Multistage Graph (Shortest Path)

A Multistage graph is a directed, weighted graph in which the nodes can be divided into a set of stages such that all edges are from a stage to next stage only (In other words there is no edge between vertices of same stage and from a vertex of current stage to previous stage).

The vertices of a multistage graph are divided into n number of disjoint subsets S = { S₁ , S_{2 ,} S_{3 ………..} S_n },  where S₁ is the source and S_n is the sink ( destination ). The cardinality of S₁ and S_n are equal to 1. i.e., |S₁| = |S_n| = 1.
We are given a multistage graph, a source and a destination, we need to find shortest path from source to destination. By convention, we consider source at stage 1 and destination as last stage.
Following is an example graph we will consider in this article :-

Now there are various strategies we can apply :-

• The Brute force method of finding all possible paths between Source and Destination and then finding the minimum. That’s the WORST possible strategy.
• Dijkstra’s Algorithm of Single Source shortest paths. This method will find shortest paths from source to all other nodes which is not required in this case. So it will take a lot of time and it doesn’t even use the SPECIAL feature that this MULTI-STAGE graph has.
• Simple Greedy Method – At each node, choose the shortest outgoing path. If we apply this approach to the example graph given above we get the solution as 1 + 4 + 18 = 23. But a quick look at the graph will show much shorter paths available than 23. So the greedy method fails !
• The best option is Dynamic Programming. So we need to find Optimal Sub-structure, Recursive Equations and Overlapping Sub-problems.

Optimal Substructure and Recursive Equation :-
We define the notation :- M(x, y) as the minimum cost to T(target node) from Stage x, Node y. 

Shortest distance from stage 1, node 0 to 
destination, i.e., 7 is M(1, 0).
// From 0, we can go to 1 or 2 or 3 to
// reach 7.              
M(1, 0) = min(1 + M(2, 1),
              2 + M(2, 2),
              5 + M(2, 3))

This means that our problem of 0 —> 7 is now sub-divided into 3 sub-problems :-

So if we have total 'n' stages and target
as T, then the stopping condition  will be :-
M(n-1, i) = i ---> T + M(n, T) = i ---> T

Recursion Tree and Overlapping Sub-Problems:-
So, the hierarchy of M(x, y) evaluations will look something like this :-

In M(i, j), i is stage number and
j is node number
                   M(1, 0)
           /          |         \                             
          /           |          \                            
       M(2, 1)      M(2, 2)        M(2, 3)
    /      \        /     \         /    \
M(3, 4)  M(3, 5)  M(3, 4)  M(3, 5) M(3, 6)  M(3, 6)
 .         .       .       .          .        .
 .         .       .       .          .        .
 .         .       .       .          .        .

So, here we have drawn a very small part of the Recursion Tree and we can already see Overlapping Sub-Problems. We can largely reduce the number of M(x, y) evaluations using Dynamic Programming.
Implementation details: 
The below implementation assumes that nodes are numbered from 0 to N-1 from first stage (source) to last stage (destination). We also assume that the input graph is multistage. 

We use top to bottom approach, and use dist[] array to store the value of overlapping sub-problem.

dist[i] will store the value of minimum distance from node i to node n-1 (target node).

Therefore, dist[0] will store minimum distance between from source node to target node.

C++

// CPP program to find shortest distance
// in a multistage graph.
#include<bits/stdc++.h>
using namespace std;

#define N 8
#define INF INT_MAX

// Returns shortest distance from 0 to
// N-1.
int shortestDist(int graph[N][N]) {

    // dist[i] is going to store shortest
    // distance from node i to node N-1.
    int dist[N];

    dist[N-1] = 0;

    // Calculating shortest path for
    // rest of the nodes
    for (int i = N-2 ; i >= 0 ; i--)
    {

        // Initialize distance from i to
        // destination (N-1)
        dist[i] = INF;

        // Check all nodes of next stages
        // to find shortest distance from
        // i to N-1.
        for (int j = i ; j < N ; j++)
        {
            // Reject if no edge exists
            if (graph[i][j] == INF)
                continue;

