Black and White Keys

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

--- 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

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
>