Breaking down fizzbuzz in (Dyalog) APL

The one-liner solution: {∊(3↑(0=3 5|⍵)∪1)/'Fizz' 'Buzz' ⍵}¨⍳20

The explanation:

For each value in [1, 2, ..., 20], find the boolean mask of “is this divisible by 3 or 5?” (vectorized).

{0=3 5|⍵}¨⍳20

┌───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┐
│0 0│0 0│1 0│0 0│0 1│1 0│0 0│0 0│1 0│0 1│0 0│1 0│0 0│0 0│1 1│0 0│0 0│1 0│0 0│0 1│
└───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┘

Union this with the value 1; if 1 doesn’t already appear, add it. This is now a mask for whether the result should be “fizz” (1 0), “buzz” (0 1), “fizzbuzz” (1 1), or the number itself (0 0 1).

{(0=3 5|⍵)∪1}¨⍳20

┌─────┬─────┬───┬─────┬───┬───┬─────┬─────┬───┬───┬─────┬───┬─────┬─────┬───┬─────┬─────┬───┬─────┬───┐
│0 0 1│0 0 1│1 0│0 0 1│0 1│1 0│0 0 1│0 0 1│1 0│0 1│0 0 1│1 0│0 0 1│0 0 1│1 1│0 0 1│0 0 1│1 0│0 0 1│0 1│
└─────┴─────┴───┴─────┴───┴───┴─────┴─────┴───┴───┴─────┴───┴─────┴─────┴───┴─────┴─────┴───┴─────┴───┘

Take 3 elements; if there are only 2, extend it to 3 (adding a 0). The combinations are then “fizz” (1 0 0), “buzz” (0 1 0), “fizzbuzz” (1 1 0), or the number itself (0 0 1).

{3↑(0=3 5|⍵)∪1}¨⍳20

┌─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┐
│0 0 1│0 0 1│1 0 0│0 0 1│0 1 0│1 0 0│0 0 1│0 0 1│1 0 0│0 1 0│0 0 1│1 0 0│0 0 1│0 0 1│1 1 0│0 0 1│0 0 1│1 0 0│0 0 1│0 1 0│
└─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┘

Filter the vector 'Fizz' 'Buzz' ⍵ based on this boolean mask.

{(3↑(0=3 5|⍵)∪1)/'Fizz' 'Buzz'⍵}¨⍳20

┌─┬─┬──────┬─┬──────┬──────┬─┬─┬──────┬──────┬──┬──────┬──┬──┬───────────┬──┬──┬──────┬──┬──────┐
│1│2│┌────┐│4│┌────┐│┌────┐│7│8│┌────┐│┌────┐│11│┌────┐│13│14│┌────┬────┐│16│17│┌────┐│19│┌────┐│
│ │ ││Fizz││ ││Buzz│││Fizz││ │ ││Fizz│││Buzz││  ││Fizz││  │  ││Fizz│Buzz││  │  ││Fizz││  ││Buzz││
│ │ │└────┘│ │└────┘│└────┘│ │ │└────┘│└────┘│  │└────┘│  │  │└────┴────┘│  │  │└────┘│  │└────┘│
└─┴─┴──────┴─┴──────┴──────┴─┴─┴──────┴──────┴──┴──────┴──┴──┴───────────┴──┴──┴──────┴──┴──────┘

and finally, enlist (flatten) the vector.

{∊(3↑(0=3 5|⍵)∪1)/'Fizz' 'Buzz'⍵}¨⍳20

┌─┬─┬────┬─┬────┬────┬─┬─┬────┬────┬──┬────┬──┬──┬────────┬──┬──┬────┬──┬────┐
│1│2│Fizz│4│Buzz│Fizz│7│8│Fizz│Buzz│11│Fizz│13│14│FizzBuzz│16│17│Fizz│19│Buzz│
└─┴─┴────┴─┴────┴────┴─┴─┴────┴────┴──┴────┴──┴──┴────────┴──┴──┴────┴──┴────┘ 

So concise.