Musical Idea 34: Unshadowing

05.31.2014 § 3 Comments

Continued from previous shadowing topic: Funky Shadowing

Compared to shadowing (see introductory idea here), unshadowing is rather complicated. This is because while every rhythm has a shadow, not every rhythm is another’s shadow. Said another way: it is always possible to iterate on evening your rhythm out a little further, but as you go the other direction — iterating on bringing out its unevenness — you will inevitably reach a point where some strikes are too close together and some intervals are too wide for one to find a rhythm for which your rhythm could be the midpoints of its intervals.

It’s also much easier to shadow a rhythm than to unshadow one. To shadow a rhythm, remember, all we have to do is find its intervals’ midpoints. But to unshadow a rhythm, I have needed to come up with formulas. The formula needed varies depending on whether the number of intervals in your rhythm is even or odd. I’m sure someone else out there has figured out an inverse midpoint formula for cyclically ordered sets and is using it for rocket science, but I couldn’t find anything, so here’s my best.

First, let’s get out of the way some trivial instances. To unshadow a rhythm that is one interval repeated over and over, it is simply itself offset by half, over and over. You can unshadow this vacuously forever. And to unshadow a rhythm that is two alternating intervals repeated over and over, it will only work if these intervals are the same, in which case this is identical to the situation where it is one interval repeated over and over. To understand why, recognize that you must center intervals in your unshadow on the strikes in your rhythm; if you imagine widening these thusly centered intervals until they meet, hoping they meet each other on both ends sameltimeously, you will find unfortunately that they meet first inside the smaller interval of the shadow and have nowhere left to go, leaving a gap centered on the bigger interval of the the shadow.

It’s when unshadowing a rhythm that is three intervals repeated in the same order over and over that formulas start to help. Let’s name the three intervals in the rhythm i1, i2, and i3, and the three intervals in the unshadowed rhythm u1, u2, and u3:


If any of u1, u2, or u3 are not positive, then your rhythm is not unshadowable. It cannot be another’s shadow. I suggest we refer to such a rhythm as a cast.

I have oriented these formulas so that u1 is centered on the strike dividing the intervals i1 and i2. Since it must be offset one way or the other, I mean, I figured that it was simpler to offset it forward rather than backward.


Unshadowing. Each interval in your original rhythm is comprised of one half of from each overlapping interval in its unshadow.

To understand why this formula works, return to the drawing example earlier in this chapter. Looking upward from each strike in the shadow rhythm we see two diverging diagonal lines and one vertical dotted line. The dotted line goes up to a midpoint of an interval in the original rhythm, and the diverging diagonal lines connect with the strikes on either side of that interval. The dotted line divides the interval in the original rhythm exactly in half, so we can refer to these as half intervals. Finally, realize that each interval in the shadow rhythm is the sum of two of these halves, but different halves, halves from neighboring intervals in the original rhythm. Thus we can say:

2It’s a basic system of equations. If you solve for one of the h’s in terms of no other h’s, but all i’s (whose values you know) then you can simply double that to get one of the u’s.

Let’s check it out on our original 2,8,4 rhythm. What rhythm, if any, is it the shadow of? -2, 6, 10. You can see that -2, 6, and 10 sum to 14, the same length as the shadow rhythm, as we’d expect — however, since one of the intervals is negative, sadly, it is not possible to unshadow it — it’s a cast. Just looking at it, it should be clear that the two strikes that are only 2 apart are the problem: the biggest area that can cover in an unshadowed rhythm is 4, and that’s counting the 2 it’s centered over; even if that was skewed entirely into the 8’s territory — the bigger one, more in need of covering — then the other strike sweeping in to cover the remaining 6 of the 8’s territory would be forced to sweep equally the other direction, 6 into 4’s territory, overextending by 2 (hence the negative 2). Though I suppose if you wanted to, you could figure out some way to make something musically interesting out of such negative situations.

