我把Project Euler中的第12题作为一个编程练习,并比较了我在C、Python、Erlang和Haskell中的实现(当然不是最优的)。为了获得更高的执行时间,我搜索第一个因数超过1000的三角形数,而不是原始问题中所述的500。

结果如下:

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:

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

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

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:

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:

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

简介:

C: 100% Python: 692% (PyPy占118%) Erlang: 436%(135%归功于RichardC) Haskell: 1421%

我认为C语言有一个很大的优势,因为它使用长来进行计算,而不是像其他三种那样使用任意长度的整数。它也不需要首先加载运行时(其他的呢?)

问题1: Erlang, Python和Haskell是否会因为使用任意长度的整数而降低速度,或者只要值小于MAXINT就不会?

问题2: 哈斯克尔为什么这么慢?是否有一个编译器标志关闭刹车或它是我的实现?(后者是很有可能的,因为Haskell对我来说是一本有七个印章的书。)

问题3: 你能否给我一些提示,如何在不改变我确定因素的方式的情况下优化这些实现?以任何方式优化:更好、更快、更“原生”的语言。

编辑:

问题4: 我的函数实现是否允许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?

虽然我不得不承认我的Haskell和Erlang知识非常有限,但我确实试图用这四种语言实现尽可能相似的相同算法。


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

当前回答

更多关于C版本的数字和解释。显然这么多年来没人这么做过。记得给这个答案点赞,这样它就可以放在最上面,让每个人都能看到和学习。

第一步:作者程序的基准

笔记本电脑的规格:

CPU i3 M380 (931 MHz -最大省电模式) 4 gb内存 Win7 64位 微软Visual Studio 2012终极版 Cygwin与gcc 4.9.3 Python 2.7.10

命令:

compiling on VS x64 command prompt > `for /f %f in ('dir /b *.c') do cl /O2 /Ot /Ox %f -o %f_x64_vs2012.exe`
compiling on cygwin with gcc x64   > `for f in ./*.c; do gcc -m64 -O3 $f -o ${f}_x64_gcc.exe ; done`
time (unix tools) using cygwin > `for f in ./*.exe; do  echo "----------"; echo $f ; time $f ; done`

.

----------
$ time python ./original.py

real    2m17.748s
user    2m15.783s
sys     0m0.093s
----------
$ time ./original_x86_vs2012.exe

real    0m8.377s
user    0m0.015s
sys     0m0.000s
----------
$ time ./original_x64_vs2012.exe

real    0m8.408s
user    0m0.000s
sys     0m0.015s
----------
$ time ./original_x64_gcc.exe

real    0m20.951s
user    0m20.732s
sys     0m0.030s

文件名为:integertype_architecture_compiler.exe

Integertype目前与原始程序相同(稍后详细介绍) 架构是x86或x64,取决于编译器设置 编译器是GCC或vs2012

第二步:调查、改进和再次基准

VS比gcc快250%。这两个编译器应该给出类似的速度。显然,代码或编译器选项有问题。让我们调查!

首先要注意的是整数类型。转换可能很昂贵,一致性对于更好的代码生成和优化很重要。所有整数都应该是相同的类型。

它现在是int和long的混合体。我们要改进这一点。使用哪种类型?最快的。必须对它们进行基准测试!

----------
$ time ./int_x86_vs2012.exe

real    0m8.440s
user    0m0.016s
sys     0m0.015s
----------
$ time ./int_x64_vs2012.exe

real    0m8.408s
user    0m0.016s
sys     0m0.015s
----------
$ time ./int32_x86_vs2012.exe

real    0m8.408s
user    0m0.000s
sys     0m0.015s
----------
$ time ./int32_x64_vs2012.exe

real    0m8.362s
user    0m0.000s
sys     0m0.015s
----------
$ time ./int64_x86_vs2012.exe

real    0m18.112s
user    0m0.000s
sys     0m0.015s
----------
$ time ./int64_x64_vs2012.exe

real    0m18.611s
user    0m0.000s
sys     0m0.015s
----------
$ time ./long_x86_vs2012.exe

