Path with smallest product of edges with weight>0

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Given a directed graph with N nodes and E edges where the weight of each edge is > 0, also given a source S and a destination D. The task is to find the path with the minimum product of edges from S to D. If there is no path from S to D, then print -1.

Examples:

Input: N = 3, E = 3, Edges = {{{1, 2}, 0.5}, {{1, 3}, 1.9}, {{2, 3}, 3}}, S = 1, and D = 3
Output: 1.5
Explanation:
The shortest path will be 1->2->3
with value 0.5*3 = 1.5

Input: N = 3, E = 3, Edges = {{{1, 2}, 0.5}, {{2, 3}, 0.5}, {{3, 1}, 0.5}}, S = 1, and D = 3
Output: cycle detected

Approach: The idea is to use the bellman ford algorithm. It is because Dijkstra’s algorithm cannot be used here as it works only with non-negative edges. It is because while multiplying values between [0-1), the product keeps decreasing indefinitely and 0 is returned finally.
Moreover, cycles need to be detected because if a cycle exists, the product of this cycle will indefinitely decrease the product to 0 and the product will tend to 0. For, simplicity, we will report such cycles.
The following steps can be followed to compute the result:

Initialize an array, dis[] with initial value as ‘inf’ except dis[S] as 1.
Run a loop from 1 – N-1. For each edge in the graph:
- dis[edge.second] = min(dis[edge.second], dis[edge.first]*weight(edge))
Run another loop for each edge in the graph, if any edge exits with (dis[edge.second] > dis[edge.first]*weight(edge)), then cycle is detected.
If dist[d] in infinity, return -1, else return dist[d].

Below is the implementation of the above approach:

C++

// C++ implementation of the approach.
#include <bits/stdc++.h>
using namespace std;
 
double inf = std::
    numeric_limits<double>::infinity();
 
// Function to return the smallest 
// product of edges
double bellman(int s, int d,
               vector<pair<pair<int, int>,
                           double> >
                   ed,
               int n)
{
    // If the source is equal
    // to the destination
    if (s == d)
        return 0;
 
    // Array to store distances
    double dis[n + 1];
 
    // Initialising the array
    for (int i = 1; i <= n; i++)
        dis[i] = inf;
    dis[s] = 1;
 
    // Bellman ford algorithm
    for (int i = 0; i < n - 1; i++)
        for (auto it : ed)
            dis[it.first.second] = min(dis[it.first.second],
                                       dis[it.first.first]
                                           * it.second);
 
    // Loop to detect cycle
    for (auto it : ed) {
        if (dis[it.first.second]
            > dis[it.first.first] * it.second)
            return -2;
    }
 
    // Returning final answer
    if (dis[d] == inf)
        return -1;
    else
        return dis[d];
}
 
// Driver code
int main()
{
 
    int n = 3;
    vector<pair<pair<int, int>, double> > ed;
 
    // Input edges
    ed = { { { 1, 2 }, 0.5 },
           { { 1, 3 }, 1.9 },
           { { 2, 3 }, 3 } };
 
    // Source and Destination
    int s = 1, d = 3;
 
    // Bellman ford
    double get = bellman(s, d, ed, n);
 
    if (get == -2)
        cout << "Cycle Detected";
    else
        cout << get;
}

Java

// Java implementation of the approach
import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
 
class Pair<K, V> 
{
    K first;
    V second;
 
    public Pair(K first, V second) 
    {
        this.first = first;
        this.second = second;
    }
}
 
class GFG{
 
static final float inf = Float.POSITIVE_INFINITY;
 
// Function to return the smallest
// product of edges
static float bellman(int s, int d, 
                     ArrayList<Pair<Pair<Integer,
                                         Integer>, Float>> ed, 
                     int n)
{
     
    // If the source is equal
    // to the destination
    if (s == d)
        return 0;
 
    // Array to store distances
    float[] dis = new float[n + 1];
 
    // Initialising the array
    Arrays.fill(dis, inf);
    dis[s] = 1;
 
    // Bellman ford algorithm
    for(int i = 0; i < n - 1; i++)
        for(Pair<Pair<Integer, Integer>, Float> it : ed)
            dis[it.first.second] = Math.min(dis[it.first.second], 
                                            dis[it.first.first] *
                                                it.second);
 
    // Loop to detect cycle
    for(Pair<Pair<Integer, Integer>, Float> it : ed) 
    {
        if (dis[it.first.second] > 
            dis[it.first.first] * 
                it.second)
            return -2;
    }
 
    // Returning final answer
    if (dis[d] == inf)
        return -1;
    else
        return dis[d];
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
     
    // Input edges
    ArrayList<Pair<Pair<Integer, 
                        Integer>, Float>> ed = new ArrayList<>(
            Arrays.asList(
                      new Pair<Pair<Integer, Integer>, Float>(
                           new Pair<Integer, Integer>(1, 2), 0.5f),
                    new Pair<Pair<Integer, Integer>, Float>(
                         new Pair<Integer, Integer>(1, 3), 1.9f),
                    new Pair<Pair<Integer, Integer>, Float>(
                         new Pair<Integer, Integer>(2, 3), 3f)));
 
    // Source and Destination
    int s = 1, d = 3;
 
