Ternary scale theorems: Difference between revisions

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== Theorem 5 (PWF scales are balanced) ==

All pairwise-well-formed scales are [[balanced]].

=== Proof ===

Let ''s'' be a PWF (thus primitive) scale. The case where ''s'' is equivalent to '''XYXZXYX''' can be manually verified, so by Theorem 4, the only remaining case is when ''s'' can be constructed by stacking two alternating sizes, '''g'''1 and '''g'''2, of ''k''-steps. We assume that ''s'' has [[step signature]] ''a'''''X''' ''b'''''Y''' ''b'''''Z''' where ''a'' is odd. This ''k'' corresponds to a class of generators of the primitive MOS ''a'''''X''' 2''b'''''W'''. This MOS is obtained from ''s'' by applying the letterwise substitution function π such that π('''X''') {{=}} '''X''' and π('''Y''') {{=}} π('''Z''') {{=}} '''W'''. Naturally, π applies to linear words, circular words, and step vector sizes. Additionally, we can choose ''k'' so that the two sizes of ''k''-steps in π(''s'') are:

Let ''s'' be a PWF (thus primitive) scale. The case where ''s'' is equivalent to '''XYXZXYX''' can be manually verified, so by Theorem 4, the only remaining case is when ''s'' can be constructed by stacking two alternating sizes, '''g'''1 and '''g'''2, of ''k''-steps. We assume that ''s'' has [[step signature]] ''a'''''X''' ''b'''''Y''' ''b'''''Z''' where ''a'' is odd. This ''k'' corresponds to a class of generators of the primitive MOS ''a'''''X''' 2''b'''''W'''. This MOS is obtained from ''s'' by applying the letterwise substitution function π such that {{nowrap|π('''X''') {{=}} '''X'''}} and {{nowrap|π('''Y''') {{=}} π('''Z''') {{=}} '''W'''}}. Naturally, π applies to linear words, circular words, and step vector sizes. Additionally, we can choose ''k'' so that the two sizes of ''k''-steps in π(''s'') are:

* the perfect generator '''g''' {{=}} ''t'''''X''' + (''k'' − ''t'')'''W''' (note (''k'' − ''t'') is odd by a previous proof), and

* the perfect generator {{nowrap|'''g''' {{=}} ''t'''''X''' + (''k'' − ''t'')'''W'''}} (note that {{nowrap|(''k'' − ''t'')}} is odd by a previous proof), and

* the imperfect generator '''i''' {{=}} (''t'' + 1)'''X''' + (''k'' − ''t'' − 1)'''W'''.

We have π('''g'''1) {{=}} π('''g'''2) ~~{{=~~}} '''g'''. Let '''h''' be the size in ''s'' such that π('''h''') {{=}} '''i'''. Hence only one ''k''-step subword ''h'' has this size in ''s''. By Theorem 1, we also may assume {{abs|'''g'''1}}'''Y''' {{=}} {{abs|'''g'''2}}'''Y''' + 1 and {{abs|'''g'''1}}'''Z''' {{=}} {{abs|'''g'''2}}'''Z''' − 1 (the other case corresponds to the opposite chirality).

We have {{nowrap|π('''g'''1) {{=}} π('''g'''2)}} = '''g'''. Let '''h''' be the size in ''s'' such that {{nowrap|π('''h''') {{=}} '''i'''}}. Hence only one ''k''-step subword ''h'' has this size in ''s''. By Theorem 1, we also may assume {{nowrap|{{abs|'''g'''1}}'''Y''' {{=}} {{abs|'''g'''2}}'''Y''' + 1}} and {{nowrap|{{abs|'''g'''1}}'''Z''' {{=}} {{abs|'''g'''2}}'''Z''' − 1}} (the other case corresponds to the opposite chirality).

As ''s'' is periodic and gcd(len(''s''), ''k'') {{=}} 1, it suffices to count letters in stacks of ''k''-steps in ''s''. The ''j''-step on any note of ''s'' can be computed by reducing a stack of ''k''-step subwords on that note, which alternate in size between '''g'''1 and '''g'''2. Note that:

As ''s'' is periodic and {{nowrap|gcd(len(''s''), ''k'') {{=}} 1}}, it suffices to count letters in stacks of ''k''-steps in ''s''. The ''j''-step on any note of ''s'' can be computed by reducing a stack of ''k''-step subwords on that note, which alternate in size between '''g'''1 and '''g'''2. Note that:

# Such a stack has a unique number ''m'' {{=}} ''m''(''j'') of ''k''-steps for a given ''j''.

# Such a stack has a unique number {{nowrap|''m'' {{=}} ''m''(''j'')}} of ''k''-steps for a given ''j''.

# Either such a stack has '''i''' as one of its ''k''-steps, or it does not.

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In summary: Let '''u'''1, '''u'''2 be the two sizes that do not include '''i''', and '''v''' be the size that does. Say that '''u'''1 has one more '''g'''1 than '''g'''2. Then

