﻿ ﻿THE PASCAL TRIANGLE AND PINGALA'S PROBLEMS - Counting for Poets - Numbers: Their Tales, Types, and Treasures

## Chapter 5: Counting for Poets

### 5.10.THE PASCAL TRIANGLE AND PINGALA'S PROBLEMS

The Pascal triangle in figure 5.9 contains the full solution to Pingala's third problem. But not only this, the solutions to the Pingala's first and second problem are also hidden in there.

Pingala's second problem states that the number of all paths leading from the top of the decision diagram (figure 5.7) to a point in line n at position k is equal to the number of verse meters with n syllables of which k are short. This number has been called B(n,k), and this is the number at position (n,k) in the Pascal triangle. In order to find the number of all meters with n syllables, we just have to sum up for this n all the numbers B(n,k) with k = 0 up to k = n. This takes into account all meters with a length of n syllables and thus solves Pingala's second problem.

Indeed, we find that the sum of the numbers in each row is just a power of 2 (see figure 5.12). For example, the sum of the numbers in row number 5 (corresponding to five syllables) is

1 + 5 + 10 + 10 + 5 + 1 = 32 = 25,

and this is precisely the result that was obtained previously by a different reasoning. Figure 5.12: Sum across lines in Meru Prastara.

Pingala's first problem asks that we determine the total number of meters with a given duration. Here duration is measured in moras, where a short syllable has one mora and a long syllable has two moras.

The solution to this problem is also hidden in the Pascal triangle. We only have to remember that all paths ending at a certain point (n,k) have precisely k short syllables and nk long syllables, hence they have the same duration. But there are other endpoints describing same duration. Consider figure 5.13, which shows some region of the decision tree. If we start at point “a,” anywhere in the decision tree, it takes a long syllable to go to “b” and two short syllables to go to “c”—two moras in both cases. Hence the points “b” and “c” correspond to verse meters with the same duration (two moras longer than “a”). By the same reasoning, we find that all the points along a “shallow diagonal” (the dashed line in figure 5.13) correspond to verse meters of the same duration. Figure 5.13: Points “b” and “c” correspond to verse meters with the same duration.

So if we want to know the total number of verse meters of a given duration, we have to sum up all entries in the Pascal triangle that sit along a shallow diagonal. This should give the solution to Pingala's first problem. Indeed, we find the Fibonacci numbers A(n) as the sums along the shallow diagonals in the Pascal triangle, as illustrated in figure 5.14. Figure 5.14: The sums along “shallow diagonals” are Fibonacci numbers.

Many other discoveries can be made within the Pascal triangle. You can see that the first diagonal, that is given by all numbers with k = 1, are just the natural numbers 1, 2, 3, 4…,

B(n,1) = n.

The second diagonal is also a sequence familiar from the previous chapter; it is the sequence of triangular numbers 1, 3, 6, 10, 15, 21…, characterized by the property

B(n + 1,2) = B(n,2) + B(n,1) = B(n,2) + n.

Moreover, the third diagonal sequence, the numbers with k = 3, are just the tetrahedral numbers 1, 4, 10, 20, and 35.

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