Ternary parallelogram scales are MOS substitution: Difference between revisions
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Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from previous lemma). If the order is ''N'' > ''n'', then by Lemma 2, {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are not disjoint, contradicting injectivity. | Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from previous lemma). If the order is ''N'' > ''n'', then by Lemma 2, {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are not disjoint, contradicting injectivity. | ||
The claim establishes that | |||
# the two other 1-step vectors are not found on the axis orthogonal to the first vector, as that would imply order ''mn'' | |||
# φ is minimally periodic under translation by ''n'''''w'''. Hence the two non-axial 1's on the grid from the 0 at the origin are translated by ''n'''''w'''. | |||
==== Template word is MOS ==== | ==== Template word is MOS ==== | ||