Maximal evenness: Difference between revisions

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

where the [[Wikipedia:Floor and ceiling functions|ceiling]] function fixes integers and rounds up non-integers to the next higher integer. It can be proven that 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 [[Wikipedia:Floor and ceiling functions|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.

It is easy to show that replacing ceil() with round() (rounding half-integers up) gives an equivalent definition; floor() does too, since ME(''n'', ''m'') is a MOS and thus [[chirality|achiral]].

== Complement of a maximally even subset is maximally even ==

Proof sketch: We may assume that gcd(''n'', ''m'') = 1; there are two cases.

# 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.

# 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 "| . |", and then collapse "||" to "|". The " "s form chunks of sizes that differ by 1 by (*).

== Concoctic scales ==

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 <math>\operatorname{ME}(n, m) = \{ m\mathbb{Z} + \lceil \frac{im}{n} \rceil : i \in \{0, ..., n-1\} \} \subseteq \mathbb{Z}/m\mathbb{Z},</math>
-where the [[Wikipedia:Floor and ceiling functions|ceiling]] function fixes integers and rounds up non-integers to the next higher integer. It can be proven that 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 [[Wikipedia:Floor and ceiling functions|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.
 It is easy to show that replacing ceil() with round() (rounding half-integers up) gives an equivalent definition; floor() does too, since ME(''n'', ''m'') is a MOS and thus [[chirality|achiral]].
+== Complement of a maximally even subset is maximally even ==
+Proof sketch: We may assume that gcd(''n'', ''m'') = 1; there are two cases.
+# 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.
+# 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 "| . |", and then collapse "||" to "|". The " "s form chunks of sizes that differ by 1 by (*).
 == Concoctic scales ==