[Haskell-beginners] Question about time consume when calculate prime numbers

[Yi Cheng]

Thanks for answering my question, but I'm still confused by some details.
I don't quite agree with you that Eratosthenes algorithm must be
implemented with a complexity of O(n^2) in space. When the n is used to
calculate the primes below it, it can be implemented in space complexity
O(n). For example, in languages, like C/C++, we can allocate a array. So I
think the the complexity of O(n^2) in space you mentioned, is the
complexity of "the beautiful code". So here's the question, can
Eratosthenes algorithm be implemented in a more gentle way?

Then I think maybe there is a more beautiful and practical way to implement
it.
One method of mine is trying to judge whether a number is a prime just by
the primes less than it, such as if the greatest common divisor of the
number and the product of the primes less than it equals to 1. But the
product of the primes is too large.

So I wander if there is a concise method to solve the problem with a faster
method. In my C++ version, the Eratosthenes is implemented in linear space
complexity, and optimize in filtering the numbers which can be divided by a
prime. This code is faster than the original algorithm implemented by me(It
was also implemented it in C++, and slower than the following code).
I know, when writing Haskell code, it would be better to forget some
experience in command-line language, but I want to know whether there is a
faster method to solve the problem.


Thank you.
Yi. Cheng

The code in my c++ version.
#include <iostream>
using namespace std;
int main(){
    int p[2000000] = {0};
    long sum = 0;
    int f = 1;
    for(long i=2; i <= 2000000; ++i){
        if(p[i] == 0){
            sum += i;
            for(long j = i * i; j < 2000000; j += i)
                p[j] = 1;
        }
    }
    cout<<sum<<endl;
    return 0;
}

On Wed, Sep 12, 2012 at 5:26 PM, Lorenzo Bolla <lbolla at gmail.com> wrote:

>
>
> On Wed, Sep 12, 2012 at 9:06 AM, Yi Cheng <chengyidna at gmail.com> wrote:
>
>> Recently, I'm trying to solve some problems in project euler using
>> haskell. When it came to problem 10, calculating the sum of all primes
>> below 20000000, I try to write a program which can generate primes.
>> In my memory Eratosthenes is faster than just whether a number can be
>> divided by the number less then the square root of it.
>> Firstly, I wrote the following programs:
>>
>> module Main where
>> isPrime x = isPrime' 3 x (round . sqrt. fromIntegral $ x)
>> isPrime' d target maxd
>>   | d > maxd = True
>>   | mod target d == 0 = False
>>   | otherwise = isPrime' (d + 2) target maxd
>>
>> main = print $ (sum (filter isPrime [3,5..2000000]) + 2)
>>
>> And it consume about 11s in my computer.
>> Then, I tried to figure out how to solve the problem by Eratosthenes, but
>> failed. Later, I find a program implemented by others, meeting my purpose
>> and I've used it to solve the problem:
>>
>> primes :: [Int]
>> primes = primes' [2..]
>>
>> primes' :: [Int] -> [Int]
>> primes' [] = []
>> primes' (n:ns) = n : primes' (filter (\v -> v `mod` n /= 0) ns)
>>
>> solve x = sum $ primes' [2..x]
>>
>> main = print $ solve 2000000
>>
>> Well, although the code is beautiful, it is slow. Even waiting for a
>> minute, no answer was printed.
>>
>> In C version, Eratosthenes is faster than the method implemented in my
>> earlier code, which only consume 0.3s(the earlier method consume 1.6s).
>>
>> So I want to know, why Eratosthenes implemented in Haskell is slow than
>> the ugly code implemented by me.
>> Could anyone tell me?
>>
>>
> Eratosthenes's complexity is O(n^2) (both space and time), whereas the
> "ugly" one has a sub-quadratic running complexity and linear in space.
>
> Try to profile them:
> $> ghc -O2 --make -prof -auto-all <filename>
> $> ./primes +RTS -p -hc
> $> hp2ps primes.hp
>
> You'll see that most of the time is spent allocating space which is never
> released.
> You could play a bit with strictness, but the main problem is the awful
> complexity of the algorithm.
>
> hth,
> L.
>
>
>
>
>> Thank you
>> Yi Cheng
>>
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>>
>>
>
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