How exactly does a Max 2 Sat reduce to a 3 Sat?

Question

I've been reading this article which tries and explains how the max 2 sat problem is essentially a 3-sat problem and is NP-hard.

However, if you see the article, I'm not able to understand why , after ci is satisfied, 7 out of 10 clauses are satisfied and if it is not satisfied, the 6 out of 10 clauses are satisfied.

Can someone explain to me in simple terms, and demystify what exactly the article wants to convey? Essentially, I have come to know that a max-2-sat problem is the same as a 3 sat problem. The question is I'm not able to understand why.

More formally, I wish to solve this problem:

Consider the problem MAX2SAT described as follows.

Given a 2-CNF (Conjunctive Normal Form) Boolean expression (with m clauses, n variables) and an integer k, Decide if there is an assignment satisfying at least 'k' of the total clauses? Compute the complexity class (P or NP or NP Complete) of the MAX2SAT with justification.

Answer 1

Well, a key misunderstanding that we should talk about first is that this is not reducing Max-2-sat to 3-sat as your title says. In order to prove something (in this case Max-2-sat) is np-hard, we must to the opposite and reduce 3-sat (or any other known np-hard problem) to it.

If we do this, then we show that IF the thing that we reduced to (max-2-sat) isn't np-hard THEN 3-sat isn't either because we could just do our reduction and solve 3-sat efficiently via max-2-sat. We know this to be impossible (3-sat is known np-complete, someone else did the hard work) so the idea that max-2-sat isn't np-hard is a contradiction.

Reducing TO 3-sat doesn't tell us much. Max-2-sat could still be easy in that case. In that case, Maybe reducing to 3-sat was just a bad idea and it's easier to just tackle it directly. It is only doing the reverse that proves what you are trying to show