Recursive structure of MOS scales: Difference between revisions

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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, so we need only show that binary generated scales are MOS. Specifically, we claim that any such scale has ''Myhill's property with respect to p'', which we define to mean that any interval class not ''p''-equivalent to 0 has ''exactly'' 2 sizes.

Suppose that such a scale ''S'' (with ''n'' ≥ 2 notes) has ''a''-many L steps and ''b''-many s steps per period ''p'', and has generator ''g''.

Suppose that such a scale ''S'' (with ''n'' ≥ 2 notes) has ''a''-many L steps and ''b''-many s steps per period ''p'', and has generator ''g''. Since ''S'' is generated, the interval sizes modulo ''p'' that occur in ''S'' are:

~~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:~~

{(−''n'' + 1)''g'', ..., −''g'', 0, ''g'', ..., (''n'' − 1)''g''},

{~~(−''n'' + 1)''g'', ..., −''g'',~~ 0, ''g'', ..., (''n'' − 1)''g''}.

and all sizes {0, ''g'', ..., (''n'' − 1)''g''} are distinct.

We thus have:

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

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''. We hence must have gcd(''a'', ''b'') = 1, since otherwise ''g'' cannot occur in the scale~~. In fact, {L, s} is another valid basis for the abelian group with basis {''p'', ''g''}, since by binarity we have ''p, g'' ∈ span(L, s)~~.

for appropriate integers ''c, d, e, f'', where |''c''| < ''n'' and |''e''| < ''n''. The following two steps depend only on "all sizes {0, ''g'', ..., (''n'' − 1)''g''} are distinct":

# By assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p'', and from generatedness we have ''ac'' + ''be'' ∈ {0, ..., ''n'' − 1}. Hence we must have ''ac'' + ''be'' = 0. Hence ''c'' = ±''b'' and ''e'' = ∓''a''.

# We hence must have gcd(''a'', ''b'') = 1, since otherwise ''g'' cannot occur in the scale.

Assume ''c'' = ''b'' and ''e'' = −''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] 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 the class of ''k''-steps has ''at least'' two sizes, the sizes must be ''j''(''k'')''g'' and (''j''(''k'') − ''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 by linear independence must be distinct from an (''n'' − ''k'')-step equivalent to a positive number of ''g'' generators; note that the latter (''n'' − ''k'')-step does occur in the "brightest" mode of ''S'', i.e. the mode with the most ''g'' generators stacked ''up'' rather than ''down'' from the tonic.

Now we assume that ''g'' and ''p'' are linearly independent. Then {L, s} is another valid basis for the abelian group with basis {''p'', ''g''}, since by binarity we have ''p, g'' ∈ span(L, s). Assume ''c'' = ''b'' and ''e'' = −''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] 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 the class of ''k''-steps has ''at least'' two sizes, the sizes must be ''j''(''k'')''g'' and (''j''(''k'') − ''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 by linear independence must be distinct from an (''n'' − ''k'')-step equivalent to a positive number of ''g'' generators; note that the latter (''n'' − ''k'')-step does occur in the "brightest" mode of ''S'', i.e. the mode with the most ''g'' generators stacked ''up'' rather than ''down'' from the tonic.

To show Myhillness with respect to ''p'' for non-linearly-independent ''p'' and ''g'', observe 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 in the previous case, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval not ''p''-equivalent to 0 has ''exactly'' 2 sizes.

To show Myhillness with respect to ''p'' for non-linearly-independent ''p'' and ''g'', we depart from fact (*) and observe 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 in the previous case, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval not ''p''-equivalent to 0 has ''exactly'' 2 sizes.

== See also ==

@@ Line 444: / Line 444: @@
 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, so we need only show that binary generated scales are MOS. Specifically, we claim that any such scale has ''Myhill's property with respect to p'', which we define to mean that any interval class not ''p''-equivalent to 0 has ''exactly'' 2 sizes.
-Suppose that such a scale ''S'' (with ''n'' ≥ 2 notes) has ''a''-many L steps and ''b''-many s steps per period ''p'', and has generator ''g''.
+Suppose that such a scale ''S'' (with ''n'' ≥ 2 notes) has ''a''-many L steps and ''b''-many s steps per period ''p'', and has generator ''g''. Since ''S'' is generated, the interval sizes modulo ''p'' that occur in ''S'' are:
-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:
+{(&minus;''n'' + 1)''g'', ..., &minus;''g'', 0, ''g'', ..., (''n'' &minus; 1)''g''},
-{(&minus;''n'' + 1)''g'', ..., &minus;''g'', 0, ''g'', ..., (''n'' &minus; 1)''g''}.
+and all sizes {0, ''g'', ..., (''n'' &minus; 1)''g''} are distinct.
 We thus have:
@@ Line 456: / Line 456: @@
 s = ''eg'' + ''fp''
-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''. We hence must have gcd(''a'', ''b'') = 1, since otherwise ''g'' cannot occur in the scale. In fact, {L, s} is another valid basis for the abelian group with basis {''p'', ''g''}, since by binarity we have ''p, g'' ∈ span(L, s).
+for appropriate integers ''c, d, e, f'', where |''c''| < ''n'' and |''e''| < ''n''. The following two steps depend only on "all sizes {0, ''g'', ..., (''n'' &minus; 1)''g''} are distinct":
+# By assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p'', and from generatedness we have ''ac'' + ''be'' ∈ {0, ..., ''n'' &minus; 1}. Hence we must have ''ac'' + ''be'' = 0. Hence ''c'' = ±''b'' and ''e'' = ∓''a''.
+# We hence must have gcd(''a'', ''b'') = 1, since otherwise ''g'' cannot occur in the scale.
-Assume ''c'' = ''b'' and ''e'' = &minus;''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] 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 the class of ''k''-steps has ''at least'' two sizes, the sizes must be ''j''(''k'')''g'' and (''j''(''k'') &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 by linear independence must be distinct from an (''n'' &minus; ''k'')-step equivalent to a positive number of ''g'' generators; note that the latter (''n'' &minus; ''k'')-step does occur in the "brightest" mode of ''S'', i.e. the mode with the most ''g'' generators stacked ''up'' rather than ''down'' from the tonic.
+Now we assume that ''g'' and ''p'' are linearly independent. Then {L, s} is another valid basis for the abelian group with basis {''p'', ''g''}, since by binarity we have ''p, g'' ∈ span(L, s). Assume ''c'' = ''b'' and ''e'' = &minus;''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] 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 the class of ''k''-steps has ''at least'' two sizes, the sizes must be ''j''(''k'')''g'' and (''j''(''k'') &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 by linear independence must be distinct from an (''n'' &minus; ''k'')-step equivalent to a positive number of ''g'' generators; note that the latter (''n'' &minus; ''k'')-step does occur in the "brightest" mode of ''S'', i.e. the mode with the most ''g'' generators stacked ''up'' rather than ''down'' from the tonic.
-To show Myhillness with respect to ''p'' for non-linearly-independent ''p'' and ''g'', observe 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 in the previous case, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval not ''p''-equivalent to 0 has ''exactly'' 2 sizes.
+To show Myhillness with respect to ''p'' for non-linearly-independent ''p'' and ''g'', we depart from fact (*) and observe 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 in the previous case, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval not ''p''-equivalent to 0 has ''exactly'' 2 sizes.
 == See also ==