Are NP-Complete languages closed under any regular operations?

Question

I have tried looking online, but I couldn't find any definitive statements. It would make sense to me that Union and Intersection of two NPC languages would produce a language not necessarily in NPC. Is it also true that NPC languages are not closed under the complement, concatenation, and kleene star operations?

Answer 1

For all of the examples in this answer, I'm taking the alphabet to be $\{0,1\}$. Note that the languages $\emptyset$ and $\{0,1\}^*$ are definitely not NP-complete.

• The class of NP-complete languages is not closed under intersection. For any NP-complete language $L$, let $L_0 = \{0w\mid w\in L\}$ and $L_1 = \{1w\mid w\in L\}$. $L_0$ and $L_1$ are both NP-complete but $L_0\cap L_1 = \emptyset$.

• The class of NP-complete languages is not closed under union. Given the NP-complete languages $L_0$ and $L_1$ from the previous part, let $L'_0 = L_0 \cup \{1w\mid w\in \{0,1\}^*\}\cup\{\varepsilon\}$ and $L'_1 = L_1\cup \{0w\mid w\in \{0,1\}^*\}\cup\{\varepsilon\}$. $L'_0$ and $L'_1$ are both NP-complete but $L'_0\cup L'_1 = \{0,1\}^*\!$.

• The class of NP-complete languages is not closed under concatenation. Consider the NP-complete languages $L'_0$ and $L'_1$ from the previous part. Since both languages contain $\varepsilon$, we have $L'_0L'_1 \supseteq L'_0\cup L'_1 = \{0,1\}^*\!$.

• The class of NP-complete languages is not closed under Kleene star. For any NP-complete language $L$, $L\cup \{0,1\}$ is NP-complete but $\big(L\cup \{0,1\}\big)^* = \{0,1\}^*\!$.

• If the class of NP-complete problems is closed under complementation, then NP = coNP. Whether this is true or not is one of the major open problems in complexity theory.

Answer 2

Take a look at the proofs for union, intersection, concatenation, and kleene star of NP languages, here. Its seems like a similar argument could be made for NP-Complete languages.

For notation let

• $A$ be a oracle that decides a known NP-Complete problem like 3-SAT. See the definition of turing reducible
• $L_1$ and $L_2$ are NP-Complete languages
• $M_1$ and $M_2$ are Turing machines that decide $L_1$ and $L_2$ using $A$.
• $L_3$ is $L_1 \cup L_2$
• $M_3$ is a turing machine that decides $L_3$

In the case of union from 1, we can create a new machine $M_3$ that decides $L_3$ by calling $M_1$ and $M_2$ as sub routines. In turn, each time $M_1$ or $M_2$ is called, $A$ is also called. So $M_3$ decides $L_3$ using $A$. By the argument from 1, the running time of $M_3$ is in P and since it uses $A$ as a subroutine, $L_3$ is NP-Complete. In other words, $L_3$ is NP-Complete for the same reason that $L_1$ and $L_2$ are NP-Complete.

The same argument can be made intersection and it looks like similar arguments could be made for concatenation, and kleene star.

In the case of compliment, it seems likely to be difficult to prove for the same reasons is difficult to prove compliment in NP.