﻿ ﻿SOLVING PINGALA'S FIRST COUNTING PROBLEM IN SPECIAL CASES - Counting for Poets - Numbers: Their Tales, Types, and Treasures

## Chapter 5: Counting for Poets

### 5.4.SOLVING PINGALA'S FIRST COUNTING PROBLEM IN SPECIAL CASES

A first step in a mathematical investigation is often to give a name to the object of interest. We are interested in the number of meters with a total duration of n moras, where n is an arbitrary natural number.

· The number of meters with a duration of n moras will be called A(n).

For a mathematician, n would indeed be an arbitrary natural number. For the sake of completeness, a mathematician would also consider meters of verses with a duration n = 1 or 2 moras, which are absolutely irrelevant for poetry. But this has the advantage that the answer is very easy to find: Obviously, A(1) = 1, because the only meter with a duration of one mora is the meter that consists of exactly one short syllable. For a meter with a duration of two moras, there are already two possibilities because it could consist of two short syllables or one long syllable. Thus, we findA(2) = 2. Let us collect our results:

A(1) = 1, A(2) = 2.

Now, one could proceed in a systematic way and try to determine all possible meters of total duration n = 3, 4, 5, and so on. For example, if n = 6 (which is still too short for meaningful poetry), one would find the following list of thirteen different meters:

 1: ¯ ¯ ¯ 6: ˘ ¯ ¯ ˘ 2: ˘ ˘ ¯ ¯ 7: ¯ ˘ ¯ ˘ 3: ˘ ¯ ˘ ¯ 8: ¯ ¯ ˘ ˘ 4: ¯ ˘ ˘ ¯ 9: ˘ ˘ ˘ ¯ ˘ 5: ˘ ˘ ˘ ˘ ¯ 10: ˘ ˘ ¯ ˘ ˘ 11: ˘ ¯ ˘ ˘ ˘ 12: ¯ ˘ ˘ ˘ ˘ 13: ˘ ˘ ˘ ˘ ˘ ˘

Table 5.1: A list of verse meters with a duration of six moras.

But one soon has the impression that this procedure is not really helpful. For larger n, the number of possibilities gets very large and unmanageable. How can one be sure not to omit one of the possible patterns? It is probably wiser to look for another way of determining A(n) for arbitrary n. As we will see, mathematicians do not always choose the direct approach. Sometimes, they attack a problem by working backward. In the next section, we give the general solution that results from this strategy.

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