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Neutral and interordinal intervals in MOS scales: Difference between revisions

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Given a tuning of a primitive (single-period) [[mos]] pattern aLbs with a > b, we may define two types of notes "in the cracks of" interval categories defined by aLbs:
Given a tuning of a primitive (single-period) [[mos]] pattern aLbs with a > b, we may define two types of notes "in the cracks of" interval categories defined by aLbs:
# Given 1 <= ''k'' <= a + b - 1, the '''neutral''' ''k''-step (abbrev. n''k''s) is the interval exactly halfway between the smaller ''k''-step and the larger ''k''-step of the mos. When the mos is generated by a (perfect) ''k''-step, this may instead be called the '''semiperfect''' ''k''-step (abbrev. sP''k''s), since it is halfway between the perfect and imperfect (either diminished or augmented, depending on whether the generator is bright or dark) ''k''-step.
# Given 1 ≤ ''k'' ≤ a + b - 1, the '''neutral''' ''k''-step (abbrev. n''k''s) is the interval exactly halfway between the smaller ''k''-step and the larger ''k''-step of the mos. When the mos is generated by a (perfect) ''k''-step, this may instead be called the '''semiperfect''' ''k''-step (abbrev. sP''k''s), since it is halfway between the perfect and imperfect (either diminished or augmented, depending on whether the generator is bright or dark) ''k''-step.
# Given 1 <= ''k'' <= a + b &minus; 2, and assuming that the larger ''k''-step < the smaller (''k'' + 1)-step, the '''interordinal''' between ''k''-steps and (''k'' + 1)-steps, denoted ''k''x(''k'' + 1)s or ''k''X(''k'' + 1)s (read "''k'' cross (''k'' + 1) step"), is the interval exactly halfway between the larger ''k''-step and the smaller (''k'' + 1)-step. The name comes from the fact that ''k''-steps in the diatonic mos are conventionally called "(''k'' + 1)ths".
# Given 1 ≤ ''k'' ≤ a + b &minus; 2, and assuming that the larger ''k''-step < the smaller (''k'' + 1)-step, the '''interordinal''' between ''k''-steps and (''k'' + 1)-steps, denoted ''k''x(''k'' + 1)s or ''k''X(''k'' + 1)s (read "''k'' cross (''k'' + 1) step"), is the interval exactly halfway between the larger ''k''-step and the smaller (''k'' + 1)-step. The name comes from the fact that ''k''-steps in the diatonic mos are conventionally called "(''k'' + 1)ths".


Given such a mos, it's easy to observe the following properties of the simplest equal tunings for the mos, due to the way they divide the small step (s) and the chroma (L &minus; s):
Given such a mos, it's easy to observe the following properties of the simplest equal tunings for the mos, due to the way they divide the small step (s) and the chroma (L &minus; s):
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To prove the theorem, we need a couple lemmas.
To prove the theorem, we need a couple lemmas.
=== Lemma 1 ===
=== Lemma 1 ===
Let n and k be integers, and let x be a real number such that kx is not an integer. Then floor((n+k)x) - floor(nx) >= floor(kx).
Let n and k be integers, and let x be a real number such that kx is not an integer. Then floor((n+k)x) - floor(nx) ≥ floor(kx).
==== Proof ====
==== Proof ====
floor((n+k)x) - floor(nx) = -1 + ceil((n+k)x) + ceil(-nx) >= ceil((n+k)x - nx) - 1 = ceil(kx) - 1 = floor(kx).
floor((n+k)x) - floor(nx) = -1 + ceil((n+k)x) + ceil(-nx) ≥ ceil((n+k)x - nx) - 1 = ceil(kx) - 1 = floor(kx).


