Neutral and interordinal intervals in MOS scales
Given a tuning of a primitive (single-period) mos pattern aLbs<E> with a > b and arbitrary equave E, we may define two types of notes "in the cracks of" interval categories defined by aLbs<E>:
- Given 1 ≤ k ≤ a + b - 1, the neutral k-step (abbrev. nks) 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. sPks), 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 − 2, and assuming that the larger k-step < the smaller (k + 1)-step, the interordinal between k-steps and (k + 1)-steps, denoted kx(k + 1)s or kX(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 − s):
- The basic equal tuning (2a + b)-ed contains neither neutrals nor interordinals.
- The monohard equal tuning (3a + b)-ed contains neutrals of that mos but not interordinals.
- The monosoft equal tuning (3a + 2b)-ed contains interordinals but not neutrals.
- 2(2a + b)-ed, twice the basic equal tuning, contains both types of intervals.
Examples
Diatonic (5L2s)
Basic 5L2s (diatonic, dia-): 12edo Parent mos: soft 2L5s (pentic, pt-) 1\24 2\24 m1dias 3\24 n1dias 4\24 M1dias == m1pts 5\24 1X2dias == n1pts 6\24 m2dias == M1pts 7\24 n2dias == 1x2pts 8\24 M2dias == d2pts 9\24 2X3dias == sP2pts 10\24 P3dias == P2pts 11\24 sP3dias 12\24 A3/d4dias == 2x3pts 13\24 sP4dias 14\24 P4dias == P3pts 15\24 4X5dias == sP3pts 16\24 m5dias == A3pts 17\24 n5dias == 3x4pts 18\24 M5dias == m4pts 19\24 5X6dias == n4pts 20\24 m6dias == M4pts 21\24 n6dias 22\24 M6dias 23\24
m-chromatic (7L5s)
Basic 7L5s (m-chromatic, mchr-): 19edo Parent mos 5L2s (diatonic, dia-) 1\38 2\38 m1s 3\38 n1s 4\38 M1s == m1dias 5\38 1x2s == n1dias 6\38 m2s == M1dias 7\38 n2s 8\38 M2s/m3s == 1x2dias 9\38 n3s 10\38 M3s == m2dias 11\38 3x4s == n2dias 12\38 m4s == M2dias 13\38 n4s 14\38 M4s/d5s == 2x3dias 15\38 sPs 16\38 P5s == P3dias 17\38 5x6s == sP3dias 18\38 m6s == A3dias 19\38 n6s == 3x4dias 20\38 M6s == d4dias 21\38 6x7s == sP4dias 22\38 P7s == P4dias 23\38 sP7s 24\38 A7s/m8s == 4x5dias 25\38 n8s 26\38 M8s == m5dias 27\38 8x9s == n5dias 28\38 m9s == M5dias 29\38 n9s 30\38 M9s/m10s == 5x6dias 31\38 n10s 32\38 M10s == m6dias 33\38 10x11s == n6dias 34\38 m11s == M6dias 35\38 n11s 36\38 M11s 37\38
Manual (4L1s)
Basic 2/1-equivalent 4L1s (manual, man-): 9edo Parent mos: soft 1L3s (antetric, att-) 1\18 2\18 d1mans 3\18 sP1mans 4\18 P1mans == P1atts 5\18 1x2mans == sP1atts 6\18 m2mans == A1atts 7\18 n2mans == 1x2atts 8\18 M2mans == m2atts 9\18 2x3mans == n2atts 10\18 m3mans == M2atts 11\18 n3mans == 2x3atts 12\18 M3mans == d3atts 13\18 3x4mans == sP3atts 14\18 P4mans == P3atts 15\18 sP4mans 16\18 A4mans 17\18
The Interordinal Theorem
The Interordinal Theorem relates the neutral (resp. interordinal) intervals of aLbs (with a > b) with the interordinal (resp. neutral) intervals of its parent mos, bL(a − b)s.
Statement
Suppose a > b and gcd(a, b) = 1.
- Every interordinal in basic aLbs<E> (an interval that is exactly halfway between the larger k-step and the smaller (k+1)-step) is a neutral or semiperfect interval in the parent mos bL(a-b)s<E>.
- Every interordinal interval in the parent mos bL(a-b)s<E> of basic aLbs<E> is a neutral or semiperfect interval in basic aLbs<E>. The number (b - 1) counts the places in 2(2a+b)edE (twice the basic mos tuning for aLbs<E>) where the parent's interordinal is two steps away from the ordinal categories.
Preliminaries for the proof
Below we assume that the equave is 2/1, but the proof generalizes to any equave.
Consider a primitive mos aLbs. Recall that (b - 1) satisfies:
(b - 1) = |(brightest mode of basic aLbs, ignoring equaves) ∩ (darkest mode of basic aLbs, ignoring equaves)| = #{k : 0 < k < a+b and larger k-step of basic aLbs = smaller (k+1)-step of basic aLbs}.
Also recall that the following are equivalent for a mos aLbs:
- a > b.
- The parent mos, which is bL(a-b)s, has steps L + s and L. In this context, since aLbs is assumed to have hardness 2/1, bL(a-b)s has hardness 3/2 thus is strictly proper.
To prove the theorem, we need a couple lemmas.
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).
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).
Discretizing Lemma
Consider an m-note 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
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.
Proof of Theorem
Let n = 2a + b (the basic edo tuning of aLbs) and suppose that m\(2n) is an interordinal (where m must be odd) between k-steps and (k+1)-steps, denoted kX(k+1)ms. Recall that:
- In basic aLbs, s = 1\n = 2\2n.
- A concrete mos tuning is improper iff its hardness is > 2/1 and the number of s steps it has is > 1.
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.
- 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):
- X = Larger (k+1)-step = (i+2)L + (k-i-1)s
- Smaller (k+1)-step = (i+1)L + (k-i)s
- Larger k-step = iL + (k-i)s
- Y = Smaller k-step = (i-1)L + (k-i+1)s
- Since the smaller k-step has two more s steps than the larger (k+1)-step, Y has two more complete chunks of L's than X:
Y=L^A sL...LsL...LsL...LsL...LsL^B
X=L^CsL...LsL...LsL^D
- 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+C+D+floor(rμ) ≤ |X| ≤ 1+C+D+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)
- (Lemma 1) ≥ (A+B)-(C+D)-1 + 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.
- To see why, suppose the difference is L (here k ≥ i ≥ 1, 0 < k < k+1 < a+b):
- 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).
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.