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Monthly Archives: April 2016
Protected: Subset Sum
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Posted in Appendix Sets and Partitions
Tagged Difficulty 2, Part, SOS, SP13
Protected: Minimum EdgeCost Flow
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Posted in Appendix Network Design
Tagged Difficulty 5, Min EdgeCost Flow, No G&J reference, uncited reduction, X3C
Monotone 3Satisfiability
I told Daniel when he gave me his Monotone Satisfiability reduction that the actual problem mentioned in G&J was Monotone 3Satisfiability. 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 nonnegated 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 Monotone 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 Monotone 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 subformula 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.
Posted in AppendixLogic
Tagged 3sat, Dan's Problems, Difficulty 4, Monotone 3SAT, Monotone Sat, uncited reduction
Protected: Kth Shortest Path
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Posted in Appendix Network Design
Tagged Difficulty 5, Hamiltonian Path, kth shortest path, ND31, Turing Reduction, uncited reduction
Monotone Satisfiability
This semester I’m doing an independent study with a student, Daniel Thornton, looking at NPComplete 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 where each clause in F contains all negated or nonnegated 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 and , such that all elements of are nonnegated 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 nonnegated 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. I like that the reduction involves manipulating the formula, instead of applying logical identities.
Posted in AppendixLogic
Tagged Dan's Problems, Difficulty 3, Monotone Sat, uncited reduction
Protected: Shortest WeightConstrained Path
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Posted in Appendix Network Design
Tagged Difficulty 5, Part, Shortest WeightConstrained Path, uncited reduction