Maximal evenness - Revision history

Inthar: /* Complement of a maximally even subset is maximally even */

2026-02-08T16:52:22Z

Complement of a maximally even subset is maximally even

← Older revision		Revision as of 16:52, 8 February 2026
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	Case 1: ME(''n'', ''m'') where {{nowrap\|''n'' < {{frac\|''m''\|2}}}}. This is a maximally even subset of Z/mZ with step sizes {{nowrap\|L > s > 1}}, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are {{nowrap\|L − 2}} and {{nowrap\|s − 2}} (0 is a valid chunk size), and the sizes form a maximally even MOS.		Case 1: ME(''n'', ''m'') where {{nowrap\|''n'' < {{frac\|''m''\|2}}}}. This is a maximally even subset of Z/mZ with step sizes {{nowrap\|L > s > 1}}, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are {{nowrap\|L − 2}} and {{nowrap\|s − 2}} (0 is a valid chunk size), and the sizes form a maximally even MOS.

	Case 2: ME(''n'', ''m'') where {{nowrap\|''n'' > {{frac\|''m''\|2}}}}. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since {{nowrap\|''n'' > {{frac\|''m''\|2}}}}) occupy a maximally even subset of ~~the slots of ME(''~~n~~'', ''m'')~~ (*). Now replace each 1 with "\|" and each 2 with "$\|".		Case 2: ME(''n'', ''m'') where {{nowrap\|''n'' > {{frac\|''m''\|2}}}}. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since {{nowrap\|''n'' > {{frac\|''m''\|2}}}}) occupy a maximally even subset of <math>\mathbb{Z}/n\mathbb{Z}</math> (*). Now replace each 1 with "\|" and each 2 with "$\|".

	(e.g.) {{nowrap\|2112111 → <nowiki>$\|\|\|$\|\|\|\|</nowiki>}}		(e.g.) {{nowrap\|2112111 → <nowiki>$\|\|\|$\|\|\|\|</nowiki>}}

Inthar: /* Complement of a maximally even subset is maximally even */

2025-08-05T23:57:26Z

Complement of a maximally even subset is maximally even

← Older revision		Revision as of 23:57, 5 August 2025
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	Case 1: ME(''n'', ''m'') where {{nowrap\|''n'' < {{frac\|''m''\|2}}}}. This is a maximally even subset of Z/mZ with step sizes {{nowrap\|L > s > 1}}, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are {{nowrap\|L − 2}} and {{nowrap\|s − 2}} (0 is a valid chunk size), and the sizes form a maximally even MOS.		Case 1: ME(''n'', ''m'') where {{nowrap\|''n'' < {{frac\|''m''\|2}}}}. This is a maximally even subset of Z/mZ with step sizes {{nowrap\|L > s > 1}}, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are {{nowrap\|L − 2}} and {{nowrap\|s − 2}} (0 is a valid chunk size), and the sizes form a maximally even MOS.

	Case 2: ME(''n'', ''m'') where {{nowrap\|''n'' > {{frac\|''m''\|2}}}}. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since {{nowrap\|''n'' > {{frac\|''m''\|2}}}} occupy a maximally even subset of the slots of ME(''n'', ''m'') (*). Now replace each 1 with "\|" and each 2 with "$\|".		Case 2: ME(''n'', ''m'') where {{nowrap\|''n'' > {{frac\|''m''\|2}}}}. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since {{nowrap\|''n'' > {{frac\|''m''\|2}}}}) occupy a maximally even subset of the slots of ME(''n'', ''m'') (*). Now replace each 1 with "\|" and each 2 with "$\|".

