
Recent Posts
 Hitting String April 26, 2017
 Bounded Post Correspondence Problem April 19, 2017
 Shortest Common Superstring April 12, 2017
 Shortest Common Supersequence April 7, 2017
 Longest Common Subsequence March 29, 2017
Recent Comments
Archives
 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
Meta
Tag Archives: Difficulty 3
Protected: Ratio Clique
Enter your password to view comments.
Posted in AppendixGraph Theory, Problems not in appendix
Tagged Difficulty 3, Difficulty 4, GT19, No G&J reference, Ratio Clique, uncited reduction
Protected: Bin Packing Take 2
Enter your password to view comments.
Posted in Appendix: Storage and Retrieval
Tagged 3Partition, Bin Packing, Difficulty 3, No G&J reference, SR1, uncited reduction
Protected: Numerical Matching With Target Sums
Enter your password to view comments.
Posted in Appendix Sets and Partitions
Tagged Difficulty 3, No G&J reference, Numerical 3Dimensional Matching, Numerical Matching With Target Sums, SP17, uncited reduction
Protected: MinMax Multicenter
Enter your password to view comments.
Posted in Appendix Network Design
Tagged Difficulty 3, Dominating Set, MinMax Multicenter, VC<=3, Vertex Cover
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 where each clause in contains all negated or nonnegated variables and the number of clauses and variables are equal, is there an assignment of the variables so that is satisfied? Our instance will have clauses and variables.
Example:
Here is an that has variables and clauses.
F =
The above may be satisfied by the following assignment:
The reduction:
We will reduce from Monotone SAT. So we are given an instance of Monotone SAT with the clauses here each clause is of the form where each clause has all negated or nonnegated variables. This is different from Monotone EQ SAT as we do not require the number of variables and clauses to be equal.
From this we must build an instance of Monotone EQ SAT.
We may transform our instance of Monotone SAT, , into one of Monotone EQ SAT by the following iterative procedure. New variables will be denoted by and new clauses by .
= ; i = 1; j = 1; While{number of clauses != number of variables}{ introduce two new variables ; If{number of variables number of clauses}{ Create the new clause ; ; ; } else { Create three new clauses: ; ; ; } ; }
The above algorithm will produce an equation that is in Monotone EQ SAT. This may be shown by induction. Notice that before the procedure if that we will add 2 new variables and 3 new clauses.
If then we will add 2 new variables but only a single new clause. Either way the difference between the number of variables and clauses, will decrease by . So in steps we will obtain an formula where = 0. Such an formula is an instance of Monotone EQ SAT.
True{Monotone SAT True{Monotone EQ SAT}
Here we assume that there is a truth assignment function that maps every variable to a truth value, such that is satisfied. Then after we preform the above algorithm we have an instance of , now our instance of will be of the form for some . Now notice that above will satisfy in and we may trivially satisfy by simply assigning all new variables to true.
This will give us a new truth assignment function that will satisfy
True{Monotone EQ SAT} True{Monotone SAT}
Here we assume that there is a truth assignment function that will satisfy then obviously as then must also satisfy .
(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.
Posted in AppendixLogic
Tagged Dan's Problems, Difficulty 3, Monotone EQ Sat, Monotone Sat
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: Biconnectivity Augmentation, Strong Connectivity Augmentation
Enter your password to view comments.
Posted in Appendix Network Design
Tagged Biconnectivity Augmentation, Difficulty 3, Hamiltonian Circuit, ND18, ND19, Strong Connectivity Augmentation
Protected: Optimum Communication Spanning Tree
Enter your password to view comments.
Posted in Appendix Network Design
Tagged Difficulty 3, ND7, Optimum Communication Spanning Tree, Shortest Total Path Length Spanning Tree, uncited reduction
Protected: Dominating Set on Bipartite Graphs
Enter your password to view comments.
Posted in Problems not in appendix
Tagged Difficulty 3, Dominating Set, Dominating Set on Bipartite Graphs, No G&J reference, uncited reduction
Protected: Graph Homomorphism
Enter your password to view comments.
Posted in AppendixGraph Theory
Tagged Difficulty 3, Difficulty 4, Graph 3coloring, Graph Contraction, Graph Homomorphism, GT52, uncited reduction