Select All Statement That are True The Vertex-Cover problem can be reduced in polynomial time to...

Select All Statement That are True

The Vertex-Cover problem can be reduced in polynomial time to the Set-Cover problem: The implication of this reduction is that if we are ever able to solve the Vertex Cover problem in polynomial time, we would also be able to solve the Set Cover problem In polynomial time.
If problem Xis NP-Hard, then it follows that X is necessarily not in NP.
Suppose X <p Y. If the size of the instance of X is n, then we can transform an instance of X into and instance of Yin time 0(n5) for some constant k.
if both SET-COVER and VERTEX-COVER are NP-complete problems. This implies that SET-COVER <p VERTEX-COVER.
Given decision problems X and Y, we define a new type of reduction X <E Y, where <= follows the same definition as seen in lecture for the polynomial time reduction but for the fact that we allow time exponential in the size of the instance of X for this instance to be transformed into an instance of Y. Then given X <E Y, If we were ever able to solve Yin polynomial time, it would imply we can solve X in polynomial time.

Solutions

Expert Solution

Among the given options, first, fourth and fifth statements are true.

The correct statements are:

(i) The Vertex-Cover problem can be reduced in polynomial time to the Set-Cover problem: The implication of this reduction is that if we are ever able to solve the Vertex Cover problem in polynomial time, we would also be able to solve the Set Cover problem In polynomial time.

(ii) If both SET-COVER and VERTEX-COVER are NP-complete problems. This implies that SET-COVER <p VERTEX-COVER.

(iii) Given decision problems X and Y, we define a new type of reduction X <E Y, where <= follows the same definition as seen in lecture for the polynomial time reduction but for the fact that we allow time exponential in the size of the instance of X for this instance to be transformed into an instance of Y. Then given X <E Y, If we were ever able to solve Yin polynomial time, it would imply we can solve X in polynomial time.

Second option is incorrect as if problem X is NP-Hard, then it follows that X ay or may not be in NP. So, we can't comment on that.

Third option is incorrect for X <p Y. If the size of the instance of X is n, then we can't transform an instance of X into an instance of Yin time O(n⁵) for some constant k.

Please comment in case of any doubt.
Please upvote if this helps.

venereology answered 1 year ago

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