            // We apply equation to
            // distance to target through j.
            // and compare with minimum distance
            // so far.
            dist[i] = min(dist[i], graph[i][j] +
                                        dist[j]);
        }
    }

    return dist[0];
}

// Driver code
int main()
{
    // Graph stored in the form of an
    // adjacency Matrix
    int graph[N][N] =
      {{INF, 1, 2, 5, INF, INF, INF, INF},
       {INF, INF, INF, INF, 4, 11, INF, INF},
       {INF, INF, INF, INF, 9, 5, 16, INF},
       {INF, INF, INF, INF, INF, INF, 2, INF},
       {INF, INF, INF, INF, INF, INF, INF, 18},
       {INF, INF, INF, INF, INF, INF, INF, 13},
       {INF, INF, INF, INF, INF, INF, INF, 2},
      {INF, INF, INF, INF, INF, INF, INF, INF}};

    cout << shortestDist(graph);
    return 0;
}

Java

// Java program to find shortest distance
// in a multistage graph.
import java.io.*;
import java.util.*;

class GFG {

    static int N = 8;
    static int INF = Integer.MAX_VALUE;

    // Returns shortest distance from 0 to
    // N-1.
    public static int shortestDist(int[][] graph)
    {

        // dist[i] is going to store shortest
        // distance from node i to node N-1.
        int[] dist = new int[N];

        dist[N - 1] = 0;

        // Calculating shortest path for
        // rest of the nodes
        for (int i = N - 2; i >= 0; i--) {

            // Initialize distance from i to
            // destination (N-1)
            dist[i] = INF;

            // Check all nodes of next stages
            // to find shortest distance from
            // i to N-1.
            for (int j = i; j < N; j++) {
                // Reject if no edge exists
                if (graph[i][j] == INF) {
                    continue;
                }

                // We apply recursive equation to
                // distance to target through j.
                // and compare with minimum distance
                // so far.
                dist[i] = Math.min(dist[i],
                                   graph[i][j] + dist[j]);
            }
        }

        return dist[0];
    }

    // Driver code
    public static void main(String[] args)
    {
        // Graph stored in the form of an
        // adjacency Matrix
        int[][] graph = new int[][] {
            { INF, 1, 2, 5, INF, INF, INF, INF },
            { INF, INF, INF, INF, 4, 11, INF, INF },
            { INF, INF, INF, INF, 9, 5, 16, INF },
            { INF, INF, INF, INF, INF, INF, 2, INF },
            { INF, INF, INF, INF, INF, INF, INF, 18 },
            { INF, INF, INF, INF, INF, INF, INF, 13 },
            { INF, INF, INF, INF, INF, INF, INF, 2 }
        };

        System.out.println(shortestDist(graph));
    }
}

// This code has been contributed by 29AjayKumar

Python3

# Python3 program to find shortest
# distance in a multistage graph.

# Returns shortest distance from
# 0 to N-1.
def shortestDist(graph):
    global INF

    # dist[i] is going to store shortest
    # distance from node i to node N-1.
    dist = [0] * N

    dist[N - 1] = 0

    # Calculating shortest path
    # for rest of the nodes
    for i in range(N - 2, -1, -1):

        # Initialize distance from
        # i to destination (N-1)
        dist[i] = INF

        # Check all nodes of next stages
        # to find shortest distance from
        # i to N-1.
        for j in range(N):

            # Reject if no edge exists
            if graph[i][j] == INF:
                continue

            # We apply recursive equation to
            # distance to target through j.
            # and compare with minimum
            # distance so far.
            dist[i] = min(dist[i],
                          graph[i][j] + dist[j])

    return dist[0]

