Ternary scale theorems: Difference between revisions

Line 53:

Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class of ''k''-steps reached by stacking ''r'' generators:

# from '''g'''1 '''g'''2 ... '''g'''1, we get a1 = (''r'' − 1)/2*g + '''g'''1 = ceil(''r''/2) '''g'''1 + floor(''r''/2) '''g'''2

# from '''g'''1 '''g'''2 ... '''g'''1, we get a1 = (''r'' − 1)/2 * '''g''' + '''g'''1 = ceil(''r''/2) '''g'''1 + floor(''r''/2) '''g'''2

# from '''g'''2 '''g'''1 ... '''g'''2, we get a2 = (''r'' − 1)/2*g + '''g'''2 = floor(''r''/2) '''g'''1 + ceil(''r''/2) '''g'''2

# from '''g'''2 '''g'''1 ... '''g'''2, we get a2 = (''r'' − 1)/2 * '''g''' + '''g'''2 = floor(''r''/2) '''g'''1 + ceil(''r''/2) '''g'''2

# from '''g'''2 (...even # of gens...) '''g'''1 '''g'''3 '''g'''1 (...even # of gens...) '''g'''2, we get a3 = (''r'' − 1)/2 '''g'''1 + (''r'' − 1)/2 '''g'''2 + '''g'''3 ≡ (''r'' − ''n''/2 − 3/2)'''g'''1 + (''r'' − ''n''/2 − 1/2)'''g'''2 mod e.

# from '''g'''1 (...odd # of gens...) '''g'''1 '''g'''3 '''g'''1 (...odd # of gens...) '''g'''1, we get a4 = (''r'' + 1)/2 '''g'''1 + (''r'' − 3)/2 '''g'''2 + '''g'''3 ≡ (''r'' − ''n''/2 − 1/2)'''g'''1 + (''r'' − ''n''/2 − 3/2)'''g'''2 mod e.

@@ Line 53: / Line 53: @@
 Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class of ''k''-steps reached by stacking ''r'' generators:
-# from '''g'''<sub>1</sub> '''g'''<sub>2</sub> ... '''g'''<sub>1</sub>, we get a<sub>1</sub> = (''r'' &minus; 1)/2*g + '''g'''<sub>1</sub> = ceil(''r''/2) '''g'''<sub>1</sub> + floor(''r''/2) '''g'''<sub>2</sub>
+# from '''g'''<sub>1</sub> '''g'''<sub>2</sub> ... '''g'''<sub>1</sub>, we get a<sub>1</sub> = (''r'' &minus; 1)/2 * '''g''' + '''g'''<sub>1</sub> = ceil(''r''/2) '''g'''<sub>1</sub> + floor(''r''/2) '''g'''<sub>2</sub>
-# from '''g'''<sub>2</sub> '''g'''<sub>1</sub> ... '''g'''<sub>2</sub>, we get a<sub>2</sub> = (''r'' &minus; 1)/2*g + '''g'''<sub>2</sub> = floor(''r''/2) '''g'''<sub>1</sub> + ceil(''r''/2) '''g'''<sub>2</sub>
+# from '''g'''<sub>2</sub> '''g'''<sub>1</sub> ... '''g'''<sub>2</sub>, we get a<sub>2</sub> = (''r'' &minus; 1)/2 * '''g''' + '''g'''<sub>2</sub> = floor(''r''/2) '''g'''<sub>1</sub> + ceil(''r''/2) '''g'''<sub>2</sub>
 # from '''g'''<sub>2</sub> (...even # of gens...) '''g'''<sub>1</sub> '''g'''<sub>3</sub> '''g'''<sub>1</sub> (...even # of gens...) '''g'''<sub>2</sub>, we get a<sub>3</sub> = (''r'' &minus; 1)/2 '''g'''<sub>1</sub> + (''r'' &minus; 1)/2 '''g'''<sub>2</sub> + '''g'''<sub>3</sub> ≡ (''r'' &minus; ''n''/2 &minus; 3/2)'''g'''<sub>1</sub> + (''r'' &minus; ''n''/2 &minus; 1/2)'''g'''<sub>2</sub> mod e.
 # from '''g'''<sub>1</sub> (...odd # of gens...) '''g'''<sub>1</sub> '''g'''<sub>3</sub> '''g'''<sub>1</sub> (...odd # of gens...) '''g'''<sub>1</sub>, we get a<sub>4</sub> = (''r'' + 1)/2 '''g'''<sub>1</sub> + (''r'' &minus; 3)/2 '''g'''<sub>2</sub> + '''g'''<sub>3</sub> ≡ (''r'' &minus; ''n''/2 &minus; 1/2)'''g'''<sub>1</sub> + (''r'' &minus; ''n''/2 &minus; 3/2)'''g'''<sub>2</sub> mod e.