Recursive structure of MOS scales: Difference between revisions

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=== Binary generated scales are MOS ===

By ''generatedness'', we mean that every interval in the scale is of the form ''jg'' + ''kp'' where ''g'' is a generator, ''p'' is the period, and ''j, k'' ∈ '''Z'''. We have shown that the result of chunking and reduction is generated and binary and the result of expanding from a reduced word is generated and binary. We need only show that binary generated scales are MOS (i.e. ~~satisfy Myhill's property~~).

By ''generatedness'', we mean that every interval in the scale is of the form ''jg'' + ''kp'' where ''g'' is a generator, ''p'' is the period, and ''j, k'' ∈ '''Z'''. We have shown that the result of chunking and reduction is generated and binary and the result of expanding from a reduced word is generated and binary. We need only show that binary generated scales are MOS (i.e. have maximum variety 2).

Suppose that such a scale ''S'' (with ''n'' ≥ 2 notes) has ''a''-many L steps and ''b''-many s steps per period ''p'', where gcd(''a'', ''b'') = 1, and has generator ''g''. First assume that ''g'' and ''p'' are linearly independent, so the step ratio of ''S'' is irrational. Since ''S'' is generated, the interval sizes modulo ''p'' that occur in ''S'' are:

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s = ''eg'' + ''fp''

for appropriate integers ''c, d, e, f'', |''c''| < ''n'', and |''e''| < ''n''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. In fact, [L, s] is another valid basis for the group generated by ''p'' and ''g'', since by binarity we have ''p, g'' ∈ span(L, s).

for appropriate integers ''c, d, e, f'', where |''c''| < ''n'' and |''e''| < ''n''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. In fact, [L, s] is another valid basis for the group generated by ''p'' and ''g'', since by binarity we have ''p, g'' ∈ span(L, s).

Assume ''c'' = ''b'' and ''e'' = −''a''. Let χ = L − s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness, binarity, and ~~linear independence of~~ ''p'' ~~and~~ ''g'', ~~the class~~ of ''k''-~~steps~~ must have ~~two~~ sizes ~~separated by~~ ''ng'', and ~~changing the k-steps outside these two sizes leaves~~ the range −(''n'' − 1)''g'', ..., 0, ..., (''n'' − 1)''g''. Thus the class of ''k''-steps has two sizes for 1 ≤ ''k'' ≤ (''n'' − 1).

Assume ''c'' = ''b'' and ''e'' = −''a''. Let χ = L − s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each ''j'' ∈ {1, ..., ''n'' − 1} there exists at most one ''k'' = ''k''(''j'') ∈ {1, ..., ''n'' − 1}</sub> such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if an interval class that has two sizes must have the sizes ''jg'' and (''j'' − ''n'')'g''; any other size must leave the range −(''n'' − 1)''g'', ..., 0, ..., (''n'' − 1)''g''. Thus the class of ''k''-steps has at most two sizes for 1 ≤ ''k'' ≤ (''n'' − 1). Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an (''n'' − ''k'')-step equivalent to ''−jg'', which must be distinct from an (''n'' − ''k'')-step equivalent to a positive number of ''g'' generators, which does occur in the brightest (most generators stacked ''up'' rather than ''down'') mode of ''S''.

To show ~~Myhillness~~ for non-linearly-independent ''p'' and ''g'', note that every ''k''-step (which is a specific linear combination of L and s) in the scale with rational step ratio is a limit point of the same linear combination of L and s in versions of the binary scale with linearly independent ''p'' and ''g'', and thus there must be ''at most'' 2 sizes for each generic interval. Since χ, which separates the two sizes, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval has ''exactly'' 2 sizes.

To show that maximum variety 2 holds for non-linearly-independent ''p'' and ''g'', note that every ''k''-step (which is a specific linear combination of L and s) in the scale with rational step ratio is a limit point of the same linear combination of L and s in versions of the binary scale with linearly independent ''p'' and ''g'', and thus there must be ''at most'' 2 sizes for each generic interval. Since χ, which separates the two sizes, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval has ''exactly'' 2 sizes.

