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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.

enter image description here


(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):

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    $\begingroup$ For non-musicians: a major chord can be represented as any set of three residue classes modulo 12 of the form $r,r+4,r+7$. ($r$ is the root of the chord, $r+4$ is a major third (four semitones) above the root, and $r+7$ is a perfect fifth (seven semitones) above the root.) For example, letting 0 represent the pitch $C$: an $A$ major chord would be $9,1,4\pmod{12}$, while a $D$ major chord would be $2,6,9\pmod{12}$. $\endgroup$ – Greg Martin Feb 12 '16 at 1:31
  • $\begingroup$ Note that for the purposes of this exercise, the notes are not "modulo", though. In the example A7 arpeggio, there are 3 notes with the name 'A' but each is different/distinct for this analysis. They are (0,5), (2,7) and (5,5). Each is the starting point of distinctly different "possible transitions" between arpeggios. $\endgroup$ – GreenAsJade Feb 12 '16 at 5:19
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    $\begingroup$ For what it's worth, I think this is a fantastic question! $\endgroup$ – pjs36 Feb 12 '16 at 5:45
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    $\begingroup$ I agree with pjs36, this is an interesting question, but I still can't quite fully understand it yet... keep trying! Hope someone understands this better than me. $\endgroup$ – Simply Beautiful Art Feb 13 '16 at 13:57
  • $\begingroup$ Try adding the music-theory tag for more attention. $\endgroup$ – Simply Beautiful Art Feb 13 '16 at 13:58
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One thing is for certain: The sequence of notes is eventually periodic. This is the case in the general situation that we throw a deterministic algorithm with bounded memory at a finite structure. Under suitable circumstances we can guarantee that the sequence is in fact periodic, namely at least if the algorithm is reversible. Depending on the two arpeggios this may or may not be the case: There might exist two (consecutive) notes in A such that the next highest note in D is the same for both (and hence we lose memory from which of the two we came). In fact, this situation occurs in the given example: The next highest note in $A$ after $N1\in D$ as well as after $N4\in D$ is $N5$. (This will always be the case for $A,D$ starting at the same $N1$ and differing somewhere). However, whether the "initial state" is already "in the period" or not is probably easiest determined by simulation ...

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  • $\begingroup$ I think that "lose memory of where we came from" is critical here. As you say: it is the case in the given A and D. Asking "is the initial state in the period" is a good alternative way of asking part of the question. For the given A and D, it is in the period for N=3 and it is not in the period for N=8. I don't know why not. $\endgroup$ – GreenAsJade Feb 14 '16 at 3:35
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    $\begingroup$ All in all, we have $2(|A|+|D|)-4$ "states" (being "at" a note going up or down, except that up/down does not apply to the top/bottom notes). For each point in $A$ we find the effect of "do $N-1$ steps in $A$, switch to $D$, do $N-1$ steps in $D$, switch to $A$". This is a (much depending on all that input data) map $f\colon A\to A$ and I can only repeat that I do not see any general method to decide the problem short of checking where the eventual period happens. Of course it suffices to check whether we get back to the initial point within at most $2|A|-2$ applications of the map $f$ ... $\endgroup$ – Hagen von Eitzen Feb 14 '16 at 14:37
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The answer is, to the question of whether you would arrive back to the start, is yes.

That is because notes are modular, in fact, they are modular $12$. You will only be able to go through $12$ unique chords before returning to the beginning.

However, if your chords increase from one to the next by $A$ half-steps, then we may also have

$$A=0\pmod{12}$$

In this scenario, we will not find $12$ unique chords because we will repeat back to the first chord before hitting $12$ unique chords.

To give a better answer, I require information on the distance between each note of each chord and whether or not the distance is constant when transitioning from one chord to the next (I cannot determine this because I am not a guitarist.)

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  • $\begingroup$ I'm not sure that notes, as defined for this purpose, are modular. In music, sometimes we consider notes modular because we arrive back at A after going A A# B C C# ... through all 12. However, in this exercise, low A (string 0 fret 5) is a different 'note' to middle A (string 2 fret 7) and different again to high A (string 5 fret 5). Instead, there are a fixed set of 'notes', determined by the choice of arpeggios. There is also a determined relationship between them, described in my footnote - you know whether a note is "up" or "down" from another based on that rule. $\endgroup$ – GreenAsJade Feb 12 '16 at 1:14
  • $\begingroup$ @GreenAsJade Well, that's more complicated. Since I don't understand frets and strings, I can't help you much further. Sorry. $\endgroup$ – Simply Beautiful Art Feb 12 '16 at 1:17
  • $\begingroup$ I think that the understanding needed is one of the rules, rather than the musical interpretation. (String,fret) is just a coordinate. An arpeggio is just a set of coordinates. There is a simple rule that governs what is the "next" coordinate from any given one. This "simplicity" is what makes me hope that there might be some sort of analytical approach. But - thanks for trying to help! $\endgroup$ – GreenAsJade Feb 12 '16 at 1:22
  • $\begingroup$ @GreenAsJade I understand what an arpeggio is, I just can't quite wrap my head around how to approach this any other way. Perhaps a conversion into a grand staff may help my understanding $\endgroup$ – Simply Beautiful Art Feb 12 '16 at 1:24
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    $\begingroup$ Done. Let me also apologise if it appeared I implied you didn't know what an arpeggio is. I wasn't trying to tell you what you know: rather, I was offering a more "mathematical definition" for "the purpose of this exercise". IE in this exercise, an "arpeggio" seems to be "simply an ordered set of notes" - with no other meaning. $\endgroup$ – GreenAsJade Feb 12 '16 at 5:39

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