;; Fizz Buzz ;; My sources ;; http://blog.codinghorror.com/why-cant-programmers-program/ ;; inform me that: ;; The majority of computer science graduates can't solve this problem: ;; Write a program that prints the numbers from 1 to 100. But for ;; multiples of three print "Fizz" instead of the number and for the ;; multiples of five print "Buzz". For numbers which are multiples of ;; both three and five print "FizzBuzz". ;; And Brother Downing of this parish, who actually hires people to ;; program in Java and Clojure learns me that he does indeed use this ;; to screen job applicants, and that most of them can't do it. ;; It is hard to read a thing like that without thinking: 'hang on, is that harder than it looks?' ;; So I did it, just to check: ;; I decided to use pull it out your ass driven development, where ;; you just pull the answer out of your ass. ;; First bit, print out the numbers from 1 to 100 (range 100) ;-> (0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 ...) ;; Bugger (range 1 101) ;-> (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 ...) ;; paranoid now (last (range 1 101)) ;-> 100 ;; Although I guess really (print (range 1 101)) is what I want ;; here. In PIOYADD, we defer this important user interface question for later. ;; Next, print 'Fizz' instead of all the multiples of three (map (= 0 #(% quot 3)) (range 1 101)) ;; ClassCastException java.lang.Boolean cannot be cast to clojure.lang.IFn clojure.core/map/fn--4207 (core.clj:2485) ;; bugger (map #(= 0 (quot % 3)) (range 1 101)) ;-> (true true false false false false false false false false false false false false false false false false false false false false false false false false false ...) ;; ok (map #(if (= 0 (quot % 3)) "Fizz" %) (range 1 101)) ;-> ("Fizz" "Fizz" 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 ...) ;; oh for fuck's sake (map #(if (= 0 (mod % 3)) "Fizz" %) (range 1 101)) ;-> (1 2 "Fizz" 4 5 "Fizz" 7 8 "Fizz" 10 11 "Fizz" 13 14 "Fizz" 16 17 "Fizz" 19 20 "Fizz" 22 23 "Fizz" 25 26 "Fizz" ...) ;; payday (map #(case (= 0 (mod % 3)) "Fizz" (= 0 (mod % 5)) "Buzz" %) (range 1 101)) ;; IllegalArgumentException No matching clause: false user/eval1234/fn--1235 (NO_SOURCE_FILE:1) ;; ok, I always screw that up (map #(cond (= 0 (mod % 3)) "Fizz" (= 0 (mod % 5)) "Buzz" %) (range 1 101)) ;; CompilerException java.lang.IllegalArgumentException: cond requires an even number of forms, compiling:(NO_SOURCE_PATH:1:7) ;; twice usually (map #(cond (= 0 (mod % 3)) "Fizz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (1 2 "Fizz" 4 "Buzz" "Fizz" 7 8 "Fizz" "Buzz" 11 "Fizz" 13 14 "Fizz" 16 17 "Fizz" 19 "Buzz" "Fizz" 22 23 "Fizz" "Buzz" 26 "Fizz" ...) ;; Bwahhahhhahhh! No, Mr Bond, I expect you to die. ;; And now, for the tricky bit of the problem, we unsheathe the ;; superweapon, copy-and-paste-driven development (map #(cond (or (= 0 (mod % 3)) (= 0 (mod % 3))) "FizzBuzz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (1 2 "FizzBuzz" 4 "Buzz" "FizzBuzz" 7 8 "FizzBuzz" "Buzz" 11 "FizzBuzz" 13 14 "FizzBuzz" 16 17 "FizzBuzz" 19 "Buzz" "FizzBuzz" 22 23 "FizzBuzz" "Buzz" 26 "FizzBuzz" ...) ;; hmmm (map #(cond (and (= 0 (mod % 3)) (= 0 (mod % 3))) "FizzBuzz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (1 2 "FizzBuzz" 4 "Buzz" "FizzBuzz" 7 8 "FizzBuzz" "Buzz" 11 "FizzBuzz" 13 14 "FizzBuzz" 16 17 "FizzBuzz" 19 "Buzz" "FizzBuzz" 22 23 "FizzBuzz" "Buzz" 26 "FizzBuzz" ...) ;; still wrong (map #(cond (and (= 0 (mod % 3)) (= 0 (mod % 5))) "FizzBuzz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (1 2 3 4 "Buzz" 6 7 8 9 "Buzz" 11 12 13 14 "FizzBuzz" 16 17 18 19 "Buzz" 21 22 23 24 "Buzz" 26 27 ...) ;; aargh, where did the fizzes go? (map #(cond (and (= 0 (mod % 3)) (= 0 (mod % 5))) "FizzBuzz" (= 0 (mod % 3)) "Fizz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (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" ...) ;; Wahey! That looks done. Three minutes. ;; And now, pretending you're not a filthy hacker driven development: (map #(cond (and (= 0 (mod % 3)) (= 0 (mod % 5))) "FizzBuzz" (= 0 (mod % 3)) "Fizz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (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" ...) (defn divides? [n m] (= 0 (mod n m))) (divides? 