Path with smallest product of edges with weight > 0

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 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 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:

  1. Initialise an array, dis[] with initial value as ‘inf’ except dis[S] as 1.
  2. 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))
  3. 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.
  4. If dist[d] in infinity, return -1, else return dist[d].

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


Output:

1.5

Time complexity: O(E*V)

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01