Schreier-Sims algorithm

This article describes an algorithm for the following problem: strong generating set-finding problem

Solution

Basic idea

The idea behind solving the order-finding problem for $G$ comes from the following descending chain of subgroups:

$G=G^{(0)}\geq G^{(1)}\geq G^{(2)}\ldots \geq G^{(n-1)}=e$

where $G^{(i)}=G\cap Stab_{Sym(n)}{1,2,3,...,i}$.

By an easy coset counting argument:

$[G^{(i)}:G^{(i+1)}]\leq n-i$

Obstacles and overcoming them

We can readily compute a system of coset representatives for $G^{(i+1)}$ in $G^{(i)}$ by using the breadth-first search. Further information: minimal Schreier system

The problem, however, is: how do we get a generating set for $G^{(i+1)}$ starting from a generating set for $G^{(i)}$, To solve this problem, we can use Schreier's lemma which provides a constructive approach to obtaining a generating set for the subgroup from a generating set for the whole group and a system of coset representatives (the more general problem is finding a generating set from a membership test).

This still leaves another problem: Each time we go downward on the chain, the size of the generating set may go up by a factor of $n$ (because the index of the subgroup is $O(n)$). Thus, we need to keep the size of the subset small at every stage. This can be done using a small generating set-finding algorithm, such as the Sims filter or the Jerrum's filter.

Flow of the algorithm

Description of the $i^{th}$ step of the algorithm:

Start with: a generating set $A_{i}$ for $G^{(i)}$

Do the following:

1. Finding coset representatives: Using a breadth-first search, find a minimal Schreier system of coset representatives for $G^{(i+1)}$ in $G^{(i)}$
2. Finding the index: From this, obtain $[G^{(i)}:G^{(i+1)}]$
3. Finding a generating set: Using the coset representatives and the generating set $A_{i}$, obtain a (possibly larger) generating set for $G^{(i+1)}$. This uses Schreier's lemma
4. Filtering the generating set: Apply a filter to trim the size, and call the filtered generating set $A_{i+1}$.

Analysis of running time

1. Finding coset representatives: This takes time $|A_{i}|n$ since that is the number of edges in the graph.
2. Finding the index: This comes for free
3. Finding a generating set: This takes time $|A_{i}|n^{2}$ since $A_{i}|n$ terms need to be computed and the computation of each requires multiplication operations (which take $O(n)$ time).
4. Filtering the generating set: This takes time dependent on the choice of filter. A naive analysis on Jerrum's filter reveals that the time taken is $O(|A_{i}|n^{5})$ because the new generating set with which we start is itself of size $|A_{i}|n$. Similarly a naive analysis of the Sims filter reveals that the time it takes is $O(|A_{i}|n^{3})$.

Now, if we apply the filter right at the start, then at every step, $|A_{i}|$ is polynomially bounded. Thus:

Total time = Time taken to apply filter at the start + $n\times$ Time taken at each stage

Thus:

• In the case of Jerrum's filter: $n^{5}|A|+O(n^{7})$
• In the case of Sims filter: $n^{2}|A|+O(n^{6})$