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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 \(i\)th 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 \[ f_k(x) = x \prod_{i=1}^{k} \left(1+\frac{1}{p_j-1}\right) \] 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 \[ 1+\sum_{j=1}^k \frac{1}{p_j} \prod_{i=1}^j \left(1+\frac{1}{p_i -1}\right) = \prod_{j=1}^k \left(1+\frac{1}{p_j-1}\right) \]


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 \(k=1\).

\begin{align*} 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{align*}

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

Proof \begin{align*} 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{align*}

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

\begin{align*} 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{align*}

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


\[ f_k(x) = O(x \log k) \]

Proof \begin{align*} 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{align*}

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.

Posted by Chao Xu on 2015-11-20.
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