Ternary scale theorems: Difference between revisions
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Depending on point (2), ''m'' may be even or odd. If ''m'' is even, then in any stack in ''S'' that has '''i''', the number of stacked '''g''' generators in the projection is odd, hence the number of non-'''X''' letters in the corresponding ''km''-step word in ''S'' is odd. Each incremental shift of the boundary that does not result in including '''i''' results in one '''Y''' being swapped for a '''Z''', or vice versa, while the number of '''X''' steps remains ''t''. In summary: | Depending on point (2), ''m'' may be even or odd. If ''m'' is even, then in any stack in ''S'' that has '''i''', the number of stacked '''g''' generators in the projection is odd, hence the number of non-'''X''' letters in the corresponding ''km''-step word in ''S'' is odd. Each incremental shift of the boundary that does not result in including '''i''' results in one '''Y''' being swapped for a '''Z''', or vice versa, while the number of '''X''' steps remains ''t''. In summary: | ||
* |'''g'''|<sub>'''X'''</sub> = |'''g'''|<sub>'''X'''</sub> = ''t'' | * |'''g'''<sub>1</sub>|<sub>'''X'''</sub> = |'''g'''<sub>2</sub>|<sub>'''X'''</sub> = ''t'' | ||
* |'''g'''<sub>1</sub>|<sub>'''Y'''</sub> = |'''g'''<sub>2</sub>|<sub>'''Z'''</sub> = ceil((''k'' − ''t'')/2), | * |'''g'''<sub>1</sub>|<sub>'''Y'''</sub> = |'''g'''<sub>2</sub>|<sub>'''Z'''</sub> = ceil((''k'' − ''t'')/2), | ||
* |'''g'''<sub>1</sub>|<sub>'''Z'''</sub> = |'''g'''<sub>2</sub>|<sub>'''Y'''</sub> = floor((''k'' − ''t'')/2), | * |'''g'''<sub>1</sub>|<sub>'''Z'''</sub> = |'''g'''<sub>2</sub>|<sub>'''Y'''</sub> = floor((''k'' − ''t'')/2), | ||