A combinatorial necklace is an arrangement of colored beads on circle (or circular string). Imagine the necklace is lying flat on a table and you can’t pick it up to turn it over. Two necklaces are considered equivalent if there’s a way to rotate one necklace so that its beads align with those of the other. (Another way to say this is that you could find positions on each necklace so that if you started at those positions and called out the bead colors in clockwise order, the two sequences of colors would be identical.)  So, how many distinct 12-bead necklaces could you create if you had an unlimited supply of black and white beads?  That’s the problem we’ll address in this post.

I’m interested in such black/white or binary necklaces because they correspond to musical constructs like chord types or scale families. If you want to know, for example, how many distinct chord types there are in 12-tone equal temperament (where a chord type may have any number of notes from 0 to 12), this turns out to be equivalent to the question we just posed about necklaces.

In mathematical discussions of musical chords and scales, when “the necklace question” comes up, it’s usually addressed in one of two ways. Some authors present an exhaustive list of necklaces/chords/scales obtained by brute force (presumably, the author wrote a computer program to generate all 2^12 binary strings and eliminate pairs of strings that represent equivalent necklaces). Other authors point out that you can obtain a count of the necklaces by invoking a piece of mathematical machinery called the Polya Enumeration Theorem.  Feed some numbers into Polya’s formula and we learn that there are 352 binary necklaces of size 12.

When I started thinking about necklaces I was looking for something that neither approach seemed to provide: I wanted to gain an intuitive understanding of how the possibilities arise. Where do those 352 necklaces “come from,” and why aren’t there more of them, or fewer of them, for that matter? Presumably, such intuition could be obtained by exploring the inner workings of the Polya Enumeration Theorem, but I was looking for a more direct path, using only elementary mathematical concepts, so I started trying to devise a method that would help me list all the necklaces by hand, without having to explore and eliminate duplicates. It turns out there are a number of algorithms for generating necklaces in the literature, including one due to Fredricksen, Kessler and Maiorana, and more recent work by Cattell, Ruskey, Sawada, and Serra. The method I came up with is different from approaches I’ve seen elsewhere, so I’ve decided to present it here in hopes that the right reader might find it interesting. In this post, I’m not aiming to present it as a general purpose algorithm for enumerating necklaces of arbitrary size (to do this would require some refinement) or to analyze its performance in contrast to the other algorithms — I’m just offering it as an interesting way to generate and/or categorize the possibilities for size-12 necklaces, a way that hopefully provides some intuition into the problem space, and that is amenable to execution by hand for those crazy enough to want to do it. That said, I’d be interested to collaborate with anyone who thinks this approach could be expanded into something more generally useful, or who could point out where it might already have been explored.

The Idea

The basic idea behind this approach is to view a size-12 binary necklace as an interleaving of two smaller necklaces built from partitions of two integers summing to 12. I’ll try to explain this by describing a process for assembling a necklace from scratch:

1. First, decide how many white beads to include in your necklace. Call this n. Obviously, the number of black beads will be 12-n. (n and 12-n are the two numbers we’ll be partitioning later.)
2. Now decide how many separate “clumps” the white beads will fall into. I’m using the term “clump” to mean a sequence or strip of beads of one color (with no beads of the other color between them). Call the number of white clumps c. It’s possible to see that the number of black clumps must be the same as the number of white clumps, unless the necklace is all-black or all-white.
3. Now decide how many beads will go into each clump. For the white beads, this is the same as choosing an integer partition of n with c parts (that is, a way of writing n as the sum of c positive integers). For the black beads, select an integer partition of 12-n with c parts.
4. Now treat each integer in the partitions from step 3 as a distinct type (or “color”) of bead, and arrange each partition into a necklace of c beads. The number of possibilities here will depend on the number of colors in each necklace, and how many beads there are of each color: later, we’ll refer to these two pieces of information together as the “necklace type.” At the end of this step, we have two necklaces, one of which will contribute all the white beads to our final necklace, and the other of which will contribute all the black beads.
5. Take the two c-bead necklaces from step 4 and interleave them. This means that you’ll create a single 2c-bead necklace by taking one bead from the first necklace, then one bead from the second, then one bead from the first, then one bead from the second, until you’ve used up all the beads in both necklaces. In doing this, you’ll need to choose a position on each necklace to start pulling the beads from, and you have to pull the beads from each necklace in a clockwise order.
6. Finally, take each integer-labeled bead in the interleaved necklace from step 5 and “expand” it into the corresponding number of black or white beads, giving a 12-bead binary necklace.

The Idea Illustrated

Here’s an example of how the process works.  In Step 1, we choose how many white beads to include in our necklace.  Let’s go with six white beads, meaning there will also be six black beads:

In Step 2, we choose a number of clumps (we’ll go with three), and in Step 3 we partition the white and black beads into that many clumps.  Here we’ll break the white beads into clumps of size 3, 2, and 1, and we’ll break the black beads into clumps of size 4, 1, and 1:

In Step 4, we form necklaces from the clumps.  There are two distinct necklaces we can make from the white clumps, and there’s only one necklace we can make from the black clumps.  The arrow indicates the white necklace that we’ll choose:

In Step 5, we interleave the white and black necklaces.  There are 3 distinct interleavings possible between the white necklace we selected above and the only black option; the arrow indicates which of those interleavings we’ll choose:

In Step 6 we expand our fused beads back into individual beads, giving this particular one of the 352 possible binary necklaces of size 12:

If you go back through these steps, it should be easy to verify that making any of our choices in a different way would necessarily have yielded a different necklace in the end.

The Idea Illustrated Backwards

Now let’s try the process in reverse. We’ll start with a binary necklace such as would be obtained in Step 6, and then we’ll break it apart into its pieces so as to unwind the previous steps.

