Hexachords in 12t-ET

If one takes the interval vectors for hexachords (there are 35 unique
ones) and classify them based on the frequency of vector values from
1 to 5 (leaving out tritones), you obtain a neat table with 5 columns
of 7 hexachord types each.

There are a few tie-breaking rules that one must impose, but they are
pretty simple.

There is one flaw: Sets P and Q in my hexachord system both appear
equally in columns 1 and 5. So you get P/Q and P/Q in those columns.
Otherwise the system is perfect.

I'll post this arrangement to my Files section.

The same works for pentachords, even though there are more sets that
require tie-breaker rules.

In my hexachord system, the sets in the seventh row are the ones that
require tie-breakers.

Here it is:

A G M1 M5 U
B1 H N1 N5 V1
B5 I O1 O5 V5
C J1 PQ PQ W
D J5 R1 R5 X
E K S1 S5 Y
F L T1 T5 Z

These columns are based on 4, 3, 1, 5 and 2 (Major third, minor
third, semitone, fourth, and whole-step) which is unfortunate
but that's how it evolved and now I am stuck with it.

The order in each column is not significant, even though there
are some interesting relationships.

You can classify based on hexachord families which are based on
A,M1,M5,Y (no tritones) and I, P, U (three tritones).

Sets B1<->B5 for example have the M5 relationship, (just swap 1 and
5 values). Reducing for this and the Z relation gives 26, which works
out nicely with our alphabet.

All the irregular sets at the bottom are Z related.

The other Z related sets are D, G, J1, J5, R1, R5, S1, S5.

I've neglected to post the interval vectors for all the sets, (I
still use John Rahn's book, but I should get something in my Files
section).

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:

Note: This will make more sense if you cross-reference it with
"My Hexachord System" on the Files Section - Paul Hj.

I've touched it up showing how every hexachord, it's complement,
which may be Z-related, inverse, and inverse complement can be
found from A, M1, M5, Y, U, P or I by holding down five notes
and varying the sixth - which I have labelled the "valiance")

Z related chords are marked with a Z, even though so far I only
list one version of a Z-related hexachord pair. (They are always
complementary in 12t-ET.) Plus I have only shown one direction of
a chord, (no inverses are shown).

I'll add in the interval vectors soon.

Paul Hj

> If one takes the interval vectors for hexachords (there are 35
unique
> ones) and classify them based on the frequency of vector values from
> 1 to 5 (leaving out tritones), you obtain a neat table with 5
columns
> of 7 hexachord types each.
>
> There are a few tie-breaking rules that one must impose, but they
are
> pretty simple.
>
> There is one flaw: Sets P and Q in my hexachord system both appear
> equally in columns 1 and 5. So you get P/Q and P/Q in those columns.
> Otherwise the system is perfect.
>
> I'll post this arrangement to my Files section.
>
> The same works for pentachords, even though there are more sets that
> require tie-breaker rules.
>
> In my hexachord system, the sets in the seventh row are the ones
that
> require tie-breakers.
>
>
> Here it is:
>
> A G M1 M5 U
> B1 H N1 N5 V1
> B5 I O1 O5 V5
> C J1 PQ PQ W
> D J5 R1 R5 X
> E K S1 S5 Y
> F L T1 T5 Z
>
> These columns are based on 4, 3, 1, 5 and 2 (Major third, minor
> third, semitone, fourth, and whole-step) which is unfortunate
> but that's how it evolved and now I am stuck with it.
>
> The order in each column is not significant, even though there
> are some interesting relationships.
>
> You can classify based on hexachord families which are based on
> A,M1,M5,Y (no tritones) and I, P, U (three tritones).
>
> Sets B1<->B5 for example have the M5 relationship, (just swap 1 and
> 5 values). Reducing for this and the Z relation gives 26, which
works
> out nicely with our alphabet.
>
> All the irregular sets at the bottom are Z related.
>
> The other Z related sets are D, G, J1, J5, R1, R5, S1, S5.
>
> I've neglected to post the interval vectors for all the sets, (I
> still use John Rahn's book, but I should get something in my Files
> section).
>

* * * * * * * * * * * * * * * * *

FUN HEXACHORD FACTS:

Of the 50 hexachord Tn/TnI types

4 can be found using only white keys
25 can be found using 1 black and 5 white keys
21 can be found using 2 black keys and 4 white keys

Of the 80 total hexachord types:

6 can be found using only white keys
46 can be found using 1 black key and 5 white keys
28 can be found using 2 black keys and 4 white keys

Of the 35 hexachord-partitions (which is also the count based
on interval vectors)

4 can be found using only white keys (in the partition that uses
the least amount of black keys)

20 can be found using 1 black key and 5 white keys (in the "lowest"
partition)

11 can be found using 2 black keys and 4 white keys (in the "lowest"
partition)

Interesting that comparing 0,2 black key counts and 1 black key
counts gives

25, 25
34, 46
15, 20

for the respective sets.

