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From this we can conclude that ''d'', ''e'', and ''f'' satisfy <math>\left|d - q_3b + q_5a\right| \leqslant 2Eq_3q_5</math>, <math>\left|e - q_3c + q_7a\right| \leqslant 2Eq_3q_7</math>, and <math>\left|f - q_5c + q_7d\right| \leqslant 2Eq_5q_7</math>. This has an interesting interpretation: since <math>\tval{1 & q_3 & q_5 & q_7} \wedge \tval{0 & a & b & c} =</math> <math>\bitval{a & b & c & q_3b - q_5a & q_3c - q_7a & q_5c - q_7b}</math>, if {{nowrap|E ≤ {{frac|1|4''q''<sub>5</sub>''q''<sub>7</sub>}}}}, then the full wedgie can be recovered from the octave equivalent (OE) portion of the wedgie simply by wedging it with <math>\tval{1 & q_3 & q_5 & q_7}</math> and rounding to the nearest integer. This is not a very serious constraint to place on relative error; it seems unlikely anyone would be interested in a temperament which did not fall well under this low standard. Hence we may compile lists of reasonable temperaments by presuming "reasonable" requires this bound to be met, searching through triples <math>\bitval{a & b & c & \ldots}</math> (note that if all of these are zero, 2 is being tempered out) up to some complexity bound, wedging with <math>\tmonzo{1 & q_3 & q_5 & q_7}</math>, and rounding, then checking if the GCD is one and the Pfaffian is zero (i.e. {{nowrap|''af'' − ''be'' + ''cd'' {{=}} 0}}). Then we may toss everthing which does not meet the bound on relative error; however, for a reasonable list we will want a tighter bound. | From this we can conclude that ''d'', ''e'', and ''f'' satisfy <math>\left|d - q_3b + q_5a\right| \leqslant 2Eq_3q_5</math>, <math>\left|e - q_3c + q_7a\right| \leqslant 2Eq_3q_7</math>, and <math>\left|f - q_5c + q_7d\right| \leqslant 2Eq_5q_7</math>. This has an interesting interpretation: since <math>\tval{1 & q_3 & q_5 & q_7} \wedge \tval{0 & a & b & c} =</math> <math>\bitval{a & b & c & q_3b - q_5a & q_3c - q_7a & q_5c - q_7b}</math>, if {{nowrap|E ≤ {{frac|1|4''q''<sub>5</sub>''q''<sub>7</sub>}}}}, then the full wedgie can be recovered from the octave equivalent (OE) portion of the wedgie simply by wedging it with <math>\tval{1 & q_3 & q_5 & q_7}</math> and rounding to the nearest integer. This is not a very serious constraint to place on relative error; it seems unlikely anyone would be interested in a temperament which did not fall well under this low standard. Hence we may compile lists of reasonable temperaments by presuming "reasonable" requires this bound to be met, searching through triples <math>\bitval{a & b & c & \ldots}</math> (note that if all of these are zero, 2 is being tempered out) up to some complexity bound, wedging with <math>\tmonzo{1 & q_3 & q_5 & q_7}</math>, and rounding, then checking if the GCD is one and the Pfaffian is zero (i.e. {{nowrap|''af'' − ''be'' + ''cd'' {{=}} 0}}). Then we may toss everthing which does not meet the bound on relative error; however, for a reasonable list we will want a tighter bound. | ||
If {{nowrap|C {{=}} | If {{nowrap|C {{=}} ‖W‖}} is the TE complexity, then the formula for the [[Tenney-Euclidean_metrics#Logflat TE badness|logflat badness]] B in the 7-limit rank-two case is particularly simple: {{nowrap|B {{=}} CE}}. If complexity is bounded by, for example, 20 (which allows for some quite complex temperaments) then since {{nowrap|E ≤ {{frac|1|4''q''<sub>5</sub>''q''<sub>7</sub>}}}}, {{nowrap|B ≤ {{frac|20|4''q''<sub>5</sub>''q''<sub>7</sub>}} {{=}} 0.767}}. This badness figure is easily met. While simply bounding complexity will lead to a finite list, the list would be enormous. An alternative is also to bound badness; for instance, we might produce a list of 7-limit rank-two temperaments with complexity less than 20 and a more reasonable badness limit, such as 0.05 or 0.06. | ||
== Reconstituting wedgies in general == | == Reconstituting wedgies in general == | ||
