I'm hoping that math has an answer to a question arising out of a musical exercise.
In music terms, the exercise is:
- Choose two arpeggios (sets of notes) of equal (or roughly equal) span (number of notes and distance between them)
- Start at the bottom of one, and play N notes going upwards until the top, then back down again
- After N, chose the very next note from the "other" arpeggio in the same direction you are going and repeat (play N notes)
- until you have done "every possible transition from one arpeggio to the other"
I am trying to figure out what characteristics of the arpeggios determine whether:
- you will in fact go through every possible transition from one arpeggio to the other?
- Whether this also means inherently that you arrive back where you started.
That, in a nutshell, is the question.
The exercise was given with arpeggios of length 10 and N=8
Here is my attempt to cast it in a "mathemetical way", escaping the musical context.
We have a an ordered set of nodes (aka notes :) ). In this case 30 of them. Their ordinality can be represented by a number - they are numbered, and "higher" means... higher in the numerical sense.
(N1, N2, ..., N30)
We take two subsets, call them 'A' and 'D'. These have equal size (in this case 10).
A: (N1, N5, N8, N11, N13, N17, N20, N23, N25, N29) D: (N1, N4, N6, N10, N13, N16, N18, N22, N25, N28)
An "up" transition means "find the next highest number in the target set to the current one". Inversely for a down transition.
We can transition inside the current set or to the "other" set.
The exercise is:
- Start in the direction 'up", at the lowest node in set A and make N transitions in set A
- If at any point there are no remaining notes upwards, reverse direction
- Having made N transitions in set A, transition in the same direction to set D
- Make N transitions in Set D, then transition to set A
... repeat until all transitions have been made.
The question is: what characteristics of the sets (A and D) determine whether you will take every possible transition?
(and thus arrive back at "the beginning")
With N=8, the sequence starts like this:
start in A : N1, N5, N8, N11, N13, N17, N20, N23, trans to D : N25, N28, N25, N22, N18, N16, N13, N10, trans to A : N8, N5, N1, N5, N8, N11 ...
and the question is "do we 'trans to X: NY' for all X in A,D and Y in nodes-in-A-and-D ?""
(And a simpler question - "do we ever arrive back at 'trans to A: N1'?")
I tried to represent this graphically. I laid out the initial set of 30 notes, then illustrated the subsets A and D. Then I drew in all the possible transitions.
In doing so I discovered that the specific sets (A and D) from the exercise have some interesting properties: they have some nodes in common, which result in one-way transitions, which I have highlighted.
(Edit: deleted detailed discussion of how I simulated this for guitar - I think it was a distraction).
I wrote a simulation of this. When I run it for N=3, what I see is that I do get back to the beginning and thus will repeat from here, but not every transition was covered.
A:N1 -> A:N5 -> A:N8 -> D:N10 -> D:N13 -> D:N16 -> A:N17 -> A:N20 -> A:N23 -> D:N25 -> D:N28 -> D:N25 -> A:N23 -> A:N20 -> A:N17 -> D:N16 -> D:N13 -> D:N10 -> A:N8 -> A:N5 -> A:N1 -> D:N4 -> D:N6 -> D:N10 -> A:N11 -> A:N13 -> A:N17 -> D:N18 -> D:N22 -> D:N25 -> A:N29 -> A:N25 -> A:N23 -> D:N22 -> D:N18 -> D:N16 -> A:N13 -> A:N11 -> A:N8 -> D:N6 -> D:N4 -> D:N1 -> A:N5 -> A:N8 -> A:N11 -> D:N13 -> D:N16 -> D:N18 -> A:N20 -> A:N23 -> A:N25 -> D:N28 -> D:N25 -> D:N22 -> A:N20 -> A:N17 -> A:N13 -> D:N10 -> D:N6 -> D:N4 -> A:N1
The missing two that this sequence does not transition to are A:N25 and D:N1.
With N=8 the sequence is much longer before it repeats, but also "worse". Worse in the sense that it does not arrive back at the beginning before repeating, and it leaves more "holes".
Is there some way to look at this, or represent the problem, such that you could deduce this outcome without simulating it?
Update : for those who can relate better to musical notation, here are the figures in musical notation.
The A arpeggio:
The D arpeggio
The simulation of N=3 run of the exercise - 3 notes per bar, alternating A7 arpeggio and D7 arpeggio (note that the generator does not carry accidentals over bars):