* {{abs|'''u'''1}}'''X''' {{=}} {{abs|'''u'''2}}'''X''' {{=}} {{abs|v}}'''X''' − 1

* {{nowrap|{{abs|'''u'''1}}'''X''' {{=}} {{abs|'''u'''2}}'''X'''}} {{nowrap|{{=}} {{abs|v}}'''X''' − 1}}

* {{abs|'''u'''1}}'''Y''' {{=}} {{abs|'''u'''2}}'''Y''' + 1

* {{nowrap|{{abs|'''u'''1}}'''Y''' {{=}} {{abs|'''u'''2}}'''Y''' + 1}}

* {{abs|'''u'''1}}'''Z''' {{=}} {{abs|'''u'''2}}'''Z''' − 1

* {{nowrap|{{abs|'''u'''1}}'''Z''' {{=}} {{abs|'''u'''2}}'''Z''' − 1}}

* {{abs|'''v'''}}'''Y''' {{=}} {{abs|'''v'''}}'''Z''' {{=}} {{abs|'''u'''1}}'''Y''' {{=}} min({{abs|'''u'''1}}'''Y''', {{abs|'''u'''2}}'''Y''').

* {{nowrap|{{abs|'''v'''}}'''Y''' {{=}} {{abs|'''v'''}}'''Z'''}} {{nowrap|{{=}} {{abs|'''u'''1}}'''Y'''}} {{nowrap|{{=}} min({{abs|'''u'''1}}'''Y''', {{abs|'''u'''2}}'''Y''')}}.

This proves that the set of ''j''-steps is balanced. When ''m'' is even, take the equave-complement of the set of ''j''-steps to reduce to the above case. {{qed}}

