The problems in the P class can be polynomially reduced to its complement and vica versa?

Question

I considered the Euler circle problem to decide this. The polynomial reduction is: I add a new vertex in the graph:

If the degree of each vertex is even, then I connect all the vertices with this new vertex
Otherwise, I only connect the vertices with an odd number of degrees to the new vertex Is this a correct reduction?

Can you clarify the question? Are you looking for a reduction from the problem of deciding whether a graph contains an Eulerian cycle to its complement, or are you interested in knowing whether all problems in NP can be reduced to their respective complements? — Steven, Jan 28 '24 at 09:30
I try to prove that every P problems can be reduced to its complements and the opposite direction is also true. For this purpose I used the Euler circuit problem. — Andrew19, Jan 28 '24 at 09:37
Does this answer your question? Does two languages being in P imply reduction to each other? — Bader Abu Radi, Jan 28 '24 at 10:31
Please, before asking a question, make sure that you search if it, or something very similar, already exists on the site. — Bader Abu Radi, Jan 28 '24 at 10:48

Steven · Accepted Answer · 2024-01-28T09:45:49.730

The claim is false. $\emptyset$ cannot be reduced to its complement $\Sigma^*$ (and vice-versa).

However, if you restrict yourself to languages $L$ such that $L \not\in \{ \emptyset, \Sigma^*\}$ then it is true that $L \le \overline{L}$. To see this let $n$ be any fixed element of $L$ and let $y$ be any fixed element of $\overline{L}$. Such elements exist by our choice of $L$.

The reduction is given by: $$ f(x) = \begin{cases} y & \mbox{if } x \in L \\ n & \mbox{if } x \not\in L \end{cases}. $$ Notice that $f$ can be computed in polynomial time since $L \in \mathsf{P}$, which means that we can decide in polynomial time whether $x \in L$.

The problems in the P class can be polynomially reduced to its complement and vica versa?

1 Answers1