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I study complexity and computation independently. I have a problem that I can not solve.

That's the problem:

For the SAT problem, there is a version in which we receive as input phrase $\varphi$ in the form of CNF and we have to decide whether there is a placement in each of the closures $\varphi$ provides at least one literal and does not provide at least one literal. (This version is called NAE-SAT, where NAE is the acronym for Not All Equal).

for example:

  • $( x_1 \vee \overline{x_2} \vee x_3 )\wedge (\overline{x_1} \vee \overline{x_2} \vee x_3) \in NAE-SAT$ because for the placement $s(x_1)=T$ , $s(x_2)=T$ and $s(x_3)=T$ it holds that in each clause there is a literal about T and a literal about F, since it will show from the shape: $( T \vee F \vee T )\wedge (F \vee F \vee T )$
  • $( x_1 \vee x_2 \vee x_3 )\wedge (\overline{x_1} \vee x_2 \vee x_3) \wedge (\overline{x_1} \vee \overline{x_2} \vee x_3) \wedge (\overline{x_1} \vee x_2 \vee \overline{x_3}) \notin NAE-SAT$. because in every placement for the phrase there will be at least one clause in which all the literals will be provided or all will not be provided, contrary to the requirements of the language.

We would like to show a reduction $ SAT \leq _p NAE-SAT $. Given phrase $\varphi = c_1 \wedge c_2 \wedge \cdots \wedge c_m$ (from the Boolean variables $x_1 , x_2 , \cdots , x_n $) as an instance for the SAT problem, we will first create the new variables $ y_1 , y_2 , \cdots , y_m , z$. Now, for each clause $C_i = (l_{i,1} \vee \cdots \vee l_{i,k_i})$ (which has $k_i \geq 3$ literals) we construct the clauses: $D_{i,1} = (l_{i,1} \vee \cdots \vee l_{i,k_i-1} \vee y_i) $ and $D_{i,2} = ( \overline{y_i} \vee l_{i,k_i} \vee z)$ . Finally, we will define the whole phrase to be $f(\varphi ) = D_{1,1} \wedge D_{1,2} \wedge D_{2,1} \wedge D_{2,2} \wedge \cdots \wedge D_{m,1} \wedge D_{m,2}$.

The question has 3 sections, which are related to each other, so I can not ask each one separately

Section A

For the phrase $\varphi = ( \overline{x_1} \vee x_2 \vee \overline{x_3} \vee x_4 )\wedge (x_1 \vee x_2 \vee x_3)$, what is the phrase FP that will be constructed by the reduction?

  1. $f(\varphi ) = ( \overline{x_1} \vee x_2 \vee \overline{x_3} \vee y_1 )\wedge (\overline{y_1} \vee x_4 \vee z) \wedge (x_1 \vee x_2 \vee y_2) \wedge (\overline{y_2} \vee x_3 \vee z)$

  2. $ f(\varphi ) = ( \overline{y_1} \vee x_4 \vee z) \wedge (x_1 \vee \overline{x_3} \vee y_2) \wedge (y_1 \vee y_2 \vee y_3) \wedge (\overline{y_3} \vee \overline{x_3} \vee z) $

  3. none of the answers is correct.

I think the answer is 1, It's not complicated.

Section B

Is the reduction, as a function of the length of the phrase, polynomial?

  1. Yes and also linear
  2. Yes but it is not linear
  3. This cannot be determined
  4. No, since the number of new variables we add may increase exponentially depending on the number of original variables
  5. none of the answers is correct.

I think the answer is yes, so 3 and 4 are certainly incorrect. The reduction takes a phrase in SAT and at most doubles it 2. If in SAT there were m clauses, at the moment there will be 2m clauses. And if in SAT there were n variables, currently in NAE-SAT there will be n + m variables. So in my opinion the correct answer is 1, but I'm not sure here.

Section C Is the reduction correct and defined on each input?

  1. Yes and yes
  2. Although the reduction is correct, in the description we do not refer to clauses with less than 3 liters. Nevertheless, it can be easily adapted to the case of 2 literals, and in the case of a single literal is simply rejected, as it will not be possible to simultaneously supply and not supply it.
  3. The reduction is incorrect and is not defined on the entire input, which is why we do not handle clauses of even length
  4. Although the reduction is defined on the entire input, it is incorrect.
  5. none of the answers is correct.

I think the reduction is correct because it is polynomial. And I think it's also defined, because of every clause in SAT you can get 2 clauses in NAE-SAT, but I'm not sure about that. I stand out between answer 1 and answer 2

The question was translated from Hebrew. So I have no source for the question.

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  • $\begingroup$ I think A & B are correct, but for C I don't think so. You have to show that $\varphi\in SAT\iff f(\varphi)\in NAE-SAT$. I have my doubts about how correct is the direction $f(\varphi)\in NAE-SAT \implies \varphi \in SAT$ $\endgroup$
    – nir shahar
    Jul 23 at 13:07
  • $\begingroup$ You can still credit the original source, where you encountered the Hebrew version. Translating doesn't prevent you from crediting the original source. $\endgroup$
    – D.W.
    2 days ago
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+50
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As you correctly spotted, the reduction can be implemented in polynomial time, and the blowup in the formula size is indeed linear.

The reduction is also correct, with the caveat mentioned in item 2. It's not trivial to show, but we'll go at it slowly.

So first, let's assume we allow this construction also for 2 literals. As for 1 literal clauses -- we can assume those don't exist, since given a formula $\phi$, any 1-literal clause immediately forces a value of one of the variables $x$. So we can start by plugging in this value, and removing the clause and all occurrences of $x$ (more precisely - if $x$ was assigned True, we can remove all clauses containing $x$, and remove $\neg x$ from all other clauses, and vice-versa if $x$ was assigned False).

So now for the main reduction. We need to show two directions of correctness. First, let's assume $\phi\in SAT$. Then it has a satisfying assignment $\pi$.

We'll construct a NAE-satisfying assignment for $f(\phi)$ (note that it's called an "assignment", not "placement"). Consider some clause $C_i$ of $\phi$. Since $\pi\models \phi$, then at least one literal in $C_i$ is assigned True. If one of the literals in $D_{i,1}$ is assigned True, then we assign $y_i$ to False and $z$ to False, and now both clauses that result from $C_i$ are NAE-satisfied. Otherwise, if all literals in $D_{i,1}$ are assigned to False, then the remaining literal, which appears in $D_{i,2}$ is assigned to True, so by assigning $y_i$ to be True (and $z$ still False), we get again a NAE-satisfying assignment.

Note that we set $z$ to False globally, as it is a single variable.

So the first direction is correct. For the converse, we start with the following observation: if $\tau$ is a NAE-assignment, then the ``inverse'' assignment is also NAE. Thus, if $f(\phi)$ has a NAE-assignment, we can assume w.l.o.g. that $z$ is assigned to False (if not, we invert the assignment). From here we make a similar deduction to the above, to obtain a satisfying assignment for $\phi$. Try it and write if you get stuck.

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