# Monthly Archives: April 2016

# Protected: Minimum Edge-Cost Flow

# 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.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 where each clause in contains all negated or non-negated variables, and each clause contains at most variables. Does there exist an assignment of the variables so that is satisfied?

**Example:**

the following assignment satisfies :

However:

And the following is in MonotoneĀ 3SAT form:

are both unsatisfiable.

**The reduction:**

In the following reduction we are given an instance of 3SAT,

. Here each clause is of the form:

where

and each is a literal of the form .

We use the following construction to build an instance of MonotoneĀ 3 SAT out of the above instance of 3SAT :

In each clause we have at most one literal, 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:

this yields a modified clause .

Now we must be able to guarantee that and are mapped to opposite truth values, so we introduce the new clause:

and conjunct it onto our old formula producing a new formula .

For example:

so we preform the substitution

so and

Now repeating this procedure will result in a new formula: .

We claim logical equivalence between the and This is semantically intuitive as the clause requires all substituted literal in to take the value opposite of this was the stipulation for the substitution initially. It is also verifiable by truth table construction for:

:

If there exists a truth assignment that satisfies , then we may extent this truth assignment to produce which will satisfy

by letting for all and letting for all .

Obviously if is satisfiable must be by the above construction of . So by the above claim we have that will satisfy .

:

Continuing from the above, if we have a truth assignment that satisfies , then by the claim above it also must satisfy . And is a sub-formula of so any truth assignment that satisfies must also satisfy .

(Back to me)

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

# Protected: Kth Shortest Path

# Monotone Satisfiability

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 where each clause in F contains all negated or non-negated variables, is there an assignment of the variables so that is satisfied?

**Example:**

the following assignment satisfies :

**The reduction:**

In the following reduction we are given an instance of SAT, with the clauses:

. Here each clause is of the form and each is a literal of the form

Now we build an instance of Monotone SAT from the instance of SAT given above:

For each we construct two new clauses andĀ , such that all elements of are non-negated literals and all terms in are negated literals with the addition of the new special term . Now let us build a new formula this is our instance of Monotone SAT, clauses are either all non-negated or negated.

:

Notice how we added the extra literal or to each of the clauses or respectfully. Now if there is an assignment that satisfies all of the clauses of then as only or 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 so such an assignment satisfies all .

:

Using an argument similar to the one above, For to be satisfied there must be at least one literal assignment say that satisfies each clause Now is in either or . This implies that at least one of or is also satisfied by , so simply assign the new term accordingly to satisfy the clause in not satisfied by

(back to me again)

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