在不吹栈的情况下生成素数

Question

我正在学习OCaml(原谅我的风格)并且我正在尝试编写一个函数来生成一个素数列表,直到某个上限.我已经设法以几种不同的方式做到这一点,所有这些都有效,直到你将它们扩展到相对较高的上限.

如何更改这些(其中任何一个)以便递归不会填满堆栈？我以为我的while循环版本会实现这一点,但显然不是!

发电机

let primes max =
  let isPrime p x =
    let hasDivisor a = (x mod a = 0) in
    not (List.exists hasDivisor p) in

  let rec generate p test =
    if test < max then
      let nextTest = test + 2 in
      if isPrime p test then generate (test :: p) nextTest
                        else generate p nextTest
    else p in

  generate [5; 3; 2] 7;;

这是我最成功的解决方案,因为它在运行时不会立即溢出堆栈primes 2000000;;.然而它只是在那里消耗CPU; 我只能假设它最终会完成!以下备选方案都存在堆栈溢出问题:

Eratosthenes的递归筛选

let primes max =
  let rec sieve found toTest =
    let h = List.hd toTest
    and t = List.tl toTest in

    let newPrimes = h :: found
    and doesntDivide x = (x mod h <> 0) in

    let nonDivisors = List.filter doesntDivide t in
      if nonDivisors = [] then newPrimes
                          else sieve newPrimes nonDivisors in

  let rec range a b =
    if a > b then []
             else a :: range (a + 1) b in

  let p = range 2 max in

  sieve [] p;;

Eratosthenes v2的递归筛选

let primes max =
  let rec sieve toTest =
    let h = List.hd toTest
    and t = List.tl toTest in
    let doesntDivide x = (x mod h <> 0) in
    let nonDivisors = List.filter doesntDivide t in
      if nonDivisors = [] then [h]
                          else (h :: sieve nonDivisors) in

  let rec range a b =
    if a > b then []
             else a :: range (a + 1) b in

  let p = range 2 max in

  sieve p;;

而Eratosthenes的Loop Sieve

let primes max =
  let rec range a b =
    if a > b then []
             else a :: range (a + 1) b in

  let tail = ref (range 2 max)
  and p    = ref [] in

  while !tail <> [] do
    let h = List.hd !tail
    and t = List.tl !tail in
    let doesntDivide x = (x mod h <> 0) in
    let newTail = ref (List.filter doesntDivide t) in

    tail := !newTail;
    p := h :: !p
  done;

  !p;;

Answer 1

发生堆栈溢出是因为您的范围函数不是尾递归.一个有效的是,例如

  let rec range store a b =
    if a > b then store
    else range (a :: store) (a + 1) b
  in

  let p = List.rev (range [] 2 max) in

使用该定义和格式行,给出

$ ocamlopt -o primes2 primes2.ml
$ ./primes2
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
...

既然你正在学习,我也会给你一些不请自来的风格评论:)

• 不要使用hd和tl.喜欢模式匹配.然后编译器可以告诉你错过的案例.例如

  让rec sieve找到toTest = let h = List.hd toTest和t = List.tl toTest in

将会

let rec sieve found = function
  | h :: t -> ...
  | [] -> Error handling...

• 不要使用x = [].使用模式修补.

  将x与|匹配 [] - > ... | h :: t - > ......

• 使用匿名函数而不是短(即<= 1行)命名的单用函数:

  let dontDivide x =(x mod h <> 0)in let nonDivisors = List.filter doesntdivide t in

  让nonDivisors = List.filter(fun x - >(x mod h <> 0))t in

• 谨慎使用命令功能.