    // Bellman ford
    float get = bellman(s, d, ed, n);
 
    if (get == -2)
        System.out.println("Cycle Detected");
    else
        System.out.println(get);
}
}
 
// This code is contributed by sanjeev2552

Python3

# Python3 implementation of the approach. 
import sys
 
inf = sys.maxsize;
 
# Function to return the smallest 
# product of edges 
def bellman(s, d, ed, n) : 
 
    # If the source is equal 
    # to the destination 
    if (s == d) :
        return 0; 
 
    # Array to store distances 
    dis = [0]*(n + 1); 
 
    # Initialising the array 
    for i in range(1, n + 1) :
        dis[i] = inf; 
         
    dis[s] = 1; 
 
    # Bellman ford algorithm 
    for i in range(n - 1) : 
        for it in ed : 
            dis[it[1]] = min(dis[it[1]], dis[it[0]] * ed[it]); 
 
    # Loop to detect cycle 
    for it in ed :
        if (dis[it[1]] > dis[it[0]] * ed[it]) :
            return -2; 
 
    # Returning final answer 
    if (dis[d] == inf) :
        return -1; 
    else :
        return dis[d]; 
 
# Driver code 
if __name__ == "__main__" : 
 
    n = 3;
     
    # Input edges 
    ed = { ( 1, 2 ) : 0.5 , 
        ( 1, 3 ) : 1.9 , 
        ( 2, 3 ) : 3 }; 
 
    # Source and Destination 
    s = 1; d = 3; 
 
    # Bellman ford 
    get = bellman(s, d, ed, n); 
 
    if (get == -2) :
        print("Cycle Detected"); 
    else :
        print(get); 
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Linq;
using System.Collections.Generic;
 
public class GFG{
 
public static float inf = 1000000000;
 
// Function to return the smallest
// product of edges
public static float bellman(int s, int d, 
                     List<KeyValuePair<KeyValuePair<int,
                                         int>, float>> ed, 
                     int n)
{
     
    // If the source is equal
    // to the destination
    if (s == d)
        return 0;
 
    // Array to store distances
    float[] dis = Enumerable.Repeat(inf, n+1).ToArray();
 
    dis[s] = 1;
 
    // Bellman ford algorithm
    for(int i = 0; i < n - 1; i++)
        foreach(KeyValuePair<KeyValuePair<int, int>, float> it in ed)
            dis[it.Key.Value] = Math.Min(dis[it.Key.Value], 
                                            dis[it.Key.Key] *
                                                it.Value);
 
    // Loop to detect cycle
    foreach(KeyValuePair<KeyValuePair<int, int>, float> it in ed) 
    {
        if (dis[it.Key.Value] > 
            dis[it.Key.Key] * 
                it.Value)
            return -2;
    }
 
    // Returning final answer
    if (dis[d] == inf)
        return -1;
    else
        return dis[d];
}
 
// Driver code
public static void Main(string[] args)
{
    int n = 3;
     
    // Input edges
    List<KeyValuePair<KeyValuePair<int, 
                        int>, float>> ed = new List<KeyValuePair<KeyValuePair<int, 
                        int>, float>>(){
                      new KeyValuePair<KeyValuePair<int, int>, float>(
                           new KeyValuePair<int, int>(1, 2), 0.5f),
                    new KeyValuePair<KeyValuePair<int, int>, float>(
                         new KeyValuePair<int, int>(1, 3), 1.9f),
                    new KeyValuePair<KeyValuePair<int, int>, float>(
                         new KeyValuePair<int, int>(2, 3), 3f)};
 
    // Source and Destination
    int s = 1, d = 3;
 
    // Bellman ford
    float get = bellman(s, d, ed, n);
 
    if (get == -2)
        Console.Write("Cycle Detected");
    else
        Console.Write(get);
}
}
 
// This code is contributed by rrrtnx.

Javascript

<script>
 
// Javascript implementation of the approach.
 
var inf = 1000000000;
 
// Function to return the smallest 
// product of edges
function bellman(s, d, ed, n)
{
    // If the source is equal
    // to the destination
    if (s == d)
        return 0;
 
    // Array to store distances
    var dis = Array(n+1).fill(inf);
 
    dis[s] = 1;
 
    // Bellman ford algorithm
    for (var i = 0; i < n - 1; i++)
    {
        for(var j =0 ; j< ed.length; j++)
        {
            dis[ed[j][0][1]] = Math.min(dis[ed[j][0][1]],
                                       dis[ed[j][0][0]]
                                           * ed[j][1]);
        }
    }
 
    // Loop to detect cycle
    for (var it in ed) {
         
        if (dis[it[0][1]]
            > dis[it[0][0]] * it[1])
            return -2;
    }
 
    // Returning final answer
    if (dis[d] == inf)
        return -1;
    else
        return dis[d];
}
 
// Driver code
var n = 3;
var ed;
// Input edges
ed = [ [ [ 1, 2 ], 0.5 ],
       [ [ 1, 3 ], 1.9 ],
       [ [ 2, 3 ], 3 ] ];
// Source and Destination
var s = 1, d = 3;
// Bellman ford
var get = bellman(s, d, ed, n);
if (get == -2)
    document.write( "Cycle Detected");
else
    document.write( get);
 
 
</script>

Output:

1.5

Time complexity: O(E*V)
Auxiliary Space: O(V).

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