Generator-offset property: Difference between revisions
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[Note: This is not true with SGA replaced with GO; [[blackdye]] is a counterexample that is MV4.] | [Note: This is not true with SGA replaced with GO; [[blackdye]] is a counterexample that is MV4.] | ||
==== Proof ==== | ==== Proof ==== | ||
Let | Let e be the equave of ''S''. | ||
Assuming SGA, we have two chains of generator g<sub>0</sub> (going right). The two cases are: | Assuming SGA, we have two chains of generator g<sub>0</sub> (going right). The two cases are: | ||
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Label the notes (1, ''j'') and (2, ''j''), 1 ≤ ''j'' ≤ (number of notes in the chain), for notes in the upper and lower chain respectively. | Label the notes (1, ''j'') and (2, ''j''), 1 ≤ ''j'' ≤ (number of notes in the chain), for notes in the upper and lower chain respectively. | ||
===== Statement (1) ===== | ===== Statement (1) ===== | ||
In case 1, let g<sub>1</sub> = (2, 1) − (1, 1), g<sub>2</sub> = (1, 2) − (2, 1), and g<sub>3</sub> = (1, 1) − (''n''/2, 2) = (−''n''/2*g<sub>1</sub> − g<sub>1</sub> − ''n''/2*g<sub>2</sub>) mod | In case 1, let g<sub>1</sub> = (2, 1) − (1, 1), g<sub>2</sub> = (1, 2) − (2, 1), and g<sub>3</sub> = (1, 1) − (''n''/2, 2) = (−''n''/2*g<sub>1</sub> − g<sub>1</sub> − ''n''/2*g<sub>2</sub>) mod e. We assume that g<sub>1</sub>, g<sub>2</sub> and e are '''Z'''-linearly independent. We have the chain g<sub>1</sub> g<sub>2</sub> g<sub>1</sub> g<sub>2</sub> ... g<sub>1</sub> g<sub>3</sub> which visits every note in ''S''. | ||
Since ''S'' is GO it is well-formed with respect to g = (g<sub>2</sub> + g<sub>1</sub>). Since g<sub>1</sub> and g<sub>2</sub> subtend the same number of steps, all multiples of the generator g must be even-steps, and those intervals that are "offset" by g<sub>1</sub> must be odd-steps. Letting ''M'' be the subset of all even-numbered notes (which are generated by g) and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to g, thus ''M'' (and its offset) must be a mos subset. Hence (g<sub>3</sub> + g<sub>1</sub>), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus g<sub>2</sub> and g<sub>3</sub> subtend the same number of steps, a fact we need in order to be able to substitute one instance or g<sub>2</sub> with g<sub>3</sub> in the next part. | Since ''S'' is GO it is well-formed with respect to g = (g<sub>2</sub> + g<sub>1</sub>). Since g<sub>1</sub> and g<sub>2</sub> subtend the same number of steps, all multiples of the generator g must be even-steps, and those intervals that are "offset" by g<sub>1</sub> must be odd-steps. Letting ''M'' be the subset of all even-numbered notes (which are generated by g) and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to g, thus ''M'' (and its offset) must be a mos subset. Hence (g<sub>3</sub> + g<sub>1</sub>), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus g<sub>2</sub> and g<sub>3</sub> subtend the same number of steps, a fact we need in order to be able to substitute one instance or g<sub>2</sub> with g<sub>3</sub> in the next part. | ||
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# from g<sub>2</sub> ... g<sub>2</sub>, we get a<sub>2</sub> = (''r'' − 1)/2*g<sub>0</sub> + g<sub>2</sub> = (''r'' − 1/2) g<sub>1</sub> + (''r'' + 1/2) g<sub>2</sub> | # from g<sub>2</sub> ... g<sub>2</sub>, we get a<sub>2</sub> = (''r'' − 1)/2*g<sub>0</sub> + g<sub>2</sub> = (''r'' − 1/2) g<sub>1</sub> + (''r'' + 1/2) g<sub>2</sub> | ||
# from g<sub>2</sub> (...even # of gens...) g<sub>1</sub> g<sub>3</sub> g<sub>1</sub> (...even # of gens...) g<sub>2</sub>, we get a<sub>3</sub> = (''r'' − 1)/2 g<sub>1</sub> + (''r'' − 1)/2 g<sub>2</sub> + g<sub>3</sub> ≡ (''r'' − ''n''/2 − 3/2)g<sub>1</sub> + (''r'' − ''n''/2 − 1/2)g<sub>2</sub> mod e. | # from g<sub>2</sub> (...even # of gens...) g<sub>1</sub> g<sub>3</sub> g<sub>1</sub> (...even # of gens...) g<sub>2</sub>, we get a<sub>3</sub> = (''r'' − 1)/2 g<sub>1</sub> + (''r'' − 1)/2 g<sub>2</sub> + g<sub>3</sub> ≡ (''r'' − ''n''/2 − 3/2)g<sub>1</sub> + (''r'' − ''n''/2 − 1/2)g<sub>2</sub> mod e. | ||
# from g<sub>1</sub> (...odd # of gens...) g<sub>1</sub> g<sub>3</sub> g<sub>1</sub> (...odd # of gens...) g<sub>1</sub>, we get a<sub>4</sub> = (''r'' + 1)/2 g<sub>1</sub> + (''r'' − 3)/2 g<sub>2</sub> + g<sub>3</sub> ≡ (''r'' − ''n''/2 − 1/2)g<sub>1</sub> + (''r'' − ''n''/2 − 3/2)g<sub>2</sub> mod | # from g<sub>1</sub> (...odd # of gens...) g<sub>1</sub> g<sub>3</sub> g<sub>1</sub> (...odd # of gens...) g<sub>1</sub>, we get a<sub>4</sub> = (''r'' + 1)/2 g<sub>1</sub> + (''r'' − 