3

Assume that Problem $A$ is polynomial-time reducible to problem $B$.

Claim 1: If problem $A$ is NP-hard then problem $B$ is NP-hard.

Claim 2: If problem $B$ is NP-hard then problem $A$ is NP-hard.

Claim 1 is obviously true, but why the second one is false?

Bader Abu Radi
  • 2,901
  • 1
  • 9
  • 27

1 Answers1

1

I am assuming that $A$ is a non-trivial problem. If $A$ is trivial, then the claims cannot be correct in their current form (make sure you understand why).

Consider an NP-complete problem $B$, and consider some non-trivial problem $A \in \text{P}$. Clearly, $A\leq_p B$. If we assume that claim 2 is correct, then as $A\leq_p B$, we get that $A$ is $\text{NP}$-hard and thus for every problem $L\in \text{NP}$, it holds that $ L\leq_p A$. Hence, as $A\in \text{P}$, we get that $L\in \text{P}$ as well, and so $\text{NP}\subseteq \text{P}$. Conversely, if we assume that $\text{NP}\subseteq \text{P}$, then as every non-trivial problem is $\text{P}$-hard, we get that every non-trivial problem $A$ is $\text{NP}$-hard. Hence, claim 2 holds trivially.

So we have in total that claim 2 is correct iff $\text{P} = \text{NP}$ which is quite open.

Bader Abu Radi
  • 2,901
  • 1
  • 9
  • 27
  • Yes, and the answer formalizes this intuition. As you know nothing about $A$ and its relation with problems in NP, then claim 2 actually suggests that every non-trivial language is NP-hard, which is the same as suggesting that P=NP. – Bader Abu Radi Jan 08 '21 at 23:57
  • Yes, NP-complete problems are NP-hard problems in NP. – Bader Abu Radi Jan 09 '21 at 00:05
  • We proved in the answer that 2 is correct iff P = NP. The P vs NP is an open problem. So the answer to your question, it is unknown whether calim 2 is correct, and knowing the status of claim 2 solves a long standing open problem. Yet, most experts believe that $\text{P} \neq \text{NP}$. Under the assumption that $\text{P} \neq \text{NP}$, the answer proves that 2 is not correct (B in the above answer can be any classic NP-complete problem). – Bader Abu Radi Jan 09 '21 at 00:21
  • Yes, the answer shows that claim 2 is equivalent to the P vs NP question. – Bader Abu Radi Jan 09 '21 at 00:24