Kite's thoughts on pergens: Difference between revisions
Wikispaces>TallKite **Imported revision 627092221 - Original comment: ** |
Wikispaces>TallKite **Imported revision 627100499 - Original comment: ** |
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<h2>IMPORTED REVISION FROM WIKISPACES</h2> | <h2>IMPORTED REVISION FROM WIKISPACES</h2> | ||
This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br> | This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br> | ||
: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-03- | : This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-03-02 03:35:48 UTC</tt>.<br> | ||
: The original revision id was <tt> | : The original revision id was <tt>627100499</tt>.<br> | ||
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For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2<span class="nowrap">⋅</span>P8 - 3<span class="nowrap">⋅</span>P5) / (3<span class="nowrap">⋅</span>2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This __is__ explicitly false, thus the comma can be found from m3/6 alone. G' is about 50¢, and the comma is 6<span class="nowrap">⋅</span>G' - m3. The comma splits both the octave and the fifth. | For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2<span class="nowrap">⋅</span>P8 - 3<span class="nowrap">⋅</span>P5) / (3<span class="nowrap">⋅</span>2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This __is__ explicitly false, thus the comma can be found from m3/6 alone. G' is about 50¢, and the comma is 6<span class="nowrap">⋅</span>G' - m3. The comma splits both the octave and the fifth. | ||
This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false | This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false. The unreduced pergen is (P8/4, M2/4), which also isn't explicitly false, thus (P8/4, P4/2) is a true double. It requires two commas, one for each fraction. The two commas must use different higher primes, e.g. 648/625 and 49/48. Thus __true doubles require commas of at least 7-limit__, whereas false doubles require only 5-limit. To summarize: | ||
* **A double-split pergen is __explicitly false__ if and only if m = |b|.** | |||
* **A double-split pergen is a __true double__ if and only if GCD (m, n) > |b|.** | |||
* **A double-split pergen is a __true double__ if and only if neither it nor its unreduced form is explicitly false.** | |||
A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively. | A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively. | ||
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The sharp symbol's ratio is always (-11,7) = 2187/2048, by definition. Looking at the table in the Applications section, the up symbol often equals only a few ratios. For most 5-limit temperaments, ^1 = 81/80. For most 2.3.7 temperaments, ^1 = 64/63. Most 11-limit temperaments use either 33/32 or 729/704. These **mapping commas** are used to map higher primes to 3-limit intervals, and are essential for notation. They also determine where a ratio "lands" on a keyboard. By definition they are a P1, and the only intervals that map to P1 are these commas and combinations of them. | The sharp symbol's ratio is always (-11,7) = 2187/2048, by definition. Looking at the table in the Applications section, the up symbol often equals only a few ratios. For most 5-limit temperaments, ^1 = 81/80. For most 2.3.7 temperaments, ^1 = 64/63. Most 11-limit temperaments use either 33/32 or 729/704. These **mapping commas** are used to map higher primes to 3-limit intervals, and are essential for notation. They also determine where a ratio "lands" on a keyboard. By definition they are a P1, and the only intervals that map to P1 are these commas and combinations of them. | ||
If a single-comma temperament uses double-pair notation, neither | If a single-comma temperament uses double-pair notation, neither accidental will equal the mapping comma. A double-comma temperament using double-pair notation may use the difference between two mapping commas, as in lemba, where ^1 equals 64/63 minus 81/80. | ||
Sometimes the mapping comma needs to be inverted. In diminished, which sets 6/5 = P8/4, ^1 = 80/81. in every temperament except those in the meantone family, the 81/80 comma is not tempered out, but it is still tempered, just like every ratio. Occasionally 81/80 is tempered so far that it becomes a descending interval. See also blackwood-like pergens below. | Sometimes the mapping comma needs to be inverted. In diminished, which sets 6/5 = P8/4, ^1 = 80/81. in every temperament except those in the meantone family, the 81/80 comma is not tempered out, but it is still tempered, just like every ratio. Occasionally 81/80 is tempered so far that it becomes a descending interval. See also blackwood-like pergens below. | ||
