User:Arseniiv/Three-gap theorem: Difference between revisions

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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.

: '''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.)''

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''.)''

: '''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.

:: 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.

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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.
-: '''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.)''
 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''.)''
+: '''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.
+:: 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.