	(e.g.) {{nowrap\|2112111 → <nowiki>$\|\|\|$\|\|\|\|</nowiki>}}		(e.g.) {{nowrap\|2112111 → <nowiki>$\|\|\|$\|\|\|\|</nowiki>}}

Sintel: remove sections about concoctic scales and calenders

2025-05-18T09:05:22Z

remove sections about concoctic scales and calenders

← Older revision		Revision as of 09:05, 18 May 2025
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	Consider the resulting binary word of "\|" and "$". The "\|"s form chunks of sizes that differ by 1 and are distributed in a MOS way by (*). The desired complement, occupied by the "$"'s, thus forms a maximally even subset.		Consider the resulting binary word of "\|" and "$". The "\|"s form chunks of sizes that differ by 1 and are distributed in a MOS way by (*). The desired complement, occupied by the "$"'s, thus forms a maximally even subset.

	~~== Concoctic scales ==~~
	The generator of a maximally even scale is a modular multiplicative inverse of it's number of notes and the EDO size. A maximal even scale whose generator is equal to it's note amount is called [[concoctic]]. Major and minor scales in standard Western music are such—the generator is a perfect fifth of 7 semitones, as inferred through Pythagorean tuning, and the scale has 7 notes in it.

	== Sound perception ==		== Sound perception ==
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	We simply merge {{nowrap\|52 & 293}} in a selected limit to get our answer. Let's say 17 limit, we get a {{nowrap\|52 & 243c}} temperament with a comma list 225/224, 715/714, 2880/2873, 22750/22627 and 60112/60025.		We simply merge {{nowrap\|52 & 293}} in a selected limit to get our answer. Let's say 17 limit, we get a {{nowrap\|52 & 243c}} temperament with a comma list 225/224, 715/714, 2880/2873, 22750/22627 and 60112/60025.

	~~Let's see what temperament does the Tabular Persian or Dee calendar offer {{nowrap\|(29 & 33)}}. In the 5-limit, we get a contorted Lala-Quinyo (553584375:536870912).~~

	~~== Sonifications ==~~
	* Maximally even heptatonic scale of [[19edo]] is the leap year arrangement of the {{w\|Hebrew calendar}}.
	* Maximally even octatonic scale of [[33edo]] is a leap year arrangement of the Dee calendar and the tabular, evened version of the {{w\|Iranian calendars\|Persian calendar}}.

	[[Category:Scale]]		[[Category:Scale]]
	[[Category:Todo:cleanup]]		[[Category:Todo:cleanup]]

ArrowHead294: /* Discovery of temperaments with a given generator */

2025-03-19T17:31:13Z

Discovery of temperaments with a given generator

← Older revision		Revision as of 17:31, 19 March 2025
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	Let's say we want to see what would repeatedly stacking 11th harmonic do well in all of 11-limit, in an EDO that presents it well.		Let's say we want to see what would repeatedly stacking 11th harmonic do well in all of 11-limit, in an EDO that presents it well.

	11/8 amounts to 17 steps of 37edo, and the solution to the problem {{nowrap\|17*''x'' (mod 1) {{=}} 37}} is 24, meaning if the generator is 11/8, we are dealing with a 24 tone maximally even scale. As such, the temperament we are looking for is 24 & 37, which can be interpreted as [[freivald]] or [[emka]].		11/8 amounts to 17 steps of 37edo, and the solution to the problem {{nowrap\|17*''x'' (mod 1) {{=}} 37}} is 24, meaning if the generator is 11/8, we are dealing with a 24 tone maximally even scale. As such, the temperament we are looking for is {{nowrap\|24 & 37}}, which can be interpreted as [[freivald]] or [[emka]].

	=== Example 3: On-request maximum evenness scales ===		=== Example 3: On-request maximum evenness scales ===
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	We simply merge {{nowrap\|52 & 293}} in a selected limit to get our answer. Let's say 17 limit, we get a {{nowrap\|52 & 243c}} temperament with a comma list 225/224, 715/714, 2880/2873, 22750/22627 and 60112/60025.		We simply merge {{nowrap\|52 & 293}} in a selected limit to get our answer. Let's say 17 limit, we get a {{nowrap\|52 & 243c}} temperament with a comma list 225/224, 715/714, 2880/2873, 22750/22627 and 60112/60025.