For some, it may be more helpful to check the intervals in ones rhythm against these criteria than to plug into formulas and hope for all positives:

3You can see that the formulas are all rotations of each other around the cyclical (since it repeats itself, as far as we care, forever) rhythm. In other words, no one interval can be greater or equal than the other two. Since 8 is clearly greater than 2 + 4, we could have known it wasn’t going to work out just by that.

For now I’m going to skip over unshadowing rhythms with four intervals, since as I mentioned earlier, these behave a little differently. To unshadow a rhythm that is five intervals, the formulas look similar to the formulas for rhythms with three — the pattern for odd numbered rhythms begins to emerge.

4Allow me to begin unpacking the pattern. If you look at the formulas for the three-rhythm unshadowing, you’ll see that each u is the sum of two positive i’s and one negative i, thus, equivalent to net one positive i. This makes sense since the resulting sum of u’s must be the same as the sum of i’s. So now if you look at the formulas for the five-rhythm, you’ll see that each u is the sum of three positive i’s and two negative i’s, thus, still equivalent to net one positive i.

But which alternating just-shy-of-half of the i’s is to be negative? Looking back at the three-rhythm, it was always the i numbered one less than the u. That is, for u2 it is i(2 – 1) = i1. For the five rhythm, it’s the same, but then also the one two less than that. For the seven and nine interval rhythms, you can continue making every second i backwards negative and the formula will work out.

Let’s test this out on a familiar rhythm, the clave son, with intervals 3, 3, 4, 2, 4, total length 16. Plugging in above, we find 0, 6, 2, 2, 6. Since 0 is not positive, unfortunately, the clave son is not unshadowable. 0 isn’t negative, either, though, which suggests that it’s on the cusp. Indeed, drawing it out, we find a similar situation as we did with the 2, 8, 4 rhythm: the interval of 2 can cover a maximum of 4 area. The clave son is symmetrical about this interval so we may as well center it. To cover the remaining 3 area on each 4, we have to use up exactly all of the space on each of the 3’s, meaning that the fifth interval is length zero, centered on the strike between the two threes. In other words, if it worked for your music, you could shadow the 4-rhythm 6, 2, 2, 6 into 3, 3, 4, 2, 4 if you just made sure that one of the strikes stayed put.

The equivalent inequality check for the five-rhythm resembles the one for the three-rhythm, though there are enough intervals in play at this point that the shorthand explanation does not carry. One need not check every sum of two intervals and confirm that they are not greater than the sum of the other three; one need only check these five permutations:

5In other words, each run of three every other i’s must be greater than the other two. Since 3 + 2 + 3 = 4 + 4, we know it can’t quite work (by the way, I’ve been finding these inequalities simply by setting the u to 0 and making sure that its formula is greater than that).

In case you get stuck, here’s the first formula of the set for a 7-rhythm:

6To unshadow an even rhythm, things seem to get simpler in some ways and more complicated in others. That we cannot use the same pattern of formulas we use for the odd numbered rhythms should be clear, because we cannot create a net one positive i by making any number of four i’s negative. When we start from scratch with our system of equations for a four-rhythm, initially we feel thwarted, since no matter what we do, all of h’s end up getting canceled out, so we’re unable to proceed from i to u. However, if you do it right, the trick is revealed:

7When you think about it, it kind of makes sense compared with the inequalities that function as checks for the odd-rhythms: it’s almost like the odds’ inequalities are their best attempt at cyclically achieving such an equality, distributing it across each near-even alternation.

The next thing I tried after figuring this out was not attempting to come up with equations. What I did first was pick a really weird rhythm that satisfied this equation and just looked at it to see if I could manually unshadow it. I tried 23, 11, 7, 19 (23 + 7 = 30 = 11 + 19). With h1 = 8, h2 = 3, h3 = 4, and h4 = 15, it worked out (23 = 15 + 8; 11 = 8 + 3; 7 = 3 + 4; 19 = 4 + 15). That leads to an unshadowed rhythm of 16, 6, 8, 30.

