Little Book of Hexachord Theory

The math is essentially completed, I am ready to make an outline and
write my little book.

This won't make any sense out of context, but it is pretty, so I
thought I would post this piece from my book:

The inner core of my hexachord system are hexachords with 1 tritone,
and 2 tritones.

Arrangement for 1 tritone:

BCD
NXL
TSR

Numbers of each, to total all expressions:

4,2,4
4,2,4
8,4,8

Now let's consider all expressions/transpositions represented
by exactly three of each I-Ching symbols (--, -o-, ---, -e-)
This is done so we can flip between realms. There are 90 exactly
for one and two tritone hexachords.

14,4,2
10,2,8
12,12,6

Sum rows: 30, 20, 40 Sum columns: 36,18,36. Take the "inflationary
difference:

10,2,8
6,0,4
4,8,8

Sum rows: 20,10,20 Sum columns: 20,10,20

Same thing for two tritone hexachords:

4,2,8
4,2,*
4,2,4

Sum rows: 16,6,10 Sum columns 12,6,12

Inflated:

12,6,24
12,6, *
12,6,12

Sum rows 42,18,30 Sum columns 36,18,36

Inflationary difference:

8,4,16
8,4,0
8,4,8

Sum rows: 28,12,20 Sum columns: 24,12,24

Now one can identify exactly 3 heptachords that cover all
two tritone hexachords (in families). With transpositions, and
flipping of symbols (change of realm) All hexachords are covered.

With one tritone hexachords, another flip allow complementation
(flip tritones) With two tritone hexachords, flipping non-tritones
will go between M5 pairs, or in the case of G/K and F/Z, will
flip between G-K and F-Z, while transposing too...

(Leaving out many details, of course)

Mapping between realms

AYME<->GKFZ
DLR<->PUI
SXC<->QWH
OVJ<->BNT

Half a deck of cards? 13 Hearts 13 Spades?

PGH

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<phjelmstad@...> wrote:
>
> The math is essentially completed, I am ready to make an outline and
> write my little book.
>
> This won't make any sense out of context, but it is pretty, so I
> thought I would post this piece from my book:
>
> The inner core of my hexachord system are hexachords with 1 tritone,
> and 2 tritones.
>
> Arrangement for 1 tritone:
>
> BCD
> NXL
> TSR
>
> Numbers of each, to total all expressions:
>
> 4,2,4
> 4,2,4
> 8,4,8
>
> Now let's consider all expressions/transpositions represented
> by exactly three from (A,C) and three from (B,D) of these I-Ching
symbols (- -, -o-, ---, -e-) depending on tritone count (always
three without a circle and three with a circle)

> This is done so we can flip between realms. There are 90 exactly
> for one and two tritone hexachords.
>
> 14,4,12
> 10,2,8
> 12,12,16
>
> Sum rows: 30, 20, 40 Sum columns: 36,18,36. Take the "inflationary
> difference:
>
> 10,2,8
> 6,0,4
> 4,8,8
>
> Sum rows: 20,10,20 Sum columns: 20,10,20
>
> Same thing for two tritone hexachords:
>
> 4,2,8
> 4,2,*
> 4,2,4
>
> Sum rows: 16,6,10 Sum columns 12,6,12
>
> Inflated:
>
> 12,6,24
> 12,6, *
> 12,6,12
>
> Sum rows 42,18,30 Sum columns 36,18,36
>
> Inflationary difference:
>
> 8,4,16
> 8,4,0
> 8,4,8
>
> Sum rows: 28,12,20 Sum columns: 24,12,24
>
> Now one can identify exactly 3 heptachords that cover all
> two tritone hexachords (in families). With transpositions, and
> flipping of symbols (change of realm) All hexachords are covered.
>
> With one tritone hexachords, another flip allow complementation
> (flip tritones) With two tritone hexachords, flipping non-tritones
> will go between M5 pairs, or in the case of G/K and F/Z, will
> flip between G and K or F and Z.

Transposing will cross various breaks, for the non-tritone part
(4 lines for 1 tritone hexachords, 2 lines for 2 tritone hexachords)
>
> (Leaving out many details, of course)
>
> Mapping between realms
>
> AYME<->GKFZ
> DLR<->PUI
> SXC<->QWH
> OVJ<->BNT
>
> Half a deck of cards? 13 Hearts 13 Spades?
>
> PGH

Corrected errors - PGH

Ready to write my book on Hexachord Theory. I really feel I have
pushed this theory about as far as you can.

My original goal was to find a nice seven by five arrangement of the
35 hexachord types (as classified by Allen Forte and John Rahn). I
have
achieved (a few) nice arrangements based on a construction I have
done this year.

From 924 Hexachords, you get 80 types based on

C4 X C3

This reduces to 50 types based on

D12

This reduces to 34 types based on

D4 X S3

The 35 types of my theory are based on

D12 X S2 (complementation)

The 26 letters of my system are based on

D4 X S3 X S2

You can reduce further with

S4 X S3 and get 12 classes

which is used in my theory here and there.

I also used my own I-Ching system, based on Tritones,

A weight system based on 0,1,2,3 weight classes where
0,1,2,3,4,5,6,7,8,9,10,11->0,1,2,3,2,1,0,1,2,3,2,1

Attic-Upper-Lower-Cellar based on Tritone count (0,1,2,3)

The new mold I use is based on Outer(Aut(S6)->M12 (S(5,6,12)
System and the Projective Space (PG(3,2)) which has 35 lines

The first 15 lines (Picture A) are obtained from the 15
points and 15 hyperplanes of (15,35,15) Projective Space (PG(3,2))

Briefly,

(15 * 3 + 10 * 2 + 1)* 2 (sets, complements) = 132 hexads

Now I can find 15 exact hexachords from the first 45 combinations
by choosing "one per line"

8 more are found from Picture B and C (4 of 10 in each)

the last 12 hexachords of the 35 hexachords are found by replacing
the redundant 12 (6 and 6) in Pictures B and C in a systematic
way (My Hexachord Extension)

Then are different ways (one is best) for arranging all this into
a seven by five grid.

Well I don't want to write the whole book on this post, but there
is quite a bit of theory, and I think I have something fairly
consistent and elegant here.

There are many interesting relationships and coincidences, some
involving S4 X S3, and how the 132 hexads of M12 are mapped
onto the whole space of hexachords (they are exactly 1/7 of the
924, and they make up exactly 60 types of the 80, for example)

There are some bookkeeping things you can do to balance things
out....The most interesting being the way you get 20 hexachords
from the first 8 (Fixed) and then 20 more from the last 12
(Replaced) of the 80 total, or rather 8 + 12 hexachords of
D12 X S2. The Z-relation plays into all this, plus, now I have
found a way to recover almost everything about a hexachord
from its interval vector, especially the four weight classes.

Please see SteinerS6New.xls for more theory and my work, and
especially the colored grids on the last tab. I will upload
to Files - Paul Hj's Stuff

Best,

PGH