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. 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)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 notice the following properties of the simplest equal tunings for the mos, due to the way they divide the small step and the chroma, respectively:

• The hard equal tuning (3a + b)-ed contains neutrals of that mos but not interordinals.
• The soft 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

The Interordinal Theorem

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.

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. 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 or semiperfect interval in basic aLbs is not always an interordinal interval in the parent mos.
2. (b - 1) counts the places in 2(2a+b)edo (twice the basic mos tuning for aLbs) where assigning interordinals to the parent mos of basic aLbs fails.

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 (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 (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.
• 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).