Re: [Algorithm] Sum of Primes < 1000000

From:
Patricia Shanahan <pats@acm.org>
Newsgroups:
comp.lang.java.programmer
Date:
Mon, 08 Jan 2007 14:25:02 GMT
Message-ID:
<2vsoh.7792$pQ3.3161@newsread4.news.pas.earthlink.net>
Simon wrote:

Patricia Shanahan schrieb:

Simon wrote:

Patricia Shanahan schrieb:

Simon wrote:
As a matter of curiosity, did you use a boolean[] or BitSet?

I started with a boolean[], and mine was slower than yours, about 200
milliseconds. It dropped to under 100 milliseconds when I switched to
BitSet. I believe the improvement is due to more efficient caching.

Well, that is strange. I have been using a boolean[], and using BitSet
actually
slows it down on my machine. I am using a Laptop, so CPU speed may
vary. Today
it is even faster than yesterday. The output was (code see below):

The most likely cause of the speed up I saw is more efficient caching,
because BitSet stores the data more densely. That is a very hardware
dependent effect, though less a matter of CPU speed than of cache
configuration.

    public void eraseComposite() {
    int prime = 2;
    while (prime < numbers.length) {
        sum += prime;
        // mark multiples of prime
        for (int i = 2 * prime; i < numbers.length; i += prime) {
                // (we could start with prime*prime here, but I didn't
bother
                // since it would be larger than Integer.MAX_INTEGER)
        numbers[i] = true;
        }
        // move forward to next prime
        prime++;
        while (prime < numbers.length-1 && (numbers[prime])) {
        prime++;
        }
    }
    }


There is one optimization you missed. When prime is greater than the
square root of numbers.length, there is no need to do the composite scan.

If a number less than 1,000,000 is composite, at least one prime factor
is less than 1000.


This is what I tried to express in the above comment. It was more laziness than
the reason stated above though that prevented me from implementing it :-)

Cheers,
Simon


Ah, I see. You were thinking of checking by asking whether the square of
the latest prime is less than numbers.length.

The way I did it, there is no possibility of overflow - I stored
Math.sqrt(numbers.length) as an integer. Inside the loop, I just
compared to the square root.

Patricia

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