=== Discretizing Lemma ===
=== Discretizing Lemma ===
Consider an m-note [[maximal evenness|maximally even]] mos of an n-equal division, and let 1 <= k <= m-1. Then a k-step of that mos is either floor(nk/m)\n or ceil(nk/m)\n.
Consider an m-note [[maximal evenness|maximally even]] mos of an n-equal division, and let 1 ≤ k ≤ m-1. Then a k-step of that mos is either floor(nk/m)\n or ceil(nk/m)\n.
==== Proof ====
==== Proof ====
The circular word formed by stacking k-steps of the mos is itself a maximally even mos, considered as a subset of a kn-note equal tuning. One step of an m'-note maximally even mos of an n'-note equal tuning is either floor(n'/m')\n'<equave> or ceil(n'/m')\n'<equave>, implying the lemma.
The circular word formed by stacking k-steps of the mos is itself a maximally even mos, considered as a subset of a kn-note equal tuning. One step of an m'-note maximally even mos of an n'-note equal tuning is either floor(n'/m')\n'<equave> or ceil(n'/m')\n'<equave>, implying the lemma.
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Part (1) takes some step size arithmetic:
Part (1) takes some step size arithmetic:
* Larger k+1-step of aLbs minus larger k-step of aLbs = Smaller k+1-step of aLbs minus smaller k-step of aLbs must be L, if 0 < k < k+1 < a+b. The reason is that larger 1-step = L, larger 2-step = LL, because there are more L’s than s’s.
* Larger k+1-step of aLbs minus larger k-step of aLbs = Smaller k+1-step of aLbs minus smaller k-step of aLbs must be L, if 0 < k < k+1 < a+b. The reason is that larger 1-step = L, larger 2-step = LL, because there are more L’s than s’s.
* Smaller k+1-step of aLbs minus larger k-step of aLbs >= 0, with =0 at improprieties. At the values of k and k+1 that are proper, this equals s.  
* Smaller k+1-step of aLbs minus larger k-step of aLbs ≥ 0, with =0 at improprieties. At the values of k and k+1 that are proper, this equals s.  
** To see why, suppose the difference is L (here k >= i >= 1, 0 < k < k+1 < a+b):
** To see why, suppose the difference is L (here k ≥ i ≥ 1, 0 < k < k+1 < a+b):
*** X = Larger (k+1)-step = (i+2)L + (k-i-1)s
*** X = Larger (k+1)-step = (i+2)L + (k-i-1)s
*** Smaller (k+1)-step = (i+1)L + (k-i)s
*** Smaller (k+1)-step = (i+1)L + (k-i)s
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*** <pre>X=L^CsL...LsL...LsL^D</pre>
*** <pre>X=L^CsL...LsL...LsL^D</pre>
** Let r be the number of complete chunks (maximal contiguous strings of L's) in X, and let μ = n/(a+b). Note that the spacings between the s steps of a mos form a maximally even mos, and that since a > b, 3/2 < μ < 2. By the Discretizing Lemma, we have (|·| denotes length):
** Let r be the number of complete chunks (maximal contiguous strings of L's) in X, and let μ = n/(a+b). Note that the spacings between the s steps of a mos form a maximally even mos, and that since a > b, 3/2 < μ < 2. By the Discretizing Lemma, we have (|·| denotes length):
*** 1+A+B+floor((r+2)μ) <= |Y| <= 1+A+B+ceil((r+2)μ)
*** 1+A+B+floor((r+2)μ) ≤ |Y| ≤ 1+A+B+ceil((r+2)μ)
*** 1+C+D+floor(rμ) <= |X| <= 1+C+D+ceil(rμ)
*** 1+C+D+floor(rμ) ≤ |X| ≤ 1+C+D+ceil(rμ)
*** -1 = |Y|-|X| >= (A+B)-(C+D) + floor((r+2)μ) - ceil(rμ)
*** -1 = |Y|-|X| ≥ (A+B)-(C+D) + floor((r+2)μ) - ceil(rμ)
*** = (A+B)-(C+D)-1 + floor((r+2)μ) - floor(rμ)
*** = (A+B)-(C+D)-1 + floor((r+2)μ) - floor(rμ)
*** (Lemma 1) >= (A+B)-(C+D)-1 + floor(2μ)
*** (Lemma 1) ≥ (A+B)-(C+D)-1 + floor(2μ)
*** Hence, (C+D)-(A+B) >= floor(2μ).
*** Hence, (C+D)-(A+B) ≥ floor(2μ).
*** Also, (C+D)-(A+B) <= 2*ceil(μ)-2 = 2(ceil(μ)-1) = 2*floor(μ). This is a contradiction: as 3/2 < μ < 2, we have floor(2μ) - 2*floor(μ) = 1.  
*** Also, (C+D)-(A+B) ≤ 2*ceil(μ)-2 = 2(ceil(μ)-1) = 2*floor(μ). This is a contradiction: as 3/2 < μ < 2, we have floor(2μ) - 2*floor(μ) = 1.  
* As s is the chroma of bL(a-b)s, it *would* be the difference between major and minor intervals in the parent mos, assuming these interval sizes (smaller k+1-step, larger k-step) occur in the parent; so kX(k+1) would become neutral or semiperfect.
* As s is the chroma of bL(a-b)s, it *would* be the difference between major and minor intervals in the parent mos, assuming these interval sizes (smaller k+1-step, larger k-step) occur in the parent; so kX(k+1) would become neutral or semiperfect.
* To show that these actually occur in bL(a-b)s, consider smaller and larger j-steps ( 1 <= j <= a-1) in the parent mos. These intervals also occur in the mos aLbs separated by s, and the number of j’s (“junctures”) that correspond to these places in aLbs is exactly a-1. Note that we are considering “junctures” between k-steps and k+1-steps in aLbs, excluding k = 0 and k = a+b-1, so the total number of “junctures” to consider is a+b-2. This proves part (1).
* To show that these actually occur in bL(a-b)s, consider smaller and larger j-steps ( 1 ≤ j ≤ a-1) in the parent mos. These intervals also occur in the mos aLbs separated by s, and the number of j’s (“junctures”) that correspond to these places in aLbs is exactly a-1. Note that we are considering “junctures” between k-steps and k+1-steps in aLbs, excluding k = 0 and k = a+b-1, so the total number of “junctures” to consider is a+b-2. This proves part (1).


Part (2) is easier to see: where basic aLbs is improper, larger k-step = smaller k+1-step, and larger k+1-step - smaller k-step = 2(L-s) = 2s = L. But the step L is not two steps in 2n-edo.
Part (2) is easier to see: where basic aLbs is improper, larger k-step = smaller k+1-step, and larger k+1-step - smaller k-step = 2(L-s) = 2s = L. But the step L is not two steps in 2n-edo.
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