Pergen names: Difference between revisions
Wikispaces>TallKite **Imported revision 624089721 - Original comment: ** |
Wikispaces>TallKite **Imported revision 624089859 - 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>2017-12-20 01: | : This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-20 01:32:05 UTC</tt>.<br> | ||
: The original revision id was <tt> | : The original revision id was <tt>624089859</tt>.<br> | ||
: The revision comment was: <tt></tt><br> | : The revision comment was: <tt></tt><br> | ||
The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br> | The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br> | ||
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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. | 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 | For example, half-octave (P8/2, P5) has generator P5, alternate generators P4 and vA1, 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 ^<span style="vertical-align: super;">6</span>dd2. | ||
There are also equivalent enharmonics, see below. | There are also equivalent enharmonics, see below. | ||
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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. | 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 | 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 = (-1,1) is a 7-semitone 5th, thus (-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 | k = 12a + 19b | ||
s = 7a + 11b | s = 7a + 11b | ||
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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']. | 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']. | ||
Gedras greatly facilitate | Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). 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. Because the enharmonic E is multiplied by 2, it will occur twice in the perchain, even thought the comma only occurs once. 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*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: | ||
<span style="display: block; text-align: center;">P1 - ^^M2=v<span style="vertical-align: super;">3</span>m3 - v4 - ^5 - ^<span style="vertical-align: super;">3</span>M6=vvm7 - P8</span><span style="display: block; text-align: center;">C - D^^=Ebv<span style="vertical-align: super;">3</span> - Fv - G^ - A^<span style="vertical-align: super;">3</span>=Bbvv - C</span> | <span style="display: block; text-align: center;">P1 -- ^^M2=v<span style="vertical-align: super;">3</span>m3 -- v4 -- ^5 -- ^<span style="vertical-align: super;">3</span>M6=vvm7 -- P8</span><span style="display: block; text-align: center;">C -- D^^=Ebv<span style="vertical-align: super;">3</span> -- Fv -- G^ -- A^<span style="vertical-align: super;">3</span>=Bbvv -- C</span> | ||
All single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [round(5/5), round(3/5)] = [1,1] = m2. The bare enharmonic is P4 - 5*m2 = [5,3] - 5*[1,1] = [5,3] - [5,5] = [0,-2] = -2*[0,1] = two descending d2's. The d2 must be upped, and E = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is: | All single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [round(5/5), round(3/5)] = [1,1] = m2. The bare enharmonic is P4 - 5*m2 = [5,3] - 5*[1,1] = [5,3] - [5,5] = [0,-2] = -2*[0,1] = two descending d2's. The d2 must be upped, and E = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is: | ||
<span style="display: block; text-align: center;">P1 - ^^m2=v<span style="vertical-align: super;">3</span>A1 - vM2 - ^m3 - ^<span style="vertical-align: super;">3</span>d4=vvM3 - P4</span><span style="display: block; text-align: center;">C - Db^^=C#v<span style="vertical-align: super;">3</span> - Dv - Eb^ - Fb^<span style="vertical-align: super;">3</span>=Evv - F</span> | <span style="display: block; text-align: center;">P1 -- ^^m2=v<span style="vertical-align: super;">3</span>A1 -- vM2 -- ^m3 -- ^<span style="vertical-align: super;">3</span>d4=vvM3 -- P4</span><span style="display: block; text-align: center;">C -- Db^^=C#v<span style="vertical-align: super;">3</span> -- Dv -- Eb^ -- Fb^<span style="vertical-align: super;">3</span>=Evv -- F</span> | ||
(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 = v<span style="vertical-align: super;">4</span>dd3, 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. | (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 = v<span style="vertical-align: super;">4</span>dd3, 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. | ||
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For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it 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 = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator 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 = ^<span style="vertical-align: super;">4</span>M2. Next, deduce the alternate generator 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 either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, thus G' = ^<span style="vertical-align: super;">5</span>M2, which happens to be P + G. Thus P = ^4M2, G = ^1, G' = ^5M2, and E = v10m2. | For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it 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 = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator 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 = ^<span style="vertical-align: super;">4</span>M2. Next, deduce the alternate generator 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 either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, thus G' = ^<span style="vertical-align: super;">5</span>M2, which happens to be P + G. Thus P = ^4M2, G = ^1, G' = ^5M2, and E = v10m2. | ||
