What are the stipulations for a twelve-tone row to create a matrix with two consistent diagonals?

Question

Consider the tone row for Chen Yi's "Near Distance" (C, D♭, B♭, E♭, E, G♭, A♭, A, B, F, G, D). The matrix for this row form is:

Notice that the NW–SE diagonal is all the same pitch/integer (here, 0). This is true for all properly constructed twelve-tone matrices. But notice that there is no such consistency in the NE–SW diagonal.

Now consider the row from the "Wittgenstein" Motet by Elisabeth Lutyens (C, B, E♭, G, A♭, E, D, G♭, F, D♭, A, B♭).

This row form is almost symmetrical; the intervals moving forwards are the exact same as the intervals moving backwards, with only one exception: the interval between the sixth and seventh members of the row is a descending major second moving forward but an ascending major second moving backwards. Because of this symmetry, which ends exactly halfway through the row form, notice that the matrix has a NE–SW diagonal that is "t" for the first half and "2" for the second half:

Thus my question: under what stipulations will a twelve-tone matrix have both of these diagonals using only their own pitch? (By "twelve tone" I mean a row that uses all twelve pitch classes once and only once.) Does this occur only when the row form is completely symmetrical, or are there outside possibilities that would also create such a matrix?

By the way, I'm using musictheory.net's matrix calculator for these matrices.

Answer 1

This is a very interesting question and ended up being a fun maths puzzle to solve. I solved numerically and found 3840 matrices which satisfy this condition.

Here is one of them:

Pat Muchmore pretty much hit the nail on the head with the necessary conditions in his answer. But I thought I'd walk through how you'd solve this mathematically.

Firstly, instead of putting note values in the 12-TET matrix, let's put interval values above the root. So instead of the first row looking like:

1, 3, 6, 10, 4, ...

We can write it as:

1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+6, ...

(Note that we are working in a mod12 system, so the last element of the second list here 1+2+3+4+6=16. But 16 mod 12 is 4, which is the last element of the first list.)

Also, let's write it more formally, and make no assumptions about values. If we start on x, in 12-TET we have the first row:

It's starts out ugly, but the general strategy is to figure out relations between these intervals to try to reduce them down. From this first row we can construct a full matrix. Here it is (with only 1st row, 1st column, and diagonals completed).

If you follow down the NE-SW diagonal from the top, a pattern emerges that when you move down you must remove the latest interval added (k), and when you move left you remove the earliest interval added (a). In other words, remove intervals in pairs: (a,k); (b,j); etc. Our goal is that every value in this diagonal is the same, so the only way to remove a pair of intervals and not change the value is if their sum is 12.

Therefore: a+k=12, b+j=12, etc. If you play around with some values, you'll see that this is exactly the palindrome behaviour observed. In fact, I rather say the condition is not that there must be a palindrome, but that pairs of outer intervals must sum to 12.

When we get to the center, we have x+f and x-f, which must be equal. The only way these can be equal in a mod12 system is when f=6. Great, now we know this interval, and we know for certain the diagonal must only contain x+6 in each cell.

We can also infer what the first element of the first row is. The top right cell is x+a+b+...+k, and this =x+6. The bottom left cell is x-a-a-...-k, and this =x+6. So we have the relation:

x+a+b+c+d+e+f+g+h+i+j+k = x-a-b-c-d-e-f-g-h-i-j-k. But we already know a+k=12, which = 0 in mod12. We can remove. Similarly for -a-k=-12=0. In fact we can do this for all out interval pairs until we have x+f = x-f = 6. This is only true when x = 0.

Once we have this, we know our diagonal contains x+f=0+6=6 in each cell. One last little bit of notation to improve on; instead of holding onto a,b,c,d,e,f,g,h,i,j and k. Let's use our knowledge of how the pairs must sum to 12 to rename each interval a,b,c,d,e,6,12-e,12-d,12-c,12-b,12-a.

The problem is now reduced to five parameters, and the current matrix row looks like:

This is ugly, until we realize we can cancel a lot of stuff (a-a=0, 12=0, etc), so we can simplify to:

Where we are now is very close to the end. These are the minimum required conditions to find a 12-TET matrix with a both diagonals constant. The goal from here is to construct a row such that each tone only appears once, and the intervalic relationships provided are observed. So long as the first interval is, for example, a=7, then the second last row element must be 6+7=1, and so on.

I couldn't come up with any further parameterizations that helped, or any more tricks. From here I brute forced it by constructing all possible rows that satisfied all the conditions.

This all moved very far away form music, but I had fun so thanks for the problem!

Answer 2

This is a fascinating problem, and there are still a few questions I hope to explore further.

In short, yes, most definitely it requires the retrograde to be a palindrome with, potentially, a reversed “seam” in the middle (more on that in a bit.)

Here’s my reasoning. Without losing generality, we can assume that the northwest corner is PC 0, and that will always fill the NW–SE diagonal. For now, let’s call the final PC of the first row x—and this is the value we want to have go down at least the first part of the NE–SW diagonal. So far, our first two rows are:

0 _ _ _ _ _ _ _ _ _ _ x
_ 0 _ _ _ _ _ _ _ _ x _

So let’s be quite general about the second note of our P0 and call it y. Here’s the thing: we know that the second row will be exactly y less than the top row. Let’s fill some of that in:

0 y _ _ _ _ _ _ _ _ _ x
-y 0 _ _ _ _ _ _ _ _ x (x-y)

Thus, the second to last note of our P0 row needs to be y larger than x. Right, because everything in the second row needs to be y less than the top row? So we have:

0 y _ _ _ _ _ _ _ _ (x+y) x
-y 0 _ _ _ _ _ _ _ _ x (x-y)

Thus, the penultimate note is precisely as much bigger than the last note as the second note was bigger than the first (and remember, we’re in a mod-12 universe, so it is perfectly general to talk about adding numbers, since any subtracting is the same as addition by the mod-12 complement).

This logic will continue the whole way through, the third note of the row will be precisely as much bigger than the first as the third-to-last will be bigger than the last. Etc.

Now, another thing I realized is that there is a way to make the NE–SW diagonal the same the whole way, but only with 6 as the pitch class. This is because the value has to be its own complement (note how, in your Lutyens example, the two PCs that form the other diagonal are t and 2—complements. This will always be the case), and only 0 and 6 have the property of being self-complementary. Additionally, (and maybe relatedly) this is also the only way to make (any) tone row a true palindrome. Since the prime order move from the sixth to seventh note has to be the same as the the retrograde move from the seventh to the sixth while somehow being an “increase” in one direction while being a “decrease” in the other. The only interval that goes to the same pitch class either up or down is 6.

Unanswered questions that I hope to revisit:

What other restrictions are there in the note content? You definitely can’t use any row that starts in 0 and ends on 6 and get this effect.

How related is this to all-combinatorial hexachords? I have some wondering that there might be a relation.

The fact that the true palindrome version must have a tritone relation between the first and last notes, and a tritone relation in the middle definitely calls all-interval rows to mind. Coincidence?

Anyway, let me know if I need to clarify.