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 == Theorem 5 (PWF scales are balanced) ==
 All pairwise-well-formed scales are [[balanced]].
 === Proof ===
-Let ''s'' be a PWF (thus primitive) scale. The case where ''s'' is equivalent to '''XYXZXYX''' can be manually verified, so by Theorem 4, the only remaining case is when ''s'' can be constructed by stacking two alternating sizes, '''g'''<sub>1</sub> and '''g'''<sub>2</sub>, of ''k''-steps. We assume that ''s'' has [[step signature]] ''a'''''X''' ''b'''''Y''' ''b'''''Z''' where ''a'' is odd. This ''k'' corresponds to a class of generators of the primitive MOS ''a'''''X''' 2''b'''''W'''. This MOS is obtained from ''s'' by applying the letterwise substitution function π such that π('''X''') {{=}} '''X''' and π('''Y''') {{=}} π('''Z''') {{=}} '''W'''. Naturally, π applies to linear words, circular words, and step vector sizes. Additionally, we can choose ''k'' so that the two sizes of ''k''-steps in π(''s'') are:
+Let ''s'' be a PWF (thus primitive) scale. The case where ''s'' is equivalent to '''XYXZXYX''' can be manually verified, so by Theorem 4, the only remaining case is when ''s'' can be constructed by stacking two alternating sizes, '''g'''<sub>1</sub> and '''g'''<sub>2</sub>, of ''k''-steps. We assume that ''s'' has [[step signature]] ''a'''''X''' ''b'''''Y''' ''b'''''Z''' where ''a'' is odd. This ''k'' corresponds to a class of generators of the primitive MOS ''a'''''X''' 2''b'''''W'''. This MOS is obtained from ''s'' by applying the letterwise substitution function π such that {{nowrap|π('''X''') {{=}} '''X'''}} and {{nowrap|π('''Y''') {{=}} π('''Z''') {{=}} '''W'''}}. Naturally, π applies to linear words, circular words, and step vector sizes. Additionally, we can choose ''k'' so that the two sizes of ''k''-steps in π(''s'') are:
-* the perfect generator '''g''' {{=}} ''t'''''X''' + (''k'' &minus; ''t'')'''W''' (note (''k'' &minus; ''t'') is odd by a previous proof), and
+* the perfect generator {{nowrap|'''g''' {{=}} ''t'''''X''' + (''k'' &minus; ''t'')'''W'''}} (note that {{nowrap|(''k'' &minus; ''t'')}} is odd by a previous proof), and
 * the imperfect generator '''i''' {{=}} (''t'' + 1)'''X''' + (''k'' &minus; ''t'' &minus; 1)'''W'''.
-We have π('''g'''<sub>1</sub>) {{=}} π('''g'''<sub>2</sub>) {{=}} '''g'''. Let '''h''' be the size in ''s'' such that π('''h''') {{=}} '''i'''. Hence only one ''k''-step subword ''h'' has this size in ''s''. By Theorem 1, we also may assume {{abs|'''g'''<sub>1</sub>}}<sub>'''Y'''</sub> {{=}} {{abs|'''g'''<sub>2</sub>}}<sub>'''Y'''</sub> + 1 and {{abs|'''g'''<sub>1</sub>}}<sub>'''Z'''</sub> {{=}} {{abs|'''g'''<sub>2</sub>}}<sub>'''Z'''</sub> &minus; 1 (the other case corresponds to the opposite chirality).
+We have {{nowrap|π('''g'''<sub>1</sub>) {{=}} π('''g'''<sub>2</sub>)}} =&nbsp;'''g'''. Let '''h''' be the size in ''s'' such that {{nowrap|π('''h''') {{=}} '''i'''}}. Hence only one ''k''-step subword ''h'' has this size in ''s''. By Theorem 1, we also may assume {{nowrap|{{abs|'''g'''<sub>1</sub>}}<sub>'''Y'''</sub> {{=}} {{abs|'''g'''<sub>2</sub>}}<sub>'''Y'''</sub> + 1}} and {{nowrap|{{abs|'''g'''<sub>1</sub>}}<sub>'''Z'''</sub> {{=}} {{abs|'''g'''<sub>2</sub>}}<sub>'''Z'''</sub> &minus; 1}} (the other case corresponds to the opposite chirality).
-As ''s'' is periodic and gcd(len(''s''), ''k'') {{=}} 1, it suffices to count letters in stacks of ''k''-steps in ''s''. The ''j''-step on any note of ''s'' can be computed by reducing a stack of ''k''-step subwords on that note, which alternate in size between '''g'''<sub>1</sub> and '''g'''<sub>2</sub>. Note that:
+As ''s'' is periodic and {{nowrap|gcd(len(''s''), ''k'') {{=}} 1}}, it suffices to count letters in stacks of ''k''-steps in ''s''. The ''j''-step on any note of ''s'' can be computed by reducing a stack of ''k''-step subwords on that note, which alternate in size between '''g'''<sub>1</sub> and '''g'''<sub>2</sub>. Note that:
-# Such a stack has a unique number ''m'' {{=}} ''m''(''j'') of ''k''-steps for a given ''j''.
+# Such a stack has a unique number {{nowrap|''m'' {{=}} ''m''(''j'')}} of ''k''-steps for a given ''j''.
 # Either such a stack has '''i''' as one of its ''k''-steps, or it does not.
@@ Line 259: / Line 260: @@
 In summary: Let '''u'''<sub>1</sub>, '''u'''<sub>2</sub> be the two sizes that do not include '''i''', and '''v''' be the size that does. Say that '''u'''<sub>1</sub> has one more '''g'''<sub>1</sub> than '''g'''<sub>2</sub>. Then
-* {{abs|'''u'''<sub>1</sub>}}<sub>'''X'''</sub> {{=}} {{abs|'''u'''<sub>2</sub>}}<sub>'''X'''</sub> {{=}} {{abs|v}}<sub>'''X'''</sub> &minus; 1
+* {{nowrap|{{abs|'''u'''<sub>1</sub>}}<sub>'''X'''</sub> {{=}} {{abs|'''u'''<sub>2</sub>}}<sub>'''X'''</sub>}} {{nowrap|{{=}} {{abs|v}}<sub>'''X'''</sub> &minus; 1}}
-* {{abs|'''u'''<sub>1</sub>}}<sub>'''Y'''</sub> {{=}} {{abs|'''u'''<sub>2</sub>}}<sub>'''Y'''</sub> + 1
+* {{nowrap|{{abs|'''u'''<sub>1</sub>}}<sub>'''Y'''</sub> {{=}} {{abs|'''u'''<sub>2</sub>}}<sub>'''Y'''</sub> + 1}}
-* {{abs|'''u'''<sub>1</sub>}}<sub>'''Z'''</sub> {{=}} {{abs|'''u'''<sub>2</sub>}}<sub>'''Z'''</sub> &minus; 1
+* {{nowrap|{{abs|'''u'''<sub>1</sub>}}<sub>'''Z'''</sub> {{=}} {{abs|'''u'''<sub>2</sub>}}<sub>'''Z'''</sub> &minus; 1}}
-* {{abs|'''v'''}}<sub>'''Y'''</sub> {{=}} {{abs|'''v'''}}<sub>'''Z'''</sub> {{=}} {{abs|'''u'''<sub>1</sub>}}<sub>'''Y'''</sub> {{=}} min({{abs|'''u'''<sub>1</sub>}}<sub>'''Y'''</sub>, {{abs|'''u'''<sub>2</sub>}}<sub>'''Y'''</sub>).
+* {{nowrap|{{abs|'''v'''}}<sub>'''Y'''</sub> {{=}} {{abs|'''v'''}}<sub>'''Z'''</sub>}} {{nowrap|{{=}} {{abs|'''u'''<sub>1</sub>}}<sub>'''Y'''</sub>}} {{nowrap|{{=}} min({{abs|'''u'''<sub>1</sub>}}<sub>'''Y'''</sub>, {{abs|'''u'''<sub>2</sub>}}<sub>'''Y'''</sub>)}}.
 This proves that the set of ''j''-steps is balanced. When ''m'' is even, take the equave-complement of the set of ''j''-steps to reduce to the above case. {{qed}}