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If the edo's notation uses ups and downs, the up symbol can often be equated to a 3-limit ratio. In 17-edo and 22-edo, ^1 = m2. In 31-edo and 43-edo it's d2. But in edos like 15, 21 and 24, in which the circle of 5ths skips some notes, there is no 3-limit ratio. The ratio depends on the JI interpretation of the edo. For 10-edo, ^1 might equal 16/15 or 12/11 or 13/12. | If the edo's notation uses ups and downs, the up symbol can often be equated to a 3-limit ratio. In 17-edo and 22-edo, ^1 = m2. In 31-edo and 43-edo it's d2. But in edos like 15, 21 and 24, in which the circle of 5ths skips some notes, there is no 3-limit ratio. The ratio depends on the JI interpretation of the edo. For 10-edo, ^1 might equal 16/15 or 12/11 or 13/12. | ||
The additional accidental's ratio can be changed by adding the edo's defining comma onto it. For Blackwood, 5-edo is defined by 256/243, and /1 = 81/80 = 16/15. | The additional accidental's ratio can be changed by adding the edo's defining comma onto it. For Blackwood, 5-edo is defined by 256/243, and /1 = 81/80 = 16/15. | ||
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The interval P8/2 has a "ratio" of the square root of 2, which equals 2<span style="vertical-align: super;">1/2</span>, and its monzo can be written with fractions as (1/2, 0). In general, the pergen (P8/m, (a,b)/n) implies P = (1/m, 0) and G = (a/n, b/n). These equations make the **pergen matrix** [(1/m 0) (a/n b/n)], which is P and G in terms of P8 and P12. Its inverse is [(m 0) (-am/b n/b)], which is P8 and P12 in terms of P and G, i.e. the square mapping. | The interval P8/2 has a "ratio" of the square root of 2, which equals 2<span style="vertical-align: super;">1/2</span>, and its monzo can be written with fractions as (1/2, 0). In general, the pergen (P8/m, (a,b)/n) implies P = (1/m, 0) and G = (a/n, b/n). These equations make the **pergen matrix** [(1/m 0) (a/n b/n)], which is P and G in terms of P8 and P12. Its inverse is [(m 0) (-am/b n/b)], which is P8 and P12 in terms of P and G, i.e. the square mapping. | ||
Because we started with a valid pergen, the square mapping must be an integer matrix. Since n/b is an integer, n must be a multiple of |b|. From this it follows that a and b must be coprime, otherwise a, b, and n could all be reduced by GCD (a,b), and the multigen could be simplified. Since GCD (a, b) = 1 and -am/b is an integer, it follows that m must be a multiple of |b| as well. Thus GCD (m,n) = |b| · GCD (m/|b|, n/|b|)= |b| · r. If r = 1, then GCD (m, n) = |b|, and vice versa, which is the proposed test for a false double. | Because we started with a valid pergen, the square mapping must be an integer matrix. Since n/b is an integer, n must be a multiple of |b|. From this it follows that a and b must be coprime, otherwise a, b, and n could all be reduced by GCD (a,b), and the multigen could be simplified. Since GCD (a, b) = 1 and -am/b is an integer, it follows that m must be a multiple of |b| as well. Thus GCD (m,n) = |b| · GCD (m/|b|, n/|b|) = |b| · r. If r = 1, then GCD (m, n) = |b|, and vice versa, which is the proposed test for a false double. | ||
Assume the pergen is a false double, and there's a comma C that splits both P8 and (a,b) appropriately. Can we prove r = 1? Let Q = the higher prime that C uses. Express P, G and C as monzos | Assume the pergen is a false double, and there's a comma C that splits both P8 and (a,b) appropriately. Can we prove r = 1? Let Q = the higher prime that C uses. Express P, G and C as monzos of the prime subgroup 2.3.Q, by expanding the 2x2 pergen matrix to a 3x3 matrix A: | ||
P = (1/m, 0, 0) | P = (1/m, 0, 0) | ||
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C = (u, v, w) | C = (u, v, w) | ||
The inverse of A expresses 2, 3 and Q in terms of P, G and C. If C is tempered out, the C column can be discarded, making the usual 3x2 period-generator mapping. However, if C is not tempered out, the inverse of A is a 3x3 period-generator-comma mapping, which is simply a change of basis. For example, 5-limit JI can be generated by 2/1, 3/2 and 81/80. | Here u, v and w are integers. If GCD (u, v, w) > 1, simplify C so that it = 1. The inverse of A expresses 2, 3 and Q in terms of P, G and C. If C is tempered out, the C column can be discarded, making the usual 3x2 period-generator mapping. However, if C is not tempered out, the inverse of A is a 3x3 period-generator-comma mapping, which is simply a change of basis. For example, 5-limit JI can be generated by 2/1, 3/2 and 81/80. | ||
