Ternary scale theorems: Difference between revisions

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Write ''s''1 for the scale word made from stacked 2-steps from the 0-degree, and let ''s''2 be as follows:

* In the singly even case, let ''s''2 be the circular word of 2-steps starting at the (''n''/2)-degree. We know that they differ only by interchanging '''y''' and '''z''', hence that they have the same period. Hence both ''s''1 and ''s''2 are primitive.

* In the doubly even case, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''2 be offset at a generator of the even-regular scale, which ~~by Theorem 6~~ we choose to have the same interval class as a bright generator of the MOS ''a'''''x''' 2''k'''''X'''. This is what induces the equality of ''s''1 and ''s''2 (in particular, the two scales have the same period, thus they are both primitive): Let ''s''''t'' be the period of the brightest mode of the template MOS, and let ''g'' be its bright generator class. Then the slice {{nowrap|''s''''t''[-''g'' +1 : 1]}} is the imperfect generator of the MOS. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''1 and ''s''2, we turn that slice into the bright generator, hence swapping ''s''''t''[-''g''] and ''s''''t''[-''g'' + 1]. Note that ''g'' must be odd since it generates a 2-period MOS. So (under 0-indexing) the first letter's index is even and the second letter's index is odd, which is what we want since the letters are within a stacked 2-step. While the generator might have to be higher by an (''n''/2)-step, that doesn't affect the parity since ''n''/2 is even.

* In the doubly even case, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''2 be offset at a generator of the even-regular scale, which we choose to have the same interval class as a bright generator of the MOS ''a'''''x''' 2''k'''''X'''. This is what induces the equality of ''s''1 and ''s''2 (in particular, the two scales have the same period, thus they are both primitive): Let ''s''''t'' be the period of the brightest mode of the template MOS, and let ''g'' be its bright generator class. Then the slice {{nowrap|''s''''t''[-''g'' +1 : 1]}} is the imperfect generator of the MOS. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''1 and ''s''2, we turn that slice into the bright generator, hence swapping ''s''''t''[-''g''] and ''s''''t''[-''g'' + 1]. Note that ''g'' must be odd since it generates a 2-period MOS. So (under 0-indexing) the first letter's index is even and the second letter's index is odd, which is what we want since the letters are within a stacked 2-step. While the generator might have to be higher by an (''n''/2)-step, that doesn't affect the parity since ''n''/2 is even.

We prove that ''s''1 and ''s''2 are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''1 and ''s''2. The word of stacked 2-steps of the template MOS (which is of the form {{nowrap|''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')}}), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from our chosen offset is also this same MOS word. Thus it remains to handle the cases (1) and (2) above. Whenever the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence (since ''n'' > 4) the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of {{nowrap|('''x'''+'''y''')('''x'''+'''z''')}}) is substituted.

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* In the singly even case, since there are evenly many slot letters in both ''s''1 and ''s''2, there are oddly many non-slot letters in both. Since ''s''1 and ''s''2 differ by interchanging '''y''' and '''z''', they have "opposite" filling letters, '''x''' + '''y''' being the opposite of '''x''' + '''z'''. This makes ''s''1 and ''s''2 opposite chiralities of an odd-regular MV3 scale.

* In the doubly even case, the number of non-slot letters in ''s''1 and ''s''2 is even, and we have a filling MOS of period 2. Since ''s''1 and ''s''2 are both primitive, they are both even-regular scales. {{Qed}}

== Theorem 7 (Ternary parallelogram scales are MOS substitution) ==

:''Main article: [[Ternary parallelogram scales are MOS substitution]]''

Ternary parallelogram scale words are [[MOS substitution]] scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.

