I am following the Barak and Arora book. I have asked this question regarding reduction from $\texttt{CKT-SAT}$ to $\texttt{3SAT}.$
Is it possible to show that there exists a reduction that satisfies the condition of asked question (in particular, it is direct reduction) and holds for all: $x\in\{0, 1\}^*$
there is an assignment to $f(x)$ which satisfies all the clauses in $f(x)$, except for a fixed number of them (which does not depend on the length of $x$) and
if $x=\langle C_n \rangle\in \texttt{CKT-SAT}$ , then from any assignment $z$ that satisfies $f(x)$, it is possible to produce an input $y$ that satisfies $C_n(y) =1?$
Where: $$\text{CKT-SAT}= \left\{\langle C_n \rangle:\text{$C_n$ is a Boolean circuit and $\exists x\in \{0,1\}^n$ s.t.$ C_n(x)=1$}\right\}.$$
1 Answer 1
The standard reduction based on the Tseitin transform does exactly that.
Recall that the reduction works as follows: given a circuit $C(\vec x)$, you introduce new variables $y_i$ for each non-input gate $g_i$ in the circuit, and let $f(\langle C\rangle)$ be a collection of 3ドル$-clauses that force each $y_i$ to be correctly computed from the variables corresponding to the input gates of $g_i$, together with the clause $y_o$ (where $g_o$ is the output gate of $C$). Thus:
Any assignment $\vec a$ to the original $\vec x$ variables uniquely extends to an assignment to the $\vec y$ variables that satisfies all clauses of $f(\langle C\rangle)$ except possibly the $y_o$ clause: $y_i$ is assigned the value of gate $g_i$ when $C$ is evaluated with the assignment $\vec a$.
In particular, $f(C)\smallsetminus\{y_o\}$ is always satisfiable.
The restriction of an assignment satisfying $f(\langle C\rangle)$ to the $\vec x$ variables is a satisfying input of $C$.