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:

USING: combinators io kernel math math.ranges memoize prettyprint
sequences ;

MEMO: p ( m n -- o )
    {
        { [ dup zero? ] [ nip ] }
        { [ 2dup = ] [ 2drop 1 ] }
        { [ 2dup < ] [ 2drop 0 ] }
        [ [ [ 1 - ] bi@ p ] [ [ - ] [ ] bi p + ] 2bi ]
    } cond ;

: row ( n -- seq ) dup [1,b] [ p ] with map ;

: .row ( n -- ) row [ pprint bl ] each nl ;

: .triangle ( n -- ) [1,b] [ .row ] each ;

: G ( n -- sum ) row sum ;

25 .triangle nl
"Sums:" print { 23 123 1234 12345 } [ dup pprint bl G . ] each