Monthly Archives: November 2019

Generalized Satisfiability

We’re starting a new section and a lot of the introductory problems have been covered elsewhere.

LO1 is Satisfiability

LO2 is 3-Satisfiability

LO3 is Not all Equal 3SAT

LO4 is One-In-Three 3SAT

LO5 is Maximum 2SAT

After that we finally get to a new problem:

The Problem: Generalized Satisfiability.  This is problem LO6 in the appendix.

The description: Given a list of positive integers k1 through km and a sequence S = <R1..Rm>, where each Ri is a set of strings of ki “T” or “F” symbols.  We’re also given a set U of variables, and a set of collections C1 through Cm.  Each Ci contains a ki-tuple of variables from U.

Can we assign a truth value (“T” or “F”) to each variable in  U such that for each tuple in Ci the truth value of each variable in the tuple matches the sequence of true and false values in Ri?

Example: Let all ki = 3, and m = 2.  R1 = {(T,T,T), (F,F,F)}  R2 = {(T,F,F). (F,T,T)}

U = {x1, x2, x3}.  C1 = (x1, x2, x3}.  C2 = {x1, x2, x3}

So, the question is: Can we assign truth values to x1 x2 and x3 so that the truth values map to a sequence in both R0 and R1?  Since we have the same sequences of the same variables in both sets, the answer is no.

But, suppose we change C2 to (x4, x2, x3).  Then we can set x1, x2, and x3 to true and x4 to false.  This will let the sequence in C1 map to the (T,T,T) sequence in R1, and the sequence in C2 map to the (F,T,T) sequence in R2.

Reduction: This reduction is in Shaefer’s paper, and it reduces from 3SAT.  The idea is that we will build 4 R sets, thinking of them as functions.  So:

R0 = {R(x,y,z) | x or y or z is true} = {(T,T,T), (T,T,F), (T,F,T), (T,F,F), (F,T,T), (F,T,F), (F,F,T)}

R1 =Same idea, but  “~x or y or z is true”

R2 = “~x or ~y or z is true”

R3 = ” ~x or ~y or ~z is true”

Once we’re given a CNF instance, we replace each clause in the formula with one of the Ri formulas.  Then build a C set that uses the variables in the clause.  So, for example, if a clause in the CNF formula was ~x2 ∨ x5 ∨ ~x7, we’d make the element in R corresponding to the clause R2, and we’d use the C formula (x2, x7, x5).  Notice that we changed the order of the variables in the C set because R2 expects its negated literals first.

Notice that we can find a way to assign the variables truth values to make each Ci map to its R set, we have satisfied the clause.  So this new construction is satisfiable if and only if the 3SATinstance was satisfiable.

Difficulty: Making the construction is probably a 5, mostly because you need a good way to explain how you’re making the R sets.  The hardest thing here though is understanding what the problem says.

Generalized Instant Insanity

The last problem in the “Games and Puzzles” section comes from a commercial puzzle (though one based on a much older puzzle).  It’s cool how these sorts of things pop up- a toy or game comes out and then people analyze it mathematically and find some neat features about it.

The problem: Generalized Instant Insanity.  This is problem GP15 in the appendix.

The description: Given a set Q of cubes, and a set C of colors we paint on the sides of each cube (and where |Q| = |C|) is there a way to arrange the cubes in a stack such that each color in C appears exactly once on each side?

Example: The Wikipedia page for the Instant Insanity game has a good description of the game and an example of the solution when Q = 4.

Reduction: Roberson and Munro reduce from “Exact Cover”.  I don’t think I’ve done this problem, but it’s basically X3C where the sizes of the sets can be any size.  (So it’s a trivial reduction from X3C).

They start by designing a graph representation of the puzzle, similar to the one shown on the Wikipedia page. Vertices correspond to colors, an edge connects two vertices if they are on opposite sides of the came cube, and the edges are labeled with the name of the cube.  The puzzle has a solution if and only if we can find two edge disjoint cycles that touch each vertex and edge label exactly once.  The two cycles correspond to the two sets of opposite faces we can see (since 2 sides of each cube are hidden by the stack).  Again, go look at the graphs in the Wikipedia article– they show this pretty nicely.

So, to perform the reduction, we’re given a set S {} and a collection T of subsets of S, and need to make a graph (which corresponds to a puzzle).  Each element in S will become a vertex in the graph.  Each of these vertices will have a self-loop and be labeled ζi (So each vertex is a different color, and each self-loop is part of a different cube- so each cube has one color on opposite sides).

Each element si inside a set Th of T will also have a vertex: vi,h. This vertex has an edge with the next largest level sj within Th, wrapping around if we’re at the last element of Th. These edges are labeled ϒh,j.  We also have an edge from sjto vh,j if sj is in Th labeled with δh,j.  We also have 2 copies of self-loops from each vh,j to itself labeled ϒh,j.

They then add some extra edges to give the graph the properties:

  • One of the self-loops from vh,i to itself has to be in one of the two cycle sets.
  • If the edge from vh,i to sj (labeled ϒh,j) is in the other cycle set, then the edge from sj to vh,j (labeled δh,j) is also in that cycle set
  • The ζi edge label has to be used, and so will have to be used on an si vertex the cycle that uses that vertex will also have to have the labels ϒh,j and δh,j for some h.
  • The loops (labeled ζj and ϒh,j) are in one cycle set and the paths (labeled ϒh,j and δh,j) are in the other.

With these properties, then solving the puzzle means we have to find h1 through hm that correspond to sets Th in T, but also where there is a path ϒh,jh,j in the graph for each sj in Th_i.  We make sure no elements in S are repeated by making sure the puzzle only uses each sj vertex once.

Difficulty: 8.  I glossed over a lot of the construction, especially the definition of a “p-selector”, which is a subgraph that helps set up a lot of the properties above.  That definition is very hard for me to follow.