# Mathematica Mathematica: Finding ideal generators as fast as possible

1. Jun 25, 2015

### jackmell

Hi,

I'd like to find ideal'' generators for integer mod groups as quickly as possible and am running into timing problems when the number is large. It's a two-step process: (1) find orbits for all combinations of a generator set, and (2) select which generators are ideal. For (1), the step that takes the longest is to locate in a list, where the orbits are for a set of generators. For example, here are small sections of the orbitList with the generator in the first column, order of generator in second, and the orbit list in third. The second table is a small section of genSet listing in each row, 4-generator pairs. The objective is to locate in orbitList, where the 4-pair generators are so that I can then obtain the orbit lists for the generator pairs.

orbitList:
$\left( \begin{array}{ccc} 5 & 6 & \{5,25,125,121,101,1\} \\ 11 & 6 & \{11,121,155,25,107,1\} \\ 13 & 2 & \{13,1\} \\ 17 & 6 & \{17,121,41,25,89,1\} \\ 19 & 6 & \{19,25,139,121,115,1\} \\ 23 & 6 & \{23,25,71,121,95,1\} \\ 29 & 2 & \{29,1\} \\ 31 & 6 & \{31,121,55,25,103,1\} \\ 37 & 6 & \{37,25,85,121,109,1\} \\ 41 & 2 & \{41,1\} \\ \end{array} \right)$
genSet:
$\left( \begin{array}{cccc} 13 & 29 & 41 & 5 \\ 13 & 29 & 41 & 11 \\ 13 & 29 & 41 & 17 \\ 13 & 29 & 41 & 19 \\ 13 & 29 & 41 & 23 \\ \end{array} \right)$
With Length[orbitList]=45, and Length[genSet]=81900 and I'm using the following code to find the locations of the 81900 generator pairs:
Code (Text):

orbitIndex=Flatten[Map[Position[orbitList[[All,1]],#,1,1]&,#]]&/@ genSet;

So for the first genrator set of (13,29,41,5), that code would return the locations in orbitList of (3,7,10,1), the locations of these generators in orbitList. To do all 81900, takes about 2.8 seconds at 2.2 GHz but that's just a small one and I'd like to optimize the code as I plan to work on larger ones too and other parts of the code takes a lot of time as well so I was wondering if someone could suggest a faster way of doing just this part first or is this as fast as I can run it?

Ok thanks for reading,
Jack

2. Jun 30, 2015