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In my logic class we started learning about the different complexity classes. In particular, we focused on the NP complexity class. A problem is in NP if it is solvable in polynomial time using a nondeterministic Turing machine. To show that a problem $L$ is NP-compelete, we have to show that: $1)$ $L$ is in NP, and $2)$ if $L$ is solvable in polynomial time, then all problems in NP can be solved in polynomial time.

In regards to $2)$ we can use reduction. So if you already know a NP-complete problem $M$, then it is enough to show that we can use a polynomial time algorithm that we can use for $L$, in order to solve $M$ in polynomial time.

I'm not sure I understand the explanation for the reduction section.

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  • $\begingroup$ So in simple terms your question is: if $L$ is NP-complete and $L\in P$, then why we can solve all problems in NP in polynomial time? $\endgroup$ – Bader Abu Radi Jan 9 at 22:51
  • $\begingroup$ @BaderAbuRadi In my script it says that "2) L is NP". And yes. That is what I would like to know along with a clear explanation of what the explanation of reduction is trying to say. $\endgroup$ – Ski Mask Jan 9 at 22:53
  • $\begingroup$ The definition of NP-completeness you give is wrong, as it would include all NP-intermediary problems too. $\endgroup$ – Arno Feb 9 at 9:18
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A problem $L$ is $\text{NP}$-complete if $L$ is in $\text{NP}$, and $L$ is $\text{NP}$-hard (that is, $A\leq_p L$ for all $A\in \text{NP}$ ). Consider the following claims.

Claim 1: if $L$ is $\text{NP}$-complete and $L\in \text{P}$, then $\text{NP} \subseteq \text{P}$ (that is, all problems in $\text{NP}$ can be solved in deterministic polynomial time).

Claim 2: If $ A \leq_p B$, then $B\in \text{P} \to A \in \text{P}$.

What you asked about is claim 1. To begin with, its correctness follows from claim 2. Indeed, if $L$ is $\text{NP}$-complete, then for every $A\in \text{NP}$, it holds that $A\leq_p L$. Thus, if we assume that $L\in \text{P}$, then we get by claim 2 that $A\in \text{P}$, and therefore $\text{NP} \subseteq \text{P}$. So what is missing for you is the proof of claim 2, and a sketch of its proof goes as follows. If $M_f$ is a TM that computes a polynomial time reduction $f$ from $A$ to $B$, and $M_B$ is a TM that decides $B$ in polynomial time, then a TM $M_A$ that decides $A$ in polynomial time operates as follows. On input $x$, $M_A$ computes $y = M_f(x)$, then $M_A$ runs $M_B$ on $y$, and answers the same. What is left to show is that $M_A$ decides $A$ in polynomial time, and I leave that to you (use the fact that |y| is polynomial, and the fact that a composition of polynomials is a polynomial).

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