Dijkstra’s shortest path with minimum edges

16 May, 2022
Prerequisite: Dijkstra’s shortest path algorithm Given an adjacency matrix graph representing paths between the nodes in the given graph. The task is to find the shortest path with minimum edges i.e. if there a multiple short paths with same cost then choose the one with the minimum number of edges.Consider the graph given below:  
There are two paths from vertex 0 to vertex 3 with the weight of 12:  
0 -> 1 -> 2 -> 30 -> 4 -> 3
0 -> 1 -> 2 -> 3
0 -> 4 -> 3
Since, Dijkstra’s algorithm is a greedy algorithm that seeks the minimum weighted vertex on every iteration, so the original Dijkstra’s algorithm will output the first path but the result should be the second path as it contains minimum number of edges.Examples:  
Input: graph[][] = { {0, 1, INFINITY, INFINITY, 10}, {1, 0, 4, INFINITY, INFINITY}, {INFINITY, 4, 0, 7, INFINITY}, {INFINITY, INFINITY, 7, 0, 2}, {10, INFINITY, INFINITY, 2, 0} }; Output: 0->4->3 INFINITY here shows that u and v are not neighborsInput: graph[][] = { {0, 5, INFINITY, INFINITY}, {5, 0, 5, 10}, {INFINITY, 5, 0, 5}, {INFINITY, 10, 5, 0} }; Output: 0->1->3  
 
Approach: The idea of the algorithm is to use the original Dijkstra’s algorithm, but also to keep track on the length of the paths by an array that stores the length of the paths from the source vertex, so if we find a shorter path with the same weight, then we will take it.Let’s follow the upper example iteration by iteration: Consider we want to find the shortest path from vertex 0 to vertex 3Initial state: The distance and the parent of all vertices are Infinity and NILL respectively, as usual. But now, we have one more array called pathlength[] that stores the length of the path from the source vertex to all of the vertices. Initially, we set all the elements of pathlength[] to 0.  
First iteration: First we seek the vertex which contains the minimum distance which is vertex 0, as shown in the image above. Then, we traverse all its neighbors which are not blackened, which are 1 and 4. Since, the distance of vertex 1 and 4 are infinity, then we reduce their weights to 1 and 10 respectively. Update the parents, and set the pathlength[] for each vertex (1 and 4) to 1 because they can be reached from source vertex by 1 edge. After that, we blacken the vertex as the original Dijkstra’s algorithm does.  
Second iteration: We continue to seek the non-blakened vertex which contains the minimum distance, which is vertex 1, and after that we reduce its neighbor’s weight to 1 + 4 = 5 and update its parent as the original Dijkstra’s algorithm does, and set its pathlength[] to 2, because it is two edges away from the source vertex. Finally, we blacken vertex 1.  
Third iteration: Again, the non-blackened vertex which contains the minimum distance is vertex 2, so we update the non-blackened neighbors. It has one non-blackened neighbor which is vertex 3. So, we update its weight from Infinity to 5 + 7 = 12 then we set its parent to 2 and set its pathlength[] to 3 because it is 3 edges sway from the source vertex. Finally, we blacken vertex 2.  
Fourth iteration: In this iteration, the algorithm acts different from the original Dijkstra’s algorithm. We seek the non-blackened vertex which contains the minimum distance which is 4. Since, the distance to vertex 3 from the source vertex is 12 (0->1->2->3) and the distance of vertex 4 plus the edge (4, 3) is 12, it means that we just found a new path to vertex 3 from the source vertex with the same weight. Then, we check if the new path is shorter (in edges) than the existing one and we take the one with the minimum edges. Finally, we blacken vertex 4.  