3 15) ;-> false ;; sigh (defn divides? [n m] (= 0 (mod m n))) (divides? 3 15) ;-> true (divides? 15 3) ;-> false ;; rah! ;; and so, behold: beauty is truth, and truth, beauty (map #(cond (divides? 15 %) "FizzBuzz" (divides? 3 %) "Fizz" (divides? 5 %) "Buzz" :else %) (range 1 101)) ;-> (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" ...) ;; finally, smugness driven development: (def fizzbuzz (map #(cond (divides? 15 %) "FizzBuzz" (divides? 3 %) "Fizz" (divides? 5 %) "Buzz" :else %) (map inc (range)))) ;; There are those who would call this 'premature abstraction', but they deserve not the names of men. (print (take 100 fizzbuzz)) ;; (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 ...) ;; And I suppose a regression test would be nice, if I'm trying to ;; give some sort of professional impression: (= (take 27 fizzbuzz) (list 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")) ;; And some paranoid checks, if I'm going to put this travesty on my blog: (count (filter #(= "Fizz" %) (take 1000 fizzbuzz))) ; -> 267 (count (filter #(= "FizzBuzz" %) (take 1000 fizzbuzz))) ;-> 66 (count (filter #(= "Buzz" %) (take 1000 fizzbuzz))) ;-> 134 (* (+ 134 66) 5) ;-> 1000 (* (+ 267 66) 3) ;-> 999 ;; bah, that's close enough for government work. I declare myself ;; done. Three minutes of flail and two minutes of tidying up and ;; checking it works. ;; So my question to the wider community is: Does my three minutes of ;; flailing trying to remember the semantics of my favourite language ;; (which I can perfectly imagine looking dreadful at an interview) ;; count as a fail or a pass? ;; In a previously unknown assembly language, but given a library to ;; print numbers on the screen, I can imagine taking half an hour to ;; get this working. Without the library, it's a research project. ;; And obviously, if I tried to do it in Haskell, I'd have to spend ;; three weeks remembering what a monad was in order to print the ;; damned thing out, but at least the type system would magically ;; guarantee the correctness of the final program. ;; In other words, is fizzbuzz really actually quite hard, or is ;; everyone out there a complete idiot?
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Tuesday, May 20, 2014
Fizz Buzz : An Interview Question
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I'm new at Clojure, but I go through this kind of "iterative unfucking up" quite often. I'm glad it's not just me. I just hope it gets better as you get better in the language...
ReplyDeleteLaughed hard, thanks for sharing
ReplyDelete; Tried to solve it using replace function e.g.
ReplyDelete(repalce {3 "Fizz" 5 "Buzz" 15 "FizzBuzz"} (take 50 (drop 1 (range))))
; So need to create the hash first. Hash
(defn hashHelp [n2 x2]
(loop [a1 (interpose x2 (take 50 (iterate #(+ n2 %) n2))) h1 {}] (let [[x y & rest1] a1] (if y (recur rest1 (assoc h1 x y)) h1)))
)
; use the replace function
(replace (merge-with #(str %1 %2) (hashHelp 3 "Fizz") (hashHelp 5 "Buzz")) (take 100 (drop 1 (range))))
You're one of the top Google search results for "fizzbuzz in clojure" (without the quotes). Thanks for posting! Relatable and educational.
ReplyDelete