# Driver code
N = 8
INF = 999999999999

# Graph stored in the form of an
# adjacency Matrix
graph = [[INF, 1, 2, 5, INF, INF, INF, INF],
         [INF, INF, INF, INF, 4, 11, INF, INF],
         [INF, INF, INF, INF, 9, 5, 16, INF],
         [INF, INF, INF, INF, INF, INF, 2, INF],
         [INF, INF, INF, INF, INF, INF, INF, 18],
         [INF, INF, INF, INF, INF, INF, INF, 13],
         [INF, INF, INF, INF, INF, INF, INF, 2]]

print(shortestDist(graph))

# This code is contributed by PranchalK

C#

// C# program to find shortest distance
// in a multistage graph.
using System;

class GFG
{
    static int N = 8;
    static int INF = int.MaxValue;

    // Returns shortest distance from 0 to
    // N-1.
    public static int shortestDist(int[,] graph) {

        // dist[i] is going to store shortest
        // distance from node i to node N-1.
        int[] dist = new int[N];

        dist[N-1] = 0;

        // Calculating shortest path for
        // rest of the nodes
        for (int i = N-2 ; i >= 0 ; i--)
        {

            // Initialize distance from i to
            // destination (N-1)
            dist[i] = INF;

            // Check all nodes of next stages
            // to find shortest distance from
            // i to N-1.
            for (int j = i ; j < N ; j++)
            {
                // Reject if no edge exists
                if (graph[i,j] == INF)
                    continue;

                // We apply recursive equation to
                // distance to target through j.
                // and compare with minimum distance
                // so far.
                dist[i] = Math.Min(dist[i], graph[i,j] +
                                            dist[j]);
            }
        }

        return dist[0];
    }

    // Driver code
    static void Main()
    {
        // Graph stored in the form of an
        // adjacency Matrix
        int[,] graph = new int[,]
          {{INF, 1, 2, 5, INF, INF, INF, INF},
           {INF, INF, INF, INF, 4, 11, INF, INF},
           {INF, INF, INF, INF, 9, 5, 16, INF},
           {INF, INF, INF, INF, INF, INF, 2, INF},
           {INF, INF, INF, INF, INF, INF, INF, 18},
           {INF, INF, INF, INF, INF, INF, INF, 13},
           {INF, INF, INF, INF, INF, INF, INF, 2}};

        Console.Write(shortestDist(graph));
    }
}

// This code is contributed by DrRoot_

Javascript

<script>

// JavaScript program to find shortest distance
// in a multistage graph.

let N = 8;
let INF = Number.MAX_VALUE;

// Returns shortest distance from 0 to
    // N-1.
function shortestDist(graph)
{
    // dist[i] is going to store shortest
        // distance from node i to node N-1.
        let dist = new Array(N);

        dist[N - 1] = 0;

        // Calculating shortest path for
        // rest of the nodes
        for (let i = N - 2; i >= 0; i--)
        {

            // Initialize distance from i to
            // destination (N-1)
            dist[i] = INF;

            // Check all nodes of next stages
            // to find shortest distance from
            // i to N-1.
            for (let j = i; j < N; j++)
            {
                // Reject if no edge exists
                if (graph[i][j] == INF)
                {
                    continue;
                }

                // We apply recursive equation to
                // distance to target through j.
                // and compare with minimum distance
                // so far.
                dist[i] = Math.min(dist[i], graph[i][j]
                        + dist[j]);
            }
        }

        return dist[0];
}

let graph = [[INF, 1, 2, 5, INF, INF, INF, INF],
         [INF, INF, INF, INF, 4, 11, INF, INF],
         [INF, INF, INF, INF, 9, 5, 16, INF],
         [INF, INF, INF, INF, INF, INF, 2, INF],
         [INF, INF, INF, INF, INF, INF, INF, 18],
         [INF, INF, INF, INF, INF, INF, INF, 13],
         [INF, INF, INF, INF, INF, INF, INF, 2]];
document.write(shortestDist(graph));


// This code is contributed by rag2127

</script>

Output

9

Time Complexity : The time complexity of the given code is O(N^2), where N is the number of nodes in the graph. This is because the code involves two nested loops that iterate over all pairs of nodes in the graph, and each iteration performs a constant amount of work (i.e., comparing and updating distances). Since the graph is represented using an adjacency matrix, accessing an element takes constant time. Therefore, the overall time complexity of the algorithm is O(N^2).

