FizzBuzz in APL

Print all the numbers from 1 to 100, but if the number is divisible by 3, print “Fizz”, if it’s divisible by 5, print “Buzz”, and if it’s divisible by both, print “FizzBuzz”

This famous problem is known as “FizzBuzz challenge”, and is a famous one in many interview questions. While it can be written easily with a few if-else, the real challenge is to write it using as few characters as possible. And what else can be a better contender for one-liners than APL?

So, let me first describe the algorithm. Forget about the all the numbers from 1 to 100 part. Suppose you have only one number, say 50. How do you proceed?

We do it in this way —

First we divide the number by 15, 3 and 5 and take the residue

      15 3 5 | 50
5 2 0

The | (residue) operator divides 50 by 15, 3 and 5 and gives the remainders as an array.

Now we find out which of these remainders is 0. So we compare the array with 0

      0=15 3 5|50
0 0 1

This comparison return an array in which a 0 corresponds to a non-zero remainder and 1 corresponds to a 0 remainder.

Now we just find out the index of 1, so that we know which of the remainders is 0. We use the dyadic form of ⍳

      (0=15 3 5|50)⍳1
3

This returns the position of 1(arrays are 1 indexed)

If 1 does not occur in the array (i.e. none of the remainders are 0), it returns 4 (length of array + 1)

If 1 occurs more than once, it returns the first position

Now we create an array of what we want to print

      'FizzBuzz' 'Fizz' 'Buzz', 50
 FizzBuzz Fizz Buzz 50

This creates an array of 4 elements.

Now we just pass the index calculated in the previous step in this array

      ('FizzBuzz' 'Fizz' 'Buzz', 50)[(0=15 3 5|50)⍳1]
 Buzz

So, if our number is divisible by 15, the index will be 1 and we will get ‘FizzBuzz’, if it’s divisible by 3, the index will be 2 and we will get ‘Buzz’ and so on. And if it’s not divisible by any one, the index will be 4, and we will get back the original number.

Now we just make it a function and call it on each number from 1 to 100. For that, we use ⍳ to generate all numbers from 1 to 100 and use ¨ to call it on each of the numbers

      {(‘FizzBuzz’ ‘Fizz’ ‘Buzz’,⍵)[(0=15 3 5|⍵)⍳1]}¨⍳100
 1 2  Fizz  4  Buzz   Fizz  7 8  Fizz   Buzz  11  Fizz  13 14  FizzBuzz  16 17 
       Fizz  19  Buzz   Fizz  22 23  Fizz   Buzz  26  Fizz  28 29  FizzBuzz  31 
      32  Fizz  34  Buzz   Fizz  37 38  Fizz   Buzz  41  Fizz  43 44  FizzBuzz  
      46 47  Fizz  49  Buzz   Fizz  52 53  Fizz   Buzz  56  Fizz  58 59 
       FizzBuzz  61 62  Fizz  64  Buzz   Fizz  67 68  Fizz   Buzz  71  Fizz  73 
      74  FizzBuzz  76 77  Fizz  79  Buzz   Fizz  82 83  Fizz   Buzz  86  Fizz  
      88 89  FizzBuzz  91 92  Fizz  94  Buzz   Fizz  97 98  Fizz   Buzz

This uses the anonymous function feature of Dyalog APL.

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