real    0m8.393s
user    0m0.015s
sys     0m0.000s
----------
$ time ./long_x64_vs2012.exe

real    0m8.440s
user    0m0.000s
sys     0m0.015s
----------
$ time ./uint32_x86_vs2012.exe

real    0m8.362s
user    0m0.000s
sys     0m0.015s
----------
$ time ./uint32_x64_vs2012.exe

real    0m8.393s
user    0m0.015s
sys     0m0.015s
----------
$ time ./uint64_x86_vs2012.exe

real    0m15.428s
user    0m0.000s
sys     0m0.015s
----------
$ time ./uint64_x64_vs2012.exe

real    0m15.725s
user    0m0.015s
sys     0m0.015s
----------
$ time ./int_x64_gcc.exe

real    0m8.531s
user    0m8.329s
sys     0m0.015s
----------
$ time ./int32_x64_gcc.exe

real    0m8.471s
user    0m8.345s
sys     0m0.000s
----------
$ time ./int64_x64_gcc.exe

real    0m20.264s
user    0m20.186s
sys     0m0.015s
----------
$ time ./long_x64_gcc.exe

real    0m20.935s
user    0m20.809s
sys     0m0.015s
----------
$ time ./uint32_x64_gcc.exe

real    0m8.393s
user    0m8.346s
sys     0m0.015s
----------
$ time ./uint64_x64_gcc.exe

real    0m16.973s
user    0m16.879s
sys     0m0.030s

整数类型是int long int32_t uint32_t int64_t和uint64_t from #include <stdint.h>

C语言中有很多整数类型,还有一些带符号/无符号的可以使用,还有编译为x86或x64的选择(不要与实际的整数大小混淆)。要编译和运行^^的版本太多了

第三步:理解数字

最终结论:

32位整数比64位整数快200% 无符号64位整数比有符号64位快25%(不幸的是,我对此没有解释)

陷阱问题:“C语言中int和long的大小是多少?” 正确答案是:C中int和long的大小没有很好的定义!

来自C规范:

Int至少是32位 Long至少是int型

从gcc手册页(-m32和-m64标志):

32位环境将int、long和指针设置为32位,并生成可在任何i386系统上运行的代码。 64位环境将int设置为32位,long设置为64位,指针设置为64位,并为AMD的x86-64架构生成代码。

来自MSDN文档(数据类型范围)https://msdn.microsoft.com/en-us/library/s3f49ktz%28v=vs.110%29.aspx:

Int, 4字节,也是有符号的 Long, 4字节,也称为Long int和带符号的Long int

总结一下:吸取的教训

32位整数比64位整数快。 标准整数类型在C和c++中都没有很好地定义,它们取决于编译器和体系结构。当你需要一致性和可预测性时,使用uint32_t整数族从#include <stdint.h>。 速度问题解决。所有其他语言都落后百分之百,C和c++又赢了!他们总是这样。接下来的改进将是使用OpenMP:D进行多线程处理

其他回答

在x86_64 Core2 Duo (2.5GHz)机器上使用GHC 7.0.3, gcc 4.4.6, Linux 2.6.29,对Haskell使用GHC -O2 - flvm - force-recomp编译,对C使用gcc -O3 -lm编译。

Your C routine runs in 8.4 seconds (faster than your run probably because of -O3) The Haskell solution runs in 36 seconds (due to the -O2 flag) Your factorCount' code isn't explicitly typed and defaulting to Integer (thanks to Daniel for correcting my misdiagnosis here!). Giving an explicit type signature (which is standard practice anyway) using Int and the time changes to 11.1 seconds in factorCount' you have needlessly called fromIntegral. A fix results in no change though (the compiler is smart, lucky for you). You used mod where rem is faster and sufficient. This changes the time to 8.5 seconds. factorCount' is constantly applying two extra arguments that never change (number, sqrt). A worker/wrapper transformation gives us:

 $ time ./so
 842161320  

 real    0m7.954s  
 user    0m7.944s  
 sys     0m0.004s  

没错,7.95秒。始终比C方案快半秒。没有- flvm标志,我仍然得到8.182秒,所以NCG后端在这种情况下也做得很好。

结论:Haskell非常棒。

生成的代码

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

factorCount' :: Int -> Int -> Int -> Int -> Int
factorCount' number sqrt candidate0 count0 = go candidate0 count0
  where
  go candidate count
    | candidate > sqrt = count
    | number `rem` candidate == 0 = go (candidate + 1) (count + 2)
    | otherwise = go (candidate + 1) count

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

main = print $ nextTriangle 1 1

编辑:现在我们已经探讨了这个问题,让我们来解决问题

问题1:erlang、python和haskell是否会因为使用 任意长度的整数,只要值更小 比MAXINT ?

