Tag Archives: 3-sat

Not All Equal 3SAT

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This is jumping ahead a little, but this is a variant of 3SAT that will be helpful in the future (actually, for the next problem)

The problem: Not-All-Equal-3SAT (NAE3SAT).  This is problem LO3 in the appendix.

The description: Just like 3SAT, except instead of requiring at least one of the literals in each clause to be true, we’re requiring at least one literal to be true and at least one literal to be false.  So we’re removing the case where all three literals can be true.

Notice that if we have an assignment of variables that is a NAE3SAT solution, then if we negate all of the variables, we still have a NAE3SAT solution

The reduction: From regular 3SAT.  We’re given a formula that is a collection of 3-literal clauses.  For each clause (x1 ∨ x2 ∨ x3) we create a new variable ci (for clause i) and add 2 clauses to our NAE3SAT instance:

(x1∨ x2 ∨ ci) and (x3 ∨ ~ci ∨ F)  (Where “F” is the constant value false.  We’ll talk more about this later)

The idea is that ci is there to “fix” the clause.  If the original clause was satisfiable because of x1 or x2, then ci can be false, and its negation can make the second clause true.  If the original clause was satisfiable because of x3, we can make ci true.  If the original clause had all three literals satisfiable, we can make ci false, and become an acceptable NAE3SAT solution.

Note that because the definition of NAE3SAT (and, for that matter, regular 3SAT) is defined in terms of collections of 3 variables, the constant value false we used above isn’t strictly legal.  We’ve really just shown “NAE3SAT with some constants replacing variables” is NP-Hard.  To show “normal” NAE3SAT is NP-Hard, we need to reduce the fixed constant version into the normal version:

Given an instance of NAE3SAT with fixed constants, we create 2 new variables xT and xF.  We replace all instances of the constant value true with xT, and all instances of the constant value false with xF.  We also add one additional clause:

(xT ∨ xT ∨ xF).

Note that this new clause is only NAESatisfiable if xT != xF.  We can’t directly bind xT to true and xF to false because if you take a NAE3SAT solution and negate all literals, you get another NAE3SAT solution.  But because of that, this is NAESatisfiable if and only if the original formula is.

Difficulty: It depends on what you give them.  If you make them do everything I did above, I’d say it’s an 8.  If you let them stop at the NAE3SAT with fixed constants, then maybe a 7.

 

Vertex Cover

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Back to the “core 6” with a classic problem from graph theory.  There are lots of similar graph problems, so it’s important to keep them all straight.

The Problem: Vertex Cover (VC)

The Definition: Given a graph G=(V,E) and a positive integer K (G&J say that K ≤ |V|, but really, if it’s bigger than that, you can have the algorithm just return “yes”).  Can I find a subset V’ of V, |V’| ≤ K, where every edge in E has at least one endpoint in V’?

Example: Hmm, I wonder if I can insert an image..

VC exampple

Ok, looks like that worked.  In the graph above, {c,d,g} is a vertex cover, because all edges have at least one endpoint in the set {c,d,g}.  As we’ll see later, a graph containing a clique (complete subgraph) of N nodes will need at have at least N-1 of those vertices in the cover.

The reduction: From 3SAT.  G&J on pages 54-56 show the construction.  We build a graph with vertices for all variables and their negation, with an edge between them.  We have three variables for each clause, connected in a triangle.  (The three variables correspond to “positions” in the clause).  Then there is an edge from each vertex corresponding to a literal (or its negation) to its corresponding “clause vertex”- the vertex that holds the position of that variable in the original formula.  K= # of variables + 2* number of clauses.

Difficulty: 7 , assuming it’s the first time a student has seen this sort of reduction.  If they have seen something similar (for example, a reduction from 3SAT to Clique), then maybe a 5.  The ideas are pretty similar.

Note: I teach my algorithms class using the Cormen book.  In it, they reduce 3SAT to Clique, proving Clique is NP-Complete, and then reduce Clique to VC. This works in the exact same way as the reduction from VC to Clique that I’ll be doing here next.

3-Dimensional Matching

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Last one for today.  I’m not sure how exactly I’ll space these out. We’ll see.

The Problem: 3-Dimensional Matching (3DM)

The Definition: Given disjoint sets W,X,Y all with the same number (q) of elements, and a set M of triples from WxXxY.  Is there a subset M’ of M of size q where no two elements of M’ agree in any coordinate?

I always have trouble picturing what this means, so here’s an example:

Let W = {1,2,3,4}, X = {10,20,30,40}, Y = {100,200,300,400}. q is 4 in this case.

M is a set of triples like (1,30,400), (2,20,100), and so on, with one element from each of W,X, and Y.

We’re asking for q elements of M that have no duplicates in any coordinate.  In other words, each of {1,2,3,4} appears in exactly one element of M’, each of {10,20,30,40} appears in exactly one element of M’, and so on.  So, a legal M’ could be:

{(1,20,300), (2,30,400), (3,40,100), (4,10,200)}

..assuming all 4 of those triples happened to be in the M we were given.

The Reduction: From 3SAT.  Pages 50-52 explain how we take the clauses and build W, X, Y, and M from them.  It’s pretty crazy-M contains 3 kinds of triples for 3 different kinds of jobs, build according to 3 different sets of rules.

Difficulty: 8.  I couldn’t imagine a student coming up with this without a lot of hand-holding.

3-Satisfiability

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The first of the “core six” problems, and the first NP-Complete problem that’s actually a reduction from a known NP-Complete problem.

The Problem: 3-Satisfiability (3SAT)

The Definition: (p.46) Given a set of variables U, and a set C of clauses, where each clause has exactly 3 elements, can we assign true/false values to the variables in U that satisfies all of the clauses in C?

(Note that just as the Satisfiability problem was really “CNF-SAT”, most people would call this problem “3-CNF-SAT”)

The Reduction: From SAT.  Pages 48-49 give a simple algorithm to convert arbitrarily-sized clauses into clauses of size 3, possibly by adding some extra variables.

Difficulty: 6.  It’s pretty fiddly to come up with the conversions in all cases.  But this really depends on the students’  familiarity with Boolean logic.

As an aside, I find myself compelled to note that while 3-CNF-SAT is NP-Complete, the similar problem of 2-CNF-SAT (where all clauses are of size 2) can be solved in polynomial time.  To me, one of the coolest things in Computer Science is how such small changes to a problem (like 3-CNF to 2-CNF, or  Fractional Knapsack to 0/1 Knapsack-where the solution space shrinks!) can drastically affect the complexity of the problem.