Space Complexity : The space complexity of the given program is O(N), where N is the number of nodes in the graph. This is because the program uses an array of size N to store the shortest distance from each node to the destination node N-1.

Algorithm

Input: A weighted multistage graph G with s and t as source and target vertices, respectively.
Output: The shortest path from s to t in G.
Set d(t) = 0 and d(v) = ? for all other vertices v in G.
For i = k-1 to 1:
a. For each vertex v in stage i:
i. Set d(v) = min(w(v, u) + d(u)) for all vertices u in stage i+1.
Return d(s) as the shortest path from s to t.
In the above algorithm, we start by setting the shortest path distance to the target vertex t as 0 and all other vertices as infinity. We then work backwards from the target vertex t to the source vertex s.
Starting from the second-to-last stage (k-1), we loop over all the vertices in that stage and update their shortest path distance based on the shortest path distances of the vertices in the next stage (i+1). We update the shortest path distance of a vertex v in stage i as the minimum of the sum of its weight w(v,u) and the shortest path distance d(u) of all vertices u in stage i+1 that are reachable from v.
After we have processed all stages and all vertices, the final shortest path distance d(s) will contain the shortest path from s to t.

Program

C++

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

const int INF = numeric_limits<int>::max();

// Function to find the shortest path using multistage graph
int multistage_shortest_path(vector<pair<int,
                             unordered_map<int, int>>>& graph,
                             int source, int target, int k) {
    // Initialize the shortest path distances
    vector<int> d(graph.size(), INF);
    d[target] = 0;

    // Loop over each stage from k-1 to 1
    for (int i = k - 1; i > 0; i--) {
        // Loop over all vertices in the current stage
        for (int v = 0; v < graph.size(); v++) {
            // Skip vertices not in the current stage
            if (graph[v].first != i) {
                continue;
            }

            // Update the shortest path distance of the current vertex
            for (const auto& u : graph[v].second) {
                d[v] = min(d[v], u.second + d[u.first]);
            }
        }
    }

    // Return the shortest path distance from source to target
    return d;
}

int main() {
    // Example graph
    vector<pair<int, unordered_map<int, int>>> graph = {
        {0, {}},
        {1, {{3, 2}, {4, 9}}},
        {1, {{3, 6}, {4, 3}}},
        {2, {{4, 1}}},
        {2, {{5, 4}}},
        {3, {{5, 7}}},
        {3, {{6, 2}}},
        {4, {{5, 1}, {6, 5}}},
        {4, {{6, 6}}},
        {5, {}},
        {5, {}},
        {6, {}},
        {6, {}}
    };

    // Find the shortest path from vertex 0 to vertex 12
    int shortest_path_distance = multistage_shortest_path(graph, 0, 12, 7);
    cout << "Shortest path distance from vertex 0 to vertex 12: " <<
      shortest_path_distance << endl;

    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;

public class MultistageShortestPath {
    static final int INF = Integer.MAX_VALUE;

    // Function to find the shortest path using multistage
    // graph
    public static int
    multistageShortestPath(List<List<Pair> > graph,
                           int source, int target, int k)
    {
        // Initialize the shortest path distances
        int[] d = new int[graph.size()];
        for (int i = 0; i < graph.size(); i++) {
            d[i] = INF;
        }
        d[target] = 0;

        // Loop over each stage from k-1 to 1
        for (int i = k - 1; i > 0; i--) {
            // Loop over all vertices in the current stage
            for (int v = 0; v < graph.size(); v++) {
                // Skip vertices not in the current stage
                if (v != i) {
                    continue;
                }

                // Update the shortest path distance of the
                // current vertex
                for (Pair u : graph.get(v)) {
                    d[v] = Math.min(d[v],
                                    u.weight + d[u.vertex]);
                }
            }
        }

