Ternary parallelogram scales are MOS substitution: Difference between revisions
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# ''Q''<sub>3</sub> = [-''m'' + 1 : 1] × [-''n'' + 1 : 1] | # ''Q''<sub>3</sub> = [-''m'' + 1 : 1] × [-''n'' + 1 : 1] | ||
# ''Q''<sub>4</sub> = [0 : ''m''] × [0 : ''n''] | # ''Q''<sub>4</sub> = [0 : ''m''] × [0 : ''n''] | ||
By the previous step, φ restricted to any ''m'' × ''n'' window in ℤ<sup>2</sup> is surjective, hence all of the four windows ''Q''<sub>1</sub>, ..., ''Q''<sub>4</sub> have 1 somewhere in them. Call these positions '''u'''<sub>1</sub>, ..., '''u'''<sub>4</sub> (note that none of them are the zero vector). Since {{Nowrap|φ((0, 0)) {{=}} 0}} by another application of the | By the previous step, φ restricted to any ''m'' × ''n'' window in ℤ<sup>2</sup> is surjective, hence all of the four windows ''Q''<sub>1</sub>, ..., ''Q''<sub>4</sub> have 1 somewhere in them. Call these positions '''u'''<sub>1</sub>, ..., '''u'''<sub>4</sub> (note that none of them are the zero vector). Since {{Nowrap|φ((0, 0)) {{=}} 0}} by another application of the previous step we have '''u'''<sub>1</sub>, ..., '''u'''<sub>4</sub> as the images of 1-step vectors of ''w''. Since ''w'' is ternary, exactly two of these vectors will be pairwise equal, say '''u'''<sub>''k''</sub> = '''u'''<sub>''l''</sub>. These four "quadrants" intersect in <math>[-m + 1 : m] \times \{0\} \cup \{0\} \times [-n + 1 : n],</math> entailing that '''u'''<sub>''k''</sub> = '''u'''<sub>''l''</sub> is on a coordinate axis (either the '''v'''-coordinate is 0 or the '''w'''-coordinate is 0 but not both). | ||
=== Lemma: If ''a'' has order > ''n'' in {{nowrap|ℤ/''mn''ℤ}}, then {{nowrap|{''a'', 2''a'', ..., (''n'' - 1)''a''}}} and {{nowrap|[-''m'' + 1 : ''m'']}} are not disjoint === | === Lemma: If ''a'' has order > ''n'' in {{nowrap|ℤ/''mn''ℤ}}, then {{nowrap|{''a'', 2''a'', ..., (''n'' - 1)''a''}}} and {{nowrap|[-''m'' + 1 : ''m'']}} are not disjoint === | ||