Asp Forum - Re: [QUIZ] FizzBuzz (#126

darren kirby

6/3/2007 4:44:00 PM

So: much like Marcel Proust biting into a madelaine, when I read this quiz on
Friday I had a rush of old memories. I have never heard of fizzbuzz being
used as a recruitment technique before, however, I am familiar with it
because back in high school we played it as a drinking game. If we were
feeling particularly hardcore we would also add in 'Bang' for multiples of 7.
I had as much as forgotten about fizzbuzz as I haven't though about it in
over twelve years or so...

Anyway, I have deviated from the official spec in a few ways here. First of
all, I saw no reason to stop at 100, so my solution will print results for
any user-specified 'n'. Rather than golfing, which I am no good at, I decided
to go with an obvious solution.

The second deviation is that because the solution is fairly short (ie: about a
paragraph) I thought it would be interesting to write solutions in a few
other languages as a point of comparison. To that end I also wrote solutions
in Python, Lua, Scheme, and plain old C. Hopefully you can forgive me some
off-topicness for posting these as well.

On with the code:

# fizz.rb
1.upto(ARGV[0].to_i) do |n|
if n % 15 == 0
print "FizzBuzz "
elsif n % 5 == 0
print "Buzz "
elsif n % 3 == 0
print "Fizz "
else print "#{n} "
end
end
puts

# fizz.py
import sys
for n in range(1,int(sys.argv[1])+1):
if n % 15 == 0:
sys.stdout.write("FizzBuzz ")
elif n % 3 == 0:
sys.stdout.write("Fizz ")
elif n % 5 == 0:
sys.stdout.write("Buzz ")
else: sys.stdout.write("%i " % n)
print

-- fizz.lua
for n=1,arg[1] do
if n % 15 == 0 then
io.write("FizzBuzz ")
elseif n % 3 == 0 then
io.write("Fizz ")
elseif n % 5 == 0 then
io.write("Buzz ")
else
io.write(n .. " ")
end
end
print()

/* fizz.c */
#include <stdio.h>

int main(int argc, char *argv[]) {
int i;
for (i = 1; i <= atoi(argv[1]); i++) {
if (i % 15 == 0)
printf("%s ", "FizzBuzz");
else if (i % 3 == 0)
printf("%s ", "Fizz");
else if (i % 5 == 0)
printf("%s ", "Buzz");
else
printf("%i ", i);
}
printf("\n");
return 0;
}

;;; fizz.scm
(define (fizz x)
(define (mod5? n)
(if (eq? (modulo n 5) 0) #t #f))
(define (mod3? n)
(if (eq? (modulo n 3) 0) #t #f))
(define (fizz-iter n x)
(cond
((> n x) (newline) #t)
((and (mod3? n) (not (mod5? n))) (display "Fizz ")
(fizz-iter (+ n 1) x))
((and (mod5? n) (not (mod3? n))) (display "Buzz ")
(fizz-iter (+ n 1) x))
((and (mod5? n) (mod3? n)) (display "FizzBuzz ")
(fizz-iter (+ n 1) x))
(else (display n) (display " ")
(fizz-iter (+ n 1) x))))
(fizz-iter 1 x))

cheers,
-d
--
darren kirby :: Part of the problem since 1976 :: http://badco...
"...the number of UNIX installations has grown to 10, with more expected..."
- Dennis Ritchie and Ken Thompson, June 1972

3 Answers

Pieter V.

6/3/2007 6:34:00 PM

Two quiz solutions here - one as per my interview situation, and one
for golfing.

