Ternary parallelogram scales are MOS substitution: Difference between revisions
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After rotating ''w'', we may assume that (0, 0) is labeled 0. The labeling now extends to a surjective homomorphism <math>\varphi: \mathbb{Z}^2\langle \mathbf{v},\mathbf{w}\rangle \to \mathbb{Z}/mn\mathbb{Z},</math> where {{nowrap|φ('''v''') {{=}} ''k''<sub>'''v'''</sub>}} and {{nowrap|φ('''w''') {{=}} ''k''<sub>'''w'''</sub>.}} φ has {{nowrap|[0 : ''m''] × [0 : ''n'']}} as a fundamental domain. | After rotating ''w'', we may assume that (0, 0) is labeled 0. The labeling now extends to a surjective homomorphism <math>\varphi: \mathbb{Z}^2\langle \mathbf{v},\mathbf{w}\rangle \to \mathbb{Z}/mn\mathbb{Z},</math> where {{nowrap|φ('''v''') {{=}} ''k''<sub>'''v'''</sub>}} and {{nowrap|φ('''w''') {{=}} ''k''<sub>'''w'''</sub>.}} φ has {{nowrap|[0 : ''m''] × [0 : ''n'']}} as a fundamental domain. | ||
==== Lemma: Any ''m'' × ''n'' window in ℤ<sup>2</sup> has the same cyclic ordering of elements under φ ==== | ==== Lemma: Any ''m'' × ''n'' window in ℤ<sup>2</sup> has the same cyclic ordering of elements under the φ-labeling ==== | ||
For any integers ''i''<sub>0</sub> and ''j''<sub>0</sub>, if {{nowrap|φ((''i''<sub>0</sub>, ''j''<sub>0</sub>)) {{=}} ''a'',}} then for any (''i'', ''j'') in [0 : ''m''] × [0 : ''n''], we have | |||
<math>\begin{align*}\varphi((i_0 + i, j_0 + j)) &= \varphi((i_0, j_0)) + \varphi((i, j)) | |||
\\ &= a + \varphi((i, j)),\end{align*}</math> | |||
as φ is a homomorphism. Hence corresponding elements of any two ''m'' × ''n'' windows get the same {{nowrap|ℤ/''mn''ℤ}} labeling modulo a shift. | |||
=== Step 2: By ternarity, exactly one of the 1-step vectors is parallel to a coordinate axis === | === Step 2: By ternarity, exactly one of the 1-step vectors is parallel to a coordinate axis === | ||