Program to Find Out if an Edge is a Part of a Minimum Spanning Tree in Python

Python Server Side Programming Programming

Suppose we have a 2D matrix named 'edges', that represents an undirected graph. Every item in the matrix 'edges' represents an edge and is of the form (u, v, w). This means nodes u and v are connected and the edge has the weight w. We also have integers a and b, that represent an edge (a,b). We have to find out if the edge (a, b) is part of a minimum spanning tree.

Note − the graph has to be connected and the edge (a, b) exists in the graph.

So, if the input is like edges =

[[0, 2, 100],
[1, 2, 200],
[1, 3, 100],
[2, 3, 300]],
a = 0
b = 2,

then the output will be True.

To solve this, we will follow these steps −

Define a function findPath() . This will take edges, a, b
- if a is same as b, then
  - return True
- if edges is empty, then
  - return False
- for each x in edges, do
  - if x[2] is same as -1, then
    - continue the iteration
  - new_a := -1
  - if x[0] is same as a, then
    - new_a := x[1]
  - otherwise when x[1] is same as a, then
    - new_a := x[0]
  - if new_a is not same as -1, then
    - delete x from edges
    - if findPath(edges, new_a, b) is non-zero, then
      - return True
    - insert x at the end of edges
  - return False

Now from the main function, do the following −

weight := edge weights of edge x from input array ‘edges’, if
- ((x[0] is same as a and x[1] is same as b) or(x[1] is same as a and x[0] is same as b))
edges := [edge x from input array edges if x[2] <weight]
return not findPath(edges, a, b)

Example

Let us see the following implementation to get a better understanding −

Live Demo

class Solution:
   def findPath(self, edges, a, b):
      if a == b:
         return True
      if not edges:
         return False
      for x in edges:
         if x[2] == -1:
            continue
         new_a = -1
         if x[0] == a:
            new_a = x[1]
         elif x[1] == a:
            new_a = x[0]
         if new_a != -1:
            edges.remove(x)
            if self.findPath(edges, new_a, b):
               return True
            edges.append(x)
      return False
   def solve(self, edges, a, b):
      weight = next(x for x in edges if (x[0] == a and x[1] == b) or (x[1] == a and x[0] == b))[ 2 ]
      edges = [x for x in edges if x[2] < weight]
      return not self.findPath(edges, a, b)

ob = Solution()
print(ob.solve([
   [0, 2, 100],
   [1, 2, 200],
   [1, 3, 100],
   [2, 3, 300]
], 0, 2))

Input

[
[0, 2, 100],
[1, 2, 200],
[1, 3, 100],
[2, 3, 300]
], 0, 2

Output

True

Arnab Chakraborty

Updated on: 23-Dec-2020

324 Views

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