Ternary parallelogram scales are MOS substitution: Difference between revisions
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Without loss of generality assume that {{nowrap|'''u'''<sub>'''y'''</sub> {{=}} (''b'', ''c''), ''c'' > 0,}} and {{nowrap|'''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 {{nowrap|[0 : ''n''],}} at any point it must follow the rule: "If the current '''w'''-coordinate + c ≥ ''n'', then move by ''c'' - ''n'' units (using the letter '''z'''). Otherwise, move by ''c'' units (using the letter '''y''')." | Without loss of generality assume that {{nowrap|'''u'''<sub>'''y'''</sub> {{=}} (''b'', ''c''), ''c'' > 0,}} and {{nowrap|'''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 {{nowrap|[0 : ''n''],}} at any point it must follow the rule: "If the current '''w'''-coordinate + c ≥ ''n'', then move by ''c'' - ''n'' units (using the letter '''z'''). Otherwise, move by ''c'' units (using the letter '''y''')." | ||
This pattern of movements is in fact the same as the one produced by taking the circular word {{nowrap|"1 1 1 ... 1 (1 - ''n'')"}} ((''n'' - 1)-many 1's) and stacking ( | This pattern of movements is in fact the same as the one produced by taking the circular word {{nowrap|"1 1 1 ... 1 (1 - ''n'')"}} ((''n'' - 1)-many 1's) and stacking (sums of) ''c''-step subwords. As there is only one bad position per period, the filling word can easily be seen to be MOS by stacking ''kc''-step subwords of the latter word for {{nowrap|2 ≤ ''k'' ≤ length - 1.}} {{Qed}} | ||
[[Category:Math]] | [[Category:Math]] | ||
[[Category:Pages with proofs]] | [[Category:Pages with proofs]] | ||
[[Category:Combinatorics on words]] | [[Category:Combinatorics on words]] | ||