在Haskell中,使用Integer比Int慢,但慢多少取决于执行的计算。幸运的是(对于64位机器)Int就足够了。出于可移植性的考虑,你可能应该重写我的代码,使用Int64或Word64 (C不是唯一的语言长)。

问题2:为什么haskell这么慢?有编译器标志吗 关闭刹车还是我的实现?(后者相当 就像haskell对我来说是一本有七个印章的书一样。) 问题3:你能给我一些建议吗 实现而不改变我确定因子的方式? 以任何方式优化:更好、更快、更“原生”的语言。

这就是我上面所回答的。答案是

0)通过-O2进行优化 1)尽可能使用快速(特别是不可装箱的)类型 2) rem not mod(一个经常被遗忘的优化)和 3)工人/包装器转换(可能是最常见的优化)。

问题4:我的功能实现是否允许LCO,因此 避免添加不必要的帧到调用堆栈?

是的,这不是问题所在。干得好,很高兴你考虑到这一点。

只是为了好玩。下面是一个更“原生”的Haskell实现:

import Control.Applicative
import Control.Monad
import Data.Either
import Math.NumberTheory.Powers.Squares

isInt :: RealFrac c => c -> Bool
isInt = (==) <$> id <*> fromInteger . round

intSqrt :: (Integral a) => a -> Int
--intSqrt = fromIntegral . floor . sqrt . fromIntegral
intSqrt = fromIntegral . integerSquareRoot'

factorize :: Int -> [Int]
factorize 1 = []
factorize n = first : factorize (quot n first)
  where first = (!! 0) $ [a | a <- [2..intSqrt n], rem n a == 0] ++ [n]

factorize2 :: Int -> [(Int,Int)]
factorize2 = foldl (\ls@((val,freq):xs) y -> if val == y then (val,freq+1):xs else (y,1):ls) [(0,0)] . factorize

numDivisors :: Int -> Int
numDivisors = foldl (\acc (_,y) -> acc * (y+1)) 1 <$> factorize2

nextTriangleNumber :: (Int,Int) -> (Int,Int)
nextTriangleNumber (n,acc) = (n+1,acc+n+1)

forward :: Int -> (Int, Int) -> Either (Int, Int) (Int, Int)
forward k val@(n,acc) = if numDivisors acc > k then Left val else Right (nextTriangleNumber val)

problem12 :: Int -> (Int, Int)
problem12 n = (!!0) . lefts . scanl (>>=) (forward n (1,1)) . repeat . forward $ n

main = do
  let (n,val) = problem12 1000
  print val

使用ghc -O3,它在我的机器上持续运行0.55-0.58秒(1.73GHz Core i7)。

C版本中一个更有效的factorCount函数:

int factorCount (int n)
{
  int count = 1;
  int candidate,tmpCount;
  while (n % 2 == 0) {
    count++;
    n /= 2;
  }
    for (candidate = 3; candidate < n && candidate * candidate < n; candidate += 2)
    if (n % candidate == 0) {
      tmpCount = 1;
      do {
        tmpCount++;
        n /= candidate;
      } while (n % candidate == 0);
       count*=tmpCount;
      }
  if (n > 1)
    count *= 2;
  return count;
}

