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Consider the following undirected graph with some edge costs missing.

Suppose the wavy edges form a Minimum Cost Spanning Tree for $G$. Then, which of the following inequalities NEED NOT hold?

  1. cost$(a, b) \geq 6$.
  2. cost$(b, e) \geq 5$.
  3. cost$(e, f) \geq 5$.
  4. cost$(a, d) \geq 4$.
  5. cost$(b, c) \geq 4$.
asked in Algorithms by Veteran (29.2k points)   | 362 views

2 Answers

+6 votes
Best answer

Now check this diagram, this is forest obtained from above given graph using Kruskal's algorithm for MST.

So according to the question edge $d-e$ has weight 5 and it is included in the formation of MST. Now if edges $b-e$ and $e-f$ has weight greater than 5 than it is not a problem for our MST because still we will get the given tree as Kruskal's algorithm takes the smallest weighted edge without forming a cycle.

Cost of edge $b-c \geq 4$ may also lead us to the same tree as above though Kruskal's algorithm will have choice between $c-f$ and $b-c$.

Now if the edge weight of $a-d$ becomes 4, it is guaranteed that Kruskal's algorithm will not select edge $d-e$  because its edge cost is 5, and hence the tree structure will change. But there can be the case where edge weight is greater than 4 and we still get the same tree (happens when $a-d \geq 5$). Because in the question they asked to point out an unnecessary condition this case is not the answer as we need $a-d \geq 5$ which implies $a-d \geq 4.$

Now notice option A. Put $a-b = 5.$ The given MST would not change.  So, this condition is not always necessary and hence is the answer..

 

Therefore option A is the answer .

 

answered by Boss (7k points)  
selected by
But the question is not "will not" but rather "need not".
How will it make difference @Arjun sir ? I just explained bit more in the last sentence .
a-d cannot be 4- your reason is correct. But that does not make D choice the answer rt?
sir in that way option A having wt >=6 is not necessary , so it need not be there . is that what you mean ?
yes - because thats what is asked in question :) For objective exam you will get negative even if your method is correct and final answer wrong :)
edited now !! Thanks a ton . Each and every word zaroori hota hain :P :)
why we would put a-b = 5?According to option A it should be >=6

There are ONLY two reasons, that Kruskal do not include any edge in MST-

  • either that edge forms cycle or
  • there is an equivalent edge weight and Kruskal choose to pick another one.


Tightest lower bound on every edge is -
$cost(b, c) \geq 4$ and every other edge cost $\geq$ 5.

+2 votes
I think answer should be D, bcoz if cost(a,d) =4, then it will come in MST( Min spanning tree). So this condition need not necessary hold.
answered by Junior (669 points)  
I think $cost(a, d) \geq 4$ is a MUST NOT condition, otherwise it will change the cost of MST when $cost(a, d) = 4$, whereas $cost(a, b) \geq 6$ is a NEED NOT condition. Along the same lines, remaining options are MUST conditions.

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