Let’s consider a different necklace from before–we’ll use a necklace which corresponds to the major scale and its modes.  (If white represents a scale tone and black represents an excluded tone, the interval pattern here starting from 12 o’clock is whole-whole-semi-whole-whole-whole-semi):

First we’ll collapse each clump, reversing Step 6.  Here we’ll use numbered beads as opposed to rectangular “jumbo” beads like we did above, but the idea is the same:

Now we’ll break the “interleaved” necklace above into two separate necklaces, reversing step 5:

Finally we can see that our two necklaces 21211 and 11111 (formed in step 4) were based on partitions 2+2+1+1+1=7 and 1+1+1+1+1=5 (chosen in step 3), and of course we must have chosen c=5 (in step 2), and n=7 (in step 1).

The Idea Iterated: Steps 1-3

Now if we want to enumerate all possible size-12 binary necklaces, we just need to iterate Steps 1 through 6, making the choices at each step in every possible way; as we do this, we’ll see that it’s not hard to size up the full range of possibilities at each step. Furthermore, we’ll be able to “save” some of the work we do in Steps 4 and 5 and reuse it when we return to those steps the next time.

To save time, observe that we never need to explore necklaces with more than 6 white beads. Once we’re done generating all the necklaces with up to 6 white beads, we can find the necklaces with more than 6 white beads by simply taking the complements all of the necklaces with fewer than 6 white beads (here, “taking the complement” means swapping white beads with black, and vice versa). So, each necklace that we visit with <6 white beads will contribute 2 to our total count (1 for the necklace itself, 1 for its complement); while each necklace with exactly 6 beads will only contribute 1 to the total count.

Now, looking at Step 1, we have the option to include 0, 1, 2, 3, 4, 5, or 6 white beads, and in Step 2 we can choose any number of clumps that’s less than or equal to the number of white beads. This table shows all of the possible partitions of n and 12-n into c parts, where 0<n<=6 and c<=n. All of the choices that can be made in Steps 1 through 3 are represented in this table.

Here’s the same information in a diagram:

The Idea Iterated: Step 4

In the previous section we looked at all the choices that can be made in Steps 1 through 3 of our procedure: all the ways of choosing a number of n of white beads, a number c of clumps, and a pair of integer partitions of n and 12-n into c parts each.  Now we’re going to consider all possible ways of taking those pairs of partitions, arranging each partition into a necklace (Step 4 of our procedure), and finally interleaving the necklaces (Step 5 of our procedure, covered in the next section).

The obvious way to proceed would be to consider all the pairs of white and black partitions that can be made from any single row of our table.  We could consider each pair separately (for example, let’s take the pair 4+1+1 and 3+2+1), create all possible necklaces for each member of that pair (eg. {411} and {321, 312}), and then create all the interleavings of those necklaces (e.g. 431211, 421113, 411312, 431112, 411213, 421311), repeating this process until we’ve exhausted all the pairs.

However, we can save work by observing that the same types of necklaces, and hence the same types of necklace pairs, occur repeatedly in the table.  Once we’ve done the work of finding all interleavings of two particular necklace types, we can reapply that work wherever we encounter those same necklace types in a partition pair.

What exactly do we mean by necklace type?  A partition like 4+1+1 contains two distinct integers (4 and 1) which we can think of as two distinct bead colors. There’s 1 bead of the first color and 2 beads of the second color. So we’ll say any necklace formed from 4+1+1 would be a necklace of type 2/1.  (We’ll adopt a convention of writing the necklace type in order starting from the most frequent to the least frequent color, so we’ll write 2/1 here instead of 1/2.)  Can you find another partition in the table that falls into the 2/1 necklace type?  The partition 3+1+1 fits the bill, as again there are two distinct integers (this time 3 and 1) and there’s 1 of one kind and 2 of the other kind. You won’t see it in this table, but 200+25+25 would be another example fitting the 2/1 necklace type.  Going back to our example, we’ll call all the interleavings formed from necklaces of 4+1+1 and 3+2+1 a type 2/1 x 1/1/1 interleaving (or a type 2/1 x 1/1/1 necklace).

Here are the last two columns from the table of partitions above, with each partition now annotated with its necklace type in red:

Looking at the table, we can see there are only 14 necklace types at play here, and each of them generates only a few simple necklace instances.  For example, if we consider the 2/1/1 necklace type, this means there are two beads of one color, call it X, one bead of another color, call it Y, and one bead of a third color, call it Z. There are only three ways of arranging these beads into distinct necklaces: XXYZ, XXZY, and XYXZ. (Here I’m representing a necklace as a string like XXYZ — you should imagine those beads on a circle from which you could also read them as XYZX, YZXX, ZXXY — all equivalent to XXYZ.) Here is a summary of the necklace types we’re dealing with, and their corresponding necklace instances in string form:

Looking back at our table of partitions, we can see that certain kinds of interleavings occur in our space of possibilities, while others don’t. For example, we’ve already seen that 4+1+1 and 3+2+1 make a type 2/1 x 1/1/1 interleaving, and so do 3+1+1 and 4+2+1.  But you won’t find any type 2/2 x 2/1/1 interleavings in the table.

The Idea Iterated: Step 5

Our next step will be to consider all of the interleaving types represented in our partition table and list all possible instances of each interleaving type.