	Let's see what temperament does the Tabular Persian or Dee calendar offer {{nowrap\|(29 & 33)}}. In the 5-limit, we get a contorted Lala-Quinyo (553584375:536870912).		Let's see what temperament does the Tabular Persian or Dee calendar offer {{nowrap\|(29 & 33)}}. In the 5-limit, we get a contorted Lala-Quinyo (553584375:536870912).

	== Sonifications ==		== Sonifications ==

ArrowHead294: /* Mathematics */

2025-03-19T15:08:21Z

Mathematics

← Older revision		Revision as of 15:08, 19 March 2025
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	Mathematically, if {{nowrap\|0 < ''n'' < ''m''}}, a ''maximally even (sub)set of size n'' in '''Z'''/''m'''''Z''' is any translate of the set		Mathematically, if {{nowrap\|0 < ''n'' < ''m''}}, a ''maximally even (sub)set of size n'' in '''Z'''/''m'''''Z''' is any translate of the set

	<math>\operatorname{ME}(n, m) = \{ m\mathbb{Z} + \ceil{\frac{im}{n}} : i \in \{0, ..., n-1\} \} \subseteq \mathbb{Z}/m\mathbb{Z},</math>		<math>\operatorname{ME}(n, m) = \left\{ m\mathbb{Z} + \ceil{\frac{im}{n}} : i \in \{0, ..., n-1\} \right\} \subseteq \mathbb{Z}/m\mathbb{Z},</math>

	where the {{w\|ceiling function}} fixes integers and rounds up non-integers to the next higher integer. It can be proven that when ''n'' does not divide ''m'', ME(''n'', ''m'') is a [[MOS scale\|MOS subset]] of '''Z'''/''m'''''Z''' where the two step sizes differ by exactly 1, and that the set of degrees where each step size occurs is itself maximally even in '''Z'''/''n'''''Z''', satisfying the informal definition above. ME(''n'', ''m'') is the lexicographically first mode among its rotations, and combined with the fact that it is a MOS, this implies that ME(''n'', ''m'') is the brightest mode in the MOS sense.		where the {{w\|ceiling function}} fixes integers and rounds up non-integers to the next higher integer. It can be proven that when ''n'' does not divide ''m'', ME(''n'', ''m'') is a [[MOS scale\|MOS subset]] of '''Z'''/''m'''''Z''' where the two step sizes differ by exactly 1, and that the set of degrees where each step size occurs is itself maximally even in '''Z'''/''n'''''Z''', satisfying the informal definition above. ME(''n'', ''m'') is the lexicographically first mode among its rotations, and combined with the fact that it is a MOS, this implies that ME(''n'', ''m'') is the brightest mode in the MOS sense.

ArrowHead294 at 15:07, 19 March 2025

2025-03-19T15:07:56Z

Show changes

Inthar: /* Complement of a maximally even subset is maximally even */

2025-01-01T15:23:38Z

Complement of a maximally even subset is maximally even

← Older revision		Revision as of 15:23, 1 January 2025
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	Proof sketch: We may assume that gcd(''n'', ''m'') = 1; there are two cases.		Proof sketch: We may assume that gcd(''n'', ''m'') = 1; there are two cases.

	Case 1: ME(n, m) where n < m/2. This is a maximally even subset of Z/mS with step sizes L > s > 1, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are L - 2 and s - 2 (0 is a valid chunk size), and the sizes form a maximally even MOS.		Case 1: ME(n, m) where n < m/2. This is a maximally even subset of Z/mZ with step sizes L > s > 1, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are L - 2 and s - 2 (0 is a valid chunk size), and the sizes form a maximally even MOS.

	Case 2: ME(n, m) where n > m/2. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since n > m/2) occupy a maximally even subset of the slots of ME(n, m) (*). Now replace each 1 with "\|" and each 2 with "$\|".		Case 2: ME(n, m) where n > m/2. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since n > m/2) occupy a maximally even subset of the slots of ME(n, m) (*). Now replace each 1 with "\|" and each 2 with "$\|".

