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Program to find out the critical and pseudo-critical edges in a graph in Python
Suppose, we are given a graph that contains n vertices numbered 0 to n -1. The graph is undirected and each edge has a weight. So given the graph, we have to find out the critical and the pseudo-critical edges in the graphs MST. An edge is called a critical edge if deletion of that edge causes the MST weight to increase. A pseudo-critical edge is an edge that can appear in all the graphs MSTs, but not all. We find out the index of the edges given the graph as input.
So, if the input is like
and number of vertices is 5, then the output will be [[], [0, 1, 2, 3, 4]] There are no critical edges in the given graph and all the edges are pseudo-critical. As all the edges have the same weight, any 3 edges from the graph will make an MST.
To solve this, we will follow these steps −
Define a function find_mst() . This will take num_vertices, graph, init := null, exl := null
Define one helper function visit() . This will take u
k[u] := True
for each v, w in of graph[u, an empty list], do
if exl and u is in exl and v is in exl, then
go for next iteration
if not k[v] is True, then
push triplet (w,u,v) into heap tmp
res := 0
k := a new list of size num_arrays containing value False
tmp := a new heap
if init is non-null, then
u := init
v := init
w := init
res := res + w
k[u] := True
k[v] := True
visit(u) or visit(v)
otherwise,
visit(0)
while tmp is not empty, do
w := pop smallest item from heap tmp
u := pop smallest item from heap tmp
v := pop smallest item from heap tmp
if k[u] and k[v] is non-zero, then
go for next iteration
res := res + w
if not k[u] is True, then
visit(u)
if not k[v] is True, then
visit(v)
return res if all of k are True, otherwise return infinity
From the main method, do the following:
graph := given graph
temp := find_mst(num_vertices, graph)
c_edge := a new list
p_edge := a new list
for i in range 0 to size of edges, do
if find_mst(num_vertices, graph, exl = edges[i, index 2 to end]) > temp, then
insert i at the end of c_edge
otherwise, if find_mst(num_vertices, graph, init = edges[i]) is same as temp, then
insert i at the end of p_edge
return [c_edge, p_edge]
Example
Let us see the following implementation to get better understanding
from heapq import heappop, heappush
def solve(num_vertices, edges):
graph = dict()
for u, v, w in edges:
graph.setdefault(u, []).append((v, w))
graph.setdefault(v, []).append((u, w))
temp = find_mst(num_vertices, graph)
c_edge, p_edge = [], []
for i in range(len(edges)):
if find_mst(num_vertices, graph, exl = edges[i][:2]) > temp:
c_edge.append(i)
elif find_mst(num_vertices, graph, init = edges[i]) == temp:
p_edge.append(i)
return [c_edge, p_edge]
def find_mst(num_vertices, graph, init = None, exl = None):
def visit(u):
k[u] = True
for v, w in graph.get(u, []):
if exl and u in exl and v in exl:
continue
if not k[v]:
heappush(tmp, (w, u, v))
res = 0
k = [False] * num_vertices
tmp = []
if init:
u, v, w = init
res += w
k[u] = k[v] = True
visit(u) or visit(v)
else:
visit(0)
while tmp:
w, u, v = heappop(tmp)
if k[u] and k[v]: continue
res += w
if not k[u]:
visit(u)
if not k[v]:
visit(v)
return res if all(k) else inf
print(solve(5, [[0,1,10],[1,2,10],[2,3,10],[3,4,10],[4,0,10]]))Input
5, [[0,1,10],[1,2,10],[2,3,10],[3,4,10],[4,0,10]]
Output
[[], [0, 1, 2, 3, 4]]
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