On 6/3/07, darren kirby <bulliver@badcomputer.org> wrote:
> So: much like Marcel Proust biting into a madelaine, when I read this quiz on
> Friday I had a rush of old memories. I have never heard of fizzbuzz being
> used as a recruitment technique before, however, I am familiar with it
> because back in high school we played it as a drinking game. If we were
> feeling particularly hardcore we would also add in 'Bang' for multiples of 7.
> I had as much as forgotten about fizzbuzz as I haven't though about it in
> over twelve years or so...
>
> Anyway, I have deviated from the official spec in a few ways here. First of
> all, I saw no reason to stop at 100, so my solution will print results for
> any user-specified 'n'. Rather than golfing, which I am no good at, I decided
> to go with an obvious solution.
>
> The second deviation is that because the solution is fairly short (ie: about a
> paragraph) I thought it would be interesting to write solutions in a few
> other languages as a point of comparison. To that end I also wrote solutions
> in Python, Lua, Scheme, and plain old C. Hopefully you can forgive me some
> off-topicness for posting these as well.
>
> On with the code:
>
> # fizz.rb
> 1.upto(ARGV[0].to_i) do |n|
> if n % 15 == 0
> print "FizzBuzz "
> elsif n % 5 == 0
> print "Buzz "
> elsif n % 3 == 0
> print "Fizz "
> else print "#{n} "
> end
> end
> puts
>
> # fizz.py
> import sys
> for n in range(1,int(sys.argv[1])+1):
> if n % 15 == 0:
> sys.stdout.write("FizzBuzz ")
> elif n % 3 == 0:
> sys.stdout.write("Fizz ")
> elif n % 5 == 0:
> sys.stdout.write("Buzz ")
> else: sys.stdout.write("%i " % n)
> print
>
> -- fizz.lua
> for n=1,arg[1] do
> if n % 15 == 0 then
> io.write("FizzBuzz ")
> elseif n % 3 == 0 then
> io.write("Fizz ")
> elseif n % 5 == 0 then
> io.write("Buzz ")
> else
> io.write(n .. " ")
> end
> end
> print()
>
> /* fizz.c */
> #include <stdio.h>
>
> int main(int argc, char *argv[]) {
> int i;
> for (i = 1; i <= atoi(argv[1]); i++) {
> if (i % 15 == 0)
> printf("%s ", "FizzBuzz");
> else if (i % 3 == 0)
> printf("%s ", "Fizz");
> else if (i % 5 == 0)
> printf("%s ", "Buzz");
> else
> printf("%i ", i);
> }
> printf("\n");
> return 0;
> }
>
> ;;; fizz.scm
> (define (fizz x)
> (define (mod5? n)
> (if (eq? (modulo n 5) 0) #t #f))
> (define (mod3? n)
> (if (eq? (modulo n 3) 0) #t #f))
> (define (fizz-iter n x)
> (cond
> ((> n x) (newline) #t)
> ((and (mod3? n) (not (mod5? n))) (display "Fizz ")
> (fizz-iter (+ n 1) x))
> ((and (mod5? n) (not (mod3? n))) (display "Buzz ")
> (fizz-iter (+ n 1) x))
> ((and (mod5? n) (mod3? n)) (display "FizzBuzz ")
> (fizz-iter (+ n 1) x))
> (else (display n) (display " ")
> (fizz-iter (+ n 1) x))))
> (fizz-iter 1 x))
>
> cheers,
> -d
> --
> darren kirby :: Part of the problem since 1976 :: http://badco...
> "...the number of UNIX installations has grown to 10, with more expected..."
> - Dennis Ritchie and Ken Thompson, June 1972
>
>

Paddy

6/10/2007 5:15:00 AM

On Jun 3, 5:43 pm, darren kirby <bulli...@badcomputer.org> wrote:
>
> On with the code:
>
> # fizz.rb
> 1.upto(ARGV[0].to_i) do |n|
> if n % 15 == 0
> print "FizzBuzz "
> elsif n % 5 == 0
> print "Buzz "
> elsif n % 3 == 0
> print "Fizz "
> else print "#{n} "
> end
> end
> puts
>
> # fizz.py
> import sys
> for n in range(1,int(sys.argv[1])+1):
> if n % 15 == 0:
> sys.stdout.write("FizzBuzz ")
> elif n % 3 == 0:
> sys.stdout.write("Fizz ")
> elif n % 5 == 0:
> sys.stdout.write("Buzz ")
> else: sys.stdout.write("%i " % n)
> print

A slight clean-up of the Python code.
The trailing comma on the print will
add a space as separator.

# fizz.py using print
import sys
for n in range(1, int(sys.argv[1])+1):
if n % 15 == 0:
print "FizzBuzz",
if n % 3 == 0:
print "Fizz",
if n % 5 == 0:
print "Buzz",
else:
print n,
print

- Paddy.

Paddy

6/10/2007 5:17:00 AM

On Jun 10, 6:14 am, Paddy3118 <paddy3...@googlemail.com> wrote:
> On Jun 3, 5:43 pm, darren kirby <bulli...@badcomputer.org> wrote:
>
>
>
>
>
> > On with the code:
>
> > # fizz.rb
> > 1.upto(ARGV[0].to_i) do |n|
> > if n % 15 == 0
> > print "FizzBuzz "
> > elsif n % 5 == 0
> > print "Buzz "
> > elsif n % 3 == 0
> > print "Fizz "
> > else print "#{n} "
> > end
> > end
> > puts
>
> > # fizz.py
> > import sys
> > for n in range(1,int(sys.argv[1])+1):
> > if n % 15 == 0:
> > sys.stdout.write("FizzBuzz ")
> > elif n % 3 == 0:
> > sys.stdout.write("Fizz ")
> > elif n % 5 == 0:
> > sys.stdout.write("Buzz ")
> > else: sys.stdout.write("%i " % n)
> > print
>
> A slight clean-up of thePythoncode.
> The trailing comma on the print will
> add a space as separator.
>
> # fizz.py using print
> import sys
> for n in range(1, int(sys.argv[1])+1):
> if n % 15 == 0:
> print "FizzBuzz",
> if n % 3 == 0:
> print "Fizz",
> if n % 5 == 0:
> print "Buzz",
> else:
> print n,
> print
>
> - Paddy.

I'm a Dork!
I posted my scriblings instead of:

# fizz.py using print
import sys
for n in range(1, int(sys.argv[1])+1):
if n % 15 == 0:
print "FizzBuzz"
elif n % 3 == 0:
print "Fizz"
elif n % 5 == 0:
print "Buzz"
else:
print n
print

comp.lang.ruby

Re: [QUIZ] FizzBuzz (#126

darren kirby

Pieter V.

Paddy

Paddy

x Login to ForumsZone