        // Return the shortest path distance from source to
        // target
        return d;
    }

    public static void main(String[] args)
    {
        // Example graph
        List<List<Pair> > graph = new ArrayList<>();
        for (int i = 0; i < 12; i++) {
            graph.add(new ArrayList<>());
        }

        graph.get(0).add(new Pair(3, 2));
        graph.get(0).add(new Pair(4, 9));
        graph.get(1).add(new Pair(3, 6));
        graph.get(1).add(new Pair(4, 3));
        graph.get(2).add(new Pair(4, 1));
        graph.get(2).add(new Pair(5, 4));
        graph.get(3).add(new Pair(5, 7));
        graph.get(3).add(new Pair(6, 2));
        graph.get(4).add(new Pair(5, 1));
        graph.get(4).add(new Pair(6, 5));
        graph.get(5).add(new Pair(6, 6));

        // Find the shortest path from vertex 0 to vertex 11
        // (changed target from 12 to 11)
        int shortestPathDistance
            = multistageShortestPath(graph, 0, 11, 7);
        System.out.println(
            "Shortest path distance from vertex 0 to vertex 11: "
            + shortestPathDistance);
    }

    static class Pair {
        int vertex;
        int weight;

        public Pair(int vertex, int weight)
        {
            this.vertex = vertex;
            this.weight = weight;
        }
    }
}

Python3

from math import inf

def multistage_shortest_path(graph, source, target, k):
    # Initialize the shortest path distances
    d = [inf] * (len(graph))
    d[target] = 0

    # Loop over each stage from k-1 to 1
    for i in range(k-1, 0, -1):
        # Loop over all vertices in the current stage
        for v in range(len(graph)):
            # Skip vertices not in the current stage
            if graph[v][0] != i:
                continue

            # Update the shortest path distance of the current vertex
            for u in graph[v][1]:
                d[v] = min(d[v], graph[v][1][u] + d[u])

    # Return the shortest path distance from source to target
    return d

# Example graph
graph = [
    (0, {}),
    (1, {3: 2, 4: 9}),
    (1, {3: 6, 4: 3}),
    (2, {4: 1}),
    (2, {5: 4}),
    (3, {5: 7}),
    (3, {6: 2}),
    (4, {5: 1, 6: 5}),
    (4, {6: 6}),
    (5, {}),
    (5, {}),
    (6, {}),
    (6, {}),
]

# Find the shortest path from vertex 0 to vertex 12
shortest_path_distance = multistage_shortest_path(graph, 0, 12, 7)
print("Shortest path distance from vertex 0 to vertex 12:", shortest_path_distance)

C#

using System;
using System.Collections.Generic;

class Program
{
    const int INF = int.MaxValue;

    // Function to find the shortest path using multistage graph
    static int MultistageShortestPath(List<Tuple<int, Dictionary<int, int>>> graph, int source, int target, int k)
    {
        // Initialize the shortest path distances
        int[] d = new int[graph.Count];
        for (int i = 0; i < graph.Count; i++)
            d[i] = INF;

        d[target] = 0;

        // Loop over each stage from k-1 to 1
        for (int i = k - 1; i > 0; i--)
        {
            // Loop over all vertices in the current stage
            for (int v = 0; v < graph.Count; v++)
            {
                // Skip vertices not in the current stage
                if (graph[v].Item1 != i)
                    continue;

                // Update the shortest path distance of the current vertex
                foreach (var u in graph[v].Item2)
                    d[v] = Math.Min(d[v], u.Value + d[u.Key]);
            }
        }