Essentially the same situation obtains for rank two temperaments in higher limits. The rule then is that if <math>E \leqslant \frac{1}{\binom{n}{3}\log_2\left(q\right)\log_2\left(p\right)}</math> then wedging <math>K = \tval{1 & \log_2\left(3\right) & \log_2\left(5\right) & \ldots & \log_2\left(p\right)}</math> with the val consisting of 0 followed by the first {{nowrap|''n'' − 1}} coefficients of the wedgie and rounding will give the wedgie, where ''p'' and ''q'' are the largest and second largest primes in the prime limit. | Essentially the same situation obtains for rank two temperaments in higher limits. The rule then is that if <math>E \leqslant \frac{1}{\binom{n}{3}\log_2\left(q\right)\log_2\left(p\right)}</math> then wedging <math>K = \tval{1 & \log_2\left(3\right) & \log_2\left(5\right) & \ldots & \log_2\left(p\right)}</math> with the val consisting of 0 followed by the first {{nowrap|''n'' − 1}} coefficients of the wedgie and rounding will give the wedgie, where ''p'' and ''q'' are the largest and second largest primes in the prime limit. | ||
More generally, we can reconstitute W by rounding {{nowrap|Y {{=}} (W ∨ 2) ∧ K}} to the nearest integer coefficients, where K is the JI point <math>\tval{1 & \log_2\left(3\right) & \log_2\left(5\right) & \ldots & \log_2\left(p\right)}</math> in unweighted coordinates. Then we have <math>\left\|\left(W - Y\right) + Y\right\| \leqslant \left\|W-Y\right\| + \left\|Y\right\|</math> by the triangle inequality, and since {{nowrap| | More generally, we can reconstitute W by rounding {{nowrap|Y {{=}} (W ∨ 2) ∧ K}} to the nearest integer coefficients, where K is the JI point <math>\tval{1 & \log_2\left(3\right) & \log_2\left(5\right) & \ldots & \log_2\left(p\right)}</math> in unweighted coordinates. Then we have <math>\left\|\left(W - Y\right) + Y\right\| \leqslant \left\|W-Y\right\| + \left\|Y\right\|</math> by the triangle inequality, and since {{nowrap|‖W − Y‖}} is bounded by the fact that W has been obtained by rounding, complexity, which is {{nowrap|‖(W − Y) + Y‖ {{=}} ‖W‖}}, can be bounded by ||Y||; which means it can be bounded by the coefficients of Y, which are those coefficients of W which can be found in W∨2 and over which we could be conducting a search. Moreover, we have from {{nowrap|Y ∧ K {{=}} ((W ∨ 2) ∧ K) ∧ K {{=}} 0}} that relative error, which is {{nowrap|‖W ∧ K‖}}, is {{nowrap|‖((W − Y) + Y) ∧ K‖ {{=}} ‖(W − Y) ∧ K‖}}, hence relative error is also bounded by the fact that {{nowrap|‖W − Y‖}} is bounded. This means that unless relative error is large, W can be recovered by rounding Y, and hence all wedgies within such a bound, which we may call ''recoverable'', can be found by a search on only some prospective coefficients. Temperaments which are not recoverable seem of little interest and may be ruled out of consideration. Search spaces for complexity measures such as [[Tenney-Euclidean_temperament_measures#TE Complexity|TE complexity]] which are defined in terms of the wedgie can be obtained by assuming all wedgie coefficients which are not being used to recover a wedgie are zero, which gives a minimum value for the complexity. In the case of rank two temperaments, an especially efficient complexity measure for such searches, and one with some other desirable properties, is [[generator complexity]]. | ||
In the particular case of the 11-limit in rank three, we have that {{nowrap|(W ∨ 2) ∧ K}} gives the full wedgie, which has ten coefficents, in terms of the first six upon rounding off. Using this for a search is less difficult than it sounds, since the complexity numbers for rank three are so much lower. If the relative error E satisifes {{nowrap|E ≤ {{frac|1|2√(5)''q''<sub>5</sub>''q''<sub>7</sub>''q''<sub>11</sub>}}}}, then the rounding off is guaranteed to lead to the correct result. This amount, 0.0099, is again easily met. | In the particular case of the 11-limit in rank three, we have that {{nowrap|(W ∨ 2) ∧ K}} gives the full wedgie, which has ten coefficents, in terms of the first six upon rounding off. Using this for a search is less difficult than it sounds, since the complexity numbers for rank three are so much lower. If the relative error E satisifes {{nowrap|E ≤ {{frac|1|2√(5)''q''<sub>5</sub>''q''<sub>7</sub>''q''<sub>11</sub>}}}}, then the rounding off is guaranteed to lead to the correct result. This amount, 0.0099, is again easily met. | ||