Program to find out the path between two vertices in a graph that has the minimum penalty (Python)

Suppose we are given an undirected, weighted graph and are asked to find out the path with the minimum possible penalty from node a to node b. The penalty of a path is the bitwise OR of the weights of all the edges in the path. So, we must find out such a 'minimum penalty' path, and if there exists no path between the two nodes, we return -1.

So, if the input is like

start (s) = 1, end (e) = 3; then the output will be 15.

There exist two paths between vertices 1 and 3. The optimal path is 1->2->3, the cost of the path is (10 OR 5) = 15.

To solve this, we will follow these steps −

• Define a function helper() . This will take G, s, e
  • v := a new set
  • c := a new list of size n initialized with value infinity
  • heap := a new heap containing pair (0, s)
  • while size of heap > 0, do
    • cst := pop the smallest item from heap
    • cur := pop the smallest item from heap
    • c[cur] := minimum of (cst, c[cur])
    • if (cst, cur) is present in v, then
      • go for the next iteration
    • if cur is same as e, then
      • return c[cur]
    • add pair (cst, cur) to v
    • for each neighbor, n_cost in G[cur], do
      • push values ((n_cost OR cst), neighbor) to heap
  • return c[e]
• G := [a new list containing n + 1 emoty lists]
• for each item in edges, do
  • u := item[0]
  • v := item[1]
  • w := item[2]
  • insert pair (v, w) at the end of G[u]
  • insert pair (u, w) at the end of G[v]
• ans := helper(G, s, e)
• return -1 if ans is same as inf otherwise return ans

Example

Let us see the following implementation to get better understanding −

import heapq
from math import inf

def helper(G, s, e):
    v = set()
    c = [inf] * len(G)
    heap = [(0, s)]
    while len(heap) > 0:
        cst, cur = heapq.heappop(heap)
        c[cur] = min(cst, c[cur])
        if (cst, cur) in v:
            continue
        if cur == e:
            return c[cur]
        v.add((cst, cur))
        for neighbor, n_cost in G[cur]:
            heapq.heappush(heap, (n_cost | cst, neighbor))
    return c[e]

def solve(n, edges, s, e):
    G = [[] for _ in range(n + 1)]
    for item in edges:
        u, v, w = map(int, item)
        G[u].append((v, w))
        G[v].append((u, w))
    ans = helper(G, s, e)
    return -1 if ans == inf else ans

print(solve(4, [(1, 2, 10), (2, 3, 5), (2, 4, 15), (1, 4, 20)], 1, 3))

Input

4, [(1, 2, 10), (2, 3, 5), (2, 4, 15), (1, 4, 20)], 1, 3

Output

15

Arnab Chakraborty

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Updated on: 06-Oct-2021

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