"Ternary groupings and subdivisions are *the beginning* of irregularity." - Are there any existing or possible other forms of irregularity? Could you give some examples?

Only binary rhythmic structures are completely regular. A single level will produce a group of 2, either rise/fall (off/on) or fall/rise (on/off). Two binary levels will produce a group of 4, three levels a group of 8, four levels a group of 16, and so on. All other rhythmic groupings are the result of the fusion of one or more levels via the process of removal (ablation). The ablation will always remove one beat from a group of 4 beats. Let's look at the four beats of a true bar. Beat 1 is the main down beat, and cannot be removed. Beat 4 is the rise preceding the down beat, and is fairly firmly attached to it. Beat 3 is the rise of the larger of the two levels. Beat 2 is kind of a "floating" beat; it can be either a rebound off of the down beat, or an anacrusis to beat 3. It is this "floating" beat that will be removed. We are left with a group of 3 consisting of beat 1, still the down beat; beat 2 (the old beat 3), the large level rise; and beat 3 (the old beat 4), the small level rise of the large level rise, which we call the "super-rise." This group of 3 is the simplest irregularity, and is the only one possible to produce from a group of 4. However, if we use 3 levels (a group of 8), we are able to produce groups of 6 (2+2+2 or 3+3), 7 (4+3 or 3+4), or 5 (2+3). The variety is the result of whether the ablation is at the mid-sized level, or the smallest of the 3 levels, or both, and, if at the smallest level, whether the ablation is complete (from both sets of 4) or partial (from only one or the other set of 4). Groups of 16 produce even more possibilities, and so on.

When we move to irregular structures, there may be examples that could lend themselves to possible alternate interpretations. For example, a group of 6 could be 3+3 or it could be 2+2+2. Usually, the harmony and melody taken together will reveal the correct interpretation, but there may be times when discrepancies in one or the other leave some doubt as to which interpretation is best. For example, going back to the "Little Drummer Boy" again, there is a possible alternative to the interpretation given below in answer to the original question. The alternative version has the 0 level at the level that was previously called level +1, and this larger level 0 consists of 3 binary cells (2+2+2), rather than 2 ternary cells (3+3) as in the original level +1.

More questions about ternary level 0: -how do you strike a ternary level 0 cell with the feet? -does "La Mer" have a ternary structure at the 0 level?

Let's take the first part of your question, Levels +1 and +2. Usually, once we have the 0 level, the rest should fall into place, as long as we understand how the levels fit together: at level 0 and greater, the rise (or rises if ternary) of the first rhythmic cell will always be an anacrusis (pick-up) to the rise of the first cell at the next larger level. The fall (beat) of the first cell will always coincide with the attack of the rise (off-beat) of the first cell at the next larger level. In Little Drummer Boy, the existence of ternary cells may make things seem a bit more complex. In this case, the first three binary cells at Level 0 form a large ternary cell at Level +1. We could designate this as 2+2+2 (three binary cells). Note that the fall of the first level 0 cell will coincide with the rise of the first level +1 cell, that the fall of the second level 0 cell will coincide with the super-rise of this same level +1 cell (which is ternary), and that the fall of the third level 0 cell will coincide with the fall of this same ternary level +1 cell. The two remaining Level 0 cells, which are ternary, form a large binary cell at +1 which we could designate 3+3 (two ternary cells). The rise of the second +1 cell coincides with the fall of the fourth level 0 cell (which is ternary). The fall of the second +1 cell of course coincides with the fall of the fifth level 0 cell (which again is ternary). Notice that the two Level +1 cells are of the same size (2+2+2 = 3+3). These two Level +1 cells are grouped into a single large cell at Level +2, which is binary. The rise of this large +2 cell coincides with the fall of the first +1 cell; the fall of the +2 cell coincides with the fall of the second +1 cell. So, to refine the diagram given above to include levels +1 and +2, we could use the following code:
\ = level 0 off-beat (rise)
\\ = level 0 off-off-beat (super-rise)
/ = level 0 beat (fall), level +1 off-beat (rise)
/ \ = level 0 beat (fall), level +1 off-off-beat (super-rise)
// = level 0 beat (fall), level +1 beat (fall), level +2 off-beat (rise)
/// = level 0 beat, level +1 beat, level +2 beat (falls)

Using the words of Little Drummer Boy, we would end up with this result (each line represents a level 0 cell):
\ Come they told me pa / rum pum pum pum
\ A new born King to see pa / \ rum pum pum pum
\ Our finest gifts we bring pa // rum pum pum pum
\ To lay before the King pa \\ rum pum pum pum, rum pum pum pum / rum pum pum pum
\ So to honor him pa \\ rum pum pum pum /// when we come

Note that it is only at the final fall where the beats of ALL LEVELS coincide.

