pe-23-2.clj

;;; Project Euler Problem 23 solution
;;; http://projecteuler.net/problem=23

(use 'clojure.test)

(def limit 28123)

(defn gen-sigma1-list
  "original implementation by @ypsilon-takai: https://gist.github.com/4284814"
  [^long size]
  (loop [i 2
         a (long-array size)]
    (if (< (* 2 i) size)
      (recur (inc i)
             (loop [j (* 2 i)
                    a a]
               (if (< j size)
                 (recur (+ j i)
                        (do (aset a j (+ (aget a j) i))
                          a))
                 a)))
      a)))

(defn gen-divisors [n] (filter #(zero? (mod n %)) (range 1 n)))
(defn abundant? [n] (< n (apply + (gen-divisors n))))

(defn gen-abundants [size]
  (let [sigma1-list (gen-sigma1-list size)]
    (filter #(< % (aget sigma1-list %))
            (range 1 size))))

(is (not-any? abundant? (range 1 12)))
(is (abundant? 12))

(defn gen-sums
  "candidates-ascending 中の2要素のすべての組み合わせについて和を計算する(thanks to @kohyama)"
  [candidates-ascending]
  (for [x candidates-ascending
        y candidates-ascending
        :while (<= y x)
        :let [sum (+ x y)]
        :while (<= sum limit)
        ]
    sum))

;; 補集合X-Aの要素の総和は、全体集合Xの要素の総和から部分集合Aの要素の総和を引いたものに等しい
;; thanks to @tnoda
(defn solve []
  (let [abundants (filter abundant? (range 12 (inc limit)))
        abundant-sums (set (gen-sums abundants))]
    (- (apply + (range (inc limit)))
       (apply + abundant-sums))))

(defn solve2 []
  (let [abundants (gen-abundants (inc limit))
        abundant-sums (set (gen-sums abundants))]
    (- (apply + (range (inc limit)))
       (apply + abundant-sums))))

;; (is (= (solve) 4179871))

;pe-23.core=> (time (solve))
;"Elapsed time: 25569.034676 msecs"
;4179871
;pe-23.core=> (time (solve2))
;"Elapsed time: 1789.932635 msecs"
;4179871

pe-23.clj

;;; Project Euler Problem 23 solution
;;; http://projecteuler.net/problem=23

(use 'clojure.test)

(def limit 28123)

(defn gen-divisors [n] (filter #(zero? (mod n %)) (range 1 n)))
(defn abundant? [n] (< n (apply + (gen-divisors n))))

(is (not-any? abundant? (range 1 12)))
(is (abundant? 12))

(defn gen-sums
  "candidates-ascending 中の2要素のすべての組み合わせについて和を計算する"
  [candidates-ascending]
  (if (empty? candidates-ascending)
    nil
    (let [ [lhs & larger-candidates] candidates-ascending ]
      (concat
        (for [rhs candidates-ascending
              :let [sum (+ lhs rhs)]
              :while (<= sum limit)
              ]
          sum)
        (lazy-seq (gen-sums larger-candidates))))))

(defn solve []
  (let [abundants (filter abundant? (range 12 (inc limit)))
        abundant-sums (set (gen-sums abundants))]
    (apply + (remove #(contains? abundant-sums %)
                   (range 1 (inc limit))))))

;; (is (= (solve) 4179871))

@tnodaさん

ただ，今回に限っては，

(- (apply + (range (inc limit))
   (apply + abundant-sums))

のほうが速かったかもしれません．

なるほど！補集合を求めてから総和を計算するのと、総和をそれぞれ計算して差を取るの、まったく等価ですね。
この性質は見落としてました。コードもシンプルになりますし、素晴らしいとおもいます。反映します。