* * * * * * * * * * * * * * * *

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
>
> Note: This will make more sense if you cross-reference it with
> "My Hexachord System" on the Files Section - Paul Hj.
>
> I've touched it up showing how every hexachord, it's complement,
> which may be Z-related, inverse, and inverse complement can be
> found from A, M1, M5, Y, U, P or I by holding down five notes
> and varying the sixth - which I have labelled the "valiance")
>
> Z related chords are marked with a Z, even though so far I only
> list one version of a Z-related hexachord pair. (They are always
> complementary in 12t-ET.) Plus I have only shown one direction of
> a chord, (no inverses are shown).
>
> I'll add in the interval vectors soon.
>
> Paul Hj
>
>
> > If one takes the interval vectors for hexachords (there are 35
> unique
> > ones) and classify them based on the frequency of vector values
from
> > 1 to 5 (leaving out tritones), you obtain a neat table with 5
> columns
> > of 7 hexachord types each.
> >
> > There are a few tie-breaking rules that one must impose, but they
> are
> > pretty simple.
> >
> > There is one flaw: Sets P and Q in my hexachord system both appear
> > equally in columns 1 and 5. So you get P/Q and P/Q in those
columns.
> > Otherwise the system is perfect.
> >
> > I'll post this arrangement to my Files section.
> >
> > The same works for pentachords, even though there are more sets
that
> > require tie-breaker rules.
> >
> > In my hexachord system, the sets in the seventh row are the ones
> that
> > require tie-breakers.
> >
> >
> > Here it is:
> >
> > A G M1 M5 U
> > B1 H N1 N5 V1
> > B5 I O1 O5 V5
> > C J1 PQ PQ W
> > D J5 R1 R5 X
> > E K S1 S5 Y
> > F L T1 T5 Z
> >
> > These columns are based on 4, 3, 1, 5 and 2 (Major third, minor
> > third, semitone, fourth, and whole-step) which is unfortunate
> > but that's how it evolved and now I am stuck with it.
> >
> > The order in each column is not significant, even though there
> > are some interesting relationships.
> >
> > You can classify based on hexachord families which are based on
> > A,M1,M5,Y (no tritones) and I, P, U (three tritones).
> >
> > Sets B1<->B5 for example have the M5 relationship, (just swap 1
and
> > 5 values). Reducing for this and the Z relation gives 26, which
> works
> > out nicely with our alphabet.
> >
> > All the irregular sets at the bottom are Z related.
> >
> > The other Z related sets are D, G, J1, J5, R1, R5, S1, S5.
> >
> > I've neglected to post the interval vectors for all the sets, (I
> > still use John Rahn's book, but I should get something in my
Files
> > section).
> >
>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> * * * * * * * * * * * * * * * * *
>
> FUN HEXACHORD FACTS:
>
> Of the 50 hexachord Tn/TnI types
>
> 4 can be found using only white keys
> 25 can be found using 1 black and 5 white keys
> 21 can be found using 2 black keys and 4 white keys
>
> Of the 80 total hexachord types:
>
> 6 can be found using only white keys
> 46 can be found using 1 black key and 5 white keys
> 28 can be found using 2 black keys and 4 white keys
>
> Of the 35 hexachord-partitions (which is also the count based
> on interval vectors)
>
> 4 can be found using only white keys (in the partition that uses
> the least amount of black keys)
>
> 20 can be found using 1 black key and 5 white keys (in the "lowest"
> partition)
>
> 11 can be found using 2 black keys and 4 white keys (in the "lowest"
> partition)
>
> Interesting that comparing 0,2 black key counts and 1 black key
> counts gives
>
> 25, 25
> 34, 46
> 15, 20
>
> for the respective sets.
>
> * * * * * * * * * * * * * * * *

Reducing down to 26, (Reducing for the "M5" relation), gives 4 sets
based on six white keys, 18 sets based on 5 white keys and one black
and 4 sets based on 4 white keys and 2 blacks. The all-white key sets
belong to the M5 family (just a coincidence, the symbol has nothing
to do with the "M5" relation) They are M5, N5, one part of R5 and
one part of S5.

