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Widest Path Problem | Practical application of Dijkstra’s Algorithm
  • Difficulty Level : Hard
  • Last Updated : 22 Jan, 2021

It is highly recommended to read Dijkstra’s algorithm using the Priority Queue first.
Widest Path Problem is a problem of finding a path between two vertices of the graph maximizing the weight of the minimum-weight edge in the path. See the below image to get the idea of the problem:
 

Graph

Practical Application Example: 
This problem is a famous variant of Dijkstra’s algorithm. In the practical application, this problem can be seen as a graph with routers as its vertices and edges represent bandwidth between two vertices. Now if we want to find the maximum bandwidth path between two places in the internet connection, then this problem can be solved by this algorithm.
How to solve this problem? 
We are going to solve this problem by using the priority queue ((|E|+|V|)log|V|) implementation of the Dijkstra’s algorithm with a slight change.
We solve this problem by just replacing the condition of relaxation in Dijkstra’s algorithm by:
 

max(min(widest_dist[u], weight(u, v)), widest_dist[v])

where u is the source vertex for v. v is the current vertex we are checking the condition.
This algorithm runs for both directed and undirected graph.
See the series of images below to get the idea about the problem and the algorithm:
The values over the edges represents weights of directed edges. 
 



DAG

We will start from source vertex and then travel all the vertex connected to it and add in priority queue according to relaxation condition. 
 

 

Now (2, 1) will pop up and 2 will be the current source vertex.
 

Now (3, 1) will pop from the queue. But as 3 does not have any connected vertex through directed edge nothing will happen. So (4, 2) will pop next.
 



Finally the algorithm stops, as there is no more elements in priority queue. 
 

The path with the maximum value of widest distance is 1-4-3 which has the maximum bottle-neck value of 2. So we end up getting widest distance of 2 to reach the target vertex 3.
Below is the implementation of the above approach:
 

CPP




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the required path
void printpath(vector<int>& parent, int vertex, int target)
{
    if (vertex == 0) {
        return;
    }
 
    printpath(parent, parent[vertex], target);
 
    cout << vertex << (vertex == target ? "\n" : "--");
}
 
// Function to return the maximum weight
// in the widest path of the given graph
int widest_path_problem(vector<vector<pair<int, int> > >& Graph,
                        int src, int target)
{
    // To keep track of widest distance
    vector<int> widest(Graph.size(), INT_MIN);
 
    // To get the path at the end of the algorithm
    vector<int> parent(Graph.size(), 0);
 
    // Use of Minimum Priority Queue to keep track minimum
    // widest distance vertex so far in the algorithm
    priority_queue<pair<int, int>, vector<pair<int, int> >,
                   greater<pair<int, int> > >
        container;
 
    container.push(make_pair(0, src));
 
    widest[src] = INT_MAX;
 
    while (container.empty() == false) {
        pair<int, int> temp = container.top();
 
        int current_src = temp.second;
 
        container.pop();
 
        for (auto vertex : Graph[current_src]) {
 
            // Finding the widest distance to the vertex
            // using current_source vertex's widest distance
            // and its widest distance so far
            int distance = max(widest[vertex.second],
                               min(widest[current_src], vertex.first));
 
            // Relaxation of edge and adding into Priority Queue
            if (distance > widest[vertex.second]) {
 
                // Updating bottle-neck distance
                widest[vertex.second] = distance;
 
                // To keep track of parent
                parent[vertex.second] = current_src;
 
                // Adding the relaxed edge in the prority queue
                container.push(make_pair(distance, vertex.second));
            }
        }
    }
 
    printpath(parent, target, target);
 
    return widest[target];
}
 
// Driver code
int main()
{
 
    // Graph representation
    vector<vector<pair<int, int> > > graph;
 
    int no_vertices = 4;
 
    graph.assign(no_vertices + 1, vector<pair<int, int> >());
 
    // Adding edges to graph
 
    // Resulting graph
    // 1--2
    // |  |
    // 4--3
 
    // Note that order in pair is (distance, vertex)
    graph[1].push_back(make_pair(1, 2));
    graph[1].push_back(make_pair(2, 4));
    graph[2].push_back(make_pair(3, 3));
    graph[4].push_back(make_pair(5, 3));
 
    cout << widest_path_problem(graph, 1, 3);
 
    return 0;
}

Python3




# Python3 implementation of the approach
 
# Function to prthe required path
def printpath(parent, vertex, target):
     
    # global parent
    if (vertex == 0):
        return
    printpath(parent, parent[vertex], target)
    print(vertex ,end="\n" if (vertex == target) else "--")
 
# Function to return the maximum weight
# in the widest path of the given graph
def widest_path_problem(Graph, src, target):
     
    # To keep track of widest distance
    widest = [-10**9]*(len(Graph))
 
    # To get the path at the end of the algorithm
    parent = [0]*len(Graph)
 
    # Use of Minimum Priority Queue to keep track minimum
    # widest distance vertex so far in the algorithm
    container = []
    container.append((0, src))
    widest[src] = 10**9
    container = sorted(container)
    while (len(container)>0):
        temp = container[-1]
        current_src = temp[1]
        del container[-1]
        for vertex in Graph[current_src]:
 
            # Finding the widest distance to the vertex
            # using current_source vertex's widest distance
            # and its widest distance so far
            distance = max(widest[vertex[1]],
                           min(widest[current_src], vertex[0]))
 
            # Relaxation of edge and adding into Priority Queue
            if (distance > widest[vertex[1]]):
 
                # Updating bottle-neck distance
                widest[vertex[1]] = distance
 
                # To keep track of parent
                parent[vertex[1]] = current_src
 
                # Adding the relaxed edge in the prority queue
                container.append((distance, vertex[1]))
                container = sorted(container)
    printpath(parent, target, target)
    return widest[target]
 
# Driver code
if __name__ == '__main__':
 
    # Graph representation
    graph = [[] for i in range(5)]
    no_vertices = 4
    # Adding edges to graph
 
    # Resulting graph
    #1--2
    #|  |
    #4--3
 
    # Note that order in pair is (distance, vertex)
    graph[1].append((1, 2))
    graph[1].append((2, 4))
    graph[2].append((3, 3))
    graph[4].append((5, 3))
 
    print(widest_path_problem(graph, 1, 3))
 
# This code is contributed by mohit kumar 29
Output: 
1--4--3
2

 

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