THE TIHAI ENDING AND ARITHMETIC SERIES
Compositional structures from Indian classical music

Tihai ending.

The word tihai comes from Indian classical music. It refers to a compostional structure of the form

A zg A zg A

where g is the length of the number of rests, or ``gap'' between the patterns A. A tihai occurs at the end of a composition or solo. For example, in my composition, I used the pattern F z5 F z5 F at the end (so g=5). Since pattern F has duration 13, the total duration of the tihai is 3x13 + 2x5 = 39 + 10 = 49.

Exercise:

A: dzdd
g = 3

Program your favorite drum to play this tihai. Compute the total duration of your tihai: ______.

Determining the gap length.

Indian classical music structure usually conforms to the rule

Every pattern starts and ends on the first beat of the measure.

To make this work for your tihai, the first thing to notice is that the tihai pattern A has to end on a drumhit. So A: dzddd is fine, but A: dzdzz won't work, because it ends on a rest.

Once you have made up a pattern, following the rule that it has to end on a drumhit, you need to determine the gap length g. In my composition, the pattern F had duration 13 and the gap length had duration 5. So the total length of the pattern

F z5 F z5 F

was 49. Notice that 49 = 48 + 1 = 4x12 + 1, so the pattern is equivalent to 4 measures plus one extra beat, which is the first note of the fifth measure. Notice also that 49 = 1 (mod 12)!

In general, the total length of your tihai should be equal to some number of measures plus one more beat. If m is the number of beats per measure, a is the length of pattern A, and g is the gap length, then the total duration of A zg A zg A is 3a + 2g. In order for the tihai to have the required number of beats, the following must be true

3a + 2g = 1 (mod m). This is called the tihai gap formula.

To compute the gap length for my tihai, I had to solve the equation

3x13 + 2g = 1 (mod 12)
39 + 2g = 1 (mod 12)
3 + 2g = 1 (mod 12) by reducing 39 mod 12.

To solve this equation, I can plug in values for g (g=1, 2, 3 ...) until I see that g=5 works, since

3 + 2x5 = 13 = 1 (mod 12).

Your equation may have no solutions, in which case you will have to try a pattern for A with a different duration. The fascinating thing is that Indian drummers have an intuitive sense for how to solve this equation, and are able to improvise tihais! I find the solution easier to visualize using graph paper, which I will demonstrate in class.

Arithmetic series.

An arithmetic series is a structure that appears often in Indian classical music. It may occur anywhere within the composition. There are two basic forms:

A zg AA zg AAA zg AAAA ... (counting up)

or

... AAAA zg AAA zg AA zg A (counting down)

The dots indicate that you can continue the pattern further if needed--for example, A zg AA zg AAA zg AAAA zg AAAAA is also an arithmetic series. The letter g indicates a fixed number that is the length of the gap between the A patterns. For example, in my composition I used

A z3 AA z3 AAA z3 AAAA

so g=3 for this composition.

This structure is called (by me) an arithmetic series because it's similar to a mathematical arithmetic series. A classic example of such a series is

1+2+3+4+5+6+7....

Example.

Counting up. Try the following in abcdrums

A: dzdd
Low Woodblock: A z5 AA z5 AAA z5 AAAA z5 AAAAA

Exercise: Find the total duration of the above compositional structure: _____

Counting down. Now try the opposite pattern:

Low Bongo: AAAAA z5 AAAA z5 AAA z5 AA z5 A

Exercise: Find the total duration of the above compositional structure: _____

Your answer for both durations should be the same. It is common for two instruments to play these patterns simultaneously with one counting up and the other counting down! Try it....

Low Woodblock: A z5 AA z5 AAA z5 AAAA z5 AAAAA
Low Bongo: AAAAA z5 AAAA z5 AAA z5 AA z5 A

The mathematics of arithmetic series. If you line up the patterns above like this

Low Woodblock: A z5 AA z5 AAA z5 AAAA z5 AAAAA
Low Bongo: AAAAA z5 AAAA z5 AAA z5 AA z5 A

you can see that there are five groups of 6 A's (reading both rows), so the total number of A's is 30, and since each instrument plays the same number of A's, each instrument plays 30 / 2 = 15 A's. There is a mathematical formula for the sum of an arithmetic series that uses this same principle

1+2+3+4+5+ ... +n = (n+1) x n / 2.

In our case n is the maximum number of A's, so n = 5, and the total number of A's is 5 x 6 / 2 = 15. The reasoning behind the formula is the same:

1+2+3+4+5
+ 5+4+3+2+1
_________
= 6+6+6+6+6 = 5x6,
so each of the rows equals half of the total, or 5x6 / 2.

Exercise: find the sum of the numbers 1 through 100 using the formula for the sum of an arithmetic series. _______

Determining the gap length.

Indian classical music structure usually conforms to the rule

Every pattern starts and ends on the first beat of the measure.

The first thing to notice is that the pattern A has to end on a drumhit. So A: dzddd is fine, but A: dzdzz won't work, because it ends on a rest.

In my composition, the pattern A had duration 4 and the gap length had duration 3. So the total length of the pattern

A z3 AA z3 AAA z3 AAAA

was 4x10 + 3x3 = 49. Notice that 49 = 48 + 1 = 4x12 + 1, so the pattern is equivalent to 4 measures plus one extra beat, which is the first note of the fifth measure. Notice also that 49 = 1 (mod 12)!

In order to make this work in general for the arithmetic series A zg AA zg AAA zg AAAA, we need to determine the gap length, g. First, write down the number of beats in your time signature ____, which I'll call m. My composition had 12 beats per measure, so m=12. Then write down the duration of your pattern A _______, which I'll call a. The gap g must be chosen so that the total duration of the arithmetic series is some number of measures plus one more beat. That is

10a + 3g = 1 (mod m). This is an arithmetic series gap formula.

For example, if my pattern A has duration a=5 and my time signature is m=7 beats per measure, then I look for solutions to the equation

10x5 + 3g = 1 (mod 7)
50 + 3g = 1 (mod 7)
1 + 3g = 1 (mod 7) because 50 = 1 (mod 7)

Now I look for value of g that make the equation true. There are either going to be multiple solutions or no solution. In this case, g=7 works, but g=14 would also work (in fact, g can be any number divisible by 7). If your equation has no solution, then you will have to change your pattern A so it has a different duration, or use a different arithmetic series (for example A zg AA zg AAA zg AAAA zg AAAAA). Note that if you use a different maximum number of A's, the formula will change--in this case, to

15a + 4g = 1 (mod m)

Exercise. Solve the equation 10a + 3g = 1 (mod m) using your values for a and m.