For simple interleaving types, the instances are obvious.  For example, let’s consider the 3×3 interleaving type.  There’s only one necklace of type 3, which we’ll write as XXX, and there’s only one way we can interleave it with another necklace of type 3, which we’ll write as AAA.  This single instance of 3 x 3 could be written as XAXAXA.  But, when the interleaving types get more complex, how can we find all the possibilities?  The brute force approach would to look at every pair of instances of the two necklace types, and apply the following procedure: keep one necklace fixed, and place the second necklace on top of it so that the beads of the second necklace fall into the gaps between the beads of the first; then rotate the second necklace clockwise in steps of one bead, until we’ve visited all possible positions around the circle.  Finally, eliminate interleavings that turn out to be rotationally equivalent.

Do we actually need to rotate the second necklace so it lands in every possible position, and then go through the elimination step?  No.  Let’s say first necklace can be decomposed into repeating blocks of a fixed size, like ABC (i.e. the entire necklace is ABCABCABC) while the second necklace cannot be decomposed into repeating blocks of a fixed size smaller than the necklace itself (say the necklace is XXXXXXXXY). You’d only need to rotate the second necklace through the length of one repeating block of the first necklace.  I will leave a discussion of this principle for a separate post.

Now, it turns out that there are 28 distinct interleaving types that occur in our partition table, listed below with all their instances, and all the partition pairs that fall into each type.  (Note: some of this table is currently TBD, but I think you’ll get the idea.)

The Idea Iterated: Step 6

The next step is to make all possible necklaces from each partition pair in the table above: we can do this by substituting the partition members into the precomputed necklaces of the appropriate type.  For example, let’s return to the pair (4+1+1, 3+2+1) that we considered before.  This pair is of type 2/1 x 1/1/1, and our table tells us the possible interleaved necklaces of that type are: XAXBYC, XCXAYB, XBXCYA, XAXCYB, XBXAYC, XCXBYA.  If we take X=1, Y=4, A=3, B=2, C=1 we get necklaces 131241, 111342, 121143, 131142, 121341, 111243. The requirement for setting up our substitutions is that if a letter X in occurs m times in each precomputed necklace, we need to map it to an integer that occurs m times in the partition it belongs to.

The final step is to unpack the integers in each necklace into the corresponding number of white or black beads.  Our necklace 131241, for example, would translate into white-black-black-black-white-black-black-white-white-white-white-black, or 100010011110.

If we go through this process for every partition pair in the table, and then add all the complements of necklaces with <6 white beads, we’ll get all 352 binary necklaces of size 12.  I might post them here at some point when I’m feeling up to it, but now you know one way to find them yourself (and if you don’t want to go through the labor, you can get your free necklaces here or here).  Well, I didn’t say this approach would make the enumeration easy or free from tedium, but I think it does provide a systematic process for doing it, along with a compact way of describing the structure of any necklace you encounter.

Recap: Intuition?

My main reason for going on this excursion was not that I had a pressing practical need to enumerate necklaces, but that I wanted a better intuitive understanding of the space of possibilities: I wanted to get a feel for how necklaces “arise” and how their structure can be analyzed.  Have I succeeded in obtaining, and hopefully sharing, such intuition?

Well, now if you see a necklace like 100110011100, you can view it as an “expansion” of the necklace 122232, which is built from the partition pair (1+2+3, 2+2+2), and you can observe that this pair is of type 1/1/1 x 3.  You could observe that there are in fact two interleavings of type 1/1/1 x 3 (they are XAYAZA and XAZAYA), and that 123222 (or 100111001100) is a “close relative” of the given necklace.  Finally you could observe that these two necklaces are special in that they happen to be the only interleavings of type 1/1/1 x 3 in the problem space.

Followup Questions

This section includes some followup questions which I’d like to address when I have time: hopefully soon, but quite possibly never.  I will be motivated to get around to them before never if you drop me a note letting me know what topics you’d like to see explored here, or how you think this post could be improved.

1. Can this approach be made into a recursive algorithm for generating binary necklaces of arbitrary size?  The catch is that in Step 4, we need to find necklaces of more than two colors; in this post where we were specifically dealing with 12-bead necklaces, it was possible to build these component necklaces by hand.  If we wanted to do it with a recursive call to the procedure, we’d need to generalize the procedure to handle necklaces with an arbitrary number of colors.  Perhaps this could be done using an encoding scheme where binary necklaces of a larger size get mapped to necklaces with fewer beads and more colors?
2. How does this approach relate to other necklace algorithms in the literature and does it offer any advantages?
3. Is it possible to find some kind of conceptual mapping between the steps of this procedure and the things that happen “inside” the Polya Enumeration Theorem.
4. How can this approach be applied in the context of combinatorial music theory, if at all?

Step 1: Create a Nonstandard Key Signature with your preferred number of subdivisions of the octave, and assign accidental symbols as you like.  The best explanation of how to do this I’ve found is in this PDF by Ere Lievonen.

Step 2: Set up the Aria player that comes with Finale to use a Scala file that defines your tuning.  For this, you’ll need a recent version of the Aria player, since Scala integration seems to be a newer feature.  The basic process is as follows:

• Set MIDI/Audio->Play Finale Through Audio Units
• Open this dialog: MIDI/Audio->Audio Units Banks & Effects
• Click the pencil icon for Bank 1 to open the Audio Unit Viewer
• Select Settings from the tabs on the right and look for the area labeled Tuning
• Import your chosen Scala file

That’s it (for this summary, at least).

Of course, if you only want to experiment with nonstandard tunings that have 12 notes per octave, you can skip Step 1 and go right to Step 2, selecting your Scale file in Aria Player.  There are a couple of Scala files included in the Aria installation, and you can download many more at the Scala site.  The neat thing is that if you do want more than 12 notes per octave, Step 1 and Step 2 work well together: if your nonstandard key signature has, say, 24 notes per octave and your Scala file has 24 notes, they’ll be matched during playback.  (This is how it should work, of course, but until I tested it tonight I wasn’t sure it actually would work.)  You can get any one of those 24 notes to play back properly by applying the appropriate accidental as you defined in Step 1.  It’s no longer necessary to use the older technique of creating pitch bend expressions.