The 18 sets based on 5 white and 1 black are:

(one part or all of) B5, C, D, E, F, G, H, J5, K, L, O5, Q, T5, V5,
W, X, Y and Z. Just by luck L1 and T1 also make it in, but they
are already represented by L5 and T5 (I could just say "L" and "T")

The 4 sets based on 4 white keys and 2 black keys are:

A,I,P and U which are precisely those sets with "limited
tranpositions" in this case, T4, T6, T6 and T2 respectively.

Once again, with Z-related sets I am taking the sides with lowest
black key count, and also taking the 5 part of a M5 relation (B5 part
of B). The reason for representing hexachord-partitions this way
relates to taking pentachords to represent septachord-pentachord
partitions, etc. Collapsing further for B5<->B1 just shows that
most sets with black count of two are really just cluster-chords
with high interval-1 counts (chromatic collections.

* * * * * * * * * * * * * * * * * * * * * *
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@> wrote:
> >
> > Note: This will make more sense if you cross-reference it with
> > "My Hexachord System" on the Files Section - Paul Hj.
> >
> > I've touched it up showing how every hexachord, it's complement,
> > which may be Z-related, inverse, and inverse complement can be
> > found from A, M1, M5, Y, U, P or I by holding down five notes
> > and varying the sixth - which I have labelled the "valiance")
> >
> > Z related chords are marked with a Z, even though so far I only
> > list one version of a Z-related hexachord pair. (They are always
> > complementary in 12t-ET.) Plus I have only shown one direction of
> > a chord, (no inverses are shown).
> >
> > I'll add in the interval vectors soon.
> >
> > Paul Hj
> >
> >
> > > If one takes the interval vectors for hexachords (there are 35
> > unique
> > > ones) and classify them based on the frequency of vector values
> from
> > > 1 to 5 (leaving out tritones), you obtain a neat table with 5
> > columns
> > > of 7 hexachord types each.
> > >
> > > There are a few tie-breaking rules that one must impose, but
they
> > are
> > > pretty simple.
> > >
> > > There is one flaw: Sets P and Q in my hexachord system both
appear
> > > equally in columns 1 and 5. So you get P/Q and P/Q in those
> columns.
> > > Otherwise the system is perfect.
> > >
> > > I'll post this arrangement to my Files section.
> > >
> > > The same works for pentachords, even though there are more sets
> that
> > > require tie-breaker rules.
> > >
> > > In my hexachord system, the sets in the seventh row are the
ones
> > that
> > > require tie-breakers.
> > >
> > >
> > > Here it is:
> > >
> > > A G M1 M5 U
> > > B1 H N1 N5 V1
> > > B5 I O1 O5 V5
> > > C J1 PQ PQ W
> > > D J5 R1 R5 X
> > > E K S1 S5 Y
> > > F L T1 T5 Z
> > >
> > > These columns are based on 4, 3, 1, 5 and 2 (Major third, minor
> > > third, semitone, fourth, and whole-step) which is unfortunate
> > > but that's how it evolved and now I am stuck with it.
> > >
> > > The order in each column is not significant, even though there
> > > are some interesting relationships.
> > >
> > > You can classify based on hexachord families which are based on
> > > A,M1,M5,Y (no tritones) and I, P, U (three tritones).
> > >
> > > Sets B1<->B5 for example have the M5 relationship, (just swap 1
> and
> > > 5 values). Reducing for this and the Z relation gives 26, which
> > works
> > > out nicely with our alphabet.
> > >
> > > All the irregular sets at the bottom are Z related.
> > >
> > > The other Z related sets are D, G, J1, J5, R1, R5, S1, S5.
> > >
> > > I've neglected to post the interval vectors for all the sets,
(I
> > > still use John Rahn's book, but I should get something in my
> Files
> > > section).
> > >
> >
>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> * * * * * * * * * * * * * * * * *
>
> FUN HEXACHORD FACTS:
>
> Of the 50 hexachord Tn/TnI types
>
> 4 can be found using only white keys
> 25 can be found using 1 black and 5 white keys
> 21 can be found using 2 black keys and 4 white keys
>
> Of the 80 total hexachord types:
>
> 6 can be found using only white keys
> 46 can be found using 1 black key and 5 white keys
> 28 can be found using 2 black keys and 4 white keys
>
> Of the 35 hexachord-partitions (which is also the count based
> on interval vectors)
>
> 4 can be found using only white keys (in the partition that uses
> the least amount of black keys)
>
> 20 can be found using 1 black key and 5 white keys (in the "lowest"
> partition)
>
> 11 can be found using 2 black keys and 4 white keys (in the "lowest"
> partition)
>
> Interesting that comparing 0,2 black key counts and 1 black key
> counts gives
>
> 25, 25
> 34, 46
> 15, 20
>
> for the respective sets.

