Ternary parallelogram scales are MOS substitution: Difference between revisions
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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). | 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 ''m'' > 1, ''n'' > 2, and ''a'' has order > ''n'' in {{nowrap|ℤ/''mn''ℤ}}, and {{nowrap|ℤ/''mn''ℤ}} is partitioned into windows | === Lemma === | ||
Apply the pigeonhole principle: we have | Statement: If ''m'' > 1, ''n'' > 2, and ''a'' has order > ''n'' in {{nowrap|ℤ/''mn''ℤ}}, and {{nowrap|ℤ/''mn''ℤ}} is partitioned into windows where one window has < 2''m'' - 1 adjacent elements and the rest of the windows have 2''m'' - 1, then {{nowrap|{0, ''a'', 2''a'', ..., (''n'' - 1)''a''}}} meets some window at least twice. | ||
Proof: Apply the pigeonhole principle: we have | |||
<math> | <math> | ||