== Open problems ==

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 Write ''s''<sub>1</sub> for the scale word made from stacked 2-steps from the 0-degree, and let ''s''<sub>2</sub> be as follows:
 * In the singly even case, let ''s''<sub>2</sub> be the circular word of 2-steps starting at the (''n''/2)-degree. We know that they differ only by interchanging '''y''' and '''z''', hence that they have the same period. Hence both ''s''<sub>1</sub> and ''s''<sub>2</sub> are primitive.
-* In the doubly even case, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''<sub>2</sub> be offset at a generator of the even-regular scale, which by Theorem 6 we choose to have the same interval class as a bright generator of the MOS ''a'''''x''' 2''k'''''X'''. This is what induces the equality of ''s''<sub>1</sub> and ''s''<sub>2</sub> (in particular, the two scales have the same period, thus they are both primitive): Let ''s''<sub>''t''</sub> be the period of the brightest mode of the template MOS, and let ''g'' be its bright generator class. Then the slice {{nowrap|''s''<sub>''t''</sub>[-''g'' +1 : 1]}} is the imperfect generator of the MOS. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''<sub>1</sub> and ''s''<sub>2</sub>, we turn that slice into the bright generator, hence swapping ''s''<sub>''t''</sub>[-''g''] and ''s''<sub>''t''</sub>[-''g'' + 1]. Note that ''g'' must be odd since it generates a 2-period MOS. So (under 0-indexing) the first letter's index is even and the second letter's index is odd, which is what we want since the letters are within a stacked 2-step. While the generator might have to be higher by an (''n''/2)-step, that doesn't affect the parity since ''n''/2 is even.
+* In the doubly even case, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''<sub>2</sub> be offset at a generator of the even-regular scale, which we choose to have the same interval class as a bright generator of the MOS ''a'''''x''' 2''k'''''X'''. This is what induces the equality of ''s''<sub>1</sub> and ''s''<sub>2</sub> (in particular, the two scales have the same period, thus they are both primitive): Let ''s''<sub>''t''</sub> be the period of the brightest mode of the template MOS, and let ''g'' be its bright generator class. Then the slice {{nowrap|''s''<sub>''t''</sub>[-''g'' +1 : 1]}} is the imperfect generator of the MOS. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''<sub>1</sub> and ''s''<sub>2</sub>, we turn that slice into the bright generator, hence swapping ''s''<sub>''t''</sub>[-''g''] and ''s''<sub>''t''</sub>[-''g'' + 1]. Note that ''g'' must be odd since it generates a 2-period MOS. So (under 0-indexing) the first letter's index is even and the second letter's index is odd, which is what we want since the letters are within a stacked 2-step. While the generator might have to be higher by an (''n''/2)-step, that doesn't affect the parity since ''n''/2 is even.
 We prove that ''s''<sub>1</sub> and ''s''<sub>2</sub> are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''<sub>1</sub> and ''s''<sub>2</sub>. The word of stacked 2-steps of the template MOS (which is of the form {{nowrap|''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')}}), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from our chosen offset is also this same MOS word. Thus it remains to handle the cases (1) and (2) above. Whenever the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence (since ''n'' > 4) the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of {{nowrap|('''x'''+'''y''')('''x'''+'''z''')}}) is substituted.
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 * In the singly even case, since there are evenly many slot letters in both ''s''<sub>1</sub> and ''s''<sub>2</sub>, there are oddly many non-slot letters in both. Since ''s''<sub>1</sub> and ''s''<sub>2</sub> differ by interchanging '''y''' and '''z''', they have "opposite" filling letters, '''x''' + '''y''' being the opposite of '''x''' + '''z'''. This makes ''s''<sub>1</sub> and ''s''<sub>2</sub> opposite chiralities of an odd-regular MV3 scale.
 * In the doubly even case, the number of non-slot letters in ''s''<sub>1</sub> and ''s''<sub>2</sub> is even, and we have a filling MOS of period 2. Since ''s''<sub>1</sub> and ''s''<sub>2</sub> are both primitive, they are both even-regular scales. {{Qed}}
+== Theorem 7 (Ternary parallelogram scales are MOS substitution) ==
+:''Main article: [[Ternary parallelogram scales are MOS substitution]]''
+Ternary parallelogram scale words are [[MOS substitution]] scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.
 == Open problems ==