User:Arseniiv/Three-gap theorem - Revision history

Arseniiv: missed italic

2026-04-15T19:06:32Z

missed italic

← Older revision		Revision as of 19:06, 15 April 2026
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	A bit less trivial is that "projecting" an element onto last row and column splits it (modulo 1): <math>\{ I_{m,N} + I_{N,n} \} = \{ (N - m + n - N) g \} = \{ (n - m) g \} = I_{m,n}</math>. The sum <math>I_{m,N} + I_{N,n}</math> is always 0 or 1 larger than <math>I_{m,n}</math>, in the former case really splitting the interval.		A bit less trivial is that "projecting" an element onto last row and column splits it (modulo 1): <math>\{ I_{m,N} + I_{N,n} \} = \{ (N - m + n - N) g \} = \{ (n - m) g \} = I_{m,n}</math>. The sum <math>I_{m,N} + I_{N,n}</math> is always 0 or 1 larger than <math>I_{m,n}</math>, in the former case really splitting the interval.

	: '''CLAIM 1a''': Two smallest intervals among the bottom-right edge <math>(m, N)</math> and <math>(N, n)</math> are the splitting of a step of a scale one generator less.		: '''CLAIM 1a''': Two smallest intervals among the bottom-right edge <math>(m, N)</math> and <math>(N, n)</math> are the splitting of a ''step'' of a scale one generator less.
	: '''CLAIM 1b''': Only one of ''m'', ''n'' can be zero, that is, at least one new step has a size that appears already at this point.		: '''CLAIM 1b''': Only one of ''m'', ''n'' can be zero, that is, at least one new step has a size that appears already at this point.
	:: ''(To be proven later.)''		:: ''(To be proven later.)''

Arseniiv at 18:46, 15 April 2026

2026-04-15T18:46:00Z

← Older revision		Revision as of 18:46, 15 April 2026
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	:: Because of the Claim 2, one of the old steps projects into a step of another old size and into one of those intervals, which depends on if its genspan was negative or positive (equivalently, if it sat on a lower or upper diagonal).		:: Because of the Claim 2, one of the old steps projects into a step of another old size and into one of those intervals, which depends on if its genspan was negative or positive (equivalently, if it sat on a lower or upper diagonal).
	: '''CLAIM 3b''': That means if new steps aren't the same size, that one is a new size.		: '''CLAIM 3b''': That means if new steps aren't the same size, that one is a new size.
	:: ''~~(To be proven later~~.~~)''~~		:: For now, a meh proof from contradiction: if the new steps are unequal and both happened before, then there should've been steps of both sizes in a scale one generator less, but there's also a third step size that was just split, so there were three step sizes in that scale, whereas we postulated just two.

Arseniiv: /* Setting */ wrong claim lol

2026-04-15T18:41:00Z

Setting: wrong claim lol

← Older revision		Revision as of 18:41, 15 April 2026
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	A bit less trivial is that "projecting" an element onto last row and column splits it (modulo 1): <math>\{ I_{m,N} + I_{N,n} \} = \{ (N - m + n - N) g \} = \{ (n - m) g \} = I_{m,n}</math>. The sum <math>I_{m,N} + I_{N,n}</math> is always 0 or 1 larger than <math>I_{m,n}</math>, in the former case really splitting the interval.		A bit less trivial is that "projecting" an element onto last row and column splits it (modulo 1): <math>\{ I_{m,N} + I_{N,n} \} = \{ (N - m + n - N) g \} = \{ (n - m) g \} = I_{m,n}</math>. The sum <math>I_{m,N} + I_{N,n}</math> is always 0 or 1 larger than <math>I_{m,n}</math>, in the former case really splitting the interval.

	: '''CLAIM 1''': ~~In case~~ <math>(m, N-1)</math> and <math>(N-1, n)</math> are ~~both steps~~ of an ''N''~~-note scale, then~~ one of ~~them, and~~ ''~~only~~'' ~~one of them gets split into two smaller intervals (steps of the (~~''N''~~+1)-note scale)~~, ~~unless 0 gets generated (~~that is, ~~<math>\{ N g \} = 0</math>)~~.		: '''CLAIM 1a''': Two smallest intervals among the bottom-right edge <math>(m, N)</math> and <math>(N, n)</math> are the splitting of a step of a scale one generator less.
			: '''CLAIM 1b''': Only one of ''m'', ''n'' can be zero, that is, at least one new step has a size that appears already at this point.
	:: ''(To be proven later.)''		:: ''(To be proven later.)''

