Integer Expression Membership

The last problem in this section!

The problem: Integer Expression Membership.  This is problem AN18 in the appendix.

The description: Given an integer expression e over the expressions ∪ and +, which work on sets of integers.   The ∪ operation unions two sets of integers together, and the + operation takes two sets of integers and creates the set of sums of elements from each of the sets.  Given an integer K, is K in the set of integers represented by e?

Example: Suppose A = {1,2,3}, B = {4,5,6}, C = {1,3,5}.

Then the expression (A∪C) would be {1,2,3,5} and the set (A∪C) + B would be {2,4,6,3,5,7,6,8,10}.  So if we were given that expression and a K of 8, a solver for the problem would say “yes”, but if given a K of 9, the solver for the problem would say “no”.

Reduction: Stockmeyer and Meyer show this in their Theorem 5.1.  They reduce from Subset Sum (they call it Knapsack, but their version of Knapsack doesn’t have weights, so it’s out Subset Sum problem).  So from a Subset sum instance (a set A = {a1, …, an} and a target B), we build an expression.  The expression is (a1 ∪ 0) + (a2 ∪ 0) + …  + (an ∪ 0).  Then we set K=B.

Notice that the set of integers generated by this expresion is the set of all sums of all subsets of A.  Asking is B is a member is just asking if there is a way to make the above expression (for each element ai, either taking it or taking 0) that adds up to exactly B.  This is exactly the Subset Sum problem.

Difficulty: 4.  The reduction itself is easy.  It might be hard to explain the problem itself though.

Unification for Finitely Presented Algebras

The example below (using Boolean Algebra) is the kind of description of an “Algebra” that I wish I had when I learned this stuff in school- we did a lot of stuff with modular arithmetic, that I think was a lot harder.

The problem: Unification for Finitely Presented Algebras.  This is problem AN17 in the appendix.

The description: Given a description of an algebra as a set G of generator symbols, a set O of operator symbols (of possibly varying dimensions), and a Γ of “defining relations” over G and O.  We’re also given two expressions x and y over G and O, and a set of variables V.   Can we find “interpretations” for x and y, called I(x) and I(y) such that I(x) and I(y) are formed by replacements of variables such that I(x) and I(y) represent the same element of the algebra?

Example: The paper by Kozen that has the reduction gives a pretty good example of an algebra: Suppose G was {0,1}, O was {∧, ∨, ¬}, and Γ was {0∧0 ≡ 0, 0 ∧ 1 ≡ 0, 1 ∧ 0 ≡ 0, 1 ∧ 1 ≡ 1, 0 ∨ 0 ≡ 0, 0 ∨ 1 ≡, 1 ∨ 0 ≡ 1, 1 v 1 ≡ 1, ¬0 ≡ 1, ¬1 ≡}.  This is the normal Boolean Algebra we see everywhere.

Now we can ask questions like: If x = a ∨ b ∨ c and y = 0 ∨ 1 ∨ d, can we find a unification of the variables that makes the equations the same?  The answer is yes, with the replacements {a/0, b/1, d/c}.

I think the  “represent the same element” part of the problem statement means that we can also take advantage of unification rules, so we can ask questions like does ~x ∨ y ∨ ~z unify to 1?

Reduction: Kozen’s paper makes the point that if we use the Boolean Algebra that we used in the example, then the Satisfiability problem can be represented in Boolean Algebra using the rules above, and the question “Is this formula satisfiable” can be recast as “Does this formula unify to true?”

He doesn’t provide the proof, though, so I hope I’m representing him correctly.

Difficulty: 4.  Once you see how the Boolean Algebra relates, it’s pretty easy.

Unification With Commutative Operators

The next problem is one of the “Private Communication” problems.  I’m going to try to take a stab at it myself.

The problem: Unification With Commutative Operators.  This is problem AN16 in the appendix.

