# A question in statistics around successive bounded operations

Hi,

I have no idea if the subject header is nonsense, but it’s all I could come up with.

A composer, Thom Johnson, in 1985 wrote a piece for piano called The Chord Catalogue (excerpts are here).

The work ostensibly–and I have no reason to doubt it–is a collection of every “chord” possible within the 13-note division of the notes including and between an octave.

Ie–and in this his numbering and conception of octave division is contrary to modern Serial notation, but:
c=1,c#=2,d=3, etc., till C above=13.

The first go-round, with one-note chords, gives exactly 13 separate chords; the last go-round gives one chord, with thirteen notes in it.

The go-rounds of the chords begins, as mentioned, with one-note chords. The next go-round is one consisting of two-note chords. Etc up to 13.

The pianist/composer claims 8178 chords once the whole shebang is played through.

Is this true?

What _I’d_like to see is figuring in all the different note-arrangements _within_each chord.

Something tells me that the procedures are simple (combinometrics is a cult in serial composition and analysis). But I certainly don’t know.

Leo

For each chord, each note can either be in it, or not in it. So you have 13 different choices between 2 alternatives each, for a total of 2^13 different chords. But that would be 8192 chords, not 8178-- I’m not sure why the discrepancy.

Also note that unless you include the empty chord (i.e., no notes being played at all), it’d only be 8191.

EDIT: Come to think of it, if you require that a chord have at least two notes in it, then you eliminate the empty chord and the 13 singleton chords, which brings you back down to the 8178 claimed.

And 3-note chords would be 3^13, etc?

What am I missing?

(BTW, including the singletons was my fault. The numbers/performance on total empty sets gets real weird, if John Cage is consulted.)

No Chronos has already done the 3 chord notes for you. The 2 to a power had nothing to do with the number of notes. 2 is the number of “states” for each note. Is it included or excluded. Think of pressing the piano keys down as being 1’s and not pressing them down as being 0s. Then you have a 13 digit binary number as he explained and there are 2[sup]13[/sup] of those, but you have to exclude the no note chord and the 13 one note chords.

No, Chronos had it right. For each chord, there are 13 notes each of which can either be in it or not, giving 2[sup]13[/sup] possibilities (of which, he then excluded the empty chord and the one-note degenerate chords). That gets you all the 3-note chords, all the 4-note chords, all the 5-note chords, etc., up to all the 13-note chords (of which there is one).

This kind of set is called a power set. Given some initial set { a, b, c, … } (in this case, a set of 13 notes that can be in a chord), list ALL the possible sub-sets of that set. For n elements in the original set, there are 2[sup]n[/sup] such sub-sets (including the empty set and the full original set itself).

ETA: Although I wonder if the composition mentioned in the OP actually contains any 13-note chords. Did the pianist have to use all ten fingers/thumbs, his nose, his tongue, and some other appendage to play such chords?

Waitaminnit! Upon re-reading the OP, it says explicitly that the composition DOES include all the one-note chords and the 13-note chords! So why the number 8178 and not 8191?

NOW I want to know how the pianist played it? (See ETA to above post.) Could a female pianist have played this composition? :eek:

According to the names of the tracks of the CD, it breaks down to 7178 as follows:

But a chord " in music is any harmonic set of two or more notes that is heard as if sounding simultaneously" by definition according to wikipedia. Most dictionaries seem to require 3 notes.

We await Mr. Johnson’s next project — which will cover all 88 keys of a standard piano.