3)/2 g<sub>2</sub> + g<sub>3</sub> ≡ (''r'' − ''n''/2 − 1/2)g<sub>1</sub> + (''r'' − ''n''/2 − 3/2)g<sub>2</sub> mod e. | ||
These are all distinct by '''Z'''-linear independence. By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g<sub>1</sub> and g<sub>2</sub> must themselves be step sizes. Thus we see that an even-length, abstractly SV3, GO scale must be of the form (xy)<sup>''r''</sup>xz. (Note that (xy)<sup>''r''</sup>xz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1). | These are all distinct by '''Z'''-linear independence. By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g<sub>1</sub> and g<sub>2</sub> must themselves be step sizes. Thus we see that an even-length, abstractly SV3, GO scale must be of the form (xy)<sup>''r''</sup>xz. (Note that (xy)<sup>''r''</sup>xz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1). | ||
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Consider the two alternants, detemperings of the generator ''i''X + ''j''W of the ''a''X 2''b''W mos ''T''(X, W) = ''S''(X, W, W) where gcd(''j'', 2''k'') = 1. | Consider the two alternants, detemperings of the generator ''i''X + ''j''W of the ''a''X 2''b''W mos ''T''(X, W) = ''S''(X, W, W) where gcd(''j'', 2''k'') = 1. | ||
'''Claim 1''': Deleting X's from the alternants of ''S'' gives every ''j''-step subword in the scale | '''Claim 1''': Deleting X's from the alternants of ''S'' gives every ''j''-step subword in the scale ''E''<sub>X</sub>(''S'')(Y, Z), the scale word obtained by deleting all X's from ''S''. | ||
Proof: Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has ''j'' + 1 W's and the perfect generator has ''j'' W's. Suppose that one ''j''-step word ''R'' on index ''p'' of | Proof: Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has ''j'' + 1 W's and the perfect generator has ''j'' W's. Suppose that one ''j''-step word ''R'' on index ''p'' of ''E''<sub>X</sub>(''S'') is "contained in" the corresponding "imperfect alternant" of ''S'', which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j'']. By this we mean that ''E''<sub>X</sub>(''I'') has ''R'' as a substring. Then ''S''[''p'' − 1: ''p'' − 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both detemperings of perfect generators of ''T'', and have one fewer step that is Y or Z. Thus the word ''I'' must both begin and end in a letter that is either Y or Z. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step we started with, with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords are all contained in "perfect" alternants; take ''q'' ≠ ''p'' to be the index of the first letter of one such ''j''-step subword (as contained in ''S'') and use S[''q'' : ''q'' + ''i'' + ''j'']. | ||
'''Claim 2''': If a binary scale ''U'' has ''b'' Y's and ''b'' Z's, gcd(''j'', 2''b'') = 1, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then ''U'' = (YZ)<sup>''b''</sup>. | '''Claim 2''': If a binary scale ''U'' has ''b'' Y's and ''b'' Z's, gcd(''j'', 2''b'') = 1, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then ''U'' = (YZ)<sup>''b''</sup>. | ||
Proof: Write u and v for the two sizes of ''j''-steps. Since gcd(''j'', 2''b'') = 1 there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd because gcd(''m'', 2''b'') = 1. Hence the scale steps of ''U'' are (uv)<sup>(''m''−1)/2</sup>u mod | Proof: Write u and v for the two sizes of ''j''-steps. Since gcd(''j'', 2''b'') = 1 there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd because gcd(''m'', 2''b'') = 1. Hence the scale steps of ''U'' are (uv)<sup>(''m''−1)/2</sup>u mod e and (vu)<sup>(''m''−1)/2</sup>v mod e, and the step sizes clearly alternate. | ||
These two claims prove that | These two claims prove that ''E''<sub>X</sub>(S) = (YZ)<sup>''b''</sup> and that the two alternants' sizes differ by replacing one Y for a Z. | ||
===== Statement (5) ===== | ===== Statement (5) ===== | ||
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# The sum of sizes of consecutive chunks of X (1st chunk with 2nd chunk, 3rd with 4th, ...) must form a mos. | # The sum of sizes of consecutive chunks of X (1st chunk with 2nd chunk, 3rd with 4th, ...) must form a mos. | ||
# If chunk sizes of a binary scale form a mos, the scale itself must be a mos; see [[Recursive structure of MOS scales]]. | # If chunk sizes of a binary scale form a mos, the scale itself must be a mos; see [[Recursive structure of MOS scales]]. | ||
Lastly, | Lastly, ''E''<sub>X</sub>(''S'') is the mos ''b''Y ''b''Z; hence ''S'' is elimination-mos. | ||
=== Proposition 2 (Odd GO scales are SGA) === | === Proposition 2 (Odd GO scales are SGA) === | ||