Ternary scale theorems: Difference between revisions

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Note: The xen term "brightest MOS word" is equivalent to "Christoffel word" in the paper, and similarly "brightest multiMOS word" is equivalent to "powers of a Christoffel word". Also see [[Glossary for combinatorics on words]] for more equivalents between xen community terms and standard academic terminology.

== Theorem 8 (Even-regular scales as (contra)interleavings) ==

Let ''s'' be a primitive even-regular scale of [[MOS substitution]] type ''a'''''x'''(''k'''''y''' ''k'''''z''') where ''a'' is even and gcd(''a'', ''k'') = 1. Let ''n'' = |''s''| = ''a'' + 2''k''.

# If ''n'' is oddly even, then ''s'' is a [[interleaving|contrainterleaving]] of the two opposite chiralities of an odd-regular scale.

# If ''n'' is evenly even and > 4, then ''s'' is an [[interleaving]] of two copies of a smaller even-regular scale.

# If ''n'' = 4, then ''s'' is an interleaving of a 2-note MOS.

=== Proof ===

Statement 3 is trivial and is included for completeness.

Consider the 2-steps of ''s''. If ''n'' > 4, the 2-step intervals of ''s'' must be:

# if ''a'' > 2''k'': '''y''' + '''z''', otherwise: 2'''x'''

# '''x''' + '''y'''

# '''x''' + '''z'''

We also know that ''s'' is of the form ''w''('''x''', '''y''', '''z''')''w''('''x''', '''z''', '''y'''). Hence the count of '''x''' + '''y''' = the count of '''x''' + '''z''' in all of ''s''.

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

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 ''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 the n/2-degree is also this same MOS word.

First we handle the ''a'' = 2''k'' case. This means that ''k'' = gcd(''a'', ''k'') = 1, and ''a'' = 2. This is the trivial ''n'' = 4 ('''xyxz''') case. Thus it remains to handle the cases (1) and (2) above. The steps of a MOS are distributed maximally evenly. Hence the positions occupied by '''x''' + '''X''' are a maximally even subset of <math>\mathbb{Z}/(n/2)\mathbb{Z},</math> and so are the positions occupied by 2'''X''' resp. 2'''x'''. 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 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 ('''x'''+'''y''')('''x'''+'''z''') is substituted.

Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there are two slot letters in the template MOS word of 2-steps. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' is even, 2''k'' is even). 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'''.

* In the oddly 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. This makes ''s''1 and ''s''2 opposite chiralities of an odd-regular MV3 scale.

* In the evenly 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.

== Open problems ==

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 Note: The xen term "brightest MOS word" is equivalent to "Christoffel word" in the paper, and similarly "brightest multiMOS word" is equivalent to "powers of a Christoffel word". Also see [[Glossary for combinatorics on words]] for more equivalents between xen community terms and standard academic terminology.
+== Theorem 8 (Even-regular scales as (contra)interleavings) ==
+Let ''s'' be a primitive even-regular scale of [[MOS substitution]] type ''a'''''x'''(''k'''''y''' ''k'''''z''') where ''a'' is even and gcd(''a'', ''k'') = 1. Let ''n'' = |''s''| = ''a'' + 2''k''.
+# If ''n'' is oddly even, then ''s'' is a [[interleaving|contrainterleaving]] of the two opposite chiralities of an odd-regular scale.
+# If ''n'' is evenly even and > 4, then ''s'' is an [[interleaving]] of two copies of a smaller even-regular scale.
+# If ''n'' = 4, then ''s'' is an interleaving of a 2-note MOS.
+=== Proof ===
+Statement 3 is trivial and is included for completeness.
+Consider the 2-steps of ''s''. If ''n'' > 4, the 2-step intervals of ''s'' must be:
+# if ''a'' > 2''k'': '''y''' + '''z''', otherwise: 2'''x'''
+# '''x''' + '''y'''
+# '''x''' + '''z'''
+We also know that ''s'' is of the form ''w''('''x''', '''y''', '''z''')''w''('''x''', '''z''', '''y'''). Hence the count of '''x''' + '''y''' = the count of '''x''' + '''z''' in all of ''s''.
+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 the scale word of stacked 2-steps from 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.
+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 ''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 the n/2-degree is also this same MOS word.
+First we handle the ''a'' = 2''k'' case. This means that ''k'' = gcd(''a'', ''k'') = 1, and ''a'' = 2. This is the trivial ''n'' = 4 ('''xyxz''') case. Thus it remains to handle the cases (1) and (2) above. The steps of a MOS are distributed maximally evenly. Hence the positions occupied by '''x''' + '''X''' are a maximally even subset of <math>\mathbb{Z}/(n/2)\mathbb{Z},</math> and so are the positions occupied by 2'''X''' resp. 2'''x'''. 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 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 ('''x'''+'''y''')('''x'''+'''z''') is substituted.
+Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there are two slot letters in the template MOS word of 2-steps. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' is even, 2''k'' is even). 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'''.
+* In the oddly 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. This makes ''s''<sub>1</sub> and ''s''<sub>2</sub> opposite chiralities of an odd-regular MV3 scale.
+* In the evenly 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.
 == Open problems ==

Ternary scale theorems: Difference between revisions

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