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

PythonServer Side ProgrammingProgramming

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
Published on 23-Dec-2020 11:08:09