3-SAT to Max-2-SAT Reduction I'm trying to find reduction from 3-SAT to Max-2-SAT, so far no luck.
Let me first describe it.

3-SAT: Given a CNF formula $\varphi$, where every clause in $\varphi$ has exactly 3 literals in it, one should determine if there exist an assignment that satisfies it.
Max-2-SAT: Given a CNF formula, where every clause in $\phi$ has exactly 2 literals in it, and a positive number $k$, one should determine if there exist an assignment that satisfies at least $k$ clauses.

Let me first show what I have tried so far.
Given $\varphi=\wedge _{i=1}^{n}C_i$ where: $C_i=(l_{i_1}\vee l_{i_2} \vee l_{i_3})$,
I set: $\phi=\wedge _{j=1}^{3n}D_i$, where: $D_i=(l_{i_1}\vee l_{i_2})\wedge(l_{i_1}\vee l_{i_3})\wedge(l_{i_2}\vee l_{i_3})$ and $k=2n$.
It's quite easy to see that this will not work...
Although, if there exist an assignment which satisfies $\varphi$ it means there exist an assignment that satisfies $k=2n$ clauses in $\phi$, the second direction is not true.
I found several reductions online (such as this, for example), but none of them were useful since in my problem, each clause in $\phi$ must have exactly two literals, where in the link above, the formula can also contain 1-length clause in it.
I could really use some help here.
Hint: Replace a singleton clause $a$ with $a \lor b$ and many copies of $\lnot b \lor c$, $\lnot b \lor \lnot c$.



Reductions without extra variables don't work even when singleton clauses are allowed. There are five types of clauses: $\ell_1,\lnot \ell_1, \ell_1 \lor \ell_2$, $\ell_1 \lor \lnot \ell_2$, $\lnot \ell_1 \lor \lnot \ell_2$. Suppose that $k$ of $\ell_1,\ell_2,\ell_2$ are true. The probability that a random clause of each type is true is:
$$
\begin{array}{c|cccc}
k&amp;0&amp;1&amp;2&amp;3\\\hline
\ell_1 &amp; 0 &amp; 1/3 &amp; 2/3 &amp; 1 \\
\lnot \ell_1 &amp; 1 &amp; 2/3 &amp; 1/3 &amp; 0 \\
\ell_1 \lor \ell_2 &amp; 0 &amp; 2/3 &amp; 1 &amp; 1\\
\ell_1 \lor \lnot \ell_2 &amp; 1 &amp; 1/3 &amp; 1/3 &amp; 1\\
\lnot \ell_1 \lor \lnot \ell_2 &amp; 1 &amp; 1 &amp; 2/3 &amp; 0
\end{array}
$$
We would like to take a convex combination of these rows which is of the form
$$ \begin{array}{cccc} \alpha &amp; \beta &amp; \beta &amp; \beta \end{array} $$
However, all such convex combinations have $\alpha = \beta$.