Defactoring: Difference between revisions
Cmloegcmluin (talk | contribs) |
Cmloegcmluin (talk | contribs) →temperament states: include DCF |
||
| Line 180: | Line 180: | ||
All considered, three different states are possible for a given temperament: | All considered, three different states are possible for a given temperament: | ||
# The RREF, IRREF, and | # The RREF, IRREF, HNF, and DCF are all the same. Example: [[Meantone_family#Meantone_.2812.2619.2C_2.3.5.29|meantone]] with all four equal to {{vector|{{map|1 0 -4}} {{map|0 1 4}}}}. | ||
# The RREF, IRREF, and HNF are all different. Example: [[Porcupine_family#Porcupine|porcupine]] with {{vector|{{map|1 0 <span><math>-\frac13</math></span>}} {{map|0 1 <span><math>\frac53</math></span>}}}}, {{vector|{{map|3 0 -1}} {{map|0 3 5}}}}, and {{vector|{{map|1 2 3}} {{map|0 3 5}}}}, respectively. | # The RREF, IRREF, and HNF are all different, but HNF is the same as DCF. Example: [[Porcupine_family#Porcupine|porcupine]] with RREF of {{vector|{{map|1 0 <span><math>-\frac13</math></span>}} {{map|0 1 <span><math>\frac53</math></span>}}}}, IRREF of {{vector|{{map|3 0 -1}} {{map|0 3 5}}}}, and HNF and DCF of {{vector|{{map|1 2 3}} {{map|0 3 5}}}}, respectively. | ||
# The IRREF and | # The IRREF, HNF, and DCF are the same, but the RREF is different. Example: [[Kleismic_family#Hanson|hanson]] with IRREF, HNF, and DCF of {{vector|{{map|1 0 1}} {{map|0 6 5}}}} but RREF of {{vector|{{map|1 0 1}} {{map|0 1 <span><math>\frac56</math></span>}}}}. | ||
A fourth state is possible but only for enfactored mappings, i.e. for temperoids. This is where the RREF and IRREF are the same, but the HNF is different. This is only possible for enfactored mappings because: | A fourth state is possible, but only for enfactored mappings, i.e. for temperoids. This is where the RREF and IRREF are the same, but the HNF is different, and the DCF is different still. This is only possible for enfactored mappings because: | ||
* Whenever the RREF and IRREF match, the IRREF has all 1's for pivots. | * Whenever the RREF and IRREF match, the IRREF has all 1's for pivots. | ||
* But in that case, the IRREF would also be the HNF, because there's only one possible nonnegative integer value less than 1, which is zero, and so this form would also satisfy the normal constraint. | * But in that case, the IRREF would also be the HNF, because there's only one possible nonnegative integer value less than 1, which is zero, and so this form would also satisfy the normal constraint. | ||
* The only way to avoid this would be to ensure that at least one pivot of the HNF was not 1, so that it could have some positive value somewhere in the entries above that pivot in its pivot column, to distinguish it from IRREF. For example, we could use {{vector|{{map|1 1 0}} {{map|0 2 1}}}}. | * The only way to avoid this would be to ensure that at least one pivot of the HNF was not 1, so that it could have some positive value somewhere in the entries above that pivot in its pivot column, to distinguish it from IRREF. For example, we could use {{vector|{{map|1 1 0}} {{map|0 2 1}}}}. | ||
* But now if we look at the RREF of this mapping with this HNF, we have to divide rows until all pivots are 1. So that second row would be changed to {{map|0 1 <span><math>\frac12</math></span>}}. So now RREF won't match IRREF, because it contains a noninteger. The only way to prevent this would be if a pivot of the HNF was not 1, but still every entry in that row was evenly divisible by that value, such as {{vector|{{map|1 1 0}} {{map|0 2 4}}}}. But now that row is enfactored by its pivot's value. | * But now if we look at the RREF of this mapping with this HNF, we have to divide rows until all pivots are 1. So that second row would be changed to {{map|0 1 <span><math>\frac12</math></span>}}. So now RREF won't match IRREF, because it contains a noninteger. The only way to prevent this would be if a pivot of the HNF was not 1, but still every entry in that row was evenly divisible by that value, such as {{vector|{{map|1 1 0}} {{map|0 2 4}}}}. But now that row is enfactored by its pivot's value. | ||
* If the mapping is enfactored, that is the case when the HNF ≠ DCF. | |||