Generator-offset property: Difference between revisions
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In case 1, let g<sub>1</sub> = (2, 1) − (1, 1), g<sub>2</sub> = (1, 2) − (2, 1), and g<sub>3</sub> = (1, 1) − (''n''/2, 2) = (−''n''/2*g<sub>1</sub> − g<sub>1</sub> − ''n''/2*g<sub>2</sub>) mod e. We assume that g<sub>1</sub>, g<sub>2</sub> and e are '''Z'''-linearly independent. We have the chain g<sub>1</sub> g<sub>2</sub> g<sub>1</sub> g<sub>2</sub> ... g<sub>1</sub> g<sub>3</sub> which visits every note in ''S''. | In case 1, let g<sub>1</sub> = (2, 1) − (1, 1), g<sub>2</sub> = (1, 2) − (2, 1), and g<sub>3</sub> = (1, 1) − (''n''/2, 2) = (−''n''/2*g<sub>1</sub> − g<sub>1</sub> − ''n''/2*g<sub>2</sub>) mod e. We assume that g<sub>1</sub>, g<sub>2</sub> and e are '''Z'''-linearly independent. We have the chain g<sub>1</sub> g<sub>2</sub> g<sub>1</sub> g<sub>2</sub> ... g<sub>1</sub> g<sub>3</sub> which visits every note in ''S''. | ||
Since ''S'' is GO it is well-formed with respect to (g<sub>2</sub> + g<sub>1</sub>), which must generate a mos subset. Hence (g<sub>3</sub> + g<sub>1</sub>), the imperfect generator of the mos generated by (g<sub>2</sub> + g<sub>1</sub>), subtends the same number of steps as (g<sub>2</sub> + g<sub>1</sub>). Thus g<sub>2</sub> and g<sub>3</sub> subtend the same number of steps. | |||
Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class (''k''-steps) reached by stacking ''r'' generators: | Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class (''k''-steps) reached by stacking ''r'' generators: | ||