	This claim together with the fact that for ''N'' = 1 steps are always (0,1) and (1,0), lets us mark steps of any such scale (and discharge split ones): one step always copies diagonally down-right and another is replaced with their difference. One of the steps is always on the right column and another is always on the bottom row, not coinciding in a 0. (That's obvious for another reason: there should be a step in each row and column because the last generated pitch is incident to two steps.)		This claim together with the fact that for ''N'' = 1 steps are always (0,1) and (1,0), lets us mark steps of any such scale (and discharge split ones): one step always copies diagonally down-right and another is replaced with their difference. One of the steps is always on the right column and another is always on the bottom row, not coinciding in a 0. (That's obvious for another reason: there should be a step in each row and column because the last generated pitch is incident to two steps.)

	: '''CLAIM 2''': Steps of a ~~''MOS~~ scale'' occupy the same wrapped diagonal, that is, for some constant ''C'', <math>(m, n)</math> is a step if and only if <math>C = (n-m) \bmod (N+1)</math>.		: '''CLAIM 2''': Steps of a scale with only two step sizes occupy the same wrapped diagonal, that is, for some constant ''C'', <math>(m, n)</math> is a step if and only if <math>C = (n-m) \bmod (N+1)</math>.
	:: ''(To be proven later from'' {#''L'' ⋅ ''L'' + #''s'' ⋅ ''s''} = 0''.)''		:: ''(To be proven later from'' {#''L'' ⋅ ''L'' + #''s'' ⋅ ''s''} = 0''.)''

			: '''CLAIM 3a''': After such a size, one of the new steps is (0, ''N'') or (''N'', 0).
			:: Because of the Claim 2, one of the old steps projects into a step of another old size and into one of those intervals, which depends on if its genspan was negative or positive (equivalently, if it sat on a lower or upper diagonal).
			: '''CLAIM 3b''': That means if new steps aren't the same size, that one is a new size.
			:: ''(To be proven later.)''

Arseniiv: list was autoreplaced wrong

2026-04-15T18:13:14Z

list was autoreplaced wrong

← Older revision		Revision as of 18:13, 15 April 2026
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	This kind of matrix looks like an overkill because all its values repeat in diagonals parallel to the main diagonal. It starts to look less so when we mark current scale steps (select a single element in each row and column, as pitches and steps alternate). Other trivial things to note for now are:		This kind of matrix looks like an overkill because all its values repeat in diagonals parallel to the main diagonal. It starts to look less so when we mark current scale steps (select a single element in each row and column, as pitches and steps alternate). Other trivial things to note for now are:

	'' matrices for smaller scale sizes are submatrices aligned at the top right;		* matrices for smaller scale sizes are submatrices aligned to the top right;
	'' likewise, after stacking another generator up, we just extend the matrix with a row and a column;		* likewise, after stacking another generator up, we just extend the matrix with a row and a column;
	'' each interval size is encountered once in the last row and column		* each interval size is encountered once in the last row and column.

	A bit less trivial is that "projecting" an element onto last row and column splits it (modulo 1): <math>\{ I_{m,N} + I_{N,n} \} = \{ (N - m + n - N) g \} = \{ (n - m) g \} = I_{m,n}</math>. The sum <math>I_{m,N} + I_{N,n}</math> is always 0 or 1 larger than <math>I_{m,n}</math>, in the former case really splitting the interval.		A bit less trivial is that "projecting" an element onto last row and column splits it (modulo 1): <math>\{ I_{m,N} + I_{N,n} \} = \{ (N - m + n - N) g \} = \{ (n - m) g \} = I_{m,n}</math>. The sum <math>I_{m,N} + I_{N,n}</math> is always 0 or 1 larger than <math>I_{m,n}</math>, in the former case really splitting the interval.

