Exercise 1.12

The following pattern of numbers is called Pascal's triangle.

        1
      1   1
    1   2   1
  1   3   3   1
1   4   6   4   1
      . . .

The numbers at the edge of the triangle are all 1, and each number inside the triangle is the sum of the two numbers above it. Write a procedure that computes elements of Pascal's triangle by means of a recursive process.

Write a procedure that computes elements of Pascal's triangle by means of a recursive process.

This sentence can interpreted to mean that we have to write a procedure to compute the entire triangle. That is not so. The task at hand is to compute a single element of the triangle, given the row and column of the element.

We are given 2 conditions:

1. The numbers at the edge of the triangle are all 1
2. each number inside the triangle is the sum of the two numbers above it.

Let us further examine the second case. Which are the two numbers above an element? The row is obvious, but what of the columns? Left-justifying the triangle, this becomes clearer:

1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
.   .   .   .   .   .

The two numbers are the elements in the previous column and that in the same column in the row above.

This lends itself to the two cases within our procedure p:

1. (p n 0) or (p n n) is \(1\)
2. (p row col) is (+ (p (- row 1) (- col 1)) (p (- row 1) col))

(define (p row col)
  (if (or (= col 0) (= row col))
      1
      (+ (p (- row 1) (- col 1)) (p (- row 1) col))))

Trying it out:

(p 0 0)

1

(p 3 2)

3

(p 4 2)

6