Date: Jun 12, 2013 2:15 AM
Author: JT
Subject: Combinatorics
On 12 Juni, 06:50, JT <jonas.thornv...@gmail.com> wrote:

> On 11 Juni, 22:23, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:

>

> > In article <22cede7a-ba1d-4e00-9829-8eef91562...@g9g2000vbl.googlegroups.com>,

>

> > JT <jonas.thornv...@gmail.com> wrote:

> > >Well 5!*3!*7 = 5040 and 7!=5040, tomorrow i give it another try.

>

> > Obviously, since 7! = 5! * 6 * 7, and 3! = 6. There is nothing

> > interesting in this - one of the numbers greater than 7 and

> > less-than-or-equal to 9 just happens to be a factorial. It's no more

> > significant than that 5040! = 5039! * 7!.

>

> > -- Richard

>

> The reason i looking for a pattern is i will try implement a

> permutation algorithm for elements 1-n.

>

> I have seen there is Steinhaus, Johnsson, Trotter algorithm that

> someone directed me to, my will probably be something similar working

> topdown from biggest to smallest using pairs.

> When looking at the pairs for the 5! permutation, it seem that each

> unique base pair 54,45,32 etc can produce 3! of subpairs.

>

> So for each of the 20 base pair

> 54,53,52,51,45,43,42,35,34,32,31,25,24,23,21,15,14,13,12,11 there is

> just 3! digits to play with, i was thinking this must be a working

> approach even for 7! or any number of permutations.

>

> First some combinatorial finding out how many ways to select 2 among

> 7, and from there is must follow that for each of those basepairs

> there can be 5! of combinatorial subpairs.

> I think that is the correct approach.

>

> 1. Find out how many ways 2 first digits can be combined within the

> the group of n digits and write out.

> 2. n-2 digit to calculate possible number of subpairs for each

> basepair, create write out subpair.

> 3. Repeat step 1 until n=0.

>

> 54 32 1

> 54 31 2

> 54 23 1

> 54 21 3

> 54 13 2

> 54 12 3

I will try to implement, in javascript.