Yahoo Tuning Groups Ultimate Backup tuning-math Hexachord Interval Vector Sums

Here are sums of the interval vector values for different sets of
hexachords:

80 total hexachords:

[216,218,216,222,216,112]

50 Tn/TnI type of hexachords (reduced for mirror image)

[135,136,135,138,135,71]

44 hexachord partitions (reduced for simple complementability)

[119,121,117,125,119,59]

35 hexachord partitions (reduced for both mirror image and
complementability)

[93,98,92,100,93,49]

This is also the count of unique interval vectors.

Next, a count based on frequency of each interval vector:

(Based on this grid)

7 7 6
8 8 14
8 8 14

80 hexachords:

1 time: 6
2 times: 21
4 times: 8

50 hexachords:

1 time: 20
2 times: 15

44 hexachords:

1 time: 26
2 times: 9

35 hexachords:

1 time: 35

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> Here are sums of the interval vector values for different sets of
> hexachords:
>
> 80 total hexachords:
>
> [216,218,216,222,216,112]
>
> 50 Tn/TnI type of hexachords (reduced for mirror image)
>
> [135,136,135,138,135,71]
>
> 44 hexachord partitions (reduced for simple complementability)
>
> [119,121,117,125,119,59]
>
> 35 hexachord partitions (reduced for both mirror image and
> complementability)
>
> [93,98,92,100,93,49]
>
> This is also the count of unique interval vectors.
>
> Next, a count based on frequency of each interval vector:
>
> (Based on this grid)
>
> 7 7 6
> 8 8 14
> 8 8 14
>
> 80 hexachords:
>
> 1 time: 6
> 2 times: 21
> 4 times: 8
>
> 50 hexachords:
>
> 1 time: 20
> 2 times: 15
>
> 44 hexachords:
>
> 1 time: 26
> 2 times: 9
>
> 35 hexachords:
>
> 1 time: 35
>

This can be derived using Polya's methods to consider modes
of limited transposition, symmetry and complementability. (Or
just add up existing interval vector tables!) Now, the 924
hexachords, before reducing for transposition, add up to

[2520,2520,2520,2520,2520,1260]

Here is the proof:

The proof that the columns of sets add up to the same thing is
actually pretty simple. For example, consider hexachords in 12t-ET.
There are 924 of these or C{12,6}. Consider the intervals in all the
hexachords. These are really just subsets (dyads). Since we are
counting only unordered pairs that would be 1-6. 1-5 occur 12 times,
and tritones only 6 times, in 12t-ET. What's neat is that
transpositionally symmetrical sets don't spoil the results! That is
because we are looking at every pair in every set. So for each 1-5,
these exist in 12 * C{10,4} sets and for tritones there are 6 *C
{10,4} sets. This is 2520 for 1-5 intervals and 1260 for tritones.
Since we are taking two away from every hexachords, when considering
each dyad, this leaves C{10,4} arrangements for the 4 notes that are
left.

A little math shows that 12 * C{10,4} = (C{12,6}*C{6,2}*2)/11 and 6 *
C{10,4} = (C{12,6}*C{6,2})/11. (C{6,2} is just the 15 intervals you
get in each of the 924 interval vectors)

Paul Hj

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

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If anyone is interested, I have posted intvecsym.xls in
my Files section (Paul Hj). It's an examination of all the
symmetries of hexachords in 12t-ET. (Modes of limited
transposition, complementability, reverse complementability,
symmetry, and their implications.)

Now, one can form a baseline based on what would seem typical,
and then find what is deforming the hexachord landscape based
on that.

Starting with C{12,6}, and being a little flexible, we get
a baseline of [210,210,210,210,210,105] (Even though no set
of vectors actually adds up to this, it's because 924=77*12,
so it's a lucky start for a baseline)

The baseline for modes of limited transposition is
[10,10,10,10,10,5]

For complementability

[26,26,26,26,26,13]

Reverse complementability

[80,80,80,80,80,40]

Symmetry

[54,54,54,54,54,27]

Add everything up and divide by 4 (in correspondence with
(80+8+20+32)/4)

[95,95,95,95,95,47.5]

Tritones here present a problem. So, consider a baseline of
96: [96,96,96,96,96,48]

The variation is [-3,2,-4,4,-3,1]

The spreadsheet shows what sets, mapping into themselves, or
their complement (reverse complement, inverse...) deform
the hexachord landscape (and the transpositions that are involved
with them).

48*11= 528
variation sum = -3
_____________

525, which is 35 * 15.

I still need to find a good way to produce a baseline of 384/4=96

Applying this to my 7 X 5 hexachord grid shows some interesting
properties.

For what it's worth, here's the baselines for different stages
of hexachords, based on the 95 baseline:

After limited transpositions:

[220,220,220,220,220,110] Variation [-4,-2,-4,+2,-4,+2]

After that and mirror inverse:

[137,137,137,137,137,68.5 (Ugh)], Variation [-2,-1,-2,+1,+2.5]

After all that and complementability, and reverse complementability:
(Need to do these together since I've reduced for mirror image
already)

complementability:

[26,26,26,26,26,13]

(Which is [123,123,123,123,123,61.5] Variation [-4,-2,-6,+2,-4,-2.5]
on its own)

Reverse complementability

[80,80,80,80,80,40] (No real value on its own)

Add together and divide by two

[53,53,53,53,53,26.5]

mix this in:

137 137 137 137 137 68.5
53 53 53 53 53 26.5

190 190 190 190 190 95

divide by two

[95, 95, 95, 95, 95, 47.5] and you get the final baseline

This has a variation of [-2, +3, -3, +5, -2, +1.5] = 2.5

47.5 * 11 = 522.5 + 2.5 = 525 (35 * 11)

But I still like 96 better, even though 95 comes slightly closer,
plus, it is based on results from the spreadsheet. I just don't like
using .5 when you don't have to.

One final note: Notice how clean "sets that map themselves at the
tritone" comes out, they are A, U, P, I+, I-, and also, 2^6
complementable sets" which are all the sets (64 of them) which map
into their own complement at the tritone). They are A, M1, M5, Y, E+
and E-). This is on the spreadsheet...

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