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Ternary parallelogram scales are MOS substitution: Difference between revisions

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→A technical lemma: The pigeonhole argument doesn't work
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# ''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 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).
=== A technical lemma ===
Statement: If ''m'' > 1, ''n'' > 2, ''a'' has order > ''n'' in {{nowrap|ℤ/''mn''ℤ}}, and {{nowrap|ℤ/''mn''ℤ}} is partitioned into bins where one bin consists of &le; 2''m'' - 1 adjacent elements and the rest of the bins consist of 2''m'' - 1, then {{nowrap|{0, ''a'', 2''a'', ..., (''n'' - 1)''a''}}} meets some bin at least twice.
Proof: Apply the pigeonhole principle: the number of bins is
<math>
\begin{align*}
\bigg\lceil \frac{mn}{2m-1} \bigg\rceil
&=\bigg\lceil \frac{n}{2-1/m} \bigg\rceil
\\ &\le \bigg\lceil \frac{2n}{3} \bigg\rceil < n.
\end{align*}</math>
The bin where two multiples of ''a'' co-occur must have two distinct multiples of ''a'', because the order of ''a'' is {{nowrap|> ''n''.}}


=== Step 4: The axial step is a MOS substitution slot letter ===
=== Step 4: The axial step is a MOS substitution slot letter ===
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