Fractions are allowed in the first two rows of A but not the 3rd row. Fractions are allowed in the last column of A-inverse, but not the first two columns. To avoid fractions in | 2 = 2/1 = P8 = (m, 0, 0) · (P, G, C) | ||
3 = 3/1 = P12 = (-am/b, n/b, 0) · (P, G, C) | |||
Q = Q/1 = ((av-bu)m/wb, -vn/wb, 1/w) · (P, G, C) | |||
Fractions are allowed in the first two rows of A but not the 3rd row. Fractions are allowed in the last column of A-inverse, but not the first two columns. To avoid fractions in those columns, A must be unimodular **//[I think, not sure, could it be i or 1/i for some integer i?]//**, and we have wb/mn = ±1, and w = ±mn/b. Substituting for w, we have: | |||
2 = 2/1 = P8 = (m, 0, 0) · (P, G, C) | 2 = 2/1 = P8 = (m, 0, 0) · (P, G, C) | ||
3 = 3/1 = P12 = (-am/b, n/b, 0) · (P, G, C) | |||
Q = Q/1 = (±(av-bu)/n, ±(-v)/m, ±b/mn) · (P, G, C) | |||
//**[Another try at it:]**// To split the 8ve into m parts, P8 ± C must be divisible by m, and both v and w must be a multiple of m. To split the multigen into n parts, M ± C must be divisible by n, and w must be a multiple of n. Thus for some nonzero integer k, w = k · LCM (m, n) = k · mn / GCD (m, n) = kmn/br. //**[end of another try]**// | |||
For v/m to be an integer, v must equal i·m for some integer i. Likewise, av-bu must equal j·n for some integer j. Thus bu = av - jn = iam - jn. Let p = m/rb and q = n/rb, where p and q are coprime integers, nonzero but possibly negative. Then m = prb and n = qrb. Substituting, we get bu = iaprb - jqrb, and u = r(iap - jq). Furthermore, v = im = iprb and w = ±mn/b = ±pqrrb. Thus u, v and w are all divisible by r. If r > 1, this contradicts the requirement that GCD (u, v, w) = 1, therefore r must be 1, and GCD (m, n) = |b|, and all false doubles pass the false-double test. | |||
Assuming r = 1, can we prove the existence of C = (u, v, w) for some prime Q? Let ¢(R) be the cents of the ratio R, and let ¢[M] be the cents of some monzo M. If we allow large commas, we can specify that Q = 5. | |||
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For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2<span class="nowrap">⋅</span>P8 - 3<span class="nowrap">⋅</span>P5) / (3<span class="nowrap">⋅</span>2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This <u>is</u> explicitly false, thus the comma can be found from m3/6 alone. G' is about 50¢, and the comma is 6<span class="nowrap">⋅</span>G' - m3. The comma splits both the octave and the fifth.<br /> | For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2<span class="nowrap">⋅</span>P8 - 3<span class="nowrap">⋅</span>P5) / (3<span class="nowrap">⋅</span>2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This <u>is</u> explicitly false, thus the comma can be found from m3/6 alone. G' is about 50¢, and the comma is 6<span class="nowrap">⋅</span>G' - m3. The comma splits both the octave and the fifth.<br /> | ||
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This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false | This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false. The unreduced pergen is (P8/4, M2/4), which also isn't explicitly false, thus (P8/4, P4/2) is a true double. It requires two commas, one for each fraction. The two commas must use different higher primes, e.g. 648/625 and 49/48. Thus <u>true doubles require commas of at least 7-limit</u>, whereas false doubles require only 5-limit. To summarize:<br /> | ||
<br /> | <ul><li><strong>A double-split pergen is <u>explicitly false</u> if and only if m = |b|.</strong></li><li><strong>A double-split pergen is a <u>true double</u> if and only if GCD (m, n) &gt; |b|.</strong></li><li><strong>A double-split pergen is a <u>true double</u> if and only if neither it nor its unreduced form is explicitly false.</strong></li></ul><br /> | ||
A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively.<br /> | A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively.<br /> | ||