Inthar: /* Complement of a maximally even subset is maximally even */

2024-12-10T13:20:42Z

Complement of a maximally even subset is maximally even

← Older revision		Revision as of 13:20, 10 December 2024
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	Case 1: ME(n, m) where n < m/2. This is a maximally even subset of Z/mS with step sizes L > s > 1, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are L - 2 and s - 2 (0 is a valid chunk size), and the sizes form a maximally even MOS.		Case 1: ME(n, m) where n < m/2. This is a maximally even subset of Z/mS with step sizes L > s > 1, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are L - 2 and s - 2 (0 is a valid chunk size), and the sizes form a maximally even MOS.

	Case 2: ME(n, m) where n > m/2. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since n > m/2) occupy a maximally even subset of the slots of ME(n, m) (*). Now ~~do the following steps:~~		Case 2: ME(n, m) where n > m/2. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since n > m/2) occupy a maximally even subset of the slots of ME(n, m) (*). Now replace each 1 with "\|" and each 2 with "$\|".
	# replace each 1 with "\|.\|" and each 2 with "\|$\|"
	~~# collapse "\|\|" to "\|"~~		(e.g.) 2112111 -> $\|\|\|$\|\|\|\|
	~~# remove all "."~~
	(e.g.) 2112111 -> \|$\|~~\|.\|\|.~~\|\|$\|~~\|.\|\|.\|\|.\| -> \|$\|.\|.\|$\|.\|.\|. -> \|$\|\|\|$~~\|\|\|

	Consider the resulting binary word of "\|" and "$". The "\|"s form chunks of sizes that differ by 1 and are distributed in a MOS way by (*). The desired complement, occupied by the "$"'s, thus forms a maximally even subset.		Consider the resulting binary word of "\|" and "$". The "\|"s form chunks of sizes that differ by 1 and are distributed in a MOS way by (*). The desired complement, occupied by the "$"'s, thus forms a maximally even subset.

Inthar: /* Complement of a maximally even subset is maximally even */

2024-12-10T11:47:39Z

Complement of a maximally even subset is maximally even

← Older revision		Revision as of 11:47, 10 December 2024
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	Proof sketch: We may assume that gcd(''n'', ''m'') = 1; there are two cases.		Proof sketch: We may assume that gcd(''n'', ''m'') = 1; there are two cases.

	Case 1: ME(n, m) where n < m/2. This is a maximally even subset of Z/nZ with step sizes L > s > 1, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are L - 2 and s - 2 (0 is a valid chunk size), and the sizes form a maximally even MOS.		Case 1: ME(n, m) where n < m/2. This is a maximally even subset of Z/mS with step sizes L > s > 1, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are L - 2 and s - 2 (0 is a valid chunk size), and the sizes form a maximally even MOS.

	Case 2: ME(n, m) where n > m/2. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since n > m/2) occupy a maximally even subset of the slots of ME(n, m) (*). Now do the following steps:		Case 2: ME(n, m) where n > m/2. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since n > m/2) occupy a maximally even subset of the slots of ME(n, m) (*). Now do the following steps:

Inthar: /* Complement of a maximally even subset is maximally even */

2024-12-10T11:25:46Z

Complement of a maximally even subset is maximally even

← Older revision		Revision as of 11:25, 10 December 2024
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	(e.g.) 2112111 -> \|$\|\|.\|\|.\|\|$\|\|.\|\|.\|\|.\| -> \|$\|.\|.\|$\|.\|.\|. -> \|$\|\|\|$\|\|\|		(e.g.) 2112111 -> \|$\|\|.\|\|.\|\|$\|\|.\|\|.\|\|.\| -> \|$\|.\|.\|$\|.\|.\|. -> \|$\|\|\|$\|\|\|

	Consider the resulting binary word of "\|" and "$". The "\|"s form chunks of sizes that differ by 1 and are distributed in a MOS way by (*). The desired complement, ~~parameterized~~ by "$", thus forms a maximally even subset.		Consider the resulting binary word of "\|" and "$". The "\|"s form chunks of sizes that differ by 1 and are distributed in a MOS way by (*). The desired complement, occupied by the "$"'s, thus forms a maximally even subset.

