Monthly Archives: December 2018

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.