Neutral and interordinal intervals in MOS scales

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:

1. 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 is instead called the semiperfect k-step (abbrev. sPkms or 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.
2. 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) or kX(k + 1), 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".

Examples

The Interordinal Theorem

Recall that the “impropriety number” (b - 1) of a primitive mos aLbs 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.

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.

Statement of the theorem

Suppose a > b and gcd(a, b) = 1.

1. The impropriety number (b - 1) of the mos aLbs counts the places in 2(2a+b)edo (twice the basic mos tuning for aLbs) where assigning interordinals to the parent mos of aLbs fails.
2. Every interordinal in basic aLbs (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, but a neutral interval in basic aLbs is not always an interordinal interval in the parent mos.

Proof

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) 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) 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 themselves form a maximally even mos, and that since a > b, 3/2 < μ < 2. We'll obtain a contradiction.
  • 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.
• 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 (2).