I am reading the reduction given by Sipser in his textbook \"Introduction to the
ID: 652890 • Letter: I
Question
I am reading the reduction given by Sipser in his textbook "Introduction to the Theory of Computation," on page 303. The reduction is:
egin{equation} 3SAT leq_p KCLIQUE end{equation}
I am really trying to understand everything formally -- putting everything in a strict logical notation helps me learn Math. To clarify, the content of this proof, has not helped me give other reductions because I don't understand one direction of the $iff$ in the logic of reductions.
In this reduction, $f$ must be s.t: egin{equation} win 3SAT iff f(w) in KCLIQUE end{equation} and $f$ computes within a polynomial number of steps of the input size. The polynomial part is easy for me to understand, so no problem here!
I see that the above logical statement is equivalent to: egin{equation} win 3SAT implies f(w) in KCLIQUE land w otin 3SAT implies f(w) otin KCLIQUEend{equation} The above just says yes-instances map to yes-instances and no-instances map to no-instances.
It appears that Sipser shows us: egin{equation} win 3SAT implies f(w) in KCLIQUE land f(w) in KCLIQUE implies win 3SATend{equation}
Which is also equivalent to the above by taking the contrapositive of the second implication.
Here is my understanding of the $implies$ direction. Given a yes-instance of $3SAT$, show that the reduction $f$ gives us a yes-instance for $KCLIQUE$. This seems completely natural.
I don't really understand the other direction -- namely, given a yes-instance of KCLIQUE we are supposed to show that we get a yes-instance of $3SAT$. However since the reduction goes from $3SAT$ to $KCLIQUE$ i.e. the domain is the language $3SAT$ and the Codomain is the language $KCLIQUE$, I don't understand how we show this.
It appears that the argument is; Our reduction has provided us this graph, from which we can create a satisfying assignment from?
Please help me understand the other direction, and thanks for your time.
Explanation / Answer
Your problem may be a slight misunderstanding.
Given a yes-instance of KCLIQUE we are supposed to show that we get a yes-instance of 3SAT.
This is almost correct; check the claimed equivalence again:
$qquaddisplaystyle w in mathrm{3SAT} iff f(w) in mathrm{KCLIQUE};. ag{1}$
For the "backwards" direction, you only need to assume a yes-instance from the intersection of the image of $f$ and KCLIQUE, i.e. show that
$qquaddisplaystyle orall w in operatorname{img}(f) cap mathrm{KCLIQUE}. f^{-1}(w) in mathrm{3SAT}$.
That means that you can assume some structure about the instances you need to prove the reverse direction for, namely that introduced by your reduction.
Another misunderstanding:
the domain is the language 3SAT and the Codomain is the language KCLIQUE
That's not true; note that (1) is (maybe implicitly) supposed to hold for all $w in Sigma^*$, that is the domain of $f$ is $Sigma^*$. All $w otin mathrm{3SAT}$ must therefore map to $f(w) otin mathrm{KCLIQUE}$ in order to fulfill (1), so the codomain also needs at least one value that is not a yes-instance of KCLIQUE.
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