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Let $L$ be a language in the class $FP$ of all polynomial-time solvable problems.

The class $FP$ is defined by having a TM $M$ s.t. for any $x$ it computes in polynomial time a $y$ s.t. $(x,y)\in L$. Note several $y$s may occur, the TM is required to return one, if one exists. It doesn't matter which one.

Now lets assume that for a certain $x_0$ we have both $y_a,y_b$. What if we define:

$$L'=L\backslash \{(x_0,y_a)\}$$

We remove one of these solutions (There may be more than two; at least two). Could we prove/disorove that $L'$ is in $FP$?

I am having a hard time here as $M$ the TM acts like a black box. I don't see how can I alter its way of work to ignore $y_a$ and therefore go to $y_b$.

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  • $\begingroup$ "for a certain $x_0$" Do you mean for all $x_0$? Otherwise, we can just define $M'$ which returns $y_b$ if the input is $x_0$ and otherwise runs $M$. $\endgroup$
    – JiK
    Commented May 27 at 13:36

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Suppose $L = \{(G_x, Y) \mid Y$ is the maximum or minimum vertex cover of $G_x\}$. This satisfies your requirements. Now let for $G_{x_0}$, we discard $Y_a$ as the maximum vertex cover (which is all vertices in it), and we are thus left with $Y_b$ the minimum vertex cover. Now as we know minimum vertex cover problem can not be solved in polytime unless $P = NP$, your TM can not compute $Y_b$ in polytime.

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