Kite's thoughts on pergens: Difference between revisions

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==Notating rank-3 pergens==

Conventional notation is generated by the octave and the 5th, and the notation (not the tuning itself) is rank-2. Each additional pair of accidentals increases the notation's rank by one, analogous to adding primes to a JI subgroup. Enharmonics are like commas in that each one reduces the notation's rank by one (assuming they are linearly independent). Obviously, the notation's rank must match the actual tuning's rank. Therefore the minimum number of enharmonics needed always equals the difference between the notation's rank and the tuning's rank. Examples:

Conventional notation is generated by the octave and the 5th, and the notation (not the tuning itself) is rank-2. Each additional pair of accidentals increases the notation's rank by one, analogous to adding primes to a JI subgroup. Enharmonics are like vanishing commas in that each one reduces the notation's rank by one (assuming they are linearly independent). Obviously, the notation's rank must match the actual tuning's rank. Therefore the minimum number of enharmonics needed always equals the difference between the notation's rank and the tuning's rank. Examples:

{| class="wikitable" style="text-align:center;"

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! | tuning's rank

! | notation

! | notation's rank<br>~~if no enharmonics~~

! | notation's rank<br>without any EIs

! | # of ~~enharmonics~~<br>needed

! | # of EIs<br>needed

! | ~~enharmonics~~

! | EIs

|-

| | 12-edo

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! | gen1

! | gen2

! | ~~enharmonic~~

! | EI

|-

| | Marvel aka Ruyoyo

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 ==Notating rank-3 pergens==
-Conventional notation is generated by the octave and the 5th, and the notation (not the tuning itself) is rank-2. Each additional pair of accidentals increases the notation's rank by one, analogous to adding primes to a JI subgroup. Enharmonics are like commas in that each one reduces the notation's rank by one (assuming they are linearly independent). Obviously, the notation's rank must match the actual tuning's rank. Therefore the minimum number of enharmonics needed always equals the difference between the notation's rank and the tuning's rank. Examples:
+Conventional notation is generated by the octave and the 5th, and the notation (not the tuning itself) is rank-2. Each additional pair of accidentals increases the notation's rank by one, analogous to adding primes to a JI subgroup. Enharmonics are like vanishing commas in that each one reduces the notation's rank by one (assuming they are linearly independent). Obviously, the notation's rank must match the actual tuning's rank. Therefore the minimum number of enharmonics needed always equals the difference between the notation's rank and the tuning's rank. Examples:
 {| class="wikitable" style="text-align:center;"
@@ Line 1,413: / Line 1,413: @@
 ! | tuning's rank
 ! | notation
-! | notation's rank<br>if no enharmonics
+! | notation's rank<br>without any EIs
-! | # of enharmonics<br>needed
+! | # of EIs<br>needed
-! | enharmonics
+! | EIs
 |-
 | | 12-edo
@@ Line 1,541: / Line 1,541: @@
 ! | gen1
 ! | gen2
-! | enharmonic
+! | EI
 |-
 | | Marvel aka Ruyoyo