Recursive structure of MOS scales: Difference between revisions

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=== Binary generated scales with #L coprime to #s within each period 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''', and that either ''g'' or ''−g'' occurs on every note. We have shown that the chunking procedure yields a scale that is generated and binary and that has gcd(#L, #s) = 1 within each period. We need only show 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.

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''', and that either ''g'' or ''−g'' occurs on every note. We have shown that the chunking procedure yields a scale that is generated and binary and that has gcd(#L, #s) = 1 within each period. We need only show that any such scale is a MOS. We claim 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''. Since ''S'' is generated, the interval sizes modulo ''p'' that occur in ''S'' are:

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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). 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 establish MOSness in the case of 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.

== See also ==

@@ Line 449: / Line 449: @@
 === Binary generated scales with #L coprime to #s within each period 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''', and that either ''g'' or ''&minus;g'' occurs on every note. We have shown that the chunking procedure yields a scale that is generated and binary and that has gcd(#L, #s) = 1 within each period. We need only show 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.
+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''', and that either ''g'' or ''&minus;g'' occurs on every note. We have shown that the chunking procedure yields a scale that is generated and binary and that has gcd(#L, #s) = 1 within each period. We need only show that any such scale is a MOS. We claim 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''. Since ''S'' is generated, the interval sizes modulo ''p'' that occur in ''S'' are:
@@ Line 468: / Line 468: @@
 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). 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 establish MOSness in the case of 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.
 == See also ==