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Consider $\Sigma=\{0,1\}$. Suppose that $L \subset \Sigma^*$ is $NP-$Complete. How can I prove that $L' = L \cup \{0,1\}$ is $NP-$Complete?

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$L$ is $\mathrm{NP}$-complete, so, for any $X\in\mathrm{NP}$, there is a many-one reduction $f_X$ from $X$ to $L$. Just modify $f_X$ so that it's a reduction to $L'$ instead. Note that $f_X$ already does the right thing unless you have some $w$ such that $f_X(w)\in\{0,1\}$.

Note that the same technique shows that $L\cup S$ is still $\mathrm{NP}$-complete for any finite $S\subseteq \Sigma^*$.

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  • $\begingroup$ If $\omega \in X$ then $f_X(\omega) \in X$. However, if $\omega \notin X$ then $f_X(\omega)$ could be $0$ or $1$. I guess that you have to assign a word that 's not in $L \cup \{0,1\}$ but you have to find that word and $f'$ have to be polynomial. $\endgroup$ – pepito grillo May 1 '19 at 13:36
  • $\begingroup$ You don't have to find that word. The language $X$ is fixed, so the reduction doesn't need to compute these words: they can be "hard-coded" into its definition. Imagine that the reduction is performed by a computer program: you might have to do a lot of work to find those strings so you can write that program, but you only have to do that once and it doesn't affect its running time $\endgroup$ – David Richerby May 1 '19 at 13:37

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