Ternary scale theorems: Difference between revisions

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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 our chosen offset 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.

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 is one slot letter in the template MOS for ''s''1 and one in the template MOS ''s''2, due to index parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps of the step size that occurs less frequently in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' and 2''k'' are both even).

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 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 our chosen offset 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.
+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 is one slot letter in the template MOS for ''s''<sub>1</sub> and one in the template MOS ''s''<sub>2</sub>, due to index parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps of the step size that occurs less frequently in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' and 2''k'' are both even).

Ternary scale theorems: Difference between revisions

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