与 Project Euler 的速度比较: c 和 Python 对 Erlang vs Haskell

Question

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. 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.

The result is the following:

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 with 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

Summary:

• C: 100%
• Python: 692% (118% with PyPy)
• Erlang: 436% (135% thanks to RichardC)
• Haskell: 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. Also it doesn't need to load a runtime first (Do the others?).

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?

Question 2: Why is Haskell so slow? Is there a compiler flag that turns off the brakes or is it my implementation? (The latter is quite probable as Haskell is a book with seven seals to me.)

Question 3: Can you offer me some hints how to optimize these implementations without changing the way I determine the factors? Optimization in any way: nicer, faster, more "native" to the language.

EDIT:

Question 4: Do my functional implementations permit LCO (last call optimization, a.k.a tail recursion elimination) and hence avoid adding unnecessary frames onto the call stack?

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.

---

Source codes used:

#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

转载于:https://stackoverflow.com/questions/6964392/speed-comparison-with-project-euler-c-vs-python-vs-erlang-vs-haskell

Answer 1

Using GHC 7.0.3, gcc 4.4.6, Linux 2.6.29 on an x86_64 Core2 Duo (2.5GHz) machine, compiling using ghc -O2 -fllvm -fforce-recomp for Haskell and gcc -O3 -lm for C.

• 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  

That's right, 7.95 seconds. Consistently half a second faster than the C solution. Without the -fllvm flag I'm still getting 8.182 seconds, so the NCG backend is doing well in this case too.

Conclusion: Haskell is awesome.

Resulting Code

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

EDIT: So now that we've explored that, lets address the questions

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?

In Haskell, using Integer is slower than Int but how much slower depends on the computations performed. Luckily (for 64 bit machines) Int is sufficient. For portability sake you should probably rewrite my code to use Int64 or Word64 (C isn't the only language with a long).

Question 2: Why is haskell so slow? Is there a compiler flag that turns off the brakes or is it my implementation? (The latter is quite probable as haskell is a book with seven seals to me.)

Question 3: Can you offer me some hints how to optimize these implementations without changing the way I determine the factors? Optimization in any way: nicer, faster, more "native" to the language.

That was what I answered above. The answer was

• 0) Use optimization via -O2
• 1) Use fast (notably: unbox-able) types when possible
• 2) rem not mod (a frequently forgotten optimization) and
• 3) worker/wrapper transformation (perhaps the most common optimization).

Question 4: Do my functional implementations permit LCO and hence avoid adding unnecessary frames onto the call stack?

Yes, that wasn't the issue. Good work and glad you considered this.