	== Concoctic scales ==		== Concoctic scales ==

← Older revision		Revision as of 16:52, 8 February 2026
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	Case 1: ME(''n'', ''m'') where {{nowrap\|''n'' < {{frac\|''m''\|2}}}}. This is a maximally even subset of Z/mZ with step sizes {{nowrap\|L > s > 1}}, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are {{nowrap\|L − 2}} and {{nowrap\|s − 2}} (0 is a valid chunk size), and the sizes form a maximally even MOS.		Case 1: ME(''n'', ''m'') where {{nowrap\|''n'' < {{frac\|''m''\|2}}}}. This is a maximally even subset of Z/mZ with step sizes {{nowrap\|L > s > 1}}, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are {{nowrap\|L − 2}} and {{nowrap\|s − 2}} (0 is a valid chunk size), and the sizes form a maximally even MOS.

	Case 2: ME(''n'', ''m'') where {{nowrap\|''n'' > {{frac\|''m''\|2}}}}. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since {{nowrap\|''n'' > {{frac\|''m''\|2}}}}) occupy a maximally even subset of ~~the slots of ME(''~~n~~'', ''m'')~~ (*). Now replace each 1 with "\|" and each 2 with "$\|".		Case 2: ME(''n'', ''m'') where {{nowrap\|''n'' > {{frac\|''m''\|2}}}}. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since {{nowrap\|''n'' > {{frac\|''m''\|2}}}}) occupy a maximally even subset of <math>\mathbb{Z}/n\mathbb{Z}</math> (*). Now replace each 1 with "\|" and each 2 with "$\|".

	(e.g.) {{nowrap\|2112111 → <nowiki>$\|\|\|$\|\|\|\|</nowiki>}}		(e.g.) {{nowrap\|2112111 → <nowiki>$\|\|\|$\|\|\|\|</nowiki>}}

← Older revision		Revision as of 23:57, 5 August 2025
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	Case 1: ME(''n'', ''m'') where {{nowrap\|''n'' < {{frac\|''m''\|2}}}}. This is a maximally even subset of Z/mZ with step sizes {{nowrap\|L > s > 1}}, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are {{nowrap\|L − 2}} and {{nowrap\|s − 2}} (0 is a valid chunk size), and the sizes form a maximally even MOS.		Case 1: ME(''n'', ''m'') where {{nowrap\|''n'' < {{frac\|''m''\|2}}}}. This is a maximally even subset of Z/mZ with step sizes {{nowrap\|L > s > 1}}, which determines the locations of step sizes of 2 in the complement. The rest of the complement's step sizes are 1. The sizes of the chunks of 1 are {{nowrap\|L − 2}} and {{nowrap\|s − 2}} (0 is a valid chunk size), and the sizes form a maximally even MOS.

	Case 2: ME(''n'', ''m'') where {{nowrap\|''n'' > {{frac\|''m''\|2}}}}. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since {{nowrap\|''n'' > {{frac\|''m''\|2}}}} occupy a maximally even subset of the slots of ME(''n'', ''m'') (*). Now replace each 1 with "\|" and each 2 with "$\|".		Case 2: ME(''n'', ''m'') where {{nowrap\|''n'' > {{frac\|''m''\|2}}}}. This has step sizes 1 and 2. The chunks of 1 (of nonzero size since {{nowrap\|''n'' > {{frac\|''m''\|2}}}}) occupy a maximally even subset of the slots of ME(''n'', ''m'') (*). Now replace each 1 with "\|" and each 2 with "$\|".

	(e.g.) {{nowrap\|2112111 → <nowiki>$\|\|\|$\|\|\|\|</nowiki>}}		(e.g.) {{nowrap\|2112111 → <nowiki>$\|\|\|$\|\|\|\|</nowiki>}}