Recursive structure of MOS scales: Difference between revisions

Inthar (talk | contribs)
Inthar (talk | contribs)
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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.
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''. First we handle the case gcd(''a'', ''b'') = 1.
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''.


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:
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''
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''. [This eliminates the case gcd(''a'', ''b'') > 1, since that would mean that ''g'' does not 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''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. We 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).


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.
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.