Ternary parallelogram scales are MOS substitution: Difference between revisions
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Delete all instances of the axial step '''u'''<sub>'''x'''</sub> and consider what the two remaining step sizes do to the '''w'''-coordinate. | Delete all instances of the axial step '''u'''<sub>'''x'''</sub> and consider what the two remaining step sizes do to the '''w'''-coordinate. | ||
Without loss of generality assume that '''u'''<sub>'''y'''</sub> = (''b'', ''c''), ''c'' > 0, and '''u'''<sub>''' | Without loss of generality assume that '''u'''<sub>'''y'''</sub> = (''b'', ''c''), ''c'' > 0, and '''u'''<sub>'''z'''</sub> = (''b'', ''c'' - ''n''). As the '''v'''-coordinates of both vectors are equal, we only need to look at the '''w'''-coordinate. Since the '''w'''-coordinate of a point must stay within [0 : ''n''], at any point it must follow the rule: "If the current '''w'''-coordinate + c ≥ ''n'', then move by ''c'' - ''n'' units (southward). Otherwise, move by ''c'' units (northward)." | ||
This pattern of movements is in fact the same as one produced by taking the circular word "1 1 1 ... 1 (1 - n)" and stacking ''c''-step subwords. As there is only one bad position per period, this word can easily be seen to be MOS by considering 2''c''-step subwords, 3''c''-step subwords, ... | This pattern of movements is in fact the same as one produced by taking the circular word "1 1 1 ... 1 (1 - n)" and stacking ''c''-step subwords. As there is only one bad position per period, this word can easily be seen to be MOS by considering 2''c''-step subwords, 3''c''-step subwords, ... | ||
[[Category:Pages with proofs]] | [[Category:Pages with proofs]] | ||