为什么对于 Euler 50 的等效解，Clojure 比 Python 慢 10 倍？

Question

我最近开始学习 Clojure，并决定练习 Euler 问题，以掌握可用的数据结构并练习递归和循环。

我尝试了各种方法来解决问题 50，但无论我做什么，都无法找到 1000000 的解决方案。在我查看了其他人的做法后，我猜想我正在做的事情也不应该永远持续下去，所以我在 Python 中输入了等效的算法，看看问题是否在于我对某些 Clojure 事物或 Java 设置缺乏理解。 Python 10 秒完成。对于 100000 以下的素数，Python 版本在 0.5 秒内完成，Clojure 在 5 秒内完成。

我发布的是专门为匹配 Python 代码而创建的 Clojure 版本。您能帮我理解为什么性能会有如此大的差异吗？我应该使用未经检查的添加、类型提示、原语（但在哪里？）还是什么？

这是 Clojure：

(defn prime? [n]
  (let [r (int (Math/sqrt n))]
    (loop [d 2]
      (cond
        (= n 1) false
        (> d r) true
        (zero? (rem n d)) false
        :other (recur (inc d))))))

(defn primes []
  (filter prime? (iterate inc 2)))


(defn cumulative-sum [s]
  (reduce 
    (fn [v, x] (conj v (+ (last v) x))) 
    [(first s)] 
    (rest s)))


(defn longest-seq-under [n]
  "Longest prime seq with sum under n"
  (let [ps (vec (take-while #(< % n) (primes))) ; prime numbers up to n
        prime-set (set ps)  ; set for testing of inclusion
        cs (cumulative-sum ps)
        cnt (count ps)
        max-len (count (take-while #(< % n) cs)) ; cannot have longer sequences
        sub-sum (fn [i j] ; sum of primes between the i-th and j-th      
                  (- (cs j) (get cs (dec i) 0)))
        seq-with-len (fn [m] ; try m length prime sequences and return the first where the sum is prime
                       (loop [i 0] ; try with the lowest sum
                         (if (> i (- cnt m)) ; there are no more elements for and m length sequence
                           nil ; could not find any
                           (let [j (+ i (dec m)) ; fix length
                                 s (sub-sum i j)]
                             (if (>= s n) ; overshoot
                               nil
                               (if (prime-set s) ; sum is prime
                                 [i (inc j)] ; we just looked for the first
                                 (recur (inc i))))))))] ; shift window
        (loop [m max-len] ; try with the longest sequence
          (if (not (zero? m))
            (let [[i j] (seq-with-len m) ]
              (if j 
                (subvec ps i j)
                (recur (dec m))))))))                    



(assert (= [2 3 5 7 11 13] (longest-seq-under 100)))

(let [s1000  (longest-seq-under 1000)]
  (assert (= 21 (count s1000)))
  (assert (= 953 (reduce + s1000))))

; (time (reduce + (longest-seq-under 100000))) ; "Elapsed time: 5707.784369 msecs"

Python 也是如此：

from math import sqrt
from itertools import takewhile

def is_prime(n) :
    for i in xrange(2, int(sqrt(n))+1) :
        if n % i == 0 :
            return False
    return True

def next_prime(n):
    while not is_prime(n) :
        n += 1
    return n

def primes() :
    i = 1
    while True :
        i = next_prime(i+1)
        yield i

def cumulative_sum(s):
    cs = []
    css = 0
    for si in s :
        css += si
        cs.append( css )
    return cs


def longest_seq_under(n) :
    ps = list(takewhile( lambda p : p < n, primes()))
    pss = set(ps)
    cs = cumulative_sum(ps)
    cnt = len(ps)
    max_len = len(list(takewhile(lambda s : s < n, cs)))

    def subsum(i, j):
        return cs[j] - (cs[i-1] if i > 0 else 0)

    def interval_with_length(m) :
        for i in xrange(0, cnt-m+1) :
            j = i + m - 1            
            sij = subsum(i,j)
            if sij >= n :
                return None, None
            if sij in pss : # prime
                return i, j+1
        return None, None

    for m in xrange(max_len, 0, -1) :
        f, t = interval_with_length(m)
        if t :
            return ps[f:t]


assert longest_seq_under(100) == [2, 3, 5, 7, 11, 13]
assert sum(longest_seq_under(1000)) == 953

# import timeit
# timeit.Timer("sum(longest_seq_under(100000))", "from __main__ import longest_seq_under").timeit(1) # 0.51235757617223499

谢谢

Answer 1

我认为速度减慢来自于迭代 longest-seq-under 中的序列的次数；每一次迭代都会造成损失。这是一个快速版本，基于您的代码和此处发布的答案的组合。请注意，primes 是惰性的，因此我们可以使用 def 与 defn 绑定它：

(defn prime? [n]
  (let [r (int (Math/sqrt n))]
    (loop [d 2]
      (cond (= n 1) false
            (> d r) true
            (zero? (rem n d)) false
            :else (recur (inc d))))))

(def primes (filter prime? (iterate inc 2)))

(defn make-seq-accumulator
  [[x & xs]]
  (map first (iterate
              (fn [[sum [s & more]]]
                [(+ sum s) more])
              [x xs])))

(def prime-sums
  (conj (make-seq-accumulator primes) 0))

(defn euler-50 [goal]
  (loop [c 1]
    (let [bots (reverse (take c prime-sums))
          tops (take c (reverse (take-while #(> goal (- % (last bots)))
                                            (rest prime-sums))))]
      (or (some #(when (prime? %) %)
                (map - tops bots))
          (recur (inc c))))))

在我的机器上，这大约需要 6 毫秒：

user> (time (euler-50 1000000))
"Elapsed time: 6.29 msecs"
997651

Answer 2

我将接受我自己的评论作为为什么 Python 有效而 Clojure 无效的问题的答案：使用向量的 last 是一种线性运算，它会阻止累积总和按照我的预期方式计算。

更新函数以使用如下所示的瞬态向量：

(defn cumulative-sum-2 [s]
  (loop [[x & xs] s
         ss 0
         acc (transient [])]
    (if x      
      (let [ssx (+ ss x)]
        (recur xs ssx (conj! acc ssx)))
      (persistent! acc))))

导致 Clojure 版本的运行时间始终只有 Python 的两倍。我有点希望 Clojure 对于相同的操作能比 Python 更快，不知道我是否仍然错过了一些东西。顺便说一句，我正在使用1.2。

谢谢