Np class of problems

Question

Are All problems in NP are known to be reducible to one another. I know if a problem X is in NP and any NP problem Y in NP is reducible to X then X is NP-complete. So by this assumption can we state that all NP problems are reducible to one another?

My question is if all NP problems are reducible to each other? — user10458501, Mar 15 '19 at 22:03
The answer is no. You can reduce an "easier" problem to a more "difficult" one but not necessarily the other way round. — Henry, Mar 16 '19 at 05:37

score 2 · Accepted Answer · answered Mar 16 '19 at 03:50

2

A decision problem C is NP-complete if:

C is in NP, and
Every problem in NP is reducible to C in polynomial time.

Source: https://en.wikipedia.org/wiki/NP-completeness

If all NP problems are reducible to one another, it would mean that all NP problems are NP complete, which we can not say since we still can't prove whether P = NP Refer to the image below for a better understanding.

answered Mar 16 '19 at 03:50

hsnsd

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NP=NP complete is fine but how is P=NP=NPComlete in this case ? – user10458501 Mar 16 '19 at 04:34
whether P=NP is an open problem still – hsnsd Mar 16 '19 at 05:34

score 0 · Answer 2 · answered Apr 23 '19 at 02:02

No, not all problems in NP are reducible to one another. Something that is in NP is different from something that is NP-hard. If something is in NP and is NP-hard, it is known as NP-complete. To be in NP, a problem needs to be able to be verified in polynomial time. An NP-hard problem is one that is accepted to be impossible to solve efficiently under know algorithms. To be NP-hard, a problem needs to be able to be reduced to another problem that is NP-hard. If both X and Y are in NP (like your example) then and Y is reducible to X, this does not mean that either Y or X are NP-hard because neither is reduced into an existing NP-hard problem.

Np class of problems

2 Answers2