Since V-1 vertices are blackened, the algorithm ends.Below is the implementation of the above approach: 
C++
Java
Python3
C#
// C++ program to find the shortest path// with minimum edges in a graph#include <iostream>using namespace std;#define INFINITY 9999#define n 5#define s 0#define d 3#define NILL -1int MinDistance(int*, int*);void PrintPath(int*, int); // Function to find the shortest path// with minimum edges in a graphvoid Dijkstra(int Graph[n][n], int _n, int _s, int _d){     int i, u, v, count;    int dist[n];    int Blackened[n] = { 0 };    int pathlength[n] = { 0 };    int parent[n];     // The parent Of the source vertex is always equal to nill    parent[_s] = NILL;     // first, we initialize all distances to infinity.    for (i = 0; i < n; i++)        dist[i] = INFINITY;     dist[_s] = 0;    for (count = 0; count < n - 1; count++) {        u = MinDistance(dist, Blackened);         // if MinDistance() returns INFINITY, then the graph is not        // connected and we have traversed all of the vertices in the        // connected component of the source vertex, so it can reduce        // the time complexity sometimes        // In a directed graph, it means that the source vertex        // is not a root        if (u == INFINITY)            break;        else {             // Mark the vertex as Blackened            Blackened[u] = 1;            for (v = 0; v < n; v++) {              //distance to v via u should be smaller and the path length should be minimum                if (!Blackened[v] && Graph[u][v]                         && dist[u] + Graph[u][v] <= dist[v]                         && pathlength[u] + 1 < pathlength[v]) {                    parent[v] = u;                    pathlength[v] = pathlength[u] + 1;                }            }        }    }     // Printing the path    if (dist[_d] != INFINITY)        PrintPath(parent, _d);    else        cout << "There is no path between vertex "             << _s << "to vertex " << _d;} int MinDistance(int* dist, int* Blackened){    int min = INFINITY, min_index, v;    for (v = 0; v < n; v++)        if (!Blackened[v] && dist[v] < min) {            min = dist[v];            min_index = v;        }    return min == INFINITY ? INFINITY : min_index;} // Function to print the pathvoid PrintPath(int* parent, int _d){    if (parent[_d] == NILL) {        cout << _d;        return;    }    PrintPath(parent, parent[_d]);    cout << "->" << _d;} // Driver codeint main(){    // INFINITY means that u and v are not neighbors.    int Graph[n][n] = { { 0, 1, INFINITY, INFINITY, 10 },                        { 1, 0, 4, INFINITY, INFINITY },                        { INFINITY, 4, 0, 7, INFINITY },                        { INFINITY, INFINITY, 7, 0, 2 },                        { 10, INFINITY, INFINITY, 2, 0 } };    Dijkstra(Graph, n, s, d);    return 0;}
// Java program to find the shortest path// with minimum edges in a graphimport java.io.*;import java.util.*; class GFG{     static int INFINITY = 9999, n = 5, s = 0, d = 3, NILL = -1;     // Function to find the shortest path    // with minimum edges in a graph    static void Dijkstra(int[][] Graph, int _n, int _s, int _d)    {         int i, u, v, count;        int[] dist = new int[n];        int[] Blackened = new int[n];        int[] pathlength = new int[n];        int[] parent = new int[n];         // The parent Of the source vertex is always equal to nill        parent[_s] = NILL;         // first, we initialize all distances to infinity.        for (i = 0; i < n; i++)            dist[i] = INFINITY;         dist[_s] = 0;        for (count = 0; count < n - 1; count++)        {            u = MinDistance(dist, Blackened);             // if MinDistance() returns INFINITY, then the graph is not            // connected and we have traversed all of the vertices in the            // connected component of the source vertex, so it can reduce            // the time complexity sometimes            // In a directed graph, it means that the source vertex            // is not a root            if (u == INFINITY)                break;            else            {                 // Mark the vertex as Blackened                Blackened[u] = 1;                for (v = 0; v < n; v++)                {                    if (Blackened[v] == 0 && Graph[u][v] != 0                        && dist[u] + Graph[u][v] < dist[v])                    {                        parent[v] = u;                        pathlength[v] = pathlength[parent[v]] + 1;                        dist[v] = dist[u] + Graph[u][v];                    }                    else if (Blackened[v] == 0 && Graph[u][v] != 0                            && dist[u] + Graph[u][v] == dist[v]                            && pathlength[u] + 1 < pathlength[v])                    {                        parent[v] = u;                        pathlength[v] = pathlength[u] + 1;                    }                }            }        }         // Printing the path        if (dist[_d] != INFINITY)            PrintPath(parent, _d);        else            System.out.println("There is not path between vertex " +                                _s + " to vertex " + _d);    }     static int MinDistance(int[] dist, int[] Blackened)    {        int min = INFINITY, min_index = -1, v;        for (v = 0; v < n; v++)            if (Blackened[v] == 0 && dist[v] < min)            {                min = dist[v];                min_index = v;            }        return min == INFINITY ? INFINITY : min_index;    }     // Function to print the path    static void PrintPath(int[] parent, int _d)    {        if (parent[_d] == NILL)        {            System.out.print(_d);            return;        }        PrintPath(parent, parent[_d]);        System.out.print("->" + _d);    }     // Driver Code    public static void main(String[] args)    {         // INFINITY means that u and v are not neighbors.        int[][] Graph = { { 0, 1, INFINITY, INFINITY, 10 },                        { 1, 0, 4, INFINITY, INFINITY },                        { INFINITY, 4, 0, 7, INFINITY },                        { INFINITY, INFINITY, 7, 0, 2 },                        { 10, INFINITY, INFINITY, 2, 0 } };        Dijkstra(Graph, n, s, d);    }} // This code is contributed by// sanjeev2552