I’ve just begun playing with this and will update this post as I learn more.

Consider the numbers arranged on a clock face.  Your glyph system should have the following properties:

1. The glyphs for two numbers that are opposite on the clock face (like 12 and 6) should have some visual qualities that bind them together as a pair.
2. If you look at the glyphs for any two numbers that are adjacent on the clock face (like 1 and 2), it should be easy to see which glyph represents the “lower” number in clockwise order, and which glyph represents the “higher” number.
3. The glyphs for any three numbers that form an even division of the clock face into three parts (like 12, 4, and 8) should have some common visual feature that makes them recognizable as a group.
4. The glyphs for any four numbers that form an even division of the clock face into four parts (like 12, 3, 6, and 9) should have some common visual feature that makes them recognizable as a group.
5. The parity of a number (whether the number is odd or even) should be somehow discernible from its glyph.

Thinking of this problem in musical terms, the glyphs could represent the 12 notes of the octave (assuming 12TET).  Looking at any glyph, you should be able to quickly discern the following things about the note it represents: which of the four augmented triads does the note belongs to?  Which of three fully-diminished seventh chords does it belong to?  What is its tritone, and what is that tritone’s glyph? What are its upper and lower chromatic neighbors, and what are their glyphs?

I haven’t solved this.  But, to jog your imagination, here’s one of many glyph systems I’ve come up with while considering the problem.  Do you think it has any nice properties?  Can you do better?

This second clip (20 minutes) was pure experiment.  It takes a little while to get into a groove, but one of my favorite parts is the craziness that ensues between 3:00 and 4:00, and resurfaces again towards the end:

Addendum

Since photography was the context that brought Dan and me together to make this music, I’ll end this post with an amazing video by Charles and Ray Eames that’s been in my mind since I watched it this morning.  Perhaps in some way it inspired the jam.  It’s a documentary about the Polaroid SX-70 camera, and it manages to capture, either directly or through its general spirit, almost everything I find inspiring about photography.  I made a lot of noise (of the “Oooohhh…. Ahhh….” variety) while watching it the first time.  Thanks to Carl Seglem for sending me the link!

After ’96 I would always check for the “Sardinia” section whenever I stopped in a record store, and if I found such a section, it was usually stocked with the recording I already owned.  Through the sort of physical-world-sleuthing that was necessary in those days, I did gradually expand my collection to five or six albums, but at some point I had to concede that there just wasn’t much to do as an American fan of Sardinian polyphony: no new releases, no local performances, and practically no one in my circle of acquaintance who had heard of it.  So I stopped checking for “news.”  But just the other day some fortuitous web browsing led me the best Sardinian-polyphony news I’ve come across in a decade.   There’s a group of American singers, Tenores de Aterúe, who have taken the adventurous step of studying and performing traditional cantu a tenore, and their footage on YouTube sounds great.  (Where’s your website, guys?)  In reading a little about their background, I enjoyed the story of a 2008 ephiphany, when one member who had dreamed of learning Sardinian music found a video of Tenores di Bitti demonstrating how each voice part in cantu a tenore sounds by itself, and then in ensemble.  This video became a sort of Rosetta stone that exposed the workings of the music and opened the path for a group of experienced but non-Sardinian musicians to actually learn it.  I encourage you check out Tenores de Aterúe on YouTube and see if they might be performing near you (I’ll be doing the same).  Here, I wanted to post that revealing Tenores di Bitti video, as it’s probably the best short introduction to cantu a tenore one can find (replete with charming features like the onlooker who appears in the background near 3:50, and of course, some great singing).
When I sent it to my classical voice teacher, she emailed back: “This just blew my mind.”  Specifically, she was amazed at how these musicians create — in a way that appears relaxed and free from vocal strain — a completely different set of sounds from those we work to produce in “mainstream” Western vocal practice.

Recall that a janya raga is one that is somehow derived from one of the 72 melakarta ragas. The 34,776 figure refers to one specific class of janya ragas: those that are created by omitting notes from the parent. In forming such “varja” ragas, we can omit up to two notes from the arohanam (ascent), the avarohanam (descent), or both. (And it’s permissible to omit different notes in the ascent from those we omit in the descent.) We can’t apply other processes of derivation like reordering notes or borrowing notes from other ragas: these processes lead to many more possibilities!

So where does the number 34,776 come from?  Well, if we decide to omit one note from the arohanam of a melakarta raga, there are 6 ways to do it: any note besides sa is fair game for omission. If we’re going to omit two notes, there are (6 choose 2) = 15 possibilities. And of course, if we omit no notes, there’s only 1 way to do that. This gives 6+15+1=22 options. The same 22 options exist for the avarohanam, giving 22*22=484 ways of omitting notes from a parent raga to create a janya. Except, we don’t want to count the case where no notes are omitted in the arohanam and the avarohaman both, because this leaves us with the original raga. So, the total number of janya possibilities for each melakarta raga is actually 484-1 = 483. Multiplying this by the total number of melakarta ragas, we get 483*72 = 34,776. But there’s a catch…

The process of omitting notes from two distinct parent ragas can give us two janya ragas with the same notes. For example, we can get S R2 G3 P D2 (Mohanam), by omitting M1 and N2 from S R2 G3 M1 P D2 N2 (Harikambhoji), or by omitting M2 and N3 from S R2 G3 M2 P D2 N3 (Kalyani). In Western terms, one would say you can get 1 2 3 5 6 (major pentatonic) by omitting 4 and b7 from 1 2 3 4 5 6 b7 (Mixolydian) or by omitting #4 and 7 from 1 2 3 #4 5 6 7 (Lydian). So, the 34,776 figure contains many janya ragas that actually have identical swaras. Now, if the parent raga (including its mood, characteristic phrases, ornamentation patterns, and additional swaras) is kept in mind when performing the derived raga, then two ragas derived from different parents might be perceived as distinct even though the derived ragas happen to have the same notes. In this way of looking at raga derviation, 34,776 is a plausible theoretical count. However, if we’re only interested in janya possibilities that are distinct in their notes, and we’re not considering attachments to parent ragas, then 34,776 is an overcount.

