Ternary parallelogram scales are MOS substitution: Difference between revisions

Inthar (talk | contribs)
Inthar (talk | contribs)
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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 {{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 &ge; ''n'', then move by ''c'' - ''n'' units (i.e. southward). Otherwise, move by ''c'' units (northward)."
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 &ge; ''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'')"}} and stacking (abelianized) ''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 &le; ''k'' &le; length - 1.}} {{Qed}}
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'')"}} and stacking (abelianized) ''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 &le; ''k'' &le; length - 1.}} {{Qed}}
[[Category:Pages with proofs]]
[[Category:Pages with proofs]]