Kite's thoughts on pergens: Difference between revisions

Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xE = P8 - mP = P8 - 5<span style="">⋅</span>M2 = [12,7] - 5<span style="">⋅</span>[2,1] = [2,2] = 2<span style="">⋅</span>[1,1] = 2<span style="">⋅</span>m2. Because x = 2, E will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2<span style="">⋅</span>m2 = d3). The enharmonic's '''count''' is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v<span style="vertical-align: super;">5</span>m2. Since P8 = 5<span style="">⋅</span>P + 2<span style="">⋅</span>E, the period must be ^^M2, to make the ups and downs come out even. The number of the period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:

Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [1,1] = m2. The bare enharmonic is P4 - 5<span style="">⋅</span>m2 = [5,3] - 5<span style="">⋅</span>[1,1] = [5,3] - [5,5] = [0,-2] = -2<span style="">⋅</span>[0,1] = two descending d2's. The d2 must be upped, and E = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5<span style="">⋅</span>G - 2<span style="">⋅</span>E, G must be ^^m2. The genchain is:

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10<span style="">⋅</span>P1 = m2. It must be downed, thus E = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10<span style="">⋅</span>G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x<span style="">⋅</span>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<span style="">⋅</span>P + 2<span style="">⋅</span>E, and P = ^<span style="vertical-align: super;">4</span>M2. Next, find the original half-4th generator. P = P8/5 = ~240¢, and G = P4/2 = ~250¢. Because P &lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^<span style="vertical-align: super;">4</span>M2 + ^1 = ^<span style="vertical-align: super;">5</span>M2. While the alternate multigen is more complex than the original multigen, the alternate generator is usually simpler than the original generator.

To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2), the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.

Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅M3 = [1,2] = dd3. So E = v<span style="vertical-align: super;">4</span>dd3, and G = ^M3. But by using double-pair notation, we can avoid that. We find P11/2, which equals two generators: P11/2 = 2⋅G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2⋅m6 = [1,0] = A1. For this second enharmonic, we use the second pair of accidentals: E' = \\A1 and 2⋅G = /m6 or \M6. The sum or difference of two enharmonic intervals is also an enharmonic: E + E' = v<span style="vertical-align: super;">4</span>\\d3 = 2·vv\m2, and E - E' = v<span style="vertical-align: super;">4</span>//ddd3 = 2·vv/d2. Thus vv\m2 and vv/d2 are equivalent enharmonics, and v\4 and v/d4 are equivalent generators. Here is the genchain:

One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. We have [3,2]/12 = [0,0] = P1, and G' = ^1 and E = v<span style="vertical-align: super;">12</span>m3. Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus E' = /<span style="vertical-align: super;">3</span>d2, and 4·G' = /m2. The period can be deduced from 4·G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4·G' = P4 - /m2 = \M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = \M3 + ^1 = ^\M3. Equivalent enharmonics are found from E + E' and E - 2·E'. Equivalent periods and generators are found from the many enharmonics, which also allow much freedom in chord spelling.

<span style="display: block; text-align: center;">P1 — \M3 — \\A5=/m6 — P8</span><span style="display: block; text-align: center;">C — E\ — Ab/ — C</span><span style="display: block; text-align: center;">P1 — ^\M3 — ^^\\A5=^^/m6=vv\M6 — ^<span style="vertical-align: super;">3</span>8=v/m9 — P11

<span style="display: block; text-align: center;">C — E^\ — Ab^^/=Avv\ — Dbv/ — F</span>

<span style="display: block; text-align: center;">P1 — ^/1=v/m2 — //m2=\M2 — ^M2=vm3 — /m3=\\M3 — ^\M3=v\4 — P4

<span style="display: block; text-align: center;">C — C^/=Dbv/ — Db//=D\ — D^=Ebv — Eb/=E\\ — E^/=Fv\ — F

<span style="display: block; text-align: center;">P1 -- ^<span style="vertical-align: super;">4</span>m3 -- v<span style="vertical-align: super;">4</span>M6 -- C

Because G is a M2 and E is an A2, the equivalent generator G - E is a descending A1. Ascending intervals that look descending can be confusing, one has to take into account the eighth-tone ups to see that Dv<span style="vertical-align: super;">3</span> -- Db^<span style="vertical-align: super;">6</span> is ascending. Double-pair notation may be preferable. This makes P = vM3, E = ^3d2, G = /m2, and E' = /4dd2.

<span style="display: block; text-align: center;">P1 -- vM3 -- ^m6 -- P8

<span style="display: block; text-align: center;">C -- Ev -- Ab^ -- C

<span style="display: block; text-align: center;">P1 -- /m2 -- //d3=\\A2 -- \M3 -- P4

<span style="display: block; text-align: center;">C -- Db/ -- Ebb//=D#\\ -- E\ -- F</span>