Consider two languages $L_1$ and $L_2$ each on the alphabet $\Sigma$. Let $f : \Sigma → \Sigma$ be a polynomial time computable bijection such that $(\forall x) [ x\in L_1$ iff $f(x) \in L_2]$. Further, let $f^{-1}$ be also polynomial time computable.
Which of the following CANNOT be true ?
(A) $L_1$ $\in P$ and $L_2$ is finite
(B) $L_1$ $\in NP$ and $L_2$ $\in P$
(C) $L_1$ is undecidable and $L_2$ is decidable
(D) $L_1$ is recursively enumerable and $L_2$ is recursive
GATECSESince, $f$ is a polynomial time computable bijection and $f^{-1}$ is also polynomial time computable, $L_1$ and $L_2$ should have the same complexity (isomorphic). This is because, given a problem for $L_1$, we can always do a polynomial time reduction to $L_2$ and vice verse. Hence, the answer is 'C', as in 'A', $L_1$ and $L_2$ can be finite, in 'B', $L_1$ and $L_2$ can be in $P$ and in 'D', $L_1$ and $L_2$ can be recursive. Only, in 'C' there is no intersection for $L_1$ and $L_2$, and hence it canʼt be true.
Alternatively, we can prove 'C' to be false as follows:
Given $L_2$ is decidable. Now, for a problem in $L_1$, we can have a $TM$, which takes an input x, calculates $f(x)$ in polynomial time, check $f(x)$ is in $L_2$ (this is decidable as $L_2$ is decidable), and if it is, then output yes and otherwise no. Thus $L_1$ must also be decidable.
Consider two languages $L_1$ and $L_2$ each on the alphabet $\Sigma$. Let $f : \Sigma → \Sigma$ be a polynomial time computable bijection such that $(\forall x) [ x\in L_1$ iff $f(x) \in L_2]$. Further, let $f^{-1}$ be also polynomial time computable.
Which of the following CANNOT be true ?
(A) $L_1$ $\in P$ and $L_2$ is finite
(B) $L_1$ $\in NP$ and $L_2$ $\in P$
(C) $L_1$ is undecidable and $L_2$ is decidable
(D) $L_1$ is recursively enumerable and $L_2$ is recursive
Since, $f$ is a polynomial time computable bijection and $f^{-1}$ is also polynomial time computable, $L_1$ and $L_2$ should have the same complexity (isomorphic). This is because, given a problem for $L_1$, we can always do a polynomial time reduction to $L_2$ and vice verse. Hence, the answer is 'C', as in 'A', $L_1$ and $L_2$ can be finite, in 'B', $L_1$ and $L_2$ can be in $P$ and in 'D', $L_1$ and $L_2$ can be recursive. Only, in 'C' there is no intersection for $L_1$ and $L_2$, and hence it canʼt be true.
Alternatively, we can prove 'C' to be false as follows:
Given $L_2$ is decidable. Now, for a problem in $L_1$, we can have a $TM$, which takes an input x, calculates $f(x)$ in polynomial time, check $f(x)$ is in $L_2$ (this is decidable as $L_2$ is decidable), and if it is, then output yes and otherwise no. Thus $L_1$ must also be decidable.