在main中使用gcc -O3 -lm将long类型更改为int类型,该程序始终在0.31-0.35秒内运行。

如果您利用第n个三角形数= n*(n+1)/2,并且n和(n+1)具有完全不同的质因数分解,则可以使两者运行得更快,因此可以将每个一半的因数数相乘,以得到整体的因数数。以下几点:

int main ()
{
  int triangle = 0,count1,count2 = 1;
  do {
    count1 = count2;
    count2 = ++triangle % 2 == 0 ? factorCount(triangle+1) : factorCount((triangle+1)/2);
  } while (count1*count2 < 1001);
  printf ("%lld\n", ((long long)triangle)*(triangle+1)/2);
}

将c代码的运行时间减少到0.17-0.19秒,它可以处理更大的搜索——大于10000个因数在我的机器上大约需要43秒。我给感兴趣的读者留下了类似的haskell加速。

问题1:erlang, python和haskell会因为使用任意长度的整数而降低速度吗?还是只要值小于MAXINT就不会?

This is unlikely. I cannot say much about Erlang and Haskell (well, maybe a bit about Haskell below) but I can point a lot of other bottlenecks in Python. Every time the program tries to execute an operation with some values in Python, it should verify whether the values are from the proper type, and it costs a bit of time. Your factorCount function just allocates a list with range (1, isquare + 1) various times, and runtime, malloc-styled memory allocation is way slower than iterating on a range with a counter as you do in C. Notably, the factorCount() is called multiple times and so allocates a lot of lists. Also, let us not forget that Python is interpreted and the CPython interpreter has no great focus on being optimized.

编辑:哦,好吧,我注意到你使用的是Python 3,所以range()不返回一个列表,而是一个生成器。在这种情况下,我关于分配列表的观点有一半是错误的:该函数只是分配范围对象,尽管效率很低,但没有分配包含很多项的列表那么低。

问题2:为什么haskell这么慢?是否有一个编译器标志关闭刹车或它是我的实现?(后者很有可能,因为haskell对我来说是一本有七个印章的书。)

你在使用Hugs吗?Hugs是一个相当慢的解释器。如果你正在使用它,也许你可以得到一个更好的GHC时间-但我只是在思考假设,这种东西,一个好的Haskell编译器做的是非常迷人的,远远超出我的理解:)

问题3:你能给我一些提示吗?如何在不改变我确定因素的方式的情况下优化这些实现?以任何方式优化:更好、更快、更“原生”的语言。

我得说你在玩一场不好笑的游戏。了解各种语言最好的部分是尽可能以不同的方式使用它们:)但我离题了,我只是对这一点没有任何建议。对不起,我希望有人能在这种情况下帮助你:)

问题4:我的函数实现是否允许LCO,从而避免在调用堆栈中添加不必要的帧?

据我所知,您只需要确保您的递归调用是返回值之前的最后一个命令。换句话说,像下面这样的函数可以使用这样的优化:

def factorial(n, acc=1):
    if n > 1:
        acc = acc * n
        n = n - 1
        return factorial(n, acc)
    else:
        return acc

然而,如果你的函数如下所示,你就不会有这样的优化,因为在递归调用之后有一个操作(乘法):

def factorial2(n):
    if n > 1:
        f = factorial2(n-1)
        return f*n
    else:
        return 1

我将操作分隔在一些局部变量中,以便明确执行哪些操作。然而,最常见的是看到这些函数如下所示,但它们对于我所说的观点是等价的:

def factorial(n, acc=1):
    if n > 1:
        return factorial(n-1, acc*n)
    else:
        return acc

def factorial2(n):
    if n > 1:
        return n*factorial(n-1)
    else:
        return 1

注意,这是由编译器/解释器来决定是否进行尾递归。例如,如果我记得很清楚,Python解释器就不会这样做(我在示例中使用Python只是因为它的语法流畅)。不管怎样,如果你发现了一些奇怪的东西,比如带两个参数的阶乘函数(其中一个参数有acc, accumulator等名称),现在你知道为什么人们这样做了:)

通过使用Haskell包中的一些函数,可以大大加快Haskell实现的速度。 在这种情况下,我使用了质数,它只是安装了'cabal安装质数';)

import Data.Numbers.Primes
import Data.List

triangleNumbers = scanl1 (+) [1..]
nDivisors n = product $ map ((+1) . length) (group (primeFactors n))
answer = head $ filter ((> 500) . nDivisors) triangleNumbers

main :: IO ()
main = putStrLn $ "First triangle number to have over 500 divisors: " ++ (show answer)