Now, as far as the feet go, we suggest that you start by visiting the section called More Footsies which will describe how to beat two levels with the feet by using ankle touches. Then we suggest that you become thoroughly familiar with the section on ternary subdivision including how to beat ternary rhythms with the feet. Then you will be ready to apply what you have learned to the above diagram of Little Drummer Boy. At Level 0 you would have a left foot (L) on every " \ " and a right foot (R) on anything containing a " / " ( / , / \ , // , /// ). For the ternary cells at Level 0, you would, in addition, touch your right foot to your left ankle (T) at \\ . If you wanted to beat level +1 with the feet (which would be quite slow), you would have L at / , T at / \ , R at // and ///. Level +2 would be two "giant steps": L at //, R at ///.

As for level -1, it should present no problems. It is completely binary. Simply begin on the right foot with the word "come," step with the left foot on "told," and continue at the same pace. I am afraid you will not get your desired beat on the last "pum" at this level. At level -2 (again starting on R but moving twice as fast as -1) you will get a left foot on "pum." At -3 you will finally get a down beat (right foot) on "pum."

Now, finally, for "La Mer." This tune, also known by the English title "Beyond the Sea," is ternary throughout at level 0. There are four measures on each rise and two measures on each fall. A similar structure is found in the Classical repertoire in the first section of the Hungarian Dance #1 by Brahms (in Gm) which also has 4 bars (of 2/4) on the rise and 2 bars on the fall. The big difference between the two is that "La mer" maintains this level 0 grouping right through to the end without the slightest deviation, and the Brahms (in the second section) abruptly switches to a much smaller level 0 with 1 bar on the rise and 1 bar on the fall. Another example from the classical repertoire that contains ternary cells at level 0 is the Minuet from Mozart's Symphony No. 40 in G Minor. It is actually ternary at both level -1 and, for most of the A section, at level 0, however, in this case, there is a binary cell mixed in with the ternary cells (to be precise, it is a group of 3+3, followed by a group of 3+2+3).

You said: "Notice that the two level +1 cells are of the same size (2+2+2=3+3)." Counting the bars, the first level +1 cell has rise (4 bars), super-rise (4 bars) and fall (6 bars) = 14 bars, the second level +1 cell has rise (6 bars) and fall (4 bars) = 10 bars. Shouldn't the rise, super-rise and fall of a ternary rhythmic cell and the rise and fall of a binary rhythmic cell have the same number of bars or beats? Where is the fault in my thinking?

The simple answer to your question is that we count the pick-ups to a cell as belonging to that cell. What we are calling the first level +1 cell, which happens to be ternary, consists of three binary level 0 cells (2+2+2). Each of these level 0 cells consists of 4 bars (2 on the rise, 2 on the fall), for a total of 12 bars. The rise of the first level 0 cell is an anacrusis, or pick-up, to the level +1 cell. Its duration is "stolen" from the last +1 fall. What we are calling the second level +1 cell, which happens to be binary, consists of two ternary level 0 cells (3+3). Each of these level 0 cells consists of 6 bars (2 on the rise, 2 on the super-rise, 2 on the fall), for a total of 12 bars. The rise and super-rise of the second level 0 cell are anacruses, or pick-ups, to the level +1 cell. Their durations are "stolen" from the previous +1 fall. We are therefore grouping according to where the BREATHS are in the music, which is BETWEEN cells and BEFORE the anacruses.

Your question also raises a more complex issue. Why are the rises, super-rises, and falls of level +1 not all of equal duration? To answer simply, we could just say that it is the presence of ternary cells at the next smallest level that creates this irregularity. If you want a more thorough answer, it will be necessary to delve into the process by which ternary structures are created. Ternary cells result from the "fusion" of two adjacent levels. For example, take two level 0 cells (rise/fall, rise/fall). These two cells belong to a larger +1 cell (rise/fall) with the rise of the first level 0 cell being a pick-up to the +1 cell. If we remove the pick-up, we are left with a group of three (rise\super-rise/fall), and we have fused levels 0 and +1 into a ternary cell which we call level 0+1. It is this transformational process of removal (sometimes called "ablation") that produces all irregular rhythmic structures from the basic and completely regular binary structures. In the previous answers regarding "The Little Drummer Boy," we simplified our terminology and referred to a level by only a single number, even when it contained ternary cells. It was hoped that this would avoid confusing the reader. If this last part about ternary cells and fusion seems confusing, don't worry; hopefully we will be able to deal with this topic in more detail in the future on the site. In the meantime, you may want to look at the CHANGING METERS page, especially in the section called TRANSFORMATIONS AT LEVEL 0.

Is subdivision in 3+2 of the 5/8 bar impossible? and why?

No, it is not impossible because small subdivisions within a bar are generally rebounding, permitting any combination. Only in large level pick-up association (grouping of bars) is the 3+2 grouping not advisable (in Fibonacci structure). If you look at Desmond's Take Five, you will see that there is never a breath after the first quarter of a bar. See Rhythm / Transformations / Graphics / Preface / Quinternary Levels. We hope this is satisfactory for now. If not, do not hesitate to write back.