Maximal evenness: Difference between revisions
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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/ | 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: | ||