Multistage Graph (Shortest Path)
A Multistage graph is a directed 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).
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 give 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.
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 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 codeint 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.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 codeN = 8INF = 999999999999# Graph stored in the form of an# adjacency Matrixgraph = [[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 : O(n2)
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