I dont think this is true, take the language with a single element L = {0}, then the complement L' would have infinitely many strings (everything that isn't 0). So if it were true that A is mapping reducible to its complement, everything in L' can be obtained via a computable function from some element in L, i.e 010 is in L' however there is no x in L that gives f(x) = 101, since theres only one element, 0, it maps to one result.

Would this be correct?

  • $\begingroup$ Here is a reduction from $L$ to its complement: map $0$ to $1$ and everything else to $0$. $\endgroup$ Jun 13, 2017 at 5:46

1 Answer 1


Let $L_1,L_2$ be two languages over $\{0,1\}$. A (computable) mapping reduction from $L_1$ to $L_2$ is a computable function $f\colon \{0,1\}^* \to \{0,1\}^*$ such that for all $x \in \{0,1\}^*$ it holds that $x \in L_1$ iff $f(x) \in L_2$.

In your example, $L_1 = \{0\}$ and $L_2 = \overline{\{0\}}$, there does exist a mapping reduction, given by $$ f(x) = \begin{cases} 1 & \text{if }x = 0, \\ 0 & \text{if }x \neq 0.\end{cases} $$ Clearly $f$ is computable, and you can check that $x \in L_1$ iff $f(x) \in L_2$.

Hence your example doesn't work. You can check, however, that $\emptyset$ and $\{0,1\}^*$ are two examples of languages not reducible to their complement. Indeed, these are the only examples, which you can try to prove.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.