Godtone (talk | contribs)
Godtone (talk | contribs)
 
(3 intermediate revisions by the same user not shown)
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* [[User:Godtone/zeta]]
* [[User:Godtone/zeta]]
Whose significance is backed up by the more psychoacoustically-informed tuning metrics I've designed (<code>optimal_edo_sequence</code>s from [[#My Python 3 code]]):
Whose significance is backed up by the more psychoacoustically-informed tuning metrics I've designed (<code>optimal_edo_sequence</code>s from [[#My Python 3 code]]):
* [[User:Godtone/optimal edo sequences]]
* [[User:Godtone/optimal edo sequences]] (and [[User:Godtone/strict_optimal_edo_sequences]])
...as well as by my own knowledge of tuning theory.
...as well as by my own knowledge of tuning theory.


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== My Python 3 code ==
== My Python 3 code ==
IMPORTANT NOTE: there seems to be a bug for subgroup mappings at the moment, pending investigation, but ideally usage of subgroups should be made far easier too:
IMPORTANT NOTE: there seems to be bugs for subgroup mappings at the moment, pending investigation, but ideally usage of subgroups should be made far easier too:
<syntaxhighlight lang="python">
<syntaxhighlight lang="python">
>>> sg = [2, 3, 7, 11, 13, 17, 19]
>>> sg = [2, 3, 7, 11, 13, 17, 19]
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return result
return result


# the length of the mediant path of whichever octave-revoicing gives the interval x the least length
def mediant_complexity(x,revoicing_octs=7):
def mediant_complexity(x,revoicing_octs=7):
mincomplexity = 2**30
mincomplexity = 2**30
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print()
print()
return spaces
return spaces
# there is only finitely many EDOs which provide some simplification of a set of intervals as contrasted to all-distinct
def efficient_edos( n, inconsistencies=0, min_simplifications=1, edos=range(1,1000) ):
if type(n)==int:
n = odd_lim(n)
elif type(n)==list and type(n[0])==int:
n = odd_lim(1,[],n)
results = []
for edo in edos:
v = edo
if type(v)==int:
v = val( lim(max([ prime_idx(len(fact(x))-1) for x in n ])), ed(edo) )
# else v is assumed to be a mapping
m = dict()
for x in n: # collect mappings of intervals
sedo = map_iv(v,x)
if sedo in m:
m[sedo].append(x)
else:
m[sedo] = [x]
if len(inconsistent_ivs_by_val(n,v)) <= inconsistencies:
if len(n) - len([ sedo for sedo in m ]) >= min_simplifications:
results.append(edo)
return results
</syntaxhighlight>
</syntaxhighlight>