Johnson’s algorithm for All-pairs shortest paths | Implementation

12 Oct, 2018
Given a weighted Directed Graph where the weights may be negative, find the shortest path between every pair of vertices in the Graph using Johnson’s Algorithm.
The detailed explanation of Johnson’s algorithm has already been discussed in the previous post.
Refer: Johnson’s algorithm for All-pairs shortest paths.
This post focusses on the implementation of Johnson’s Algorithm.
Algorithm:
Let the given graph be G. Add a new vertex s to the graph, add edges from new vertex to all vertices of G. Let the modified graph be G’.Run Bellman-Ford algorithm on G’ with s as source. Let the distances calculated by Bellman-Ford be h[0], h[1], .. h[V-1]. If we find a negative weight cycle, then return. Note that the negative weight cycle cannot be created by new vertex s as there is no edge to s. All edges are from s.Reweight the edges of original graph. For each edge (u, v), assign the new weight as “original weight + h[u] – h[v]”.Remove the added vertex s and run Dijkstra’s algorithm for every vertex.
Let the given graph be G. Add a new vertex s to the graph, add edges from new vertex to all vertices of G. Let the modified graph be G’.
Run Bellman-Ford algorithm on G’ with s as source. Let the distances calculated by Bellman-Ford be h[0], h[1], .. h[V-1]. If we find a negative weight cycle, then return. Note that the negative weight cycle cannot be created by new vertex s as there is no edge to s. All edges are from s.
Reweight the edges of original graph. For each edge (u, v), assign the new weight as “original weight + h[u] – h[v]”.
Remove the added vertex s and run Dijkstra’s algorithm for every vertex.
Example:Let us consider the following graph.
We add a source s and add edges from s to all vertices of the original graph. In the following diagram s is 4.
We calculate the shortest distances from 4 to all other vertices using Bellman-Ford algorithm. The shortest distances from 4 to 0, 1, 2 and 3 are 0, -5, -1 and 0 respectively, i.e., h[] = {0, -5, -1, 0}. Once we get these distances, we remove the source vertex 4 and reweight the edges using following formula. w(u, v) = w(u, v) + h[u] – h[v].
Since all weights are positive now, we can run Dijkstra’s shortest path algorithm for every vertex as source.
Below is the implementation of the above approach
# Implementation of Johnson's algorithm in Python3  # Import function to initialize the dictionaryfrom collections import defaultdictMAX_INT = float('Inf')  # Returns the vertex with minimum # distance from the sourcedef minDistance(dist, visited):      (minimum, minVertex) = (MAX_INT, 0)    for vertex in range(len(dist)):        if minimum > dist[vertex] and visited[vertex] == False:            (minimum, minVertex) = (dist[vertex], vertex)      return minVertex    # Dijkstra Algorithm for Modified # Graph (removing negative weights)def Dijkstra(graph, modifiedGraph, src):      # Number of vertices in the graph    num_vertices = len(graph)      # Dictionary to check if given vertex is     # already included in the shortest path tree    sptSet = defaultdict(lambda : False)      # Shortest distance of all vertices from the source    dist = [MAX_INT] * num_vertices      dist[src] = 0      for count in range(num_vertices):          # The current vertex which is at min Distance         # from the source and not yet included in the         # shortest path tree        curVertex = minDistance(dist, sptSet)        sptSet[curVertex] = True          for vertex in range(num_vertices):            if ((sptSet[vertex] == False) and                (dist[vertex] > (dist[curVertex] +                 modifiedGraph[curVertex][vertex])) and                (graph[curVertex][vertex] != 0)):                                  dist[vertex] = (dist[curVertex] +                                modifiedGraph[curVertex][vertex]);      # Print the Shortest distance from the source    for vertex in range(num_vertices):        print ('Vertex ' + str(vertex) + ': ' + str(dist[vertex]))  # Function to calculate shortest distances from source# to all other vertices using Bellman-Ford algorithmdef BellmanFord(edges, graph, num_vertices):      # Add a source s and calculate its min    # distance from every other node    dist = [MAX_INT] * (num_vertices + 1)    dist[num_vertices] = 0      for i in range(num_vertices):        edges.append([num_vertices, i, 0])      for i in range(num_vertices):        for (src, des, weight) in edges:            if((dist[src] != MAX_INT) and                     (dist[src] + weight < dist[des])):                dist[des] = dist[src] + weight      # Don't send the value for the source added    return dist[0:num_vertices]  # Function to implement Johnson Algorithmdef JohnsonAlgorithm(graph):      edges = []      # Create a list of edges for Bellman-Ford Algorithm    for i in range(len(graph)):        for j in range(len(graph[i])):              if graph[i][j] != 0:                edges.append([i, j, graph[i][j]])      # Weights used to modify the original weights    modifyWeights = BellmanFord(edges, graph, len(graph))      modifiedGraph = [[0 for x in range(len(graph))] for y in                    range(len(graph))]      # Modify the weights to get rid of negative weights    for i in range(len(graph)):        for j in range(len(graph[i])):              if graph[i][j] != 0:                modifiedGraph[i][j] = (graph[i][j] +                         modifyWeights[i] - modifyWeights[j]);      print ('Modified Graph: ' + str(modifiedGraph))      # Run Dijkstra for every vertex as source one by one    for src in range(len(graph)):        print ('\nShortest Distance with vertex ' +                        str(src) + ' as the source:\n')        Dijkstra(graph, modifiedGraph, src)  # Driver Codegraph = [[0, -5, 2, 3],          [0, 0, 4, 0],          [0, 0, 0, 1],          [0, 0, 0, 0]]  JohnsonAlgorithm(graph)
Modified Graph: [[0, 0, 3, 3], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]

Shortest Distance with vertex 0 as the source:

Vertex 0: 0
Vertex 1: 0
Vertex 2: 0
Vertex 3: 0

Shortest Distance with vertex 1 as the source:

Vertex 0: inf
Vertex 1: 0
Vertex 2: 0
Vertex 3: 0

Shortest Distance with vertex 2 as the source:

Vertex 0: inf
Vertex 1: inf
Vertex 2: 0
Vertex 3: 0

Shortest Distance with vertex 3 as the source:

Vertex 0: inf
Vertex 1: inf
Vertex 2: inf
Vertex 3: 0

Time Complexity: The time complexity of the above algorithm is  as Dijkstra’s Algorithm takes  for adjacency matrix. Note that the above algorithm can be made more efficient by using adjacency list instead of the adjacency matrix to represent the Graph.
Algorithms-Graph Shortest Paths Quiz
Graph Shortest Paths Quiz
Picked
Technical Scripter 2018
Data Structures
Graph
Technical Scripter
Data Structures
Graph
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Advantages and Disadvantages of Linked List
Data Structures | Array | Question 2
Amazon Interview Experience for SDE-II
Data Structures | Queue | Question 2
Data Structures | Hash | Question 5
Breadth First Search or BFS for a Graph
Depth First Search or DFS for a Graph
Dijkstra's shortest path algorithm | Greedy Algo-7
Prim’s Minimum Spanning Tree (MST) | Greedy Algo-5
Find if there is a path between two vertices in a directed graph