Reducing to 26 (reducing for the "M5" relation, so that B1 and B5 are
represented by B5 only) you get:

Zero black: 4
One black: 18
Two black: 4

The four with zero blacks are M5, N5, R5, S5 which is the M5 family
(just a coincidence, nothing to do with the "M5" relation)

The 18 with one black are:

B5,C,D,E,F,G,H,J5,K,L,O5,Q,R5,S5,T5,V5,W,X,Y, and Z, (even though
the other side of L can also qualify, plus both sides of L1, and T1,
plus one side of D, F, G, R1, Z qualify.)

The four with two blacks are A, I, P and U, which are precisely
the hexachords with "limited" transposition, in this case
T4, T6, T6, and T2 respectively.

> * * * * * * * * * * * * * * * *
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@> wrote:
> >
> > Note: This will make more sense if you cross-reference it with
> > "My Hexachord System" on the Files Section - Paul Hj.
> >
> > I've touched it up showing how every hexachord, it's complement,
> > which may be Z-related, inverse, and inverse complement can be
> > found from A, M1, M5, Y, U, P or I by holding down five notes
> > and varying the sixth - which I have labelled the "valiance")
> >
> > Z related chords are marked with a Z, even though so far I only
> > list one version of a Z-related hexachord pair. (They are always
> > complementary in 12t-ET.) Plus I have only shown one direction of
> > a chord, (no inverses are shown).
> >
> > I'll add in the interval vectors soon.
> >
> > Paul Hj
> >
> >
> > > If one takes the interval vectors for hexachords (there are 35
> > unique
> > > ones) and classify them based on the frequency of vector values
> from
> > > 1 to 5 (leaving out tritones), you obtain a neat table with 5
> > columns
> > > of 7 hexachord types each.
> > >
> > > There are a few tie-breaking rules that one must impose, but
they
> > are
> > > pretty simple.
> > >
> > > There is one flaw: Sets P and Q in my hexachord system both
appear
> > > equally in columns 1 and 5. So you get P/Q and P/Q in those
> columns.
> > > Otherwise the system is perfect.
> > >
> > > I'll post this arrangement to my Files section.
> > >
> > > The same works for pentachords, even though there are more sets
> that
> > > require tie-breaker rules.
> > >
> > > In my hexachord system, the sets in the seventh row are the
ones
> > that
> > > require tie-breakers.
> > >
> > >
> > > Here it is:
> > >
> > > A G M1 M5 U
> > > B1 H N1 N5 V1
> > > B5 I O1 O5 V5
> > > C J1 PQ PQ W
> > > D J5 R1 R5 X
> > > E K S1 S5 Y
> > > F L T1 T5 Z
> > >
> > > These columns are based on 4, 3, 1, 5 and 2 (Major third, minor
> > > third, semitone, fourth, and whole-step) which is unfortunate
> > > but that's how it evolved and now I am stuck with it.
> > >
> > > The order in each column is not significant, even though there
> > > are some interesting relationships.
> > >
> > > You can classify based on hexachord families which are based on
> > > A,M1,M5,Y (no tritones) and I, P, U (three tritones).
> > >
> > > Sets B1<->B5 for example have the M5 relationship, (just swap 1
> and
> > > 5 values). Reducing for this and the Z relation gives 26, which
> > works
> > > out nicely with our alphabet.
> > >
> > > All the irregular sets at the bottom are Z related.
> > >
> > > The other Z related sets are D, G, J1, J5, R1, R5, S1, S5.
> > >
> > > I've neglected to post the interval vectors for all the sets,
(I
> > > still use John Rahn's book, but I should get something in my
> Files
> > > section).
> > >
> >
>