The Gateway to Computer Science Excellence

+7 votes

Yes I think the claim is true as

If Edge weights are not distinct then we can have multiple Spanning trees

Suppose Consider an example

here we choose edges with weight 1 then with 2 then with 3

In case of 3 if we consider both edges whose weight is 3 the a cycle is formed so we consider only 1 edge by this we can have 2 Spanning trees with same weights

Spanning tree 1 :

Spanning Tree 2:

Let me know if I'm wrong

0 votes

Yes, I agree with the claim made and for its proof,

Taking an instance where we have more than one similar edge weights could be the graph (not neccessarily) on which applying kruskal's algo could give us more than one M.S.T's .

NOTE:- The claim holds false everytime we have distinct edge weights.

Let me know if I was wrong.

Taking an instance where we have more than one similar edge weights could be the graph (not neccessarily) on which applying kruskal's algo could give us more than one M.S.T's .

NOTE:- The claim holds false everytime we have distinct edge weights.

Let me know if I was wrong.

0 votes

It depends on whether edge weight is distinct or not.

Case 1: Edge weight is distinct.

MST is unique so Kruskal's algorithm (as its always works correctly to find MST) cannot give multiple MST.

Case 2: Edge weight not distinct

Multiple MST of same weight are possible. In this case Kruskal's algorithm may return different MST.

Good Read:

https://en.wikipedia.org/wiki/Kruskal%27s_algorithm#Proof_of_correctness

- All categories
- General Aptitude 1.9k
- Engineering Mathematics 7.6k
- Digital Logic 2.9k
- Programming and DS 4.9k
- Algorithms 4.4k
- Theory of Computation 6.2k
- Compiler Design 2.1k
- Databases 4.1k
- CO and Architecture 3.4k
- Computer Networks 4.2k
- Non GATE 1.4k
- Others 1.5k
- Admissions 595
- Exam Queries 573
- Tier 1 Placement Questions 23
- Job Queries 72
- Projects 18

50,834 questions

57,838 answers

199,510 comments

108,350 users