Tag Archives: Monotone Sat

Non-Circular Satisfiability

The next reduction uses a new variant of Satisfiability, so I figured it would be worth making a separate post for it.

The problem: Non-Circular Satisfiability.  This problem is not in the appendix.

The description: A CNF-Satisfiability clause is mixed if it contains both variables and their negations.  A Satisfiability formula is non-circular if each variable occurs in a mixed clause at most once.  Given a non-circular satisfiability formula, is it satisfiable?

Example: This is actually a more general version of Monotone Satisfiability– in Monotone Sat, no mixed clauses can exist.  So all Monotone formulas are also Non-Circular.

Notice that we’re not necessarily restricting things to 3Sat clauses- we can have more or less than 3 variables per clause.  So here is a (satisfiable) instance:

F = (x1 ∨ x2) ∧ (~x1 ∨ ~x2) ∧ (~x1 ∨ x2)

This formula has just one mixed clause (the third one) and is satisfiable if x1 is false and x2 is true.  The formula becomes unsatisfiable if we add the clause (x1 ∨ ~x2), but then we would have both variables be in a second mixed clause.

Reduction: The easy way is to just do this by restriction from Montone Sat.  (Given a monotone sat formula, it’s automatically non-circular, so we’re done).  But if we want an actual reduction, we can use the paper by Papadimitriou, Bernstein, and Rothnie which has the next two reductions in G&J and uses this problem for the first one.  They also show the reduction for this problem in their paper and reduce from regular CNF-SAT.  So we have a formula F.  For each variable x in F, suppose x appears m times in F. Create m new variables x1 through xm.  The first instance of x in F will be replaced by x1, the second by ~x2, the third by x3, and so on down.  We then need to basically add the rules x1 ≡~x2 ≡ x3 …  (forcing each new literal we replaced x with to all have the same truth values).  in CNF form, that’s equivalent to the clauses: (x1 ∨ x2) ∧ (~x2 ∨ ~x3) ∧ (x3 ∨ x4) .. and so on down.  This means our new formula is equivalent to F (and is satisfiable when F is).  Each of these new clauses is non-mixed.  The only other occurrence of each xi variable is the one time it replaces x in the original F, which may or may not be a mixed clause, but either way, that means each variable appears in at most one mixed clause.

Difficulty: 3.  This is about as hard as the Monotone Sat reduction.

Monotone EQSat

Today I’m posting 2 problems relating to my student Dan Thornton’s independent study work.  I’m out of town this week and next, so I have these posts set to automatically go up on Tuesday afternoon.

Dan’s independent study was based on the “Automaton Identification” problem, but to show that reduction, he needs to use a variant of 3SAT, which he shows here:

The problem: Monotone EQ SAT. This is a specific instance of Monotone SAT.

The description:
We are given a conjunction of clauses F = \wedge_{i=1}^{l} C_{i} where each clause in F contains all negated or non-negated variables z_{j} and the number of clauses and variables are equal, is there an assignment of the variables so that F is satisfied? Our instance will have l clauses and variables.

Example:
Here is an F that has 4 variables and 4 clauses.

F =( \neg x_{1} \vee \neg x_{2} \vee \neg x_{3} \vee \neg x_{4} ) \wedge ( x_{1} \vee x_{2} \vee x_{4} ) \wedge ( \neg x_{2} \vee \neg x_{3} \vee \neg x_{4} ) \wedge ( x_{1} )

The above F may be satisfied by the following assignment:
x_{1} \rightarrow True
x_{2} \rightarrow False
x_{3} \rightarrow False
x_{4} \rightarrow True

The reduction:
We will reduce from Monotone SAT. So we are given an instance of Monotone  SAT with the clauses F = \wedge_{i=1}^{n} C_{i} here each clause is of the form C_{i} = (x_{i1} \vee ... \vee x_{ik_i} ) where each clause has all negated or non-negated variables. This is different from Monotone EQ SAT as we do not require the number of variables and clauses to be equal.
From this F we must build an instance of Monotone EQ SAT.
We may transform our instance of Monotone SAT, F, into one of Monotone EQ SAT by the following iterative procedure. New variables will be denoted by z'_{j} and new clauses by C'_{i}.

 F' = F;
 i = 1;
 j = 1;
 While{number of clauses != number of variables}{
   introduce two new variables z'_{j} \ , \ z'_{j+1};
   If{number of variables < number of clauses}{
     Create the new clause C'_{i} = (z'_{j} \vee z'_{j+1});
     F' = F' \wedge C'_{i};
     i = i+1;
   }
   else {
     Create three new clauses:
      C'_{i} = z'_{j} \ ,
      C'_{i+1} = z'_{j+1} \ ,
      C'_{i + 2} = (z'_{j} \vee z'_{j+1} );
     F' = F' \wedge C'_{i} \wedge C'_{i+1} \wedge C'_{i+2};
     i = i + 3;
   }
 j = j+2;
 }

The above algorithm will produce an equation F' that is in Monotone EQ  SAT. This may be shown by induction. Notice that before the procedure if number \ of \ variables > number \ of \ clauses that we will add 2 new variables and 3 new clauses.

If number \ of \ variables < number \ of \ clauses then we will add 2 new variables but only a single new clause. Either way the difference between the number of variables and clauses, dif =| number \ of \ clauses \ - \ number \ of \ variables | will decrease by 1. So in O(dif) steps we will obtain an formula where dif = 0. Such an formula is an instance of Monotone EQ SAT.

True{Monotone SAT \Rightarrow True{Monotone EQ SAT}
Here we assume that there is a truth assignment function \theta: z_{j} \rightarrow \lbrace True,False \rbrace that maps every variable to a truth value, such that F is satisfied. Then after we preform the above algorithm we have an instance of F', now our instance of F' will be of the form F \wedge ( \wedge_{i=1}^{k} C'_{i}) for some k \geq 1. Now notice that \theta above will satisfy F in F' and we may trivially satisfy \wedge_{i=1}^{k} C'_{i} by simply assigning all new variables z'_{j} to true.
This will give us a new truth assignment function \theta' that will satisfy F'

True{Monotone EQ SAT} \Rightarrow True{Monotone SAT}
Here we assume that there is a truth assignment function theta' that will satisfy F' then obviously as F' = F \wedge ( \wedge_{i=1}^{k} C'_{i}) then theta' must also satisfy F.

(Back to me)

Difficulty: 3.  The logical manipulations aren’t hard, but it is possible to mess them up.  For example, it’s important that the algorithm above reduces the difference in variables and clauses by 1 each iteration.  If it can reduce by more, you run the risk of skipping over the EQ state.

Monotone 3-Satisfiability

I told Daniel when he gave me his Monotone Satisfiability reduction that the actual problem mentioned in G&J was Monotone 3-Satisfiability.  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 Monotone  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 Monotone  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

This semester I’m doing an independent study with a student, Daniel Thornton, looking at NP-Complete problems.  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.  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} and  \ 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.  I like that the reduction involves manipulating the formula, instead of applying logical identities.