Enumerating the janya possibilities that are note-wise distinct is more complicated. The rest of the post describes an approach I came up with when I first started thinking about the problem.  After I published the post, I received a comment from Narayana Santhanam pointing out that the technique of generating functions from enumerative combinatorics is another, more compact approach to counting janyas; and, in searching further, I found a 2002 paper by K. Balasubramanian that uses this approach.  Finally, I learned that P. Sriram and V. N. Jambunathan had investigated this question and published similar findings in a 1991 paper in the Journal of the Madras Music Academy.  (See the notes and comments at the end of this post for more details.)  Although the approach I present here is more verbose than some of the others, I hope it will be interesting to readers looking for insight into how all the possibilities arise.

In what follows, I assume R2=G1, R3=G2, D2=N1, and D3=N2 even though one could say those swaras would be performed with different ornamentation and/or intonation.  In this count, for example, an arohanam of S R2 M1 P D2 S will be considered equivalent to S G1 M1 P D2 as since we treat R2=G1.

To organize our count, we’ll consider how many notes occur in the union of the ascent and the descent of a janya raga. This union might have 5, 6 or 7 notes.

If the union contains 5 notes, then the ascent and descent must contain those same 5 notes. To find the number of janya ragas of this kind, we need to consider all the ways of choosing 4 notes from 11 possibilities without violating the melakarta rules. Note that we’re choosing 4 notes out of 11, not 5 out of 12, because one note, sa, is mandated. Unfortunately, we can’t use a simple formula for (11 choose 4) because this would count “illegal” cases where there are two ma’s, or where there are more than two notes betwen sa and ma.  One way to count only the legal possibilities is to divide the scale above sa into four sections. The first section contains ri and ga. The second section contains ma. The third section contains pa. The fourth section contains da and ni. Now there are 6 ways of filling the first section (i.e. ri and ga can be R1/G1, R1/G2, R1/G3, R2/G2, R2/G3, or R3/G3), two ways of filling the second section (i.e. ma can be M1 or M2), one way of filling third section (pa is always Pa), and six ways of filling the fourth section.  (This is how we get 6*2*1*6 = 72 melakarta ragas.) If we omit some notes, then one or more of the scale sections will no longer be full. (In particular, if the first section has only one note in it, there are 4 possibilities for its identity: R1, R2=G1, R3=G2, or G3.) If we write 2|1|1|2 to indicate a scale with all sections full, then here are the possibilities distributions of sections sizes for a five-note derived scale where the sections are not all full: 2|1|1|0, 2|0|1|1, 2|0|0|2, 2|1|0|1, 1|1|1|1, 1|0|1|2, 1|1|0|2, 0|1|1|2. Taking this list and substituting the sections sizes with the number of ways of filling each section to the designated size, we get the following count: 6*2*1*1 + 6*1*1*4 + 6*1*1*6 + 6*2*1*4 + 4*2*1*4 + 4*1*1*6 + 4*2*1*6 + 1*2*1*6 = 236.

Now if the union of the ascent and descent contains 6 notes, the respective sizes of the ascent/descent might be 6/6, 5/6, 6/5, or 5/5. The 6/6 case is similar to above: we know the ascent and descent use the same six notes, and we need to consider the number of ways of choosing 5 out of 11 notes without violating the melakarta rules. Following the logic above, the possible distributions of section sizes are 1|1|1|2, 2|0|1|2, 2|1|0|2, and 2|1|1|1. This gives 4*2*1*6 + 6*1*1*6 + 6*2*1*6 + 6*2*1*4 = 204 possibilities. In the 5/6 or 6/5 case, we can see that side containing 5 notes must be a proper subset of the side containing 6. The number of possibilities here can be calculated as (# ways of choosing 5 out of 11 notes legally)*(# ways of removing one of those 5 notes to create the smaller side) = 204*5 = 1020. Since we can make the smaller side the ascent or the descent, we multiply the last figure by two, giving 2040. Finally, in the 5/5 case, we can see that the ascent and descent must share precisely 4 notes including sa (each side contains a note that’s not in the other, creating a union of size 6). The number of possibilities can be calculated as (# ways of choosing 5 out of 11 notes legally, to create the 6-note union including Sa)*(# ways of picking 3 notes from those 5 to be shared by the ascent and descent)*(# ways of picking one of the remaining 2 notes to be in the ascent, while the other goes in the descent) = 204*(5 choose 3)*2 = 4080.

