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?

Answer 1

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 $x \in L$.