It was so easy to figure this out, that I couldn’t quite believe I had just chanced upon the solution taking my first crack at it. So I tried it another way. With h1 = 10, h2 = 1, h3 = 6, and h4 = 13, it worked out too (23 = 13 + 10; 11 = 10 + 1, 7 = 1 + 6; 19 = 6 + 11). And with h1 = 9, h2 = 2, h3 = 5, and h4 = 14. I realized that as long as you had set i1 + i3 = i2 + i4, anything went. Sweet!

However, now I had another problem. How to pick the “best” unshadow? Well, I suppose in the spirit of the challenge of continuing unshadowing until a cast is reached, I should pick the unshadow which is itself unshadowable, that is, which itself satisfies i1 + i3 = i2 + i4; in other words, u1 + u3 = u2 + u4 (and, naturally, h1 + h3 = h2 + h4).

So after throwing these two equations into the mix, we have enough information to come up with these formulas for the u’s of a 4-rhythm:

8Some notes on this set of formulas:

  • You may notice that unlike in the odd rhythm formulas, where every i is needed, here there is one less of them needed; in this case, for each u, the partner of its numerically corresponding i is omitted.
    • Let’s call adjacent intervals neighbors, and opposite intervals partners — partners like in a four-person card game  and your teammate sits across from you, and since they’re together on the same side of that critical equality.
    • This is because of that alternation equality again, i1 + i3 = i2 + i4 : if we ever know three intervals, then the fourth one is determined; and thus one of the intervals gets to be left out of each equation.
  • Another notable thing about these even rhythms is that there are countless different ways besides these to phrase these equations; the challenge was finding the most concise and instructive set of them.
    • Part of the issue seems to be that they work directionally.
      • By that I mean that while your intuition might say that the rhythm is symmetrical and there’s no reason I couldn’t switch between the two neighboring intervals, the trick is that you have to be consistent with which side of neighbor you pick for it to work.
    • The other part of the issue is that it seems like you can pointlessly scale the coefficients on the terms in the numerator.
      • I suppose you could do this for the odd rhythms, too, but for some reason when I was trying to derive these formulas it kept escalating, coming out like this until I figured out how to pare it back down.
      • In other words, you can add an alternating +n -n +n -n to the coefficients.
        • Including the zero.
        • Aligned so that negatives continue to alternate.
        • And so that a descending/ascending pattern is found in the absolute values of the coefficients (an aesthetically more pleasing pattern, but more computationally cumbersome).
        • For example, choosing n as 3, you could find u1 as –i1 + 4i2 – 3i3 + 2i4
        • All that’s important is that at the end of the day you net one i.

At a certain point, even rhythms with even numbers of intervals which have special formulas for unshadowed rhythms that are supposedly unshadowable find their non-unshadowable casts. This means, weirdly enough, when you hit this cast, that the previous unshadowing lied… or something.

For example, start with 2, 3, 5, 4. Take it to 5/2, 3/2, 9/2, 11/2. Then to 9/2, 1/2, 5/2, 13/2. But then you get 15/2, 3/2, -1/2, 11/2. Well… hard to say about lying, since you can still unshadow this to 17, 1, 1, 9, but that’s not further unshadowable.

The formula for telling when the process breaks is found again just by setting u’s to 0:


In other words, no interval can be greater or equal to its partner plus twice its neighbor in one direction, or said another equivalent way, no interval can be less than its partner minus twice its neighbor in the other direction. As I mentioned earlier, these even rhythm formula sets work directionally, so you don’t have to satisfy both sets at once; I understand this to be the equivalent of allowing the shadowing to push its weight around in one direction, as opposed to unnecessarily forcing it to distribute it evenly, centered on the original rhythm.

The pattern for 6-rhythms is:

10You can see there’s this interesting counting down then back up pattern on the coefficients. At this point you might notice how the even rhythm pattern resembles the odd rhythm pattern in that it’s the i numbered just before the u which is negative, and counting back it’s every other i that’s negative, skipping the one that’s zeroed out.

And for 8-rhythms, I believe you should start being able to induce this:


Next shadowing topic: More Shadowing Techniques


§ 3 Responses to Musical Idea 34: Unshadowing

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