## Grade School Multiplication

### November 18, 2011

We begin with the function that displays a single multiplication. We find it easiest to plan everything first, calculating the width of the output display in the variable `len`

and the number of lines in the body of the multiplication in the variable `lines`

:

`(define (mult x y)`

(let* ((ys (digits y))

(len (length (digits (* x y))))

(lines (length (filter positive? ys))))

(display-right x len)

(display-right y len)

(display (repeat len #\-)) (newline)

(when (< 1 lines)

(let loop ((ys (reverse ys)) (i 0) (z 1))

(when (pair? ys)

(if (zero? (car ys))

(loop (cdr ys) i (* z 10))

(begin (display-right (* x (car ys) z) (- len i))

(loop (cdr ys) (+ i (ilog10 z) 1) 1)))))

(display (repeat len #\-)) (newline))

(display-right (* x y) len)))

We used several utility functions that aren’t shown here, though all are at http://programmingpraxis.codepad.org/AMItBoYw: `digits`

, which returns a list of the digits in a number, is from the Standard Prelude; `repeat`

returns a string consisting of *n* repetitions of the character *c*; `display-right`

displays a number *n* right-justified in a string of width *len*, and appends a newline; and `ilog10`

returns the number of digits in the input number *n*, minus one.

Function `problem-a`

takes a *filename* and outputs the requested multiplications:

`(define (problem-a filename)`

(with-input-from-file filename

(lambda ()

(let loop ((x (read)) (y (read)) (i 1))

(when (and (positive? x) (positive? y))

(display "Problem ") (display i) (newline)

(mult x y)

(loop (read) (read) (+ i 1)))))))

To solve the problem, say `problem-a multiply.in`

.

Pages: 1 2

My quickly hacked together version in Python. Would love to see more elegant versions, but right now can’t think of one.

minor correction: instead of writing my expand_string() function, I could have used something like:

sigh… correction on my correction: I should leave the string conversion in :)

”’

My python solution

”’

i1=int(raw_input(‘enter first no’))

i2=int(raw_input(‘enter second no’))

print i1,’\n’,i2

print ‘————‘

t=i2

cnt=0

x=0

while t!=0:

x=(t%10)*i1

if cnt==0 and x!=0:

print x

elif x!=0:

print x*(cnt*10)

t=t/10

cnt+=1

print ‘————‘

print i1*i2

A Clojure solution: