Author Archives: Sean T. McCulloch

About Sean T. McCulloch

I'm an Associate Professor of Computer Science at Ohio Wesleyan University

Monotone 3-Satisfiability

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I told Daniel when he gave me his Monotone Satisfiability reduction that the actual problem mentioned in G&J was Monotone 3-Satisfiability.Ā  So he went off and did that reduction too.
The Problem:
Monotone 3 SAT. This is a more restrictive case of Monotone SAT

The Description:
Given an formula of clauses F' = \wedge_{i=1}^{n} C'_{i} where each clause in F' contains all negated or non-negated variables, and each clause C_{i} contains at most 3 variables. Does there exist an assignment of the variables so that F' is satisfied?

Example:

\\ F_{1} = (x_{1} \vee x_{3}) \wedge \\ (\neg x_{2} \vee \neg x_{3} \vee \neg x_{4}) \wedge \\ (x_{3} \vee x_{2} \vee x_{4}) \wedge \\ ( \neg x_{3} \vee \neg x_{5} \vee \neg x_{1})
the following assignment satisfies F'_{1}:
\\ x_{1} \mapsto True\\ x_{2} \mapsto False\\ x_{3} \mapsto True\\ x_{4} \mapsto True\\ x_{5} \mapsto False
However:
\\ F_{2} = (\neg x_{1} \vee \neg x_{2} \vee \neg x_{3}) \wedge \\ (x_{1} \vee \neg x_{2} \vee \neg x_{3}) \wedge\\ (\neg x_{1} \vee x_{2} \vee \neg x_{3})\wedge \\ (\neg x_{1} \vee \neg x_{2} \vee x_{3})\wedge\\ (x_{1} \vee x_{2} \vee \neg x_{3})\wedge\\ (\neg x_{1} \vee x_{2} \vee x_{3})\wedge\\ (x_{1} \vee \neg x_{2} \vee x_{3})\wedge\\ (x_{1} \vee x_{2} \vee x_{3})
And the following is F_{2}' in MonotoneĀ  3SAT form:
\\ F_{2}' = (\neg x_{1} \vee \neg x_{2} \vee \neg x_{3}) \wedge \\ (\neg y_{1} \vee \neg x_{2} \vee \neg x_{3}) \wedge\\ (\neg x_{1} \vee \neg y_{2} \vee \neg x_{3})\wedge \\ (\neg x_{1} \vee \neg x_{2} \vee \neg y_{3})\wedge \\ (x_{1} \vee x_{2} \vee y_{3})\wedge\\ (y_{1} \vee x_{2} \vee x_{3})\wedge\\ (x_{1} \vee y_{2} \vee x_{3})\wedge\\ (x_{1} \vee x_{2} \vee x_{3}) \wedge \\ (y_{1} \vee x_{1}) \wedge (\neg y_{1} \vee \neg x_{1}) \wedge \\ (y_{1} \vee x_{2}) \wedge (\neg y_{2} \vee \neg x_{2})\wedge \\ (y_{1} \vee x_{3}) \wedge (\neg y_{3} \vee \neg x_{3})
are both unsatisfiable.

The reduction:
In the following reduction we are given an instance of 3SAT,
F = \wedge_{i=1}^{n} C_{i}. Here each clause is of the form:
C_{i} = x_{i1} \vee ... \vee x_{ik_i} where
k_{i} < 4
and each x_{ik_i} is a literal of the form \neg z_{l} \ or \ z_{l} .
We use the following construction to build an instance of MonotoneĀ  3 SAT out of the above instance of 3SAT :
In each clause C_{i} we have at most one literal, z_{l} \ or \ \neg z_{l} that is not of the same parity as the rest of the literals in the clause. For every such literal, we may preform the following substitution:
z_{l} \rightarrow \neg y_{l} \ or \ \neg z_{l} \rightarrow y_{l} this yields a modified clause C'_{i}.
Now we must be able to guarantee that z_{l} and y_{l} are mapped to opposite truth values, so we introduce the new clause:
C''_{i} \ = \ ( z_{l} \vee y_{l}) \wedge ( \neg z_{l} \vee \neg y_{l}) and conjunct it onto our old formula F producing a new formula F'.

