# Divide and conquer over cyclic groups

Recently we found a divide and conquer algorithm over \Z_m. We have an efficient algorithm for the case where all the elements S are in \Z_m^*, the set of units.

Our divide and conquer algorithm basically partition the numbers by throwing them into different subgroups. Assume m has distinct prime factors p_1 < \ldots < p_k, and for simplicity, define p_0=1. We partition S to S_0,\ldots,S_k, where S_i contain all the elements that's divisible by p_i but not p_j for any j>i. We recursively apply our algorithm to each S_i/p_i in \Z_{m/p_i} and combine the solution.

This gives us a recurrence relation, and the crucial part of the recurrence involves the following function.

Let p_i be the ith smallest prime number, and p_0 defined as 1. If we know that f_0(x) = x and f_k(x) = \sum_{i=0}^k f_i(x/p_i), then we can show that \displaystyle f_k(x) = x \prod_{i=1}^{k} \left(1+\frac{1}{p_j-1}\right) by induction.

First, we would need a small lemma.

Let a_1,\ldots,a_k be real numbers such that non of them is 0 or 1, then \displaystyle 1+\sum_{j=1}^k \frac{1}{p_j} \prod_{i=1}^j \left(1+\frac{1}{p_i -1}\right) = \prod_{j=1}^k \left(1+\frac{1}{p_j-1}\right)

Proof by induction, basically \frac{1}{x} (1+\frac{1}{x-1})=\frac{1}{x-1} for any x\neq 0,1, so this is true when k=1.

\displaystyle \begin{aligned} 1+\sum_{j=1}^k \frac{1}{p_j} \prod_{i=1}^j \left(1+\frac{1}{p_i -1}\right) &= \prod_{j=1}^{k-1}\left(1+\frac{1}{p_j-1}\right) + \frac{1}{p_k}\prod_{j=1}^{k}\left(1+\frac{1}{p_j-1}\right)\\ &= \left(1+\frac{1}{p_k} \left(1 + \frac{1}{p_{k-1}}\right) \right)\prod_{j=1}^{k-1}\left(1+\frac{1}{p_j-1}\right)\\ &= \left(1+\frac{1}{p_k -1}\right)\prod_{j=1}^{k-1}\left(1+\frac{1}{p_j-1}\right)\\ \end{aligned}

\displaystyle f_k(x) = x \prod_{i=1}^{k} \left(1+\frac{1}{p_i-1}\right)

\displaystyle \begin{aligned} f_k(x) &= \sum_{i=0}^k f_i(x/p_i)\\ f_k(x)-f_k(x/p_i) &= \sum_{i=0}^{k-1} f_i(x/p_i)\\ &= x \sum_{i=0}^{k-1} \frac{1}{p_i} \prod_{j=1}^i \left(1+\frac{1}{p_j-1}\right)\\ &= x \left(1 + \sum_{i=1}^{k-1} \frac{1}{p_i} \prod_{j=1}^i \left(1+\frac{1}{p_j-1}\right)\right)\\ &= x \prod_{i=1}^{k-1} \left(1+\frac{1}{p_i-1}\right)\\ \end{aligned}

We can substitute f_k(x) = x \prod_{i=1}^{k} \left(1+\frac{1}{p_i-1}\right), and see the result matches.

\displaystyle \begin{aligned} x\left(1-\frac{1}{p_i}\right) \prod_{i=1}^{k} \left(1+\frac{1}{p_i-1}\right) = x \prod_{i=1}^{k-1} \left(1+\frac{1}{p_i-1}\right) \end{aligned}

It's also useful to bound f_k(x).

\displaystyle f_k(x) = O(x \log k)

\displaystyle \begin{aligned} f_k(x) &= x \prod_{i=1}^{k} \left(1+\frac{1}{p_i-1}\right)\\ &\leq x \exp \left( \sum_{i=1}^k \frac{1}{p_i-1}\right)\\ &\leq x \exp \left( 1 + \sum_{i=1}^k \frac{1}{p_i}\right)\\ &\leq x \exp \left( \log \log (k \log k) + A \right)\\ &= O(x \log k) \end{aligned}

It uses the facts on sum of reciprocals of the primes.

Apply this to the actual algorithm running time analysis, we would get a O(\log \log m) blow up of the running time for the \Z_m^* algorithm.