
Recent Posts
 Generalized Satisfiability November 22, 2019
 Generalized Instant Insanity November 8, 2019
 Crossword Puzzle Construction October 25, 2019
 Square Tiling October 10, 2019
 LeftRight Hackenbush for Redwood Furniture September 27, 2019
Recent Comments
 Frank Ciceri on How to get the archive password
 Frank Ciceri on How to get the archive password
 Sean T. McCulloch on How to get the archive password
 Frank Ciceri on How to get the archive password
 Sean T. McCulloch on Hamiltonian Path
Archives
 November 2019
 October 2019
 September 2019
 August 2019
 July 2019
 June 2019
 May 2019
 April 2019
 March 2019
 February 2019
 January 2019
 December 2018
 November 2018
 October 2018
 September 2018
 August 2018
 July 2018
 June 2018
 May 2018
 April 2018
 March 2018
 February 2018
 January 2018
 December 2017
 November 2017
 October 2017
 September 2017
 August 2017
 July 2017
 June 2017
 May 2017
 April 2017
 March 2017
 February 2017
 January 2017
 December 2016
 November 2016
 October 2016
 September 2016
 August 2016
 July 2016
 June 2016
 May 2016
 April 2016
 March 2016
 February 2016
 January 2016
 December 2015
 November 2015
 October 2015
 September 2015
 August 2015
 July 2015
 June 2015
 May 2015
 April 2015
 March 2015
 February 2015
 January 2015
 December 2014
 November 2014
 October 2014
 September 2014
 August 2014
 July 2014
 June 2014
Categories
 Algebra and Number Theory
 Appendix Algebra and Number Theory
 Appendix Automata and Language Theory
 Appendix Games and Puzzles
 Appendix Mathematical Programming
 Appendix Network Design
 Appendix Program Optimization
 Appendix Sets and Partitions
 AppendixGraph Theory
 AppendixLogic
 Appendix: Sequencing and Scheduling
 Appendix: Storage and Retrieval
 Chapter 3 Exercises
 Core Problems
 Overview
 Problems not in appendix
 Uncategorized
Meta
Monthly Archives: April 2016
Protected: Subset Sum
Enter your password to view comments.
Posted in Appendix Sets and Partitions
Tagged Difficulty 2, Part, SOS, SP13
Protected: Minimum EdgeCost Flow
Enter your password to view comments.
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
Enter your password to view comments.
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
Enter your password to view comments.
Posted in Appendix Network Design
Tagged Difficulty 5, Part, Shortest WeightConstrained Path, uncited reduction