loading...

re: Challenge: find 'Kaprekar numbers' VIEW POST

FULL DISCUSSION
 

Time to get my repl up and running. Not very clean, but gets the job done. Here's some Clojure:

(ns kaprekar)

(defn is-kap? [n]
  (let [square (* n n)
        ss (str square)
        length (count ss)
        ss (if (odd? length) (str "0" ss) ss)
        length (count ss)
        half-length (/ length 2)
        first-half (bigint (subs ss 0 half-length))
        second-half (bigint (subs ss half-length))]
    (= n (+ first-half second-half))))

(defn find-kaps []
  (loop [i 1 found 0]
    (if (is-kap? i)
      (do
        (println i)
        (when (< found 7)
          (recur (inc i) (inc found))))
      (recur (inc i) found))))

Thanks to Richard Orelup' tip to zero-pad the number if its length is odd.

 

Taking inspiration from Thomas Much's answer, I refactored my code to be more functional-esque and idiomatic:

(ns kaprekar)

(defn is-kap? [n]
  (let [square (* n n)
        ss (str square)
        length (count ss)
        ss (if (odd? length) (str "0" ss) ss)
        length (count ss)
        half-length (/ length 2)
        first-half (bigint (subs ss 0 half-length))
        second-half (bigint (subs ss half-length))]
    (= n (+ first-half second-half))))

(defn find-kaps []
  (->>
    (iterate inc 1)
    (filter is-kap?)
    (take 8)
    (run! println)))
Code of Conduct Report abuse