The description: Given a set V of variables, a set C of constants, a set of n ordered pairs (ei, fi) (1 ≤ i ≤ n) representing “expressions”.  Each expression is defined recursively as either a variable from V, a constant from C, or (e+f) where e and f are expressions.  Can we create a way to assign each variable v in V a variable-free expression I(v) such that, if I(e) denotes the expression obtained by making all such replacements of variables in an expression e, that I(ei) ≡ I(fi) for all i?  We define e ≡ f as either:

  • e = f  (the only possible way if e or f is a constant)
  • If e = (a+b) and f = (c + d) then either (a ≡ c and b ≡ d) or (a ≡ d and b ≡ c)

Example:  Suppose C was {0,1}, and V was (x,y,z}.  We will have pairs ((x+y), (0+1)) and ((y + z), (0 + 0)).  Then substituting x with 1, and y and z with 0 makes the substituted pairs equivalent.

Reduction: I think this can be done with One-In-Three 3SAT.  The SAT instance defines variables x1..xk.  Each variable xi in the SAT instance will correspond to 2 variables in the set V in our unification instance: xi and ~xi.  Each clause in the SAT instance (l1, l2, l3) will turn into the expression (e,f) where the “e side” is (l1+l2+l3), where each li is the variable corresponding to the literal (i.e., either some xj or some ~xj).  The “f side” of all expressions is (1+0+0), which corresponds to exactly one of the literals being true.

We also need to add some extra expressions to make sure that we set each xi and ~xi to opposite signs.  So each variable xi in the SAT instance gets a pair whose “e side” is (xi + ~xi), and whose “f side” is 0+1.

I think that the unification instance we’ve constructed has a solution if and only if we can find a way to:

  1. Set exactly one literal in each clause to true
  2. Set exactly one of xi and ~xi to true, for all xi

..which is exactly when the original One-In-Three 3SAT instance was solvable.

I’ll admit to being a little worried about whether this is correct, because the comment in G&J talked about there being at most 7 constants or variables in an expression, and I did it with just 3.  Maybe it’s because I’m using One-In-Three 3SAT instead of regular 3SAT, or maybe I’m missing something.

Difficulty: 4 if I’m right.  I think starting from this version of 3SAT is a big hint.

Equilibrium Point

This next reduction is confusing to me, and I wonder if it’s because there is a typo in the paper.

The problem: Equilibrium Point.  This is problem AN15 in the appendix.

The description: Given a set X = {x1..xn} of variables, a collection F = {F1..Fn} of integer product polynomials over the variables, and a set of “ranges” M = {M1..Mn} of subsets of the integers.  Can we find a sequence Y = {y1..yn}, where each yi ∈ Mi, and for all y ∈ Mi, Fi(y1, y2,…yi-1, yi, yi+1, …yn) ≥ Fi(y1, y2,…yi-1, y, yi+1, …yn)?

Example: This concept of “Equilibrium point” is best through of from the perspective of Game Theory.  The functions F are the utility functions for each player.  The sequence Y is the set of choices each player makes.  We are asking whether we can find a set of values in Y where any player i changing their yi value to something else will not improve their personal Fi score.

So the classic “Prisoner Dilemma” problem can be represented in these terms: There are 2 players, so n is 2.  Each range is with in {0,1}, where 0 means “stay silent” and 1 means “betray”.  F1 is defined by a table:

Player 2 stays silent Player 2 betrays
Player 1 stays silent -1 -3
Player 1 betrays 0 -2

F2 is defined similarly (the 0 and -3 scores switch places).

Notice that if we chose y1=y2 = 0 (both sides stay silent).  Then F1(0,0)= F2=(0,0) = -1.  But this is < F1(1,0), where player 1 betrays.  So this is not an equilibrium point.

y1=y2=1 is an equilibrium point, where both functions return -2.  Any player changing their choice from 1 to 0 will see their F function go from -2 to -3.

Reduction: Sahni does this in his “Computationally Related Problems” paper that we’ve used in the past to do some graph reductions.  This reduction is from 3SAT.   I’ll just say now that he could have avoided a lot of manipulation if he’d have used One-In-Three 3Sat.  From a 3SAT instance, we build a game where there is one clause for each player, and the range of choices for each player is between {0,1}.  The goal is to make a function fi for a clause Ci that is 1 iff the corresponding clause is true.  I’ll skip over the manipulations he does because he’s using a harder SAT problem than he needs to.