        // Return the shortest path distance from source to target
        return d;
    }

    static void Main(string[] args)
    {
        // Example graph
        List<Tuple<int, Dictionary<int, int>>> graph = new List<Tuple<int, Dictionary<int, int>>>
        {
            new Tuple<int, Dictionary<int, int>>(0, new Dictionary<int, int>()),
            new Tuple<int, Dictionary<int, int>>(1, new Dictionary<int, int> { { 3, 2 }, { 4, 9 } }),
            new Tuple<int, Dictionary<int, int>>(1, new Dictionary<int, int> { { 3, 6 }, { 4, 3 } }),
            new Tuple<int, Dictionary<int, int>>(2, new Dictionary<int, int> { { 4, 1 } }),
            new Tuple<int, Dictionary<int, int>>(2, new Dictionary<int, int> { { 5, 4 } }),
            new Tuple<int, Dictionary<int, int>>(3, new Dictionary<int, int> { { 5, 7 } }),
            new Tuple<int, Dictionary<int, int>>(3, new Dictionary<int, int> { { 6, 2 } }),
            new Tuple<int, Dictionary<int, int>>(4, new Dictionary<int, int> { { 5, 1 }, { 6, 5 } }),
            new Tuple<int, Dictionary<int, int>>(4, new Dictionary<int, int> { { 6, 6 } }),
            new Tuple<int, Dictionary<int, int>>(5, new Dictionary<int, int>()),
            new Tuple<int, Dictionary<int, int>>(5, new Dictionary<int, int>()),
            new Tuple<int, Dictionary<int, int>>(6, new Dictionary<int, int>()),
            new Tuple<int, Dictionary<int, int>>(6, new Dictionary<int, int>())
        };

        // Find the shortest path from vertex 0 to vertex 12
        int shortestPathDistance = MultistageShortestPath(graph, 0, 12, 7);
        Console.WriteLine("Shortest path distance from vertex 0 to vertex 12: " + shortestPathDistance);
    }
}

Javascript

function multistage_shortest_path(graph, source, target, k) {
    // Initialize the shortest path distances
    let d = Array(graph.length).fill(Infinity);
    d[target] = 0;

    // Loop over each stage from k-1 to 1
    for (let i = k - 1; i > 0; i--) {
        // Loop over all vertices in the current stage
        for (let v = 0; v < graph.length; v++) {
            // Skip vertices not in the current stage
            if (graph[v][0] != i) {
                continue;
            }

            // Update the shortest path distance of the current vertex
            for (let u in graph[v][1]) {
                d[v] = Math.min(d[v], graph[v][1][u] + d[u]);
            }
        }
    }

    // Return the shortest path distance from source to target
    return d;
}

// Example graph
let graph = [
    [0, {}],
    [1, {3: 2, 4: 9}],
    [1, {3: 6, 4: 3}],
    [2, {4: 1}],
    [2, {5: 4}],
    [3, {5: 7}],
    [3, {6: 2}],
    [4, {5: 1, 6: 5}],
    [4, {6: 6}],
    [5, {}],
    [5, {}],
    [6, {}],
    [6, {}],
];

// Find the shortest path from vertex 0 to vertex 12
let shortest_path_distance = multistage_shortest_path(graph, 0, 12, 7);
console.log("Shortest path distance from vertex 0 to vertex 12:", shortest_path_distance);

Output

Shortest path distance from vertex 0 to vertex 12: inf

In the above code, the graph variable represents a multistage graph with 13 vertices and 7 stages. Each tuple in the graph list contains a vertex’s stage number and a dictionary of its adjacent vertices and their weights.

We call the multistage_shortest_path function with the graph variable, the source vertex index (0), the target vertex index (12), and the number of stages (7). The function returns the shortest path distance from the source to the target vertex, which is printed to the console.

Time and Auxiliary Space

The time complexity of the multistage graph shortest path algorithm depends on the number of vertices and the number of stages in the graph. The outer loop iterates over the stages, which takes O(k) time. The inner loop iterates over the vertices in each stage, and for each vertex, it examines its adjacent vertices. Since the graph is represented as an adjacency list, this takes O(E) time, where E is the number of edges in the graph. Therefore, the total time complexity of the algorithm is O(kE).

The space complexity of the algorithm is O(V), where V is the number of vertices in the graph. This is because we store the shortest path distances for each vertex in a list of size V. Additionally, we store the graph as an adjacency list, which also requires O(V) space.