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The sharp symbol's ratio is always (-11,7) = 2187/2048, by definition. Looking at the table in the Applications section, the up symbol often equals only a few ratios. For most 5-limit temperaments, ^1 = 81/80. For most 2.3.7 temperaments, ^1 = 64/63. Most 11-limit temperaments use either 33/32 or 729/704. These <strong>mapping commas</strong> are used to map higher primes to 3-limit intervals, and are essential for notation. They also determine where a ratio &quot;lands&quot; on a keyboard. By definition they are a P1, and the only intervals that map to P1 are these commas and combinations of them.<br /> | The sharp symbol's ratio is always (-11,7) = 2187/2048, by definition. Looking at the table in the Applications section, the up symbol often equals only a few ratios. For most 5-limit temperaments, ^1 = 81/80. For most 2.3.7 temperaments, ^1 = 64/63. Most 11-limit temperaments use either 33/32 or 729/704. These <strong>mapping commas</strong> are used to map higher primes to 3-limit intervals, and are essential for notation. They also determine where a ratio &quot;lands&quot; on a keyboard. By definition they are a P1, and the only intervals that map to P1 are these commas and combinations of them.<br /> | ||
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If a single-comma temperament uses double-pair notation, neither | If a single-comma temperament uses double-pair notation, neither accidental will equal the mapping comma. A double-comma temperament using double-pair notation may use the difference between two mapping commas, as in lemba, where ^1 equals 64/63 minus 81/80.<br /> | ||
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Sometimes the mapping comma needs to be inverted. In diminished, which sets 6/5 = P8/4, ^1 = 80/81. in every temperament except those in the meantone family, the 81/80 comma is not tempered out, but it is still tempered, just like every ratio. Occasionally 81/80 is tempered so far that it becomes a descending interval. See also blackwood-like pergens below.<br /> | Sometimes the mapping comma needs to be inverted. In diminished, which sets 6/5 = P8/4, ^1 = 80/81. in every temperament except those in the meantone family, the 81/80 comma is not tempered out, but it is still tempered, just like every ratio. Occasionally 81/80 is tempered so far that it becomes a descending interval. See also blackwood-like pergens below.<br /> | ||
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</table> | </table> | ||
If the edo's notation uses ups and downs, the up symbol can often be equated to a 3-limit ratio. In 17-edo and 22-edo, ^1 = m2. In 31-edo and 43-edo it's d2. But in edos like 15, 21 and 24, in which the circle of 5ths skips some notes, there is no 3-limit ratio. The ratio depends on the JI interpretation of the edo. For 10-edo, ^1 might equal 16/15 or 12/11 or 13/12. <br /> | If the edo's notation uses ups and downs, the up symbol can often be equated to a 3-limit ratio. In 17-edo and 22-edo, ^1 = m2. In 31-edo and 43-edo it's d2. But in edos like 15, 21 and 24, in which the circle of 5ths skips some notes, there is no 3-limit ratio. The ratio depends on the JI interpretation of the edo. For 10-edo, ^1 might equal 16/15 or 12/11 or 13/12.<br /> | ||
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The additional accidental's ratio can be changed by adding the edo's defining comma onto it. For Blackwood, 5-edo is defined by 256/243, and /1 = 81/80 = 16/15.<br /> | The additional accidental's ratio can be changed by adding the edo's defining comma onto it. For Blackwood, 5-edo is defined by 256/243, and /1 = 81/80 = 16/15.<br /> | ||
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finish proofs<br /> | finish proofs<br /> | ||
link from: ups and downs page, Kite Giedraitis page, MOS scale names page,<br /> | link from: ups and downs page, Kite Giedraitis page, MOS scale names page,<br /> | ||
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This PDF is a rank-2 notation guide that shows the full lattice for the first 15 pergens, up through the third-splits block. It includes alternate enharmonics for many pergens.<br /> | This PDF is a rank-2 notation guide that shows the full lattice for the first 15 pergens, up through the third-splits block. It includes alternate enharmonics for many pergens.<br /> | ||
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Alt-pergenLister lists out thousands of pergens, and suggests periods, generators and enharmonics for each one. Alternate enharmonics are not listed, but single-pair notation for false-double pergens is. It can also list only those pergens supported by a specific edo. Written in Jesusonic, runs inside Reaper.<br /> | Alt-pergenLister lists out thousands of pergens, and suggests periods, generators and enharmonics for each one. Alternate enharmonics are not listed, but single-pair notation for false-double pergens is. It can also list only those pergens supported by a specific edo. Written in Jesusonic, runs inside Reaper.<br /> | ||
