Let $A$ and $B$ be finite alphabets and let $\#$ be a symbol outside both $A$ and $B$. Let $f$ be a total function from $A^*$ to $B^*$. We say $f$ is computable if there exists a Turing machine $M$ which given an input $x$ in $A^*$, always halts with $f(x)$ on its tape. Let $L_f$ denote the language $\{x\#f(x) \mid x \in A^*\}$. Which of the following statements is true:

We analyze the equivalence between $f$ being computable and $L_f$ being recursive (decidable):

1. If $f$ is computable $\implies L_f$ is recursive:

   • Given an input string $w$, a Turing machine first verifies if $w$ is of the form $x \# y$ where $x \in A^*$ and $y \in B^*$. If not, it rejects.

   • Since $f$ is a total computable function, the machine can compute $f(x)$ on a separate track/tape, which is guaranteed to halt.

   • It then compares $f(x)$ with $y$. If they are identical, it accepts; otherwise, it rejects.

   • Since this procedure halts on all inputs, $L_f$ is decidable (recursive).

2. If $L_f$ is recursive $\implies f$ is computable:

   • Given an input $x \in A^*$, we can compute $f(x)$ by enumerating all strings $y \in B^*$ in standard canonical/lexicographical order: $y_1, y_2, y_3, \dots$.

   • For each candidate $y_i$, we run the decider for $L_f$ on $x \# y_i$.

   • Because $f$ is a total function, there exists exactly one string $y = f(x)$, and the decider will accept in finite time for that $y$.

   • The machine then outputs this $y$ and halts. Thus, $f$ is computable.

Hence, $f$ is computable if and only if $L_f$ is recursive, which corresponds to option (A).