41 lines
2.1 KiB
APL
41 lines
2.1 KiB
APL
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∇ sv ← fizzbuzz n; t;d
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[1] ⍝⍝ Solve the popular 'fizzbuzz' problem in APL.
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[2] ⍝⍝ \param n - highest number to compute (≥0)
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[3] ⍝⍝ \returns sv - a vector of strings representing the fizzbuzz solution for ⍳n
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[4] ⍝⍝ (note we return a string vector to avoid a mixed-type result; remove the
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[5] ⍝⍝ ⍕ function from the (⍕t[⍵]) term to see the difference).
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[6] ⍝⍝⍝⍝
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[7] t←⍳n ⍝ the sequence 1..n itself which we'll pick from
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[8] ⍝ ... or the words 'fizz', 'buzz', 'fizzbuzz' depending on
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[9] ⍝ ... divisibility by 3 and/or 5
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[10] ⍝⎕←t ⍝ (Uncomment to see during call)
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[11]
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[12] d←1+(+⌿ ⊃ {((0=3|⍵)) (2×(0=5|⍵))} ⍳n)
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[13] ⍝ || || | | | ↓↓
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[14] ⍝ || || | | | ⍳n: generate range (1..n)
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[15] ⍝ || || | ↓.....................↓ ↓↓
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[16] ⍝ || || | A dfn (lambda) taking its right arg (⍵, ⍳n here) to compute two boolean
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[17] ⍝ || || | vectors(v12): divisibility by 3 and 5, respectively, for each of ⍳n
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[18] ⍝ || || ↓
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[19] ⍝ || || ⊃: Disclose ('lift-up' and pad w/zeros) the 'ragged' matrix of vectors (v12)
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[20] ⍝ || || holding divisibility by 3 and 5 of each ⍳n
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[21] ⍝ || ↓↓
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[22] ⍝ || +⌿: Sum (v12) row-wise to count divisibility (0=neither 3 nor 5, 1=3, 2=3 and 5)
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[23] ⍝ ↓↓
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[24] ⍝ 1+: Add one to (v12) to make them 1-based for indexing below:
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[25] ⍝⎕←d
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[26]
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[27] sv ← { ((⍕t[⍵]) 'Fizz' 'Buzz' 'FizzBuzz') [d[⍵]]}¨ ⍳n
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[28] ⍝ | | | | | |
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[29] ⍝ | | | ↓....↓ |
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[30] ⍝ | |................................↓ idx |
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[31] ⍝ | ( lookup output vector ) |
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[32] ⍝ ↓...........................................↓
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[33] ⍝ A dfn (lambda) taking as its right arg (⍵) ⍳n and using the 'each' (¨)
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[34] ⍝ operator to apply the lambda to each (idx) of ⍳n.
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[35]
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[36] ⍝⍝ USAGE
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[37] ⍝⍝ ⎕ ← ,fizzbuzz 15
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[38] ⍝ 1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz
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∇
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