Problem Statement

This task is a variation of the short story by Arthur C. Clarke. (Solvers should be aware of the consequences of completing this task.) In detail, to specify what is meant by a   “name”:

Display the first 25 rows of a number triangle which begins: Where row

n

{\displaystyle n}

corresponds to integer

n

{\displaystyle n}

,   and each column

C

{\displaystyle C}

in row

m

{\displaystyle m}

from left to right corresponds to the number of names beginning with

C

{\displaystyle C}

. A function

G ( n )

{\displaystyle G(n)}

should return the sum of the

n

{\displaystyle n}

-th   row. Demonstrate this function by displaying:

G ( 23 )

{\displaystyle G(23)}

,

G ( 123 )

{\displaystyle G(123)}

,

G ( 1234 )

{\displaystyle G(1234)}

,   and

G ( 12345 )

{\displaystyle G(12345)}

.
Optionally note that the sum of the

n

{\displaystyle n}

-th   row

P ( n )

{\displaystyle P(n)}

is the     integer partition function. Demonstrate this is equivalent to

G ( n )

{\displaystyle G(n)}

by displaying:

P ( 23 )

{\displaystyle P(23)}

,

P ( 123 )

{\displaystyle P(123)}

,

P ( 1234 )

{\displaystyle P(1234)}

,   and

P ( 12345 )

{\displaystyle P(12345)}

.

If your environment is able, plot

P ( n )

{\displaystyle P(n)}

against

n

{\displaystyle n}

for

n

1 … 999

{\displaystyle n=1\ldots 999}

.

Let's start with the solution:

(define (nine-billion-names row column)
   (cond
      ((<= row 0) 0)
      ((<= column 0) 0)
      ((< row column) 0)
      ((= row 1) 1)
      (else
         (let ((addend (nine-billion-names (- row 1) (- column 1)))
               (augend (nine-billion-names (- row column) column)))
	         (+ addend augend)))))
 
(define (print-row row)
   (for-each (lambda (x)
         (display (nine-billion-names row x))
         (display " "))
      (iota row 1)))
 
(define (print-triangle rows)
   (for-each (lambda (x)
         (print-row x)
         (print))
      (iota rows 1)))
 
(print-triangle 25)