# Python program to find the shortest path# with minimum edges in a graphdef Dijkstra(Graph, _s, _d):    row = len(Graph)    col = len(Graph[0])    dist = [float("Inf")] * row    Blackened =[0] * row    pathlength =[0] * row    parent = [-1] * row    dist[_s]= 0    for count in range(row-1):        u = MinDistance(dist, Blackened)         # if MinDistance() returns INFINITY, then the graph is not        # connected and we have traversed all of the vertices in the        # connected component of the source vertex, so it can reduce        # the time complexity sometimes        # In a directed graph, it means that the source vertex        # is not a root        if u == float("Inf"):            break        else:             # Mark the vertex as Blackened            Blackened[u]= 1        for v in range(row):            if Blackened[v]== 0 and Graph[u][v] and dist[u]+Graph[u][v]<dist[v]:                parent[v]= u                pathlength[v]= pathlength[parent[v]]+1                dist[v]= dist[u]+Graph[u][v]            elif Blackened[v]== 0 and Graph[u][v] and dist[u]+Graph[u][v]== dist[v] and pathlength[u]+1<pathlength[v]:                parent[v]= u                pathlength[v] = pathlength[u] + 1    if dist[_d]!= float("Inf"):         # Printing the path        PrintPath(parent, _d)    else:        print("There is no path between vertex ", _s, "to vertex ", _d) # Function to print the pathdef PrintPath(parent, _d):    if parent[_d]==-1:        print(_d,end='')        return    PrintPath(parent, parent[_d])    print("->", _d,end='')def MinDistance(dist, Blackened):    min = float("Inf")    for v in range(len(dist)):        if not Blackened[v] and dist[v]<min:            min = dist[v]            Min_index = v    return float("Inf") if min == float("Inf") else Min_index # Driver code# float("Inf") means that u and v are not neighborsGraph =[[0, 1, float("Inf"), float("Inf"), 10],       [1, 0, 4, float("Inf"), float("Inf")],       [float("Inf"), 4, 0, 7, float("Inf")],       [float("Inf"), float("Inf"), 7, 0, 2],       [10, float("Inf"), float("Inf"), 2, 0]]Dijkstra(Graph, 0, 3)
// C# program to find the shortest path// with minimum edges in a graphusing System;class GFG{   static int INFINITY = 9999, n = 5, s = 0, d = 3, NILL = -1;   // Function to find the shortest path  // with minimum edges in a graph  static void Dijkstra(int [,] Graph, int _n, int _s, int _d)  {     int i, u, v, count;    int[] dist = new int[n];    int[] Blackened = new int[n];    int[] pathlength = new int[n];    int[] parent = new int[n];     // The parent Of the source vertex is always equal to nill    parent[_s] = NILL;     // first, we initialize all distances to infinity.    for (i = 0; i < n; i++)      dist[i] = INFINITY;     dist[_s] = 0;    for (count = 0; count < n - 1; count++)    {      u = MinDistance(dist, Blackened);       // if MinDistance() returns INFINITY, then the graph is not      // connected and we have traversed all of the vertices in the      // connected component of the source vertex, so it can reduce      // the time complexity sometimes      // In a directed graph, it means that the source vertex      // is not a root      if (u == INFINITY)        break;      else      {         // Mark the vertex as Blackened        Blackened[u] = 1;        for (v = 0; v < n; v++)        {          if (Blackened[v] == 0 && Graph[u,v] != 0              && dist[u] + Graph[u,v] < dist[v])          {            parent[v] = u;            pathlength[v] = pathlength[parent[v]] + 1;            dist[v] = dist[u] + Graph[u,v];          }          else if (Blackened[v] == 0 && Graph[u,v] != 0                   && dist[u] + Graph[u,v] == dist[v]                   && pathlength[u] + 1 < pathlength[v])          {            parent[v] = u;            pathlength[v] = pathlength[u] + 1;          }        }      }    }     // Printing the path    if (dist[_d] != INFINITY)      PrintPath(parent, _d);    else      Console.Write("There is not path between vertex " +                    _s + " to vertex " + _d);  }   static int MinDistance(int[] dist, int[] Blackened)  {    int min = INFINITY, min_index = -1, v;    for (v = 0; v < n; v++)      if (Blackened[v] == 0 && dist[v] < min)      {        min = dist[v];        min_index = v;      }    return min == INFINITY ? INFINITY : min_index;  }   // Function to print the path  static void PrintPath(int[] parent, int _d)  {    if (parent[_d] == NILL)    {      Console.Write(_d);      return;    }    PrintPath(parent, parent[_d]);    Console.Write("->" + _d);  }   // Driver Code  public static void Main()  {     // INFINITY means that u and v are not neighbors.    int [,] Graph = { { 0, 1, INFINITY, INFINITY, 10 },                     { 1, 0, 4, INFINITY, INFINITY },                     { INFINITY, 4, 0, 7, INFINITY },                     { INFINITY, INFINITY, 7, 0, 2 },                     { 10, INFINITY, INFINITY, 2, 0 } };    Dijkstra(Graph, n, s, d);  }} // This code is contributed by jana_sayantan.
0->4->3
 
Time Complexity: O(V^2) where V is the number of vertices and E is the number of edges.Auxiliary Space: O(V + E) 
sanjeev2552
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amartyaghoshgfg
abhishekdutta
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