The last situation is where the union of the ascent and descent contains 7 notes. Here, the respective sizes of the ascent/descent might be 5/5, 6/5, 5/6, 6/6, 5/7, 7/5, 6/7, or 7/6.  In the 5/5 case, the intersection must contain 3 notes. The number of possibilities is: (# of ways of choosing 6 out of 11 notes following the melakarta rules, to create a 7-note parent raga including Sa)*(# ways of choosing 2 out of those 6, to create a 3-note intersection including Sa)*(# ways of picking two from outside the intersection to be in the ascent, while the other 2 go in the descent) = 72*(6 choose 2)*(4 choose 2) = 72*15*6 = 6480. In the 5/6 or 6/5 case, the intersection is size 4, and we can choose 3 out of six notes to form it. By similar logic to the previous case, the number of possibilities is 72*(6 choose 3)*(3 choose 2) = 72*20*3 = 4320 for each case by itself.  In the 6/6 case the intersection must be size 5, and the number of possibilities is 72*(6 choose 4)*2 = 72*15*2 = 2160. In the 5/7 and 7/5 cases, the smaller side of the scale must be a proper subset of the larger one, and we can form the smaller side by omitting two out of six notes from the larger. This gives 72*(6 choose 2) = 1080 possibilities for each case. And in the 6/7 and 7/6 cases, the smaller side again must be a proper subset of the larger one, so the number of possibilities is 72*(6 choose 1) = 432 for each case.

Here’s a recap of the cases we considered and the possibilities that exist in each case:

Union size 5.  Case 5/5: 236 possibilities.

Union size 6.  Case 6/6: 204 possibilities. Cases 5/6 and 6/5: 2040 possibilities together. Case 5/5: 4080 possibilities.

Union size 7.  Case 5/5: 6480 possibilities. Cases 5/6 and 6/5: 8640 possibilities together. Case 6/6: 2160 possibilities. Cases 5/7 and 7/5: 2160 possibilities together. Cases 6/7 and 7/6: 864 possibilities together.

The grand total is 236+204+2040+4080+6480+8640+2160+2160+864= 26864.

Now, how can we be sure that 26864 correct? In fact, I made many calculation errors as I was working through this the first time, leaving me in a state of great suspicion.  However, after I spotted and corrected my mistakes, I gained further confidence by writing a Groovy script that generates all distinct janya possibilities by brute force, and also yields… 26864!  (Warning: this is script-style code, meant for one-time use and not designed to be maintainable or easy to read, but it gets the job done.)

// this script is written in Groovy 2.1.0

// generate all melakarta ragas, representing each
// raga as a list of 11 ones and zeros
melakartas = []
for (riGa in [1,1,0,0].permutations()) {
    for (ma in [0,1].permutations()) {
        for (daNi in [1,1,0,0].permutations()) {
            melakartas.add(riGa + ma + [1] + daNi)
        }
    }
}
assert melakartas.size()==72

// generate all ways of deleting up to two notes from
// the ascent and descent of a raga
deletions = ([0..5,0..5].combinations()).collect( {it as Set} ) as Set
deletions.add([])
deletionPairs = [deletions,deletions].combinations()
deletionPairs.remove([[],[]])

// generate all possible janyas as ascending/descending scale pairs,
// by applying all possible deletions to each melakarta raga;
// keep the janyas in a set so that identical janyas will
// not be double-counted
janyas = [] as Set
for (mela in melakartas) {
    def noteIndices = (0..(mela.size()-1)).findAll({mela[it]==1})
    for (deletionPair in deletionPairs) {
        def ascending = mela.clone()
        def descending = mela.clone()
        deletionPair[0].each { ascending[noteIndices[it]]=0 }
        deletionPair[1].each { descending[noteIndices[it]]=0 }
        janyas.add([ascending, descending])
    }
}

println "Number of Distinct Janyas: ${janyas.size()}"

Notes:

How many Janya ragas have actually been named and used in Carnatic music?  I came across Raga Pravaham, a wonderful reference by musician, composer and musicologist D. Pattammal—the work includes over 5000 named ragas.  At the end of the text, there is an appendix that places the number of janya possibilities at 126936, without eliminating duplicates.  This count is so much larger than 34776 because it uses a looser definition of permissible janyas in which up to five notes may be deleted from the arohanam, avarohanam, or both.

See the comments by Narayana Santhanam below for an approach to counting janya ragas using generating functions.  Also see Combinatorial Enumeration of Ragas by K. Balasubramanian, published in Journal of Integer Sequences Vol. 5, 2002 and How Many Janya Ragas Are There? by P. Sriram and V. N. Jambunathan in the Journal of the Madras Music Academy, 1991, pp. 144-155.  See also a discussion of this topic at rasikas.org.

Major Pentatonic as a Scale Complement

One of the neat the things about standard major pentatonic scale, which uses degrees {1, 2, 3, 5, 6}, is that it can be seen as the complement of a full 7-note major scale: it contains all the notes that some major scale excludes.  For example, the black keys on the piano–all the notes that are not in C major–form Gb major pentatonic.  In general, if you take all the notes that a major scale excludes, you’ll have the major pentatonic scale rooted at its tritone.  We can visualize this with a clock-style diagram of the octave.  Here, the numbers are not hours but degrees of the major scale, and the red dots represent notes outside the scale.  In the leftmost diagram, the polygon formed by connecting the red dots represents the pentatonic scale rooted at #4 of the parent scale.  Transposing the scale up a semitone (rotating the polygon 30 degrees clockwise) gives the major pentatonic rooted at 5 of the parent scale.  Transposing it down a semitone gives the major pentatonic rooted at 4 of the parent scale.  And transposing it by a tritone (rotating it 180 degrees) gives the pentatonic rooted at 1 of the parent scale.  Notice how the pentatonics rooted at 1, 4, and 5 exist entirely within the parent scale, which is to say that none of the polygon vertices fall on any of the red dots.  The green circle indicates the root of the pentatonic scale, and the dotted line inside the polygon is its axis of symmetry.

