Tag Archives: Difficulty 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 \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:

\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 \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 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 \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 \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.

 

Non-Divisibility of a Product Polynomial

Now back to the problem we skipped over last week.

The problem: Non-Divisibility of a Product Polynomial.  This is problem AN6 in the appendix.

The description: Just like Non-Trivial Greatest Common Divisor, we’re given a set of m sequences of pairs of  integers (so sequence Ai is <(ai[1],bi[1]) …(ai[k],bi[k])>), where each b element is ≥ 0.  We’re also given an integer N.

Just like Non-Trivial Greatest Common Divisor, we build the polynomials \sum_{j=1}^k a_i[j]*z^{b_i[j]}, for each i from 1 to m.  But now we want to multiply them all together.  If we do, is the product not divisible by zN-1?

Example: Suppose A1 was <(1,3), (2,2), (-1,1), (-2,0)> and A2 was <(5,4), (3,2), (7,0)>.  Then A1‘s polynomial is z3 + 2z2-z-2, and A2‘s polynomial is 5z4+3z2+7.  The product of these is: 5z7+10z6 -2z5 – 4z4 + 4z3 + 8z2-7z-14.  If N=2, then it turns out that z2-1 is a factor (it’s actualy a factor of A1), so the decision problem would say “No”.

Reduction: The reduction for this is again by Plaisted, using his polynomials that come from logical statements. For each clause C of a 3SAT formula, you build the polynomial PC for that clause, and from that QC = (xN-1)/PC  He mentions in his paper that it is “easy to verify” (but doesn’t actually prove it) that The product of all of these QC‘s / (xN-1) is analytic iff S is inconsistent.  I think the reason is that QC represents in his polynomial the ways to assign variables to make a formula false. So if all assignments make it false, the product is equal to all assignments.  I still don’t quite see how the divisibility follows.

Difficulty: 8.  It’s about as hard as the last one.

Non-Trivial Greatest Common Divisor

We’ll do this next one out of order because I think this order is better explained by the paper.

The problem: Non-Trivial Greatest Common Divisor.  This is problem AN7 in the appendix.

The description: Given a set of m sequences of pairs of  integers (so sequence Ai is <(ai[1],bi[1]) …(ai[k],bi[k])>), where each b element is ≥ 0.  I’m pretty sure k can be different in each sequence.  If we build the polynomials \sum_{j=1}^k a_i[j]*z^{b_i[j]}, for each i from 1 to m, does the greatest common divisor of those polynomials have degree greater than 0?

Example: I think the hardest part of this description is picturing the polynomials, so here’s a simple example showing how that works.  Suppose m=2, and A1 was <(20,3),(4,2),(30,1)> and A2 was <(2,4),(8,1)>

(As an aside, I wish G&J didn’t use A for both the name of a sequence and lowercase a for the name of one of the entries of each pair.  They’re not related.)

The polynomial corresponding to A1 is 20x3 + 4x2+30x, and the polynomial corresponding to A2 is 2x4 +8x.  The polynomial 2x is a divisor of both, and I think it’s the GCD.  (If not, the GCD is larger, and so also has degree >0).  So, this problem instance would have the answer “yes”.

Reduction:

This reduction is also by Plaisted, who did last week’s reduction, but is in his 1984 paper.  His reduction is from 3SAT.  In the paper, he gives an algorithm that turns a logical formula into a polynomial.  The polynomial p that gets created has the property that p(z) = 0 if and only if z is an M-th root of unity (a solution to zM=1).  This leads to a way of setting the truth values of the logical formula: if ω is the root of unity, set variable Pj to true if ω^{M/q_j}=1, where qj is the jth prime number.  It turns out if you do that, you make the entire logical formula true.

So, for our problem, we’re handed a 3SAT instance with n variables.  Let M be the product of the first n primes.  The formula is satisfiable iff we can make all clauses true, which happens iff there is an Mth root of unity ω such that the polynomial on each clause is 0 (so we make a polynomial for each clause in the SAT formula). He states an “easy” identity (without proof) that the polynomial of A∧B is the gcd of the polynomials of A and B.  So therefore if ω exists for all polynomials, the gcd has degree greater than 0.

Difficulty: 8.  This is a little easier to follow because he states the properties of his polynomial up front.  I’d like to see more proofs of his “easy” properties and to be honest, I’m not sure how the last step happens (how having ω exist means that the gcd has degree > 0).  I also don’t know how you’d ask students to come up with these polynomials on their own.

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