Define hi(x’) to be 2* the product of all of the fi(x’) values (for some literal x’.  If x;’ is a positive literal, use the variable.  If it’s a negated literal, use 1- the variable).  F1 (x’) = h1(x’) for all players.  This means that if the formula was satisfiable, everyone could score 2, but if it wasn’t, they’d always score 0.

Now it gets weird.  We’re going to set up a second game G, a 2 player game with no equilibrium point, then define a second payoff function for our original game F2 where F2 (x) = the g function of x for the first 2 players, but 0 for everyone else.

The paper says that the actual payoff for the actual game we’re creating is: F(X) = F1(x) + F2(x) * 2 – F1(x)

The “2” is a payout of 2 for all players- since the above depends on matrix math, it’s an nx1 vector of all 2’s. This formula is very weird to me because the F1 and -F1 should cancel out.  This is where I think there might be a typo.  I’m pretty convinced there is a typo on the previous page where he was building his little fi function (he uses a + where there should be a -).  I’m thinking that there are missing parentheses in this formula, and it should be F(X) = F1(x)+F2(x)*(2-F1(x))

Now two things can happen.  If the formula was satisfiable, then F1(x) is all 2’s, and that is the max payout for everyone and is an equilibrium point.  If the formula was not satisfiable, then F1(x) is all 0’s, and so the scores in the F2 part influence the score for F, but the F2 part has no equilibrium, so F doesn’t either.

Difficulty: 8.  I think I found the typo though.

Permanent Evaluation

Going back to the problem we skipped over last week.

The problem: Permanent Evaluation.  This is problem AN13 in the appendix.

The description: Given an nxn matrix M of 0’s and 1’s, and a positive integer K, is the permanent of M equal to K?

Example: The permanent of M = \displaystyle \sum_\sigma \prod_{i=1}^n A_{i,\sigma(i)}

That is, for each permutation of the columns, we multiply down the main diagonal.  The sum of all of those products is the permanent.

1 2 3
4 5 6
7 8 9

..then the permanent is 1*5*9 + 1*6*8 + 2*4*9 + 2*6*7 + 3*5*7 + 3*4*8 = 450

Of course, we’re looking at 0/1 matrices, so I think what we’re really asking is how many permutations of the columns have 1’s on the main diagonal.

(Wrong) Reduction: If I’m right above that all we’re doing is counting how many ways there are 1’s in the main diagonal, then this becomes a pretty easy Turing Reduction from Hamiltonian Path.  Given an adjacency matrix of a graph, we want to know if the permanent of the adjacency matrix is 1 or more.  (Any Hamiltonian Path will give a permutation that multiplies to 1, and any permutation of vertices that is not a Hamiltonian Path multiplies to 0).   Given how complicated the “actual” reduction is, I’m a little worried that I missed something, though.

Edit on 1/21: This isn’t right.  The problem is that while you’re premuting the columns, you’re not permuting the rows.  So if we permute the column to be the second vertex in the Hamiltonian Path, the second row is still the vertices adjacent to vertex #2 (which might not be the second vertex in the path).

That’s a shame.  I wonder if there is a way to manipulate the problem to make it work this way anyway.

(Correct) Reduction:

The reduction by Valiant that G&J point you to uses 3SAT, He shows that if you have a formula F, and define t(F) to be 2x the number of literals in F – the number of clauses in F, then there is some function f,  computable by a deterministic Turing Machine  in polynomial time, that maps a formula to a matrix.  (The matrix has entries in {-1..3}, he does another transformation later to convert it to a 0/1 matrix).  The permanent of that matrix = 4^{t(F)} * s(F) , where s(F) is the number of ways to satisfy F.

Since one of the consequences of Cook’s Theorem is that we can take an arbitrary non-deterministic Turing Machine and turn it into a Satisfiability formula, we get the reduction.

