Talk:K*N subgroups: Difference between revisions

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= ARCHIVED WIKISPACES DISCUSSION BELOW =
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'''All discussion below is archived from the Wikispaces export in its original unaltered form.'''
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== example ==
== Termination condition ==
Hello Gene Ward Smith,


can you please give a simple example? Thanks :)
<q>When two successive values i and i+1 lead to the same normal list</q>


- '''xenwolf''' May 29, 2011, 12:42:46 PM UTC-0700
This may cause termination too early. For example... 5-limit 3*24 subgroup of 24edo, I want a result "2.3.125 subgroup", but no additional base at i=2, cause terminating the algorithm and result in "2.3 subgroup". Simply it should be do up to i=k.
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--[[User:Dummy index|Dummy index]] ([[User talk:Dummy index|talk]]) 13:29, 13 June 2024 (UTC)
Basically, you take the val of k*N-edo, get the primes with mappings divisible by k (integer steps of N-edo), and for those that don't, find some power, product, or quotient that does. For the 2*N subgroup you'll usually square the first prime mapped to an odd number of 2*N-edo steps and multiply the prime by any other prime mapped to an odd number of steps of 2*N-edo.


- '''bootmii''' December 22, 2016, 04:30:17 PM UTC-0800
Hmm, subgroup may be extended by this → <q>add to that a basis for the commas of the k*N patent val</q>
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Commas are all unison, 0 is divisible by k, not intrusive.
Ah, before that, 5-limit 24edo is enfactored... Sorry.
--[[User:Dummy index|Dummy index]] ([[User talk:Dummy index|talk]]) 14:55, 13 June 2024 (UTC)
 
New example, 5-limit 9*19 subgroup of 19edo. Because 3 and 5 are equally flattened on 19edo, patent val of 171edo maps these to 1\171 sharpened than intervals of 19edo, respectively. i=1 and i=2 result in intervals {{(}}1/1{{)}}(*2^n), its subgroup is 2-limit. Add commas (such as unimodular basis) then we gain only 2.3^171.3^8*5 subgroup. In this case we need at least i=5, {{(}}1/1, 3^5*5^4, 3^4*5^5{{)}} then we gain 2.3^9.3^8*5 subgroup. And of course, in 3-limit, we need i=9 to gain 2.3^9 subgroup. --[[User:Dummy index|Dummy index]] ([[User talk:Dummy index|talk]]) 14:07, 23 June 2024 (UTC)

Latest revision as of 14:07, 23 June 2024

This page also contains archived Wikispaces discussion.

Termination condition

When two successive values i and i+1 lead to the same normal list

This may cause termination too early. For example... 5-limit 3*24 subgroup of 24edo, I want a result "2.3.125 subgroup", but no additional base at i=2, cause terminating the algorithm and result in "2.3 subgroup". Simply it should be do up to i=k. --Dummy index (talk) 13:29, 13 June 2024 (UTC)

Hmm, subgroup may be extended by this → add to that a basis for the commas of the k*N patent val

Commas are all unison, 0 is divisible by k, not intrusive. Ah, before that, 5-limit 24edo is enfactored... Sorry. --Dummy index (talk) 14:55, 13 June 2024 (UTC)

New example, 5-limit 9*19 subgroup of 19edo. Because 3 and 5 are equally flattened on 19edo, patent val of 171edo maps these to 1\171 sharpened than intervals of 19edo, respectively. i=1 and i=2 result in intervals {1/1}(*2^n), its subgroup is 2-limit. Add commas (such as unimodular basis) then we gain only 2.3^171.3^8*5 subgroup. In this case we need at least i=5, {1/1, 3^5*5^4, 3^4*5^5} then we gain 2.3^9.3^8*5 subgroup. And of course, in 3-limit, we need i=9 to gain 2.3^9 subgroup. --Dummy index (talk) 14:07, 23 June 2024 (UTC)