def floyd_warshall(graph: nx.DiGraph) -> list:
    nodes_to_index = {node: i for i, node in enumerate(graph.nodes)}
    distances = [[float('inf') for _ in range(len(graph.nodes))] for _ in range(len(graph.nodes))]
    #set distance for directly connected nodes
    for edge in graph.edges:
        distances[nodes_to_index[edge[0]]][nodes_to_index[edge[1]]] = graph.edges[edge]['weight']
    #vertices are distance zero to themselves
    for vertex in graph.nodes:
        distances[nodes_to_index[vertex]][nodes_to_index[vertex]] = 0
    for k in range(len(graph.nodes)):
        for i in range(len(graph.nodes)):
            for j in range(len(graph.nodes)):
	            #repeatedly lower distance if shorter path exists
                distances[i][j] = min(distances[i][j], distances[i][k] + distances[k][j])
    return distances

Python 3 (NetworkX) - Leo's Method - DP iterative approach

def floyd_warshall(graph: nx.DiGraph):
    # Let V be the number of nodes in the graph:
    V = graph.number_of_nodes()
    
    # Create a 2-dimensional distance array.
    # This acts like an adjacency matrix. The
    # element at distance[i][j] means the shortest
    # distance between vertices i and j.
    dist = [[float("inf") for _ in range(V)] for _ in range(V)]
    
    # Initialise the leading diagonal as 0. This is because
    # the minimum distance from i to i is 0.
    for i in range(V):
        dist[i][i] = 0
        
    # Initialise initial distances. For each edge pair, the 
    # initial shortest distance between them will just be 
    # the weight.
    for u, v, data in graph.edges(data=True):
        dist[u][v] = data["weight"]
    
    for k in range(V):
        for j in range(V):
            for i in range(V):
                if dist[i][j] > dist[i][k] + dist[k][j]:
                    dist[i][j] = dist[i][k] + dist[k][j]
                    
    return dist
    
distances = floyd_warshall(G)
for i in range(len(distances)):
    for j in range(len(distances)):
        print(distances[i][j], end=" ")
    print()

Python 3 (NetworkX) - Leo's Method - DP recursion/cache approach

from functools import cache
 
@cache
def floyd_warshall_subproblem(i: int, j: int, k: int):
    if k == 0:
        has_direct_edge = G.has_edge(i, j)
        return G.get_edge_data(i, j)["weight"] if has_direct_edge else float("inf")
    
    dist_i_to_k = floyd_warshall_subproblem(i, k, k - 1)
    dist_k_to_j = floyd_warshall_subproblem(k, j, k - 1)
    dist_without_k = floyd_warshall_subproblem(i, j, k - 1)
    
    return min(dist_without_k, dist_i_to_k + dist_k_to_j)
 
# Initialize a 2D array representing distance
V = G.number_of_nodes()
dist = [[float("inf") for _ in range(V)] for _ in range(V)]
 
# Iterate and populate all minimum distances
for i in range(V):
    for j in range(V):
        dist[i][j] = floyd_warshall_subproblem(i, j, V - 1)
 
for i in range(len(dist)):
    for j in range(len(dist)):
        print(distances[i][j], end=" ")
    print()

Proof

Proof

Assuming all the correct pre-conditions exist for the Floyd Warshall All Pairs Shortest Path algorithm the following is a proof by induction. 

All the nodes in the graph are given consecutive and unique integer identifiers and a matrix of |V| by |V| is used to maintain the solution. 

After the initialisation when k=0 we have the dist[u,v] initialised as a simple adjacency matrix showing the direct edge weights between any pairs of nodes u and v. All other node pairs without direct edges are initialised with infinity. 

Base Case: When k=1 if going via node k=1 is less than the current distance between nodes numbered i and j, which is i⇒1⇒j will be reduce the distance by since the condition dist[i,1]+dist[1,j] < dist[i,j] will be true since nodes i and j do not have a direct edge connecting them, but are connected via node k=1. 

Inductive Step: Suppose for (k-1) passes all distances via nodes 1..(k-1) have been updated correctly for the outer loop of the triple nested loop in the FW algorithm and that dist[i,j] contains the shortest distance from node I to node j where the intermediate vertices lie within $\{1,\dots ,k-1\}$ of the graph. 

On the kth iteration of the outer triple nested loop of the FW algorithm there are two possibilities. 

• Case 1: dist[i,j] will remain unchanged as going via node k if a path exists is not cheaper, so the cheaper distance is retained 

• Case 2: dist[i,j] will reduce in distance as going via node k is cheaper i⇒k, k⇒j than what was already explored with i⇒(k-1),(k-1)⇒j 

As k is incremented from 1 to |V| we have then relaxed all the edges from i to j via node k where a path exists. We explored all the i⇒k, k⇒j paths in the graph, so FW gives the correct answer for the whole graph G.