Ternary parallelogram scales are MOS substitution: Difference between revisions
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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 (i.e. southward). Otherwise, move by ''c'' units (northward)." | 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 (i.e. southward). Otherwise, move by ''c'' units (northward)." | ||
This pattern of movements is in fact the same as the 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 ''kc''-step subwords for 2 ≤ ''k'' ≤ length - 1. | This pattern of movements is in fact the same as the 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 ''kc''-step subwords for 2 ≤ ''k'' ≤ length - 1. {{Qed}} | ||
[[Category:Pages with proofs]] | [[Category:Pages with proofs]] | ||