764 views

2 Answers

1 1 vote

FIRST APPROACH: DRAW MST AND CHECK

SECOND APPROACH:

A-B,A-D,B-E ARE SELECTED USING KRUSKAL SO EDGES B-D AND D-E CANT BE SELECTED AS THEY WILL LEAD TO CYCLE. WE HAVE REMAINING 2 EDGES TO FULFILL MST CONDITION.

WE CAN SELECT 2 EDGES FROM BC,EF,CF,EC. SO 4C2 WAYS=6

BUT CHOOSING EDGES BC AND EC CAN LEAD TO CYCLE. SO TOTAL MST=6-1=5

Position:
Show:

Related questions

2 2 votes
0 0 answers
1.2k
1.2k views
1 1 vote
1 1 answer
2.3k
2.3k views
Mak Indus asked Nov 6, 2018
2,272 views
. Assume an array A[1….n] has n-elements, and every element of an array is less than or equal to n. An element is said to be majority element, if it occurred in more than...
0 0 votes
0 0 answers
327
327 views
Miku221 asked Sep 3, 2024
327 views
Given a connected graph has $N$ vertices and $M$ edges. Each node $i$ has the weight value $w_i$. Define the strength of a path from $s$ to $t$ is the maximum weight of a...
1 1 vote
1 answers 1 answer
1.3k
1.3k views
Debargha Mitra Roy asked Aug 25, 2024
1,270 views
Which of the statement is/are correct?(a) First edge added by Kruskal’s algorithm can be the last edge added by prim’s algorithm(b) In a graph, if one raises the length o...