# APL — solving Fizz Buzz

Note: see the rest of the series so far.

This is a clas­sic inter­view ques­tion, and after some exper­i­men­ta­tion I end­ed up with the fol­low­ing solu­tion in APL:

$F \ \iota 100$

=> 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

The answer looks cor­rect, though this is per­haps not the most straight for­ward solu­tion, so let’s walk through what’s hap­pen­ing here:

• first turn the input array $\omega$ into booleans to deter­mine if they’re divis­i­ble by 3 $0=3 | \omega$ or 5 $0=5 | \omega$

$0 = 3 | \omega$ : 0 0 1 0 0 1 0 0 1 0 0 1 …

$0 = 5 | \omega$ : 0 0 0 0 1 0 0 0 0 1 0 0 …

• assum­ing that $\omega$ has length of 100, then trans­form the boolean arrays from the pre­vi­ous step so that 1s becomes 101s (for those divis­i­ble by 3) and 102s (for those divis­i­ble by 5)

$(1 + \rho \omega) \times 0 = 3 | \omega$ : 0 0 101 0 0 101 0 0 101 0 0 101…

$(2 + \rho \omega) \times 0 = 5 | \omega$ : 0 0 0 0 102 0 0 0 0 102 0 0 0…

• add the two arrays togeth­er, so that those divis­i­ble by 3 has val­ue 101, divis­i­ble by 5 has val­ue 102 and those divis­i­ble by 15 has val­ue 203

0 0 101 0 102 101 0 0 101 102 0 101 0 0 203 0 0 101 …

• apply min­i­mum $\lfloor$ to this array with 103 being the operand, so now num­bers divis­i­ble by 15 has val­ue 103 instead of 203

0 0 101 0 102 101 0 0 101 102 0 101 0 0 103 0 0 101 …

• apply max­i­mum $\lceil$ to this array with the input array $\omega$ to end up with the fol­low­ing

1 2 101 4 102 101 7 8 101 102 11 101 13 14 103 16 17 101

• use this array as index for an array that is made up of the input array $\omega$ with ‘Fizz’ ‘Buzz’ ‘FizzBuzz’ added to the end at index posi­tion 101, 102 and 103 respec­tive­ly