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Let $G = (V,E)$ be an undirected connected simple (i.e., no parallel edges or self-loops) graph with the weight function $w: E \rightarrow \mathbb{R}$ on its edge set. Let $w(e_{1}) < w(e_{2}) < · · · < w(e_{m})$, where $E = \left\{e_{1}, e_{2}, . . . , e_{m}\right\}$. Suppose $T$ is a minimum spanning tree of $G$. Which of the following statements is FALSE?

  1. The tree $T$ has to contain the edge $e_{1}$.
  2. The tree $T$ has to contain the edge $e_{2}$.
  3. The minimum weight edge incident on each vertex has to be present in $T$.
  4. $T$ is the unique minimum spanning tree in $G$.
  5. If we replace each edge weight $w_{i} = w(e_{i})$ by its square $w^{2}_{i}$ , then $T$ must still be a minimum spanning tree of this new instance.
asked in Algorithms by Veteran (39.7k points) 253 1302 1929 | 421 views

Point to be noticed --> 'Edges weights belongs to real number'

1 Answer

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Answer is E .  The catch here is Edges weights belongs to real number . Therefore  edge weight can be negative . In that case the minimum spanning tree may be different .



(Here every edge weight is distinct, therefore MST is unique. You do it using any algo.)

Option A is True. If we apply kruskal's algorithm then it will choose $e_1$

Option B is True. If we apply kruskal's algorithm then it will also choose $e_2$, and 2 edges can not forms a cycle. ($e_3$ is not guaranteed in MST, as it may form cycle.)

Option C is also true. If we apply prims also on any vertex (say u) then it chooses minimum

weight edge incident on vertex u.

Option D is true. Because every edge weight is distinct. 

answered by Boss (7.1k points) 7 29 78
edited by
whats the meaning of c) option
if their is min cost edge from every vertex  then it always be in min cost spanning tree.

each vertex connected via one min cost edge.
ok , suppose there is pentagon and weight is in this way , that if consider min weighted edge then there will be cycle so i will left it ... so vertex having min edge but i did not consider it ...

so whats the actual meaning ...??

read definition all edge weight are distinct .


node 1 connect with 2 edge with edge weight 1,2,3 then choose 1 weight  edge

node 2 connect with 2 edge with edge weight 2,3 ,4 then choose 2 weight  edge

i considered as distinct .... by the way i got what are you saying ....thanks now clear... :)


If the min cost edge of any vertex forms a cycle then we don't consider it in MST. So option c is not true always.

I agree with option E

Here edge weights are distinct and using prims algo you can't skip min cost edge in a cycle....iinstead you will remove higher cost edges
Yes got it... Thnks

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