FizzBuzz
​

​Tom Scott has a great video about the FizzBuzz interview question used by Google and others. 

The programming problem is to print out numbers within a loop, except if the number is divisible by 3 then print "Fizz" and if it's divisible by 5 print "Buzz". And if it's divisible by both, print "FizzBuzz". Sounds easy, but it requires a little thought. ​

Here's our solution in Forth:

: fizz  ( n -- rem , divisible by 3? )
   3 mod
   dup 0= if 
      ." Fizz" 
   then
;

: buzz  ( n -- rem , divisible by 5? )
   5 mod
   dup 0= if 
      ." Buzz" 
   then
;

: fizzbuzz ( n -- , fizzbuzz )
  dup
  for
    cr
    dup i -
    dup fizz
    over buzz
    * if        \ if neither remainders are 0, print number
      . else drop
    then 
  next 
  drop
;

To test it:

20 fizzbuzz
​100 fizzbuzz

How long does it take to run on a Scamp?

ticks 100 fizzbuzz ticks - abs .

gives a result of 15 ms.

​In comparison, Fizzbuzz solutions in python, running on a PC, have been benchmarked between 200 ms and 400 ms. 

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