<span style="display: block; text-align: center;">P1 - ^<span style="vertical-align: super;">4</span>M2=v<span style="vertical-align: super;">6</span>m3 - vvP4 - ^^P5 - ^<span style="vertical-align: super;">6</span>M6=v<span style="vertical-align: super;">4</span>m7 - P8</span><span style="display: block; text-align: center;">C - D^<span style="vertical-align: super;">4</span>=Ebv<span style="vertical-align: super;">6</span> - Fvv - G^^ - A^<span style="vertical-align: super;">6</span>=Bbv<span style="vertical-align: super;">4</span> - C</span><span style="display: block; text-align: center;">P1 - ^<span style="vertical-align: super;">5</span>M2=v<span style="vertical-align: super;">5</span>m3 - P4</span><span style="display: block; text-align: center;">C - D^<span style="vertical-align: super;">5</span>=Ebv<span style="vertical-align: super;">5</span> - F</span> | |||
<span style="display: block; text-align: center;">P1 - ^<span style="vertical-align: super;">4</span>M2=v<span style="vertical-align: super;">6</span>m3 - vvP4 - ^^P5 - ^<span style="vertical-align: super;">6</span>M6=v<span style="vertical-align: super;">4</span>m7 - P8</span><span style="display: block; text-align: center;">C - D^<span style="vertical-align: super;">4</span>=Ebv<span style="vertical-align: super;">6</span> - Fvv - G^^ - A^<span style="vertical-align: super;">6</span>=Bbv<span style="vertical-align: super;">4</span> - C</span> | |||
<span style="display: block; text-align: center;">P1 - ^<span style="vertical-align: super;">5</span>M2=v<span style="vertical-align: super;">5</span>m3 - P4</span><span style="display: block; text-align: center;">C - D^<span style="vertical-align: super;">5</span>=Ebv<span style="vertical-align: super;">5</span> - F</span> | |||
To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2. | To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2. | ||
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Unreducing replaces the generator with an alternate generator. Any number of periods plus or minus a single generator makes an <strong>alternate</strong> generator. A generator or period plus or minus any number of enharmonics makes an <strong>equivalent</strong> 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.<br /> | Unreducing replaces the generator with an alternate generator. Any number of periods plus or minus a single generator makes an <strong>alternate</strong> generator. A generator or period plus or minus any number of enharmonics makes an <strong>equivalent</strong> 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.<br /> | ||
<br /> | <br /> | ||
For example, (P8/2, P5) has generator P5, alternate | For example, half-octave (P8/2, P5) has generator P5, alternate generators P4 and vA1, 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 ^<span style="vertical-align: super;">6</span>dd2.<br /> | ||
<br /> | <br /> | ||
There are also equivalent enharmonics, see below.<br /> | There are also equivalent enharmonics, see below.<br /> | ||
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The <strong>keyspan</strong> 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 <strong>stepspan</strong> 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.<br /> | The <strong>keyspan</strong> 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 <strong>stepspan</strong> 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.<br /> | ||
<br /> | <br /> | ||
Every 3-limit interval can be uniquely expressed as the combination of a keyspan and a stepspan. This combination is called a <strong>gedra</strong>, analogous to a monzo, but written in brackets not parentheses: 3/2 is a 7-semitone 5th, thus | Every 3-limit interval can be uniquely expressed as the combination of a keyspan and a stepspan. This combination is called a <strong>gedra</strong>, analogous to a monzo, but written in brackets not parentheses: 3/2 = (-1,1) is a 7-semitone 5th, thus (-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:<br /> | ||
k = 12a + 19b<br /> | k = 12a + 19b<br /> | ||
s = 7a + 11b<br /> | s = 7a + 11b<br /> | ||
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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'].<br /> | 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'].<br /> | ||
<br /> | <br /> | ||
Gedras greatly facilitate | Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). 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 <u>always</u> 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. Because the enharmonic E is multiplied by 2, it will occur twice in the perchain, even thought the comma only occurs once. 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*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:<br /> | ||
<span style="display: block; text-align: center;">P1 - ^^M2=v<span style="vertical-align: super;">3</span>m3 - v4 - ^5 - ^<span style="vertical-align: super;">3</span>M6=vvm7 - P8</span><span style="display: block; text-align: center;">C - D^^=Ebv<span style="vertical-align: super;">3</span> - Fv - G^ - A^<span style="vertical-align: super;">3</span>=Bbvv - C</span><br /> | <span style="display: block; text-align: center;">P1 -- ^^M2=v<span style="vertical-align: super;">3</span>m3 -- v4 -- ^5 -- ^<span style="vertical-align: super;">3</span>M6=vvm7 -- P8</span><span style="display: block; text-align: center;">C -- D^^=Ebv<span style="vertical-align: super;">3</span> -- Fv -- G^ -- A^<span style="vertical-align: super;">3</span>=Bbvv -- C</span><br /> | ||