== See also ==

@@ Line 442: / Line 442: @@
 === Binary generated scales are MOS ===
-By ''generatedness'', we mean that every interval in the scale is of the form ''jg'' + ''kp'' where ''g'' is a generator, ''p'' is the period, and ''j, k'' ∈ '''Z'''. We have shown that the result of chunking and reduction is generated and binary and the result of expanding from a reduced word is generated and binary. We need only show that binary generated scales are MOS (i.e. satisfy Myhill's property).
+By ''generatedness'', we mean that every interval in the scale is of the form ''jg'' + ''kp'' where ''g'' is a generator, ''p'' is the period, and ''j, k'' ∈ '''Z'''. We have shown that the result of chunking and reduction is generated and binary and the result of expanding from a reduced word is generated and binary. We need only show that binary generated scales are MOS (i.e. have maximum variety 2).
 Suppose that such a scale ''S'' (with ''n'' ≥ 2 notes) has ''a''-many L steps and ''b''-many s steps per period ''p'', where gcd(''a'', ''b'') = 1, and has generator ''g''. First assume that ''g'' and ''p'' are linearly independent, so the step ratio of ''S'' is irrational. Since ''S'' is generated, the interval sizes modulo ''p'' that occur in ''S'' are:
@@ Line 454: / Line 454: @@
 s = ''eg'' + ''fp''
-for appropriate integers ''c, d, e, f'', |''c''| < ''n'', and |''e''| < ''n''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. In fact, [L, s] is another valid basis for the group generated by ''p'' and ''g'', since by binarity we have ''p, g'' ∈ span(L, s).
+for appropriate integers ''c, d, e, f'', where |''c''| < ''n'' and |''e''| < ''n''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. In fact, [L, s] is another valid basis for the group generated by ''p'' and ''g'', since by binarity we have ''p, g'' ∈ span(L, s).
-Assume ''c'' = ''b'' and ''e'' = &minus;''a''. Let χ = L &minus; s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness, binarity, and linear independence of ''p'' and ''g'', the class of ''k''-steps must have two sizes separated by ''ng'', and changing the k-steps outside these two sizes leaves the range &minus;(''n'' &minus; 1)''g'', ..., 0, ..., (''n'' &minus; 1)''g''. Thus the class of ''k''-steps has two sizes for 1 ≤ ''k'' ≤ (''n'' &minus; 1).
+Assume ''c'' = ''b'' and ''e'' = &minus;''a''. Let χ = L &minus; s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each ''j'' ∈ {1, ..., ''n'' &minus; 1} there exists at most one ''k'' = ''k''(''j'') ∈ {1, ..., ''n'' &minus; 1}</sub> such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if an interval class that has two sizes must have the sizes ''jg'' and (''j'' &minus; ''n'')'g''; any other size must leave the range &minus;(''n'' &minus; 1)''g'', ..., 0, ..., (''n'' &minus; 1)''g''. Thus the class of ''k''-steps has at most two sizes for 1 ≤ ''k'' ≤ (''n'' &minus; 1). Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an (''n'' &minus; ''k'')-step equivalent to ''&minus;jg'', which must be distinct from an (''n'' &minus; ''k'')-step equivalent to a positive number of ''g'' generators, which does occur in the brightest (most generators stacked ''up'' rather than ''down'') mode of ''S''.
-To show Myhillness for non-linearly-independent ''p'' and ''g'', note that every ''k''-step (which is a specific linear combination of L and s) in the scale with rational step ratio is a limit point of the same linear combination of L and s in versions of the binary scale with linearly independent ''p'' and ''g'', and thus there must be ''at most'' 2 sizes for each generic interval. Since χ, which separates the two sizes, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval has ''exactly'' 2 sizes.
+To show that maximum variety 2 holds for non-linearly-independent ''p'' and ''g'', note that every ''k''-step (which is a specific linear combination of L and s) in the scale with rational step ratio is a limit point of the same linear combination of L and s in versions of the binary scale with linearly independent ''p'' and ''g'', and thus there must be ''at most'' 2 sizes for each generic interval. Since χ, which separates the two sizes, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval has ''exactly'' 2 sizes.
 == See also ==