Pergen names: Difference between revisions

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<h2>IMPORTED REVISION FROM WIKISPACES</h2>

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: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-17 04:42:52 UTC</tt>.

: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-17 05:48:37 UTC</tt>.

: The original revision id was <tt>~~623937571~~</tt>.

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||= (P8/2, M2/4) ||= half-octave, quarter-tone ||= (-17,2,0,0,4) ||= large quadruple jade ||= large quadruple jade ||= Lj4T ||

||= (P8, P12/5) ||= fifth-twelfth ||= (-10,-1,5) ||= magic ||= large quintuple yellow ||= Ly5T ||

(P8/2, P4/2) is called half-everything because the ~~3/2~~ fifth, the ~~9/8 major 2nd~~, ~~and~~ every ~~other~~ 3-limit interval is split in half.

(P8/2, P4/2) is called half-everything because the fifth is also split in half, and since every 3-limit interval can be expressed as the sum of octaves and fifths, every single 3-limit interval is also split in half.

The color name indicates the amount of splitting: deep splits something into two parts, triple into three parts, etc. The multi-gen is usually some voicing of the 4th or 5th, but can be any 3-limit interval, as in the second to last example

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=__Derivation__=

For any comma containing primes 2 and 3, let M = the GCD of all the monzo's exponents other than the 2-exponent, and let N = the GCD of all its higher-prime exponents. The comma will split the octave into M parts, and if N > M, it will split some other 3-limit interval into N parts. Therefore a comma with only two primes, one of which is 2, always splits the octave (unless the other prime's exponent is ±1, e.g. 32/31). And a comma with only one higher prime will always split something, unless that prime's exponent is ±1.

For any comma containing primes 2 and 3, let m = the GCD of all the monzo's exponents other than the 2-exponent, and let n = the GCD of all its higher-prime exponents. The comma will split the octave into m parts, and if n > m, it will split some other 3-limit interval into n parts. Therefore a comma with only two primes, one of which is 2, always splits the octave (unless the other prime's exponent is ±1, e.g. 32/31). And a comma with only one higher prime will always split something, unless that prime's exponent is ±1.

In a multi-comma rank-2 or higher temperament, it's possible that one comma will contain only the 1st and 2nd primes, and the 2nd prime is directly related to the 1st prime. If this happens, the multi-gen must use the 1st and 3rd primes. If the 3rd prime is also directly related, the 4th prime is used, and so forth.

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G = (-y, x) / xz + nP = (-y, x) / xz + nP8/x = (nz - y, x) / xz

**Rank-2 pergen from the [(x, 0), (y, z)] matrix: {P8/x, (nz-y, x)/xz}**

|| **Rank-2 pergen from the [(x, 0), (y, z)] matrix: {P8/x, (nz-y, x)/xz}** ||

Rank-3 example: Breedsmic is 2.3.5.7 with 2401/2400 = (-5,-1,-2,4) tempered out. [[http://x31eq.com/cgi-bin/rt.cgi?ets=130_171_270&limit=7|x31.com]] gives us this matrix:

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||~ pergen ||~ enharmonic

interval(s) ||~ equiva-

lence(s) ||~ ~~split interval(s)~~ ||~ ~~genchain~~(s) and

lence(s) ||~ period and

~~perchains~~(s) ||~ examples ||~ compatible edos

generator ||~ perchain(s) and

genchains(s) ||~ examples ||~ compatible edos

(12-31 only) ||

||= (P8, P5) ||= none ||= none ||= none ||= C - G ||= meantone ||= 12, 13b, 14*, 15*, 16,

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If the pergen is not explicitly false, put the pergen in its **unreduced** form, which is always explicitly false if the pergen is a false double. The unreduced form replaces the generator with the difference between the period and the generator: (P8/m, M/n) = (P8/m, P8/m - M/n) = (P8/m, (n*P8 - m*M)/n*m). The new multigen is the product of the outer elements (P8 and n) minus the product of the inner elements (m and M), divided by the product of the fractions (m and n). Invert the new multi-gen if descending, and reduce if possible. The new multi-gen will have either a larger fraction, or a larger size in cents, or both, hence the name unreduced.

For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced ~~form has~~ (2*P8 - 3*P5)/(3*2) = m3/6, and the ~~new~~ pergen is (P8/3, m3/6). This __is__ explicitly false, thus the comma can be found from m3/6 alone. R is about 50¢, and the comma is 6*R - 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*P8 - 3*P5)/(3*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. R is about 50¢, and the comma is 6*R - m3. The comma splits both the octave and the fifth.