All single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [round(5/5), round(3/5)] = [1,1] = m2. The bare enharmonic is P4 - 5*m2 = [5,3] - 5*[1,1] = [5,3] - [5,5] = [0,-2] = -2*[0,1] = two descending d2's. The d2 must be upped, and E = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is:<br /> | All single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [round(5/5), round(3/5)] = [1,1] = m2. The bare enharmonic is P4 - 5*m2 = [5,3] - 5*[1,1] = [5,3] - [5,5] = [0,-2] = -2*[0,1] = two descending d2's. The d2 must be upped, and E = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is:<br /> | ||
<br /> | <br /> | ||
<span style="display: block; text-align: center;">P1 - ^^m2=v<span style="vertical-align: super;">3</span>A1 - vM2 - ^m3 - ^<span style="vertical-align: super;">3</span>d4=vvM3 - P4</span><span style="display: block; text-align: center;">C - Db^^=C#v<span style="vertical-align: super;">3</span> - Dv - Eb^ - Fb^<span style="vertical-align: super;">3</span>=Evv - F</span><br /> | <span style="display: block; text-align: center;">P1 -- ^^m2=v<span style="vertical-align: super;">3</span>A1 -- vM2 -- ^m3 -- ^<span style="vertical-align: super;">3</span>d4=vvM3 -- P4</span><span style="display: block; text-align: center;">C -- Db^^=C#v<span style="vertical-align: super;">3</span> -- Dv -- Eb^ -- Fb^<span style="vertical-align: super;">3</span>=Evv -- F</span><br /> | ||
(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 = v<span style="vertical-align: super;">4</span>dd3, 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.<br /> | (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 = v<span style="vertical-align: super;">4</span>dd3, 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.<br /> | ||
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For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it 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 = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator 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 = ^<span style="vertical-align: super;">4</span>M2. Next, deduce the alternate generator 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 either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, thus G' = ^<span style="vertical-align: super;">5</span>M2, which happens to be P + G. Thus P = ^4M2, G = ^1, G' = ^5M2, and E = v10m2.<br /> | For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it 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 = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator 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 = ^<span style="vertical-align: super;">4</span>M2. Next, deduce the alternate generator 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 either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, thus G' = ^<span style="vertical-align: super;">5</span>M2, which happens to be P + G. Thus P = ^4M2, G = ^1, G' = ^5M2, and E = v10m2.<br /> | ||
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<span style="display: block; text-align: center;">P1 - ^<span style="vertical-align: super;">4</span>M2=v<span style="vertical-align: super;">6</span>m3 - vvP4 - ^^P5 - ^<span style="vertical-align: super;">6</span>M6=v<span style="vertical-align: super;">4</span>m7 - P8</span><span style="display: block; text-align: center;">C - D^<span style="vertical-align: super;">4</span>=Ebv<span style="vertical-align: super;">6</span> - Fvv - G^^ - A^<span style="vertical-align: super;">6</span>=Bbv<span style="vertical-align: super;">4</span> - C</span><span style="display: block; text-align: center;">P1 - ^<span style="vertical-align: super;">5</span>M2=v<span style="vertical-align: super;">5</span>m3 - P4</span><span style="display: block; text-align: center;">C - D^<span style="vertical-align: super;">5</span>=Ebv<span style="vertical-align: super;">5</span> - F</span><br /> | |||
<span style="display: block; text-align: center;">P1 - ^<span style="vertical-align: super;">4</span>M2=v<span style="vertical-align: super;">6</span>m3 - vvP4 - ^^P5 - ^<span style="vertical-align: super;">6</span>M6=v<span style="vertical-align: super;">4</span>m7 - P8</span><span style="display: block; text-align: center;">C - D^<span style="vertical-align: super;">4</span>=Ebv<span style="vertical-align: super;">6</span> - Fvv - G^^ - A^<span style="vertical-align: super;">6</span>=Bbv<span style="vertical-align: super;">4</span> - C</span> | |||
<span style="display: block; text-align: center;">P1 - ^<span style="vertical-align: super;">5</span>M2=v<span style="vertical-align: super;">5</span>m3 - P4</span><span style="display: block; text-align: center;">C - D^<span style="vertical-align: super;">5</span>=Ebv<span style="vertical-align: super;">5</span> - F</span> | |||
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To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.<br /> | To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.<br /> | ||
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