6,438 views

If all the edge weights of an undirected graph are positive, then any subset of edges that connects all the vertices and has minimum total weight is a

1. Hamiltonian cycle
2. grid
3. hypercube
4. tree

### 1 comment

MST → TREE

1. Hamiltonian cycle $\Rightarrow$ This is a cycle. A cycle will not only connect all vertices, it will have $1$ extra edge than necessary. So I can just remove that edge & get better cost "subset of edges" which connect all vertices. So, this is FALSE.
2. grid $\Rightarrow$ A grid graph has cycles and so this is FALSE for same reason as option A.
3. Hypercube $\Rightarrow$ A hypercube graph also has cycles. So, this also is FALSE.
4. Tree $\Rightarrow$ This is answer. We need to have Minimum spanning tree to be exact.
"If all the edge weights of an undirected graph are positive, then any subset of edges that connects all the vertices and has minimum total weight is a Minimum Spanning Tree". !

(D) is TRUE.

I agree with the answer but only for this question...

coz "tree" in general does not touch all the vertices..ri8 ? Tree is a subgraph which has no cycle..not necessarily contains "ALL" vertices..

@Nitesh here we are asked about a subset of EDGES and not Vertices. So to have all vertices and subset of edges so that weight is minimum, we definitely need to have a tree, an MST to be precise.

Target : Get a subset of edges such that weight is minimum and it connects all vertices.

Since, all weights are positive and we need a subset with minimum weight, then we should avoid cycles, coz we should have only a single path between any pair of vertices in the graph. Having a cycle will unnecessarily add up edge weight and will do no help in accomplishing the target.

This rules out all options A, B and C.

Hence, answer = option D

Tree is a minimally connected Graph.

So, Option (D) is Correct.

I think it is (D)

### 1 comment

is Hamiltonian Cycle not Possible?why?