But I guess what I am saying is, is it not interesting that the
Steiner hexads comprise exactly 1/7 in each black white category?

Colors: 0B6W,1B5W,2B4W,3B3W,4B2W,5B1W
Totals 7,105,350,350,105,7
Steiner 1,15,50,50,35,1

--- In tuning-math@yahoogroups.com, "Paul H" <phjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul H" <phjelmstad@> wrote:
> >
> > Have been studying Myhill's Property and also Rothenberg Scales.
> >
> > Interesting that the diatonic and the pentatonic scales have
> > Myhill's property. So I thought I would study hexachords based
> > on black and white keys. Subsets of each have the property,
> > combined together, i don't know....
> >
> > Here is how all 924 hexachords break down
> >
> > White Keys Black Keys Hexachords
> >
> > 6 0 7
> > 5 1 105
> > 4 2 350
> > 3 3 350
> > 2 4 105
> > 1 5 7
> >
> > Hold C fixed
> >
> > 6 0 6
> > 5 1 75
> > 4 2 200
> > 3 3 150
> > 2 4 30
> > 1 5 1
> >
> > Notice subtracting Column Three Lower from Column Three Upper
> produces
> >
> > 1,30,150,200,75,6 which is just the lower set reversed. Also
> > 1+30+200=6+75+150=231 which is exactly half of 462 and 1/4 of 924..
> >
> > Now I have found that all 80 hexachord types can be expressed
> > with no more than 2 black keys.
> >
> > Now to study Rothenberg Scales and their hexachord subsets.
> >
> > PGH
>
> Now, of course the combinations for white and black keys are based on
> 7 * 1, 21 * 5, 35 * 10, 35 * 10, 21 * 5, 7 * 1 which is simple
> combinations. This gives
>
> 7,105,350,350,105,7. Now I was excited to find that the Steiner
> System I mention in my other post (Rothenberg Scales) is literally
> 1/7 of this in every combination: 1,15,50,50,15,1.
>
> I studied this a bit, and found that of course, of the 66 Steiner
> Hexads and their complements, all five blacks are used in each row.
> (If a hexad has 2 blacks, its complement will have 3, etc)
>
> Therefore 66*5=330 blacks used. Now of course one would expect
> equal usage, and its true, each black is used 66 times.
>
> Now its possible to derive an algorithm which will split the 924
> hexads into seven systems. (The first a Steiner System as given
> and the last six systems a Sextuple Steiner System, and possibly
> six independent systems?) Now for example with no blacks,
> we have one in SS, and of course 7 possible.
>
> With one black there are 15 sets, and of course each black is
> used 3 times (3 3 3 3 3) and in the Septuple system (all hexads)
> it's clear that each black is used 21 times, for each 21 white
> 5-sets (Binom(7,5)). One might expect that there are just 3 white
> 5-sets used, which is 1/7 of the 21 in the full system, so
> that you obtain a one-to-one with each black, for each one, but
> it is more subtle, there are 15 white 5-sets, and the 6 not
> used determine the key. It gets more elaborate with 3W 3B but
> the same principles apply.
>
> Finding Seven systems will determine a matching to the 35
> lines of PG(3,2), which is used to give the construction of the
> Steiner System (5,6,12) using Picture A in 3 combinations
> (15*3) and Pictures B and C (10 + 10) and a base set (012345)
> this makes 66, with their complements makes 132 hexads.
>
> I also feel that this will match up one to one with the 35
> hexachord types based on interval vector, that is D12 X S2
> where S2 is 2-complementation.
>
> PGH
>