For example:
C_{i} \ = \ (z_{l_1} \vee z_{l_2} \vee \neg z_{l_3}) so we preform the substitution
\neg z_{l_3} \rightarrow y_{l_3}
so C'_{i} \ = \ (z_{l_1} \vee z_{l_2} \vee y_{l_3}) and C''_{i} \ = \ (z_{l_3} \vee y_{l_3}) \wedge ( \neg z_{l_3} \vee \neg y_{l_3})

Now repeating this procedure will result in a new formula: F' = (\wedge_{i=1}^{n} C'_{i}) \wedge (\wedge_{k=1}^{m} C''_{k}).
We claim logical equivalence between the C_{i} \wedge C''_{i} and C'_{i} \wedge C''_{i} This is semantically intuitive as the C''_{i} clause requires all substituted literal y_{l} in C'_{i} to take the value opposite of z_{l} this was the stipulation for the substitution initially. It is also verifiable by truth table construction for:
\\ (z_{l_1} \vee z_{l_2} \vee \neg z_{l_3}) \wedge (z_{l_3} \vee y_{l_3}) \wedge ( \neg z_{l_3} \vee \neg y_{l_3}) \Leftrightarrow \\ (z_{l_1} \vee z_{l_2} \vee y_{l_3}) \wedge (z_{l_3} \vee y_{l_3}) \wedge ( \neg z_{l_3} \vee \neg y_{l_3})

True_{3SAT} \Rightarrow True_{Monotone \ 3 \ SAT}:
If there exists a truth assignment \phi_{F} that satisfies F, then we may extent this truth assignment to produce \phi_{G} which will satisfy
G = F \wedge (\wedge_{k=1}^{m} C''_{k}) by letting \phi_{G} (z_{l}) = \phi_{F} (x_{l}) for all l and letting \phi_{G}(y_{l}) = \neg \phi_{F}(z_{l}) for all l.
Obviously if F is satisfiable G must be by the above construction of \phi_{G}. So by the above claim we have that \phi_{G} will satisfy F'.
True_{Monotone \ 3 \ SAT} \Rightarrow True_{3SAT}:
Continuing from the above, if we have a truth assignment \phi_{F'} that satisfies F', then by the claim above it also must satisfy G. And F is a sub-formula of G so any truth assignment that satisfies G must also satisfy F.

(Back to me)

Difficulty: 4, since it’s a little harder than the regular Monotone Sat one.

Monotone Satisfiability

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This semester I’m doing an independent study with a student, Daniel Thornton, looking at NP-Complete problems.Ā  He came up with a reduction for Monotone Satisfiability, and since I hadn’t gotten to that problem yet, I told him if he wrote it up, I’d post it.

So, here it is.Ā  Take it away, Daniel!

The Problem: Monotone SAT. This is mentioned in problem LO2 in the book.

The description:
Given an set of clauses F' = \wedge_{i=1}^{n} C_{i} where each clause in F contains all negated or non-negated variables, is there an assignment of the variables so that F' is satisfied?

Example:
F' = (x_{1} \vee x_{3} \vee x_{4}) \wedge (\neg x_{2} \vee \neg x_{3} \vee \neg x_{4}) \wedge (x_{3} \vee x_{2} \vee x_{4})
the following assignment satisfies F':
x_{1} \mapsto False
x_{2} \mapsto False
x_{3} \mapsto False
x_{4} \mapsto True

The reduction:
In the following reduction we are given an instance of SAT, with the clauses:
F = \wedge_{i=1}^{n} C_{i}. Here each clause is of the form C_{i} = x_{i1} \vee x_{i2} \vee ... \vee x_{ik_i}and each x_{ij} is a literal of the form \neg z_{ij} \ or \ z_{ij}
Now we build an instance of Monotone SAT from the instance of SAT given above:
For each C_{i} we construct two new clauses \\ C'_{2i} = z_{i l_1} \vee ... \vee z_{il_k } \vee z'_{i} andĀ  \ C'_{2i-1}= \neg z_{il_k+1} \vee ... \vee \neg z_{il_m} \vee \neg z'_{i}, such that all elements of C'_{2i} are non-negated literals and all terms in C'_{2i-1} are negated literals with the addition of the new special term z'_{i}. Now let us build a new formula F' = \wedge_{i'=1}^{2n} C'_{i'} this is our instance of Monotone SAT, clauses are either all non-negated or negated.

True_{Monotone SAT} \Rightarrow True_{SAT}:
Notice how we added the extra literal z'_{i} or \neg z'_{i} to each of the clauses C'_{2i} or C'_{2i-1} respectfully. Now if there is an assignment that satisfies all of the clauses of F' then as only C_{2i} or C_{2i-1} may be satisfied by the appended extra literal, one of the clauses must be satisfied by it’s other literals. These literals are also in C_{i} so such an assignment satisfies all C_{i} \in F.

True_{SAT} \Rightarrow True_{Monotone SAT}:
Using an argument similar to the one above, For F to be satisfied there must be at least one literal assignment say z_{iy} that satisfies each clause C_{i} Now z_{iy} is in either C'_{2i} or C'_{2i-1}. This implies that at least one of C'_{2i} or C'_{2i -1} is also satisfied by z_{iy}, so simply assign the new term z'_{i} accordingly to satisfy the clause in F' not satisfied by z_{iy}

(back to me again)

Difficulty: 3.Ā  I like that the reduction involves manipulating the formula, instead of applying logical identities.