与Project Euler的速度比较：C vs Python vs Erlang vs Haskell

问题：

I have taken Problem #12 from Project Euler as a programming exercise and to compare my (surely not optimal) implementations in C, Python, Erlang and Haskell. 我将Project Euler中的问题＃12作为编程练习并比较了我在C，Python，Erlang和Haskell中的（当然不是最优的）实现。 In order to get some higher execution times, I search for the first triangle number with more than 1000 divisors instead of 500 as stated in the original problem. 为了获得更高的执行时间，我搜索第一个三角形数字，其中有超过1000个除数而不是原始问题中所述的500。

The result is the following: 结果如下：

C: C：

lorenzo@enzo:~/erlang$ gcc -lm -o euler12.bin euler12.c
lorenzo@enzo:~/erlang$ time ./euler12.bin
842161320

real    0m11.074s
user    0m11.070s
sys 0m0.000s

Python: 蟒蛇：

lorenzo@enzo:~/erlang$ time ./euler12.py 
842161320

real    1m16.632s
user    1m16.370s
sys 0m0.250s

Python with PyPy: Python与PyPy：

lorenzo@enzo:~/Downloads/pypy-c-jit-43780-b590cf6de419-linux64/bin$ time ./pypy /home/lorenzo/erlang/euler12.py 
842161320

real    0m13.082s
user    0m13.050s
sys 0m0.020s

Erlang: 二郎：

lorenzo@enzo:~/erlang$ erlc euler12.erl 
lorenzo@enzo:~/erlang$ time erl -s euler12 solve
Erlang R13B03 (erts-5.7.4) [source] [64-bit] [smp:4:4] [rq:4] [async-threads:0] [hipe] [kernel-poll:false]

Eshell V5.7.4  (abort with ^G)
1> 842161320

real    0m48.259s
user    0m48.070s
sys 0m0.020s

Haskell: 哈斯克尔：

lorenzo@enzo:~/erlang$ ghc euler12.hs -o euler12.hsx
[1 of 1] Compiling Main             ( euler12.hs, euler12.o )
Linking euler12.hsx ...
lorenzo@enzo:~/erlang$ time ./euler12.hsx 
842161320

real    2m37.326s
user    2m37.240s
sys 0m0.080s

Summary: 摘要：

C: 100% C：100％
Python: 692% (118% with PyPy) Python：692％（使用PyPy时为118％）
Erlang: 436% (135% thanks to RichardC) Erlang：436％（135％归功于RichardC）
Haskell: 1421% 哈斯克尔：1421％

I suppose that C has a big advantage as it uses long for the calculations and not arbitrary length integers as the other three. 我认为C有一个很大的优势，因为它使用long进行计算而不是任意长度整数作为其他三个。 Also it doesn't need to load a runtime first (Do the others?). 此外，它不需要首先加载运行时（其他人？）。

Question 1: Do Erlang, Python and Haskell lose speed due to using arbitrary length integers or don't they as long as the values are less than MAXINT ? 问题1： Erlang，Python和Haskell是否由于使用任意长度整数而失去速度，或者只要值小于MAXINT就不会失败？

Question 2: Why is Haskell so slow? 问题2：为什么Haskell这么慢？ Is there a compiler flag that turns off the brakes or is it my implementation? 是否有编译器标志关闭刹车或是我的实施？ (The latter is quite probable as Haskell is a book with seven seals to me.) （后者非常可能，因为Haskell是一本带有七个印章的书。）

Question 3: Can you offer me some hints how to optimize these implementations without changing the way I determine the factors? 问题3：您能否提供一些提示，如何优化这些实现而不改变我确定因素的方式？ Optimization in any way: nicer, faster, more "native" to the language. 以任何方式进行优化：更好，更快，更“本地”的语言。

EDIT: 编辑：

Question 4: Do my functional implementations permit LCO (last call optimization, aka tail recursion elimination) and hence avoid adding unnecessary frames onto the call stack? 问题4：我的功能实现是否允许LCO（最后调用优化，也就是尾递归消除），从而避免在调用堆栈中添加不必要的帧？

I really tried to implement the same algorithm as similar as possible in the four languages, although I have to admit that my Haskell and Erlang knowledge is very limited. 我真的试图在四种语言中尽可能地实现相同的算法，尽管我必须承认我的Haskell和Erlang知识非常有限。

Source codes used: 使用的源代码：

#include <stdio.h>
#include <math.h>

int factorCount (long n)
{
    double square = sqrt (n);
    int isquare = (int) square;
    int count = isquare == square ? -1 : 0;
    long candidate;
    for (candidate = 1; candidate <= isquare; candidate ++)
        if (0 == n % candidate) count += 2;
    return count;
}

int main ()
{
    long triangle = 1;
    int index = 1;
    while (factorCount (triangle) < 1001)
    {
        index ++;
        triangle += index;
    }
    printf ("%ld\n", triangle);
}

#! /usr/bin/env python3.2

import math

def factorCount (n):
    square = math.sqrt (n)
    isquare = int (square)
    count = -1 if isquare == square else 0
    for candidate in range (1, isquare + 1):
        if not n % candidate: count += 2
    return count

triangle = 1
index = 1
while factorCount (triangle) < 1001:
    index += 1
    triangle += index

print (triangle)

-module (euler12).
-compile (export_all).

factorCount (Number) -> factorCount (Number, math:sqrt (Number), 1, 0).

factorCount (_, Sqrt, Candidate, Count) when Candidate > Sqrt -> Count;

factorCount (_, Sqrt, Candidate, Count) when Candidate == Sqrt -> Count + 1;

factorCount (Number, Sqrt, Candidate, Count) ->
    case Number rem Candidate of
        0 -> factorCount (Number, Sqrt, Candidate + 1, Count + 2);
        _ -> factorCount (Number, Sqrt, Candidate + 1, Count)
    end.

nextTriangle (Index, Triangle) ->
    Count = factorCount (Triangle),
    if
        Count > 1000 -> Triangle;
        true -> nextTriangle (Index + 1, Triangle + Index + 1)  
    end.

solve () ->
    io:format ("~p~n", [nextTriangle (1, 1) ] ),
    halt (0).

factorCount number = factorCount' number isquare 1 0 - (fromEnum $ square == fromIntegral isquare)
    where square = sqrt $ fromIntegral number
          isquare = floor square

factorCount' number sqrt candidate count
    | fromIntegral candidate > sqrt = count
    | number `mod` candidate == 0 = factorCount' number sqrt (candidate + 1) (count + 2)
    | otherwise = factorCount' number sqrt (candidate + 1) count

nextTriangle index triangle
    | factorCount triangle > 1000 = triangle
    | otherwise = nextTriangle (index + 1) (triangle + index + 1)

main = print $ nextTriangle 1 1