The actual construction of that function f is complicated.  Given a formula, we construct a graph and use the matrix as the adjacency matrix of the graph.  The variables, literals, and clauses get mapped to subgraphs.

Difficulty: If my way was right, I’d give it a 4- I think it’s easier than most Turing Reductions.  The Valiant construction is an 8.

Cosine Product Integration

I’m back from break, and we’ll go a little out of order at first since this is the last problem that’s similar to the ones we’ve been working on.

The problem: Cosine Product Integration.  This is problem AN14 in the appendix.

The description: Given a sequence of n integers (a_i..a_n), does \displaystyle \int_0^{2\Pi} (\prod_{i=1}^n cos(a_i\theta)) d\theta = 0?

Example:

Suppose our sequence was (1,2,3).  Then our integral is:

\displaystyle \int_o^{2\Pi}(\theta*2\theta*3\theta)d\theta

This is just 6\theta^3, which integrates to \frac{6x^4}{4}, which over the interval 0..2\pi is 24\pi^4

Reduction: This one is in Plaisted’s 1976 paper. In it, he notes that if you look at the function \displaystyle \prod_{j=1}^{k} (x^{a_j}+x^{-a_j}) the constant term in the power series expansion of that product is 0 if and only if the cosine integral is 0.  I have trouble seeing that myself.  The cooler thing is that you can make that constant term 0 if and only if you can take the sequence of a_i elements and partition them into 2 sets with the same sum.  So we can take an instance of the Partition problem, use the elements of the set as the elements of our sequence and then feed them into the product formulas.

Difficulty: It really depends on how easily you can see the connection between the cosine integral and the product formula (and, of course, how easily you could have thought of it). I find it hard, so I’m giving it an 8.

Periodic Solution Recurrence Relation

Probably the last post of the year- enjoy the holidays, everyone!

The problem: Periodic Solution Recurrence Relation.  This is problem AN12 in the appendix.

The description: Given a set of m ordered pairs (c_1,b_1) through (c_m, b_m with each b_i >0, can we find a sequence a_0 though a_{n-1} of integers, such that if we build the infinite sequence \displaystyle a_i = \sum^m_{j-1} c_j*a_{i-b_j} is periodic: that is, a_i \equiv a_{i (mod \: n)} for all i?

Example: Here’s a super simple example: m=2 and the pairs are (1,1) and (2,2).  This gives us the recurrence a_i = a_{i-1}  + 2a_{i-2}.  If we start with 1,1, this gives the sequence 1,1,3,5,11,21,43,…  which is periodic mod 10 (the last digit always repeats 1,1,3,5)

Reduction: This shows up in Plaisted’s 1984 paper.  He mentions it as a corollary to his Theorem 5.1 which showed that Non-Trivial Greatest Common Divisor and Root of Modulus 1 were NP-Complete.  Similar to the Root of Modulus 1 problem, we build a polynomial from a set of clauses that has 0’s on the unit circle.  The polynomial also has a leading coefficient of 1.  This means, apparently, that the recurrence relation corresponding to the polynomial has a periodic solution if and only if the polynomial has a root on the complex unit circle, which only happens if the original 3SAT formula was satisfiable.

Difficulty: 8.

Number of Roots for a Product Polynomial

The problem: Number of Roots for a Product Polynomial.  This is problem AN11 in the appendix.

The description: Given a set of sequences A1 through Am , each Ai containing a sequence of k pairs (a_i[1],b_i[1]) through (a_i[k],b_i[k]) , and an integer K.  If we build a polynomial for each Ai by \displaystyle \sum_{j=1}^k a_i[j]*z^{b_i[j]}, and then multiply all of those polynomials together, does the resulting product polynomial have less than K complex roots?

Example:  Suppose A1 was <(1,2), (2,1), (1,0)>, A2 was <(3,3), (2,2), (1,1), (0,0)>, and A3 was <(5,1), (7,0)>.  These represent the polynomials x2+2x+1, 3x3 + 2x2 + x, and 5x+7.  (I’m pretty sure it’s ok for the sequences to be of different length, because we could always add (0,0) pairs to shorter sequences).  This multiplies out to 15 x6 + 61 x5 + 96 x4+ 81 x3 +  50 x2 + 26x +7, which has 4 distint complex roots, according to Mathematica.