Arseniiv: Created page with "This is my attempt at proving the wikipedia:three-gap theorem| with added MOS-related extended facts all with a single construction, so to speak. Right now (2026-04-15) I'm looking at the interval matrix, reordered to reflect the generator chain. == Preliminaries == TODO: continued fractions are bound to get used, write on their matrix composition representation using $S = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ and $A = \begin{bmatrix} 1 & 1..."$

2026-04-15T18:08:53Z

Created page with "This is my attempt at proving the wikipedia:three-gap theorem with added MOS-related extended facts all with a single construction, so to speak. Right now (2026-04-15) I'm looking at the interval matrix, reordered to reflect the generator chain. == Preliminaries == TODO: continued fractions are bound to get used, write on their matrix composition representation using <math>S = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}</math> and <math>A = \begin{bmatrix} 1 & 1..."

New page

This is my attempt at proving the [[wikipedia:three-gap theorem|three-gap theorem]] with added MOS-related extended facts all with a single construction, so to speak.

Right now (2026-04-15) I'm looking at the interval matrix, reordered to reflect the generator chain.

== Preliminaries ==

TODO: continued fractions are bound to get used, write on their matrix composition representation using <math>S = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}</math> and <math>A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}</math>.

== Setting ==

Here we as usual for this topic work in logarithmic interval domain. Period is just 1, and generator size is ''g'' ∈ (0; 1). It's stacked ''N'' ≥ 1 times to get ''N'' + 1 notes <math>a_m = \{ m g \}, m\in 0..N</math> where {''x''} denotes the fractional part of ''x'' (equivalently, ''x'' mod 1), though it may be useful to replace 0 with 1 in some cases (hopefully not). Note that <math>a_m</math> go in generator chain order, not in the circular order of their pitch classes.

Denote an interval from <math>a_m</math> upwards to <math>a_n</math> as <math>(m, n)</math>. Let a ''genchain interval matrix'' <math>I_{m,n} = \{ a_n - a_m \} = \{ (n - m) g \}</math> contain sizes of each interval.

Example for 12edo diatonic and ''N'' = 3: <math>\frac1{12} \begin{bmatrix} 0 & 7 & 2 & 9 \\ 5 & 0 & 7 & 2 \\ 10 & 5 & 0 & 7 \\ 3 & 10 & 5 & 0 \end{bmatrix}</math>.

This kind of matrix looks like an overkill because all its values repeat in diagonals parallel to the main diagonal. It starts to look less so when we mark current scale steps (select a single element in each row and column, as pitches and steps alternate). Other trivial things to note for now are:

'' matrices for smaller scale sizes are submatrices aligned at the top right;
'' likewise, after stacking another generator up, we just extend the matrix with a row and a column;
'' each interval size is encountered once in the last row and column

A bit less trivial is that "projecting" an element onto last row and column splits it (modulo 1): <math>\{ I_{m,N} + I_{N,n} \} = \{ (N - m + n - N) g \} = \{ (n - m) g \} = I_{m,n}</math>. The sum <math>I_{m,N} + I_{N,n}</math> is always 0 or 1 larger than <math>I_{m,n}</math>, in the former case really splitting the interval.

: '''CLAIM 1''': In case <math>(m, N-1)</math> and <math>(N-1, n)</math> are both steps of an ''N''-note scale, then one of them, and ''only'' one of them gets split into two smaller intervals (steps of the (''N''+1)-note scale), unless 0 gets generated (that is, <math>\{ N g \} = 0</math>).
:: ''(To be proven later.)''

This claim together with the fact that for ''N'' = 1 steps are always (0,1) and (1,0), lets us mark steps of any such scale (and discharge split ones): one step always copies diagonally down-right and another is replaced with their difference. One of the steps is always on the right column and another is always on the bottom row, not coinciding in a 0. (That's obvious for another reason: there should be a step in each row and column because the last generated pitch is incident to two steps.)

: '''CLAIM 2''': Steps of a ''MOS scale'' occupy the same wrapped diagonal, that is, for some constant ''C'', <math>(m, n)</math> is a step if and only if <math>C = (n-m) \bmod (N+1)</math>.
:: ''(To be proven later from'' {#''L'' ⋅ ''L'' + #''s'' ⋅ ''s''} = 0''.)''