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Red indicates problems. Generators of 50¢ or less are in red. Enharmonics of a 3rd or more are in red. Screenshots of the first 38 pergens:<br /> | Red indicates problems. Generators of 50¢ or less are in red. Enharmonics of a 3rd or more are in red. Screenshots of the first 38 pergens:<br /> | ||
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Listing all valid pergens is not a trivial task, like listing all valid edos or all valid MOS scales. Not all combinations of octave fractions and multigen fractions make a valid pergen. The search for rank-2 pergens can be done by looping through all possible square mappings [(x, y), (0, z)], and using the formula (P8/x, (i·z - y, x) / xz). While x is always positive and z is always nonzero, y can take on any value. For any x and z, y can be constrained to produce a reasonable cents value for 3/1. Let T be the tempered twefth 3/1. The mapping says T = y·P + z·G = y·P8/x + z·G. Thus y = x·(T/P8 - z·G/P8). We adopt the convention that G is less than half an octave. We constrain T so that the 5th is between 600¢ and 800¢, which certainly includes anything that sounds like a 5th. Thus T is between 3/2 and 5/3 of an octave. We assume that if the octave is stretched, the ranges of T and G will be stretched along with it. The outer ranges of y can now be computed, using the floor function to round down to the nearest integer, and the ceiling function to round up:<br /> | Listing all valid pergens is not a trivial task, like listing all valid edos or all valid MOS scales. Not all combinations of octave fractions and multigen fractions make a valid pergen. The search for rank-2 pergens can be done by looping through all possible square mappings [(x, y), (0, z)], and using the formula (P8/x, (i·z - y, x) / xz). While x is always positive and z is always nonzero, y can take on any value. For any x and z, y can be constrained to produce a reasonable cents value for 3/1. Let T be the tempered twefth 3/1. The mapping says T = y·P + z·G = y·P8/x + z·G. Thus y = x·(T/P8 - z·G/P8). We adopt the convention that G is less than half an octave. We constrain T so that the 5th is between 600¢ and 800¢, which certainly includes anything that sounds like a 5th. Thus T is between 3/2 and 5/3 of an octave. We assume that if the octave is stretched, the ranges of T and G will be stretched along with it. The outer ranges of y can now be computed, using the floor function to round down to the nearest integer, and the ceiling function to round up:<br /> | ||
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The interval P8/2 has a &quot;ratio&quot; of the square root of 2, which equals 2<span style="vertical-align: super;">1/2</span>, and its monzo can be written with fractions as (1/2, 0). In general, the pergen (P8/m, (a,b)/n) implies P = (1/m, 0) and G = (a/n, b/n). These equations make the <strong>pergen matrix</strong> [(1/m 0) (a/n b/n)], which is P and G in terms of P8 and P12. Its inverse is [(m 0) (-am/b n/b)], which is P8 and P12 in terms of P and G, i.e. the square mapping.<br /> | The interval P8/2 has a &quot;ratio&quot; of the square root of 2, which equals 2<span style="vertical-align: super;">1/2</span>, and its monzo can be written with fractions as (1/2, 0). In general, the pergen (P8/m, (a,b)/n) implies P = (1/m, 0) and G = (a/n, b/n). These equations make the <strong>pergen matrix</strong> [(1/m 0) (a/n b/n)], which is P and G in terms of P8 and P12. Its inverse is [(m 0) (-am/b n/b)], which is P8 and P12 in terms of P and G, i.e. the square mapping.<br /> | ||
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Because we started with a valid pergen, the square mapping must be an integer matrix. Since n/b is an integer, n must be a multiple of |b|. From this it follows that a and b must be coprime, otherwise a, b, and n could all be reduced by GCD (a,b), and the multigen could be simplified. Since GCD (a, b) = 1 and -am/b is an integer, it follows that m must be a multiple of |b| as well. Thus GCD (m,n) = |b| · GCD (m/|b|, n/|b|)= |b| · r. If r = 1, then GCD (m, n) = |b|, and vice versa, which is the proposed test for a false double.<br /> | Because we started with a valid pergen, the square mapping must be an integer matrix. Since n/b is an integer, n must be a multiple of |b|. From this it follows that a and b must be coprime, otherwise a, b, and n could all be reduced by GCD (a,b), and the multigen could be simplified. Since GCD (a, b) = 1 and -am/b is an integer, it follows that m must be a multiple of |b| as well. Thus GCD (m,n) = |b| · GCD (m/|b|, n/|b|) = |b| · r. If r = 1, then GCD (m, n) = |b|, and vice versa, which is the proposed test for a false double.<br /> | ||