§

Major Vs. Minor Pentatonic: Comparing Definitions

If the major pentatonic is built from degrees {1, 2, 3, 5, 6} it’s natural to wonder why we define the minor pentatonic as {1, b3, 4, 5, b7}.  Why not simply flatten 3 and 6 in the major pentatonic formula so that it fits inside a minor scale: {1, 2, b3, 5, b6}?  I don’t know the history of how the major and minor pentatonics evolved in various cultures over time, but I will comment on some abstract reasons why {1, b3, 4, 5, b7} might be more versatile as a scale than {1, 2, b3, 5, b6}.  First of all, it’s convenient to define our minor pentatonic so that it turns out as a mode of our major pentatonic.  This lets us shift from major to relative minor while staying in the same pool of notes.  Indeed C-major pentatonic (C D E G A) and the standard A-minor pentatonic (A C D E G) are modes of each other, or reorderings of the same set of notes.  But if we were to use the formula {1, 2, b3, 5, b6} to get an alternative A-minor pentatonic, it would be A B C E F, which doesn’t fully overlap with C-major pentatonic.  Another notable thing about the standard major and minor pentatonics are that they have no semitones, no tritones, and lots of perfect fourths and fifths, making them very “user-friendly” scales.  The same is not true of {1, 2, b3, 5, b6}, which has two semitones (2-b3 and 5-b6), one tritone (2-b6), and fewer perfect fourths and fifths.  This alternate scale might still be great to make music with, but it lacks the many of the convenient properties of the standard minor pentatonic.

§

Melodic Minor Pentatonic

With these observations in mind, let’s look at the question of how to create a pentatonic scale from the 7-note jazz melodic minor scale.  In sacrificing two notes, we want to preserve as much of the character of melodic minor as possible.  So, which two notes can we afford to lose?  We should probably keep 1 and 5!  To have a minor sound, we need to keep b3.  To have something different from a plain old minor scale, we need keep at least one of 6 or 7.  Now, as far as I know, there isn’t one universally accepted way to make the remaining choices; that’s to say, the idea of a melodic minor pentatonic hasn’t been standardized in the same way that the major and natural minor pentatonics have been.  I’ve seen melodic minor pentatonic defined alternatively as {1, 2, b3, 5, 6}, as {1, b3, 4, 5, 7}, and as {1, b3, 4, 5, 6}.  What I’d like to do here is explore some interesting qualities of that last definition, {1, b3, 4, 5, 6}.  First of all, it’s the only one of the lot that has no semitones.  In fact it’s a member of one of only three possible pentatonic families with that property (see my post on anhemitonic pentatonics).  Second, it turns out to be a complement of the 7-note melodic minor scale, in the same way that the major pentatonic is a complement of the 7-note major scale!  And third, you can actually transpose the whole thing up a whole-step while remaining inside the 7-note melodic minor parent scale.  To experiment with this, start by playing C-melodic-minor: C D Eb F G A B. Now make up some phrases using {1, b3, 4, 5, 6} or C Eb F G A.  Now transpose your phrases up a whole-step, so you’re using the same {1, b3, 4, 5, 6} pattern rooted at D.  This will give you the notes D F G A B.  The nice thing is that after this whole-step transposition, you’re still playing inside the parent scale of C-melodic minor.  Now try doing that with the other versions of the melodic minor pentatonic and let me know how it works out.

These diagrams show how the {1, b3, 4, 5, 6} pentatonic pattern originates as the complement of a parent melodic minor scale.  If we take the excluded notes of the parent scale (i.e. the red dots) and transpose them down a semitone, we get our {1, b3, 4, 5, 6} pentatonic rooted at 1 of the parent scale.  If we transpose the excluded notes up a semitone, we get the same pentatonic pattern rooted at 2, and still fitting entirely inside the parent.

§

Mix and Match

Notice that {1, b3, 4, 5, 6} is also a subset of the Dorian mode of the major scale, so we could call it a “Dorian pentatonic” as well as a “melodic minor pentatonic.”  There is a difference in how it functions in those two context though.  If you’re thinking of 7-note Dorian as your parent scale, you can only form this pentatonic on the Dorian root: any attempt to transpose it will put you outside the Dorian note set.  However, if you’re thinking of 7-note melodic minor as the parent, then as we just saw, you can transpose this pentatonic between 1 and 2.  It’s also worth noting that the standard minor pentatonic fits inside the full melodic minor scale–you can play it on 2–but you can’t transpose it anywhere else.  For example, if C-melodic-minor is your parent scale, you can play D minor pentatonic (D F G A C) and remain inside the parent.

In the diagram on the left below, you’ll see how the melodic minor pentatonic (as we’ve defined it) fits inside a major scale.  Here the pentatonic root is at positioned 2 of the major scale (or at 1 of the major scale’s Dorian mode).  In the diagram on the right, you’ll see how the standard minor pentatonic pattern fits inside a melodic minor scale.  Again, the pentatonic root is positioned at 2.  If we think of the same shape as representing a major pentatonic, its root would positioned at 4 of the enclosing scale.

If our scale has no semitones, then there must be an excluded note between any two included ones.  We’ll represent these as red dots on the pentagon’s edges:

Without making any choices to “customize” our scale, we’ve already accounted for ten notes: there must be five excluded notes somewhere between the five included notes.  We only have two notes in the octave left to play with, and all we can do is decide which edge or edges of the pentagon to put them on.  Since we know that each edge has at least one excluded note, we might as well remove the corresponding red dots from our diagram and start with blank edges.  (All we’ll be doing from now on is adding more red dots, and the order in which we place multiple red dots on the same edge isn’t significant, so the first five red dots have no bearing on our upcoming choices.)