Reduction: This is another one that uses Plaisted’s 1977 paper.  (It’s problem P4).  He builds the polynomials PC and QC in the same way that he did in the reduction for Non-Divisibility of a Product Polynomial.  One of the statements that he says is “easy to verify” is that The product of the Q polynomials for each clause has N (for us, K) zeroes in the complex plane if and only if the original 3SAT formula was inconsistent.

Difficulty: I’m giving all of these problems based on the polynomials that come from a formula an 8.

Root of Modulus 1

After taking a week off for Thanksgiving, we move on to another equation problem.

The problem: Root of Modulus 1.  This is problem AN10 in the appendix.

The description: Given a set of ordered pairs (a_1,b_1) through (a_n, b_n) of integers, each b_i is non-negative.  Can we find a complex number q where \mid q \mid= 1 such that \displaystyle \sum_{i=1}^n a_i * q^{b_i} =0?

Example: It was hard for me to come up with an interesting example (where q is not just 1 or i), so thanks to this StackOverflow post for giving me something I could use.

Let our ordered pairs be (5,2), (-6,1), and (5,0).  This gives us the polynomial 5x2-6x+5.  Plugging these into the quadratic formula get us the roots \frac{3}{5} \pm  \frac{4}{5} i, which is on the complex unit circle.

Reduction: This one is again from Plaisted’s 1984 paper.  It again uses his polynomial that we’ve seen in some other problems (most recently Non-Divisibility of a Product Polynomial).  So again, we start with a 3SAT instance and build the polynomial.  He starts by showing that if you have a polynomial with real coefficients p(z), then p(z)*p(1/z) is a real, non-negative polynomial on the complex unit circle, and it has zeros on the unit circle exactly where p(z) does.

Then, we can do this for the sum of the polynomials made out of each clause, which means that this new polynomial has 0’s on the unit circle exactly where the original one did.  Which means it has a 0 on the complex unit circle if and only if the formula was consistent.

Difficulty: 8.  I’m starting to appreciate the coolness of turning a formula into a polynomial, and how it makes a lot of problems easier.  I just wish it was clearer to see how it all works.

 

Algebraic Equations over GF[2]

AN8 is Quadratic Diophantine Equations.

The problem: Algebraic Equations over GF[2].  This is problem AN9 in the appendix.

The description: Given a set P of m polynomials over n variables (x1 through xn) where each polynomial is the sum of terms that is either 1 or the product of distinct xi, can we find a value ui for each xi in the range {0,1} that make each polynomial 0, if we define 1+1=0, and 1*1 = 1?

Example: It helps to think of GF[2] as a boolean logic world, where + is XOR and * is AND.  So, suppose we have three variables, and the polynomials:

  • P1 = 1 + x1x2 + x2x3
  • P2 = x1 + x1x2x3

..Then setting x1=0, x2=1, x3=1 makes both polynomials 0.

Reduction: G&J say that Fraenkel and Yesha use X3C, but the paper I found uses 3SAT.  We’re given an equation that has n variables and m clauses.  The variables of our polynomials will be the same variables in the 3SAT instance.  For each clause, we build a polynomial by:

  • Replacing a negated literal (~x) with the equation 1 + x.  (Remember, + means XOR in this system)
  • Replacing an OR clause (A ∨ B) with the equation A+ B +A*B
  • Xoring the whole above thing with 1.

Notice that the first replacement makes ~x have the opposite truth value of x, the second replacement rule is logically equivalent to A∨B, and the third part makes the polynomial 0 if and only if the clause evaluated to 1.  So the polynomial is 0 if and only if the clause is satisfiable.  So all polynomials are 0 if and only if the all clauses are satisfiable.

Difficulty: 5.  This is easy to follow.  It’s a little tricky to make students come up with the equivalence rules above, but I think if you can explain it right, it’s not that bad.