This creates 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. Its unreduced form has (2*P8 - 4*P4)/(2*4) = (2*M2)/8, which reduces to M2/4. 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.

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) can be constructed from 128/125 and 49/48, which split the octave and the 4th respectively.

Unreducing replaces the generator with an alternate generator. Any number of periods plus or minus a single generator makes an **alternate** generator. A generator or period plus or minus any number of enharmonics makes an **equivalent** generator or period. An equivalent generator is always the same size in cents, since the enharmonic is always 0¢. An equivalent generator is the same interval, merely notated differently.

For example, (P8/2, P5) has generator P5, alternate generator P4, period vA4, and equivalent period ^d5. (P8, P5/2) has generator ^m3 and equivalent generator vM3. Of the two equivalent generators, "the" generator is always the smaller one (smaller degree, or if the degrees are the same, more diminished). This is because the equivalent generator can be more easily found by adding the enharmonic, rather than subtracting it. For example, ^m3 + vvA1 = vM3 is an easier calculation than vM3 - vvA1 = ^m3. This is particularly true with complex enharmonics like ^6dd2.

==Finding a notation for a pergen==

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There are multiple notations for a given pergen, depending on the enharmonic interval(s). Preferably, the enharmonic's degree will be a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting. If it's a unison, it will always be an A1. (P1 would be pointless, d1 would be inverted to A1, and AA1 would be split into two A1's.) If it's a 2nd, preferably it will be a m2 or a d2 or a dd2, and not a M2 or an A2. There is an easy method for finding such a pergen, if one exists. First, some basic concepts:

The **keyspan** of an interval is the number of keys or frets or semitones that the interval spans in 12-edo. Most musicians know that a minor 2nd is one key or fret and a major 2nd is two keys or frets. The keyspans of larger intervals aren't as well known. The concept can easily be expanded to other edos, but we'll assume 12-edo for now. The **stepspan** of an interval is simply the degree minus one. M2, m2, A2 and d2 all have a stepspan of 1. P4 and ~~A4 both~~ have stepspan 3. The stepspan can be thought of as the 7-edo keyspan. Again, this concept can be expanded to include pentatonicism, etc., but we'll assume heptatonicism for now.

The **keyspan** of an interval is the number of keys or frets or semitones that the interval spans in 12-edo. Most musicians know that a minor 2nd is one key or fret and a major 2nd is two keys or frets. The keyspans of larger intervals aren't as well known. The concept can easily be expanded to other edos, but we'll assume 12-edo for now. The **stepspan** of an interval is simply the degree minus one. M2, m2, A2 and d2 all have a stepspan of 1. P5, d5 and A5 all have stepspan 4. The stepspan can be thought of as the 7-edo keyspan. Again, this concept can be expanded to include pentatonicism, octotonicism, etc., but we'll assume heptatonicism for now.

Every 3-limit interval can be uniquely expressed as the combination of a keyspan and a stepspan. This combination is called a **gedra**, analogous to a monzo, but written in brackets not parentheses: 3/2 is a 7-semitone 5th, thus 3/2 = (-1,1) = [7,4]. 9/8 = (-3,2) = [2,1] = a 2-semitone 1-step interval. The octave 2/1 = [12,7]. For any 3-limit interval (a,b), there is a unique gedra [k,s], and vice versa:

Every 3-limit interval can be uniquely expressed as the combination of a keyspan and a stepspan. This combination is called a **gedra**, analogous to a monzo, but written in brackets not parentheses: 3/2 is a 7-semitone 5th, thus 3/2 = (-1,1) = [7,4]. 9/8 = (-3,2) = [2,1] = a 2-semitone 1-step interval. The octave 2/1 = [12,7]. For any 3-limit interval with a monzo (a,b), there is a unique gedra [k,s], and vice versa:

k = 12a + 19b

s = 7a + 11b

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b = 7a - 12b

Gedras can be manipulated exactly like monzos. Just as (a,b) + (a',b') = (a+a',b+b'), likewise [k,s] + [k',s'] = [k+k',s+s']. If (a,b) is a stack of n intervals, with (a,b) = n(a',b') = (na', nb'), then [k,s] = n[k',s'] = [nk',ns'].

Gedras can be manipulated exactly like monzos. Just as (a,b) + (a',b') = (a+a',b+b'), likewise [k,s] + [k',s'] = [k+k',s+s']. If (a,b) is a stack of n intervals, with (a,b) = n(a',b') = (na', nb'), and if (a,b) = [k,s], then [k,s] = n[k',s'] = [nk',ns'].

A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. One fifth of an octave [12,7] is approximately [round(12/5), round(7/5)] = [2,1] = (-3,2) = M2. This approximation will __always__ produce an enharmonic interval with the smallest possible keyspan and stepspan, which is usually the preferred enharmonic. In this example, the bare enharmonic is the difference between P8 and 5*M2, which is [12,7] - 5*[2,1] = [12,7] - [10,5] = [2,2] = 2*[1,1] = 2*m2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the ~~number of periods~~, thus E = v5m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. Equipped with the period and the enharmonic, the perchain is easily found:

Gedras greatly facilitate find a pergen's period, generator and/or enharmonic. A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. For example, consider (P8/5, P5). One fifth of an octave [12,7] is approximately [round(12/5), round(7/5)] = [2,1] = (-3,2) = M2. This approximation will __always__ produce an enharmonic interval with the smallest possible keyspan and stepspan, which is usually the preferred enharmonic. In this example, the bare enharmonic is the difference between P8 and 5*M2, which is [12,7] - 5*[2,1] = [12,7] - [10,5] = [2,2] = 2*[1,1] = 2*m2. "2*" indicates that the enharmonic will occur twice in the perchain. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v5m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. Equipped with the period and the enharmonic, the perchain is easily found:

(P8/5, P5) perchain: P1 - ^^M2=v3m3 - v4 - ^5 - ^3M6=vvm7 - P8, or C - D^^=Ebv3 - Fv - G^ - A^3=Bbvv - C

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(P8, P4/5) genchain: P1 - ^^m2=v3A1 - vM2 - ^m3 - ^3d4=vvM3 - P4, or C - Db^^=C#v3 - Dv - Eb^ - Fb^3=Evv - F

(P8, P5/2) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4*G = [1,2] = dd3. It must be downed, thus E = ~~v4dd3~~, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because ~~it makes it easier to find~~ the ~~alternate generator G' = G + E~~. ~~Otherwise G' would be G -E~~, ~~and it's harder to subtract than add~~. ~~For example (P8, P5/2) has G = ^m3 and E = vvA1~~

(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4*G = [1,2] = dd3. It must be downed, thus E = v4dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.

To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multi-gen as before. Then deduce the period from the enharmonic.

To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multi-gen as before. Then deduce the period from the enharmonic. An alternate generator can be found by adding or subtracting periods and possibly inverting.

For example, (P8/5, P4/2) unreduces to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, we'll address that later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^4M2. Our small generator G can generate m2, but not P4. ~~Any number of periods plus or minus a single generator makes~~ an alternate generator. We can deduce ~~the 4th's generator~~ G' as we did for the period: ~~the 4th~~ plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is 1 or -1. We choose -1, to get a smaller G', ~~to which E can be added~~. 2*G' = P4 - E, and G' = ^~~5M2~~, which happens to be P + G.

For example, (P8/5, P4/2) unreduces to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, we'll address that later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^4M2. Our small generator G can generate m2, but not P4. For that we need an alternate generator G'. We can deduce G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is 1 or -1. We choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, and G' = ^5M2, which happens to be P + G.

(P8/5, P4/2): P = ^~~4M2~~, G = ^1, G' = ^~~5M2~~, E = ~~v10m2~~

(P8/5, P4/2): P = ^4M2, G = ^1, G' = ^5M2, E = v10m2

perchain: P1 - ^4M2=v6m3 - vvP4 - ^^P5 - ^6M6=v4m7 - P8, or C - D^4=Ebv6 - Fvv - G^^ - A^6=Bbv4 - C

genchain: P1 - ^5M2=v5m3 - P4, or C - D^5=Ebv5 - F

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</table>

(P8/2, P4/2) is called half-everything because the ~~3/2~~ fifth, the ~~9/8 major 2nd~~, ~~and~~ every ~~other~~ 3-limit interval is split in half.

(P8/2, P4/2) is called half-everything because the fifth is also split in half, and since every 3-limit interval can be expressed as the sum of octaves and fifths, every single 3-limit interval is also split in half.

The color name indicates the amount of splitting: deep splits something into two parts, triple into three parts, etc. The multi-gen is usually some voicing of the 4th or 5th, but can be any 3-limit interval, as in the second to last example

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<h1 id="toc1"><a name="Derivation"></a>Derivation</h1>

For any comma containing primes 2 and 3, let M = the GCD of all the monzo's exponents other than the 2-exponent, and let N = the GCD of all its higher-prime exponents. The comma will split the octave into M parts, and if N &gt; M, it will split some other 3-limit interval into N parts. Therefore a comma with only two primes, one of which is 2, always splits the octave (unless the other prime's exponent is ±1, e.g. 32/31). And a comma with only one higher prime will always split something, unless that prime's exponent is ±1.

For any comma containing primes 2 and 3, let m = the GCD of all the monzo's exponents other than the 2-exponent, and let n = the GCD of all its higher-prime exponents. The comma will split the octave into m parts, and if n &gt; m, it will split some other 3-limit interval into n parts. Therefore a comma with only two primes, one of which is 2, always splits the octave (unless the other prime's exponent is ±1, e.g. 32/31). And a comma with only one higher prime will always split something, unless that prime's exponent is ±1.

In a multi-comma rank-2 or higher temperament, it's possible that one comma will contain only the 1st and 2nd primes, and the 2nd prime is directly related to the 1st prime. If this happens, the multi-gen must use the 1st and 3rd primes. If the 3rd prime is also directly related, the 4th prime is used, and so forth.

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G = (-y, x) / xz + nP = (-y, x) / xz + nP8/x = (nz - y, x) / xz

Rank-2 pergen from the [(x, 0), (y, z)] matrix: {P8/x, (nz-y, x)/xz}

<tr>

<td>Rank-2 pergen from the [(x, 0), (y, z)] matrix: {P8/x, (nz-y, x)/xz}

</td>

</tr>

</table>

Rank-3 example: Breedsmic is 2.3.5.7 with 2401/2400 = (-5,-1,-2,4) tempered out. <a class="wiki_link_ext" href="http://x31eq.com/cgi-bin/rt.cgi?ets=130_171_270&amp;limit=7" rel="nofollow">x31.com</a> gives us this matrix:

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lence(s)

</th>

<th>~~split interval(s)~~

<th>period and

generator

</th>

<th>~~genchain~~(s) and

<th>perchain(s) and

~~perchains~~(s)

genchains(s)

</th>

<th>examples

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If the pergen is not explicitly false, put the pergen in its unreduced form, which is always explicitly false if the pergen is a false double. The unreduced form replaces the generator with the difference between the period and the generator: (P8/m, M/n) = (P8/m, P8/m - M/n) = (P8/m, (n*P8 - m*M)/n*m). The new multigen is the product of the outer elements (P8 and n) minus the product of the inner elements (m and M), divided by the product of the fractions (m and n). Invert the new multi-gen if descending, and reduce if possible. The new multi-gen will have either a larger fraction, or a larger size in cents, or both, hence the name unreduced.

For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced ~~form has~~ (2*P8 - 3*P5)/(3*2) = m3/6, and the ~~new~~ pergen is (P8/3, m3/6). This is explicitly false, thus the comma can be found from m3/6 alone. R is about 50¢, and the comma is 6*R - 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*P8 - 3*P5)/(3*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. R is about 50¢, and the comma is 6*R - m3. The comma splits both the octave and the fifth.

This creates 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. Its unreduced form has (2*P8 - 4*P4)/(2*4) = (2*M2)/8, which reduces to M2/4. 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.

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) can be constructed from 128/125 and 49/48, which split the octave and the 4th respectively.

Unreducing replaces the generator with an alternate generator. Any number of periods plus or minus a single generator makes an alternate generator. A generator or period plus or minus any number of enharmonics makes an equivalent generator or period. An equivalent generator is always the same size in cents, since the enharmonic is always 0¢. An equivalent generator is the same interval, merely notated differently.

For example, (P8/2, P5) has generator P5, alternate generator P4, period vA4, and equivalent period ^d5. (P8, P5/2) has generator ^m3 and equivalent generator vM3. Of the two equivalent generators, &quot;the&quot; generator is always the smaller one (smaller degree, or if the degrees are the same, more diminished). This is because the equivalent generator can be more easily found by adding the enharmonic, rather than subtracting it. For example, ^m3 + vvA1 = vM3 is an easier calculation than vM3 - vvA1 = ^m3. This is particularly true with complex enharmonics like ^6dd2.

<h2 id="toc7"><a name="Further Discussion-Finding a notation for a pergen"></a>Finding a notation for a pergen</h2>

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There are multiple notations for a given pergen, depending on the enharmonic interval(s). Preferably, the enharmonic's degree will be a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting. If it's a unison, it will always be an A1. (P1 would be pointless, d1 would be inverted to A1, and AA1 would be split into two A1's.) If it's a 2nd, preferably it will be a m2 or a d2 or a dd2, and not a M2 or an A2. There is an easy method for finding such a pergen, if one exists. First, some basic concepts:

The keyspan of an interval is the number of keys or frets or semitones that the interval spans in 12-edo. Most musicians know that a minor 2nd is one key or fret and a major 2nd is two keys or frets. The keyspans of larger intervals aren't as well known. The concept can easily be expanded to other edos, but we'll assume 12-edo for now. The stepspan of an interval is simply the degree minus one. M2, m2, A2 and d2 all have a stepspan of 1. P4 and ~~A4 both~~ have stepspan 3. The stepspan can be thought of as the 7-edo keyspan. Again, this concept can be expanded to include pentatonicism, etc., but we'll assume heptatonicism for now.

The keyspan of an interval is the number of keys or frets or semitones that the interval spans in 12-edo. Most musicians know that a minor 2nd is one key or fret and a major 2nd is two keys or frets. The keyspans of larger intervals aren't as well known. The concept can easily be expanded to other edos, but we'll assume 12-edo for now. The stepspan of an interval is simply the degree minus one. M2, m2, A2 and d2 all have a stepspan of 1. P5, d5 and A5 all have stepspan 4. The stepspan can be thought of as the 7-edo keyspan. Again, this concept can be expanded to include pentatonicism, octotonicism, etc., but we'll assume heptatonicism for now.

Every 3-limit interval can be uniquely expressed as the combination of a keyspan and a stepspan. This combination is called a gedra, analogous to a monzo, but written in brackets not parentheses: 3/2 is a 7-semitone 5th, thus 3/2 = (-1,1) = [7,4]. 9/8 = (-3,2) = [2,1] = a 2-semitone 1-step interval. The octave 2/1 = [12,7]. For any 3-limit interval (a,b), there is a unique gedra [k,s], and vice versa:

Every 3-limit interval can be uniquely expressed as the combination of a keyspan and a stepspan. This combination is called a gedra, analogous to a monzo, but written in brackets not parentheses: 3/2 is a 7-semitone 5th, thus 3/2 = (-1,1) = [7,4]. 9/8 = (-3,2) = [2,1] = a 2-semitone 1-step interval. The octave 2/1 = [12,7]. For any 3-limit interval with a monzo (a,b), there is a unique gedra [k,s], and vice versa:

k = 12a + 19b

s = 7a + 11b

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b = 7a - 12b

Gedras can be manipulated exactly like monzos. Just as (a,b) + (a',b') = (a+a',b+b'), likewise [k,s] + [k',s'] = [k+k',s+s']. If (a,b) is a stack of n intervals, with (a,b) = n(a',b') = (na', nb'), then [k,s] = n[k',s'] = [nk',ns'].

Gedras can be manipulated exactly like monzos. Just as (a,b) + (a',b') = (a+a',b+b'), likewise [k,s] + [k',s'] = [k+k',s+s']. If (a,b) is a stack of n intervals, with (a,b) = n(a',b') = (na', nb'), and if (a,b) = [k,s], then [k,s] = n[k',s'] = [nk',ns'].

A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. One fifth of an octave [12,7] is approximately [round(12/5), round(7/5)] = [2,1] = (-3,2) = M2. This approximation will always produce an enharmonic interval with the smallest possible keyspan and stepspan, which is usually the preferred enharmonic. In this example, the bare enharmonic is the difference between P8 and 5*M2, which is [12,7] - 5*[2,1] = [12,7] - [10,5] = [2,2] = 2*[1,1] = 2*m2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the ~~number of periods~~, thus E = v5m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. Equipped with the period and the enharmonic, the perchain is easily found:

Gedras greatly facilitate find a pergen's period, generator and/or enharmonic. A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. For example, consider (P8/5, P5). One fifth of an octave [12,7] is approximately [round(12/5), round(7/5)] = [2,1] = (-3,2) = M2. This approximation will always produce an enharmonic interval with the smallest possible keyspan and stepspan, which is usually the preferred enharmonic. In this example, the bare enharmonic is the difference between P8 and 5*M2, which is [12,7] - 5*[2,1] = [12,7] - [10,5] = [2,2] = 2*[1,1] = 2*m2. &quot;2*&quot; indicates that the enharmonic will occur twice in the perchain. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v5m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. Equipped with the period and the enharmonic, the perchain is easily found:

(P8/5, P5) perchain: P1 - ^^M2=v3m3 - v4 - ^5 - ^3M6=vvm7 - P8, or C - D^^=Ebv3 - Fv - G^ - A^3=Bbvv - C

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