So how many different ways are there of positioning our two remaining dots on blank edges of a pentagon?  We’re looking for arrangements that are distinct in the sense that they can’t be rotated to match each other (in musical terms, this is the same as requiring that two scales can’t be transposed so their notes coincide).  There are only three distinct arrangements:

For each of these arrangements, we can create five modes, depending on which white dot we choose as the root.  An edge with one red dot means there will be two excluded notes between the white endpoints (remember that we dropped one of those excluded notes to simplify the diagram), which means the endpoints will be separated by three semitones or a minor third (m3).  Similarly, an edge with two red dots represents a major third (M3).  An edge with no red dots represents a whole tone (M2).  Walking around the pentagon clockwise, you can extract the interval pattern for the scale as a sequence of whole tones, minor thirds, and major thirds.

Our first template (two red dots together on one edge) represents the interval pattern M2-M2-M3-M2-M2 and generates all of the five-note subsets of the whole-tone scale.  This is fascinating material, but not what comes to everyone’s mind they hear the word “pentatonic.”  Our second template (red dots on adjacent edges) represents the interval pattern m3-M2-M2-M2-m3 and also creates some uncommon pentatonics, although one of these (1, b3, 4, 5, 6) can be seen as a subset of the Dorian mode of the major scale, or as a subset of the melodic minor scale.  The third template (red dots separated by one edge on one side, and two on another) represents the interval pattern M2-M2-m3-M2-m3 and generates the familiar major and minor pentatonics.

The diagram below includes all fifteen possibilities.  The five modes of each of the three template are arranged around the points of three large pentagons.  As you move clockwise around one of the large pentagons, you’ll see that the root note in the small pentagon (the white dot labeled 1) shifts around, while the red dots remains in place.  The remaining vertices in each small pentagon are labeled as scale degrees relative to the given root.

[Note: I've proofread this diagram once, but please let me know if you spot any errors in the scale-degree labeling.]

Here’s a spin on the classic Circle Of Fifths from music theory.  The circle is usually presented with note names and key signatures for the 12 major (and corresponding relative minor) keys in the diatonic system, arranged clockwise in ascending perfect fifths.  What I’ve done here is to create a Venn diagram showing the relationship between any three adjacent keys in the Circle of Fifths, with special attention to the pentatonic subsets of each key.  As a musician, I’ve been working with the Circle of Fifths for a long time, but this diagram shows some aspects of the circle that only just dawned on me–I hope this way of seeing things might be interesting to others too.

All notes are labeled with respect to an arbitrary major key that we’ll treat as the home key or I.  Its notes are 1, 2, 3, 4, 5, 6, 7.  Stepping counterclockwise in the Circle of Fifths would take us to IV and stepping clockwise would take us to V.  To label the notes of those keys with respect to the home key, we also need b7 (flat seven) and #4.  You’ll see that the notes of all three keys appear in the diagram, left to right, in ascending fifths: b7, 4, 1, 5, 2, 6, 3, 7, #4.

Notice that all the notes of the home key appear inside the large blue oval (it contains everything but b7 and #4).  Within that oval, there are three circles representing the three major pentatonic scales based on IV, I and V.  The yellow circle on the left, containing {4, 1, 5, 2, 6}, represents IV-pentatonic.  (To play these notes as a pentatonic scale, you’d order them as 4, 5, 6, 1, 2.)  The red circle on the right, containing {5, 2, 6, 3, 7}, represents V-pentatonic (which you’d order as 5, 6, 7, 2, 3).  The circle wedged between the yellow and blue circles contains {1, 5, 2, 6, 3}: this is I-pentatonic, which you’d play as 1, 2, 3, 5, 6.

That’s a bunch of facts I just hadn’t realized before drawing this diagram: if you take C major as your home key, you can find the notes of F-pentatonic and G-pentatonic within it.  In fact, C major can be seen as the union of F- and G-pentatonic.  (Another way to say this is that if you take the pentatonics based on any two keys a whole tone apart, their union gives you the major key that falls right in between them on the Circle of Fifths.)  Why is that interesting?  Well, for one thing, it shows that if you stick to pentatonic scales, you can modulate between F, C, and G, all while staying within the seven notes of C major.  When I hear “modulation” I usually think either “accidentals” or “change of key signature,” but if we’re modulating between closely related pentatonics we can do it all inside the confines of one seven-note major scale.  How do you modulate between I-pentatonic and V-pentatonic?  Look at the diagram: you’re moving from the middle circle to the red circle on the right.  Just stop playing 1 and start playing 7.  Another way to say this is that shifting the root of a major pentatonic scale down a semitone gives you new pentatonic rooted a fifth above the original root. (In the seven note world, you modulate up a fifth by raising 4 to #4, which becomes the leading tone of the new key.  In the pentatonic world, there are no leading tones!  You modulate up a fifth by lowering the root!)

The full seven-note major keys based on IV and V are represented by the two large gray circles in the diagram.  The gray circle on the left represents IV and contains the yellow circle as its pentatonic subset–its two additional notes are the tritone b7-3.  The large gray circle on the right represents V and contains the red circle as its pentatonic subset–its two additional notes are the tritone 1-#4.  Notice that the small middle circle encloses all the notes in the intersection of the two large gray circles: {1, 5, 2, 6, 3}. What this means is that I-pentatonic consists of the notes in common between the two heptatonic major scales at IV and V.

It’s also interesting to observe that the notes 5, 2, and 6 (which we could interpret as a quartal chord on 6, or a sus2 chord on 5, or a sus4 chord on 2) are the most extensively shared notes in the diagram, being the intersection of I-pentatonic, IV-pentatonic, and V-pentatonic.

Here’s a look at some of the pieces you’ll find in the full diagram.  You can think of this as F major on the left, with its pentatonic in yellow; G major on the right, with its pentatonic in red, and C major in the middle, shown as the union of F-pentatonic and G-pentatonic–all labeled from the perspective of C.

Here’s the main diagram again, with each circle labeled as Heptatonic or Pentatonic: