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2. Consider the following problem: Input: graph G, integer k Question: is it possible to partition vertices of G into k disjoint independent sets? Is this problem polynomial or NP-complete? Explain your answer
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Answer #1

Given trouble is NP-complete.

To show any problem is thought to be NP-complete.

It has to create happy two conditions:

1) The clique difficulty is in NP

2) There exists an NP-complete reducible to a clique trouble.

1)consider a place with n number of rudiments has 2^n number of subsets.

graph with n amount of vertices has 2^n number of achievable subgraphs.for checking clique in each sub graph takes exponential occasion by deterministic mechanism.

That income it is not in P

2) reduce a 3SAT predicament to the NP form.A 3 SAT predicament has a solution ,if the resultant vertices form a faction.

That means we are capable to reduce the 3SAT crisis to clique problem, then clique problem is NP-complete.

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