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Assume the pergen is a false double, and there's a comma C that splits both P8 and (a,b) appropriately. Can we prove r = 1? Let Q = the higher prime that C uses. Express P, G and C as monzos | Assume the pergen is a false double, and there's a comma C that splits both P8 and (a,b) appropriately. Can we prove r = 1? Let Q = the higher prime that C uses. Express P, G and C as monzos of the prime subgroup 2.3.Q, by expanding the 2x2 pergen matrix to a 3x3 matrix A:<br /> | ||
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P = (1/m, 0, 0)<br /> | P = (1/m, 0, 0)<br /> | ||
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C = (u, v, w)<br /> | C = (u, v, w)<br /> | ||
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The inverse of A expresses 2, 3 and Q in terms of P, G and C. If C is tempered out, the C column can be discarded, making the usual 3x2 period-generator mapping. However, if C is not tempered out, the inverse of A is a 3x3 period-generator-comma mapping, which is simply a change of basis. For example, 5-limit JI can be generated by 2/1, 3/2 and 81/80.<br /> | Here u, v and w are integers. If GCD (u, v, w) &gt; 1, simplify C so that it = 1. The inverse of A expresses 2, 3 and Q in terms of P, G and C. If C is tempered out, the C column can be discarded, making the usual 3x2 period-generator mapping. However, if C is not tempered out, the inverse of A is a 3x3 period-generator-comma mapping, which is simply a change of basis. For example, 5-limit JI can be generated by 2/1, 3/2 and 81/80.<br /> | ||
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Fractions are allowed in the first two rows of A but not the 3rd row. Fractions are allowed in the last column of A-inverse, but not the first two columns. To avoid fractions in | 2 = 2/1 = P8 = (m, 0, 0) · (P, G, C)<br /> | ||
3 = 3/1 = P12 = (-am/b, n/b, 0) · (P, G, C)<br /> | |||
Q = Q/1 = ((av-bu)m/wb, -vn/wb, 1/w) · (P, G, C)<br /> | |||
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Fractions are allowed in the first two rows of A but not the 3rd row. Fractions are allowed in the last column of A-inverse, but not the first two columns. To avoid fractions in those columns, A must be unimodular <strong><em>[I think, not sure, could it be i or 1/i for some integer i?]</em></strong>, and we have wb/mn = ±1, and w = ±mn/b. Substituting for w, we have:<br /> | |||
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2 = 2/1 = P8 = (m, 0, 0) · (P, G, C)<br /> | 2 = 2/1 = P8 = (m, 0, 0) · (P, G, C)<br /> | ||
3 = 3/1 = P12 = (-am/b, n/b, 0) · (P, G, C)<br /> | |||
Q = Q/1 = (±(av-bu)/n, ±(-v)/m, ±b/mn) · (P, G, C)<br /> | |||
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<em><strong>[Another try at it:]</strong></em> To split the 8ve into m parts, P8 ± C must be divisible by m, and both v and w must be a multiple of m. To split the multigen into n parts, M ± C must be divisible by n, and w must be a multiple of n. Thus for some nonzero integer k, w = k · LCM (m, n) = k · mn / GCD (m, n) = kmn/br. <em><strong>[end of another try]</strong></em><br /> | |||
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For v/m to be an integer, v must equal i·m for some integer i. Likewise, av-bu must equal j·n for some integer j. Thus bu = av - jn = iam - jn. Let p = m/rb and q = n/rb, where p and q are coprime integers, nonzero but possibly negative. Then m = prb and n = qrb. Substituting, we get bu = iaprb - jqrb, and u = r(iap - jq). Furthermore, v = im = iprb and w = ±mn/b = ±pqrrb. Thus u, v and w are all divisible by r. If r &gt; 1, this contradicts the requirement that GCD (u, v, w) = 1, therefore r must be 1, and GCD (m, n) = |b|, and all false doubles pass the false-double test.<br /> | |||
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Assuming r = 1, can we prove the existence of C = (u, v, w) for some prime Q? Let ¢(R) be the cents of the ratio R, and let ¢[M] be the cents of some monzo M. If we allow large commas, we can specify that Q = 5.<br /> | |||
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