计时:

您的原始程序:

PS> measure-command { bin\012_slow.exe }

TotalSeconds      : 16.3807409
TotalMilliseconds : 16380.7409

改进的实现

PS> measure-command { bin\012.exe }

TotalSeconds      : 0.0383436
TotalMilliseconds : 38.3436

正如你所看到的,在同一台机器上,这台机器运行38毫秒,而你的机器运行16秒:)

编译命令:

ghc -O2 012.hs -o bin\012.exe
ghc -O2 012_slow.hs -o bin\012_slow.exe

c++ 11, < 20ms for me -在这里运行它

我理解您想要一些技巧来帮助提高您的语言特定知识,但由于这里已经很好地介绍了这一点,我想我应该为那些可能看过您的问题的mathematica注释等并想知道为什么这段代码如此之慢的人添加一些上下文。

这个答案主要是为了提供上下文,希望能够帮助人们更容易地评估您的问题/其他答案中的代码。

这段代码只使用了一些(丑陋的)优化,与所使用的语言无关,基于:

每个三角数的形式都是n(n+1)/2 N和N +1是互质 除数的数量是一个乘法函数

#include <iostream>
#include <cmath>
#include <tuple>
#include <chrono>

using namespace std;

// Calculates the divisors of an integer by determining its prime factorisation.

int get_divisors(long long n)
{
    int divisors_count = 1;

    for(long long i = 2;
        i <= sqrt(n);
        /* empty */)
    {
        int divisions = 0;
        while(n % i == 0)
        {
            n /= i;
            divisions++;
        }

        divisors_count *= (divisions + 1);

        //here, we try to iterate more efficiently by skipping
        //obvious non-primes like 4, 6, etc
        if(i == 2)
            i++;
        else
            i += 2;
    }

    if(n != 1) //n is a prime
        return divisors_count * 2;
    else
        return divisors_count;
}

long long euler12()
{
    //n and n + 1
    long long n, n_p_1;

    n = 1; n_p_1 = 2;

    // divisors_x will store either the divisors of x or x/2
    // (the later iff x is divisible by two)
    long long divisors_n = 1;
    long long divisors_n_p_1 = 2;

    for(;;)
    {
        /* This loop has been unwound, so two iterations are completed at a time
         * n and n + 1 have no prime factors in common and therefore we can
         * calculate their divisors separately
         */

        long long total_divisors;                 //the divisors of the triangle number
                                                  // n(n+1)/2

        //the first (unwound) iteration

        divisors_n_p_1 = get_divisors(n_p_1 / 2); //here n+1 is even and we

        total_divisors =
                  divisors_n
                * divisors_n_p_1;

        if(total_divisors > 1000)
            break;

        //move n and n+1 forward
        n = n_p_1;
        n_p_1 = n + 1;

        //fix the divisors
        divisors_n = divisors_n_p_1;
        divisors_n_p_1 = get_divisors(n_p_1);   //n_p_1 is now odd!

        //now the second (unwound) iteration

        total_divisors =
                  divisors_n
                * divisors_n_p_1;

        if(total_divisors > 1000)
            break;

        //move n and n+1 forward
        n = n_p_1;
        n_p_1 = n + 1;

        //fix the divisors
        divisors_n = divisors_n_p_1;
        divisors_n_p_1 = get_divisors(n_p_1 / 2);   //n_p_1 is now even!
    }

    return (n * n_p_1) / 2;
}

int main()
{
    for(int i = 0; i < 1000; i++)
    {
        using namespace std::chrono;
        auto start = high_resolution_clock::now();
        auto result = euler12();
        auto end = high_resolution_clock::now();

        double time_elapsed = duration_cast<milliseconds>(end - start).count();

        cout << result << " " << time_elapsed << '\n';
    }
    return 0;
}

我的台式机平均花费19毫秒,笔记本电脑平均花费80毫秒,这与我在这里看到的大多数其他代码相差甚远。毫无疑问,还有许多优化方法可用。