MC *_*tch 7 java arrays primes multithreading sieve-of-eratosthenes
I am trying to create a fast prime generator in Java. It is (more or less) accepted that the fastest way for this is the segmented sieve of Eratosthenes: https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes. Lots of optimizations can be further implemented to make it faster. As of now, my implementation generates 50847534 primes below 10^9 in about 1.6 seconds, but I am looking to make it faster and at least break the 1 second barrier. To increase the chance of getting good replies, I will include a walkthrough of the algorithm as well as the code.
Still, as a TL;DR, I am looking to include multi-threading into the code
For the purposes of this question, I want to separate between the 'segmented' and the 'traditional' sieves of Eratosthenes. The traditional sieve requires O(n) space and therefore is very limited in range of the input (the limit of it). The segmented sieve however only requires O(n^0.5) space and can operate on much larger limits. (A main speed-up is using a cache-friendly segmentation, taking into account the L1 & L2 cache sizes of the specific computer). Finally, the main difference that concerns my question is that the traditional sieve is sequential, meaning it can only continue once the previous steps are completed. The segmented sieve however, is not. Each segment is independent, and is 'processed' individually against the sieving primes (the primes not larger than n^0.5). This means that theoretically, once I have the sieving primes, I can divide the work between multiple computers, each processing a different segment. The work of eachother is independent of the others. Assuming (wrongly) that each segment requires the same amount of time t to complete, and there are k segments, One computer would require total time of T = k * t, whereas k computers, each working on a different segment would require a total amount of time T = t to complete the entire process. (Practically, this is wrong, but for the sake of simplicity of the example).
This brought me to reading about multithreading - dividing the work to a few threads each processing a smaller amount of work for better usage of CPU. To my understanding, the traditional sieve cannot be multithreaded exactly because it is sequential. Each thread would depend on the previous, rendering the entire idea unfeasible. But a segmented sieve may indeed (I think) be multithreaded.
Instead of jumping straight into my question, I think it is important to introduce my code first, so I am hereby including my current fastest implementation of the segmented sieve. I have worked quite hard on it. It took quite some time, slowly tweaking and adding optimizations to it. The code is not simple. It is rather complex, I would say. I therefore assume the reader is familiar with the concepts I am introducing, such as wheel factorization, prime numbers, segmentation and more. I have included notes to make it easier to follow.
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
public class primeGen {
public static long x = (long)Math.pow(10, 9); //limit
public static int sqrtx;
public static boolean [] sievingPrimes; //the sieving primes, <= sqrtx
public static int [] wheels = new int [] {2,3,5,7,11,13,17,19}; // base wheel primes
public static int [] gaps; //the gaps, according to the wheel. will enable skipping multiples of the wheel primes
public static int nextp; // the first prime > wheel primes
public static int l; // the amount of gaps in the wheel
public static void main(String[] args)
{
long startTime = System.currentTimeMillis();
preCalc(); // creating the sieving primes and calculating the list of gaps
int segSize = Math.max(sqrtx, 32768*8); //size of each segment
long u = nextp; // 'u' is the running index of the program. will continue from one segment to the next
int wh = 0; // the will be the gap index, indicating by how much we increment 'u' each time, skipping the multiples of the wheel primes
long pi = pisqrtx(); // the primes count. initialize with the number of primes <= sqrtx
for (long low = 0 ; low < x ; low += segSize) //the heart of the code. enumerating the primes through segmentation. enumeration will begin at p > sqrtx
{
long high = Math.min(x, low + segSize);
boolean [] segment = new boolean [(int) (high - low + 1)];
int g = -1;
for (int i = nextp ; i <= sqrtx ; i += gaps[g])
{
if (sievingPrimes[(i + 1) / 2])
{
long firstMultiple = (long) (low / i * i);
if (firstMultiple < low)
firstMultiple += i;
if (firstMultiple % 2 == 0) //start with the first odd multiple of the current prime in the segment
firstMultiple += i;
for (long j = firstMultiple ; j < high ; j += i * 2)
segment[(int) (j - low)] = true;
}
g++;
//if (g == l) //due to segment size, the full list of gaps is never used **within just one segment** , and therefore this check is redundant.
//should be used with bigger segment sizes or smaller lists of gaps
//g = 0;
}
while (u <= high)
{
if (!segment[(int) (u - low)])
pi++;
u += gaps[wh];
wh++;
if (wh == l)
wh = 0;
}
}
System.out.println(pi);
long endTime = System.currentTimeMillis();
System.out.println("Solution took "+(endTime - startTime) + " ms");
}
public static boolean [] simpleSieve (int l)
{
long sqrtl = (long)Math.sqrt(l);
boolean [] primes = new boolean [l/2+2];
Arrays.fill(primes, true);
int g = -1;
for (int i = nextp ; i <= sqrtl ; i += gaps[g])
{
if (primes[(i + 1) / 2])
for (int j = i * i ; j <= l ; j += i * 2)
primes[(j + 1) / 2]=false;
g++;
if (g == l)
g=0;
}
return primes;
}
public static long pisqrtx ()
{
int pi = wheels.length;
if (x < wheels[wheels.length-1])
{
if (x < 2)
return 0;
int k = 0;
while (wheels[k] <= x)
k++;
return k;
}
int g = -1;
for (int i = nextp ; i <= sqrtx ; i += gaps[g])
{
if(sievingPrimes[( i + 1 ) / 2])
pi++;
g++;
if (g == l)
g=0;
}
return pi;
}
public static void preCalc ()
{
sqrtx = (int) Math.sqrt(x);
int prod = 1;
for (long p : wheels)
prod *= p; // primorial
nextp = BigInteger.valueOf(wheels[wheels.length-1]).nextProbablePrime().intValue(); //the first prime that comes after the wheel
int lim = prod + nextp; // circumference of the wheel
boolean [] marks = new boolean [lim + 1];
Arrays.fill(marks, true);
for (int j = 2 * 2 ;j <= lim ; j += 2)
marks[j] = false;
for (int i = 1 ; i < wheels.length ; i++)
{
int p = wheels[i];
for (int j = p * p ; j <= lim ; j += 2 * p)
marks[j]=false; // removing all integers that are NOT comprime with the base wheel primes
}
ArrayList <Integer> gs = new ArrayList <Integer>(); //list of the gaps between the integers that are coprime with the base wheel primes
int d = nextp;
for (int p = d + 2 ; p < marks.length ; p += 2)
{
if (marks[p]) //d is prime. if p is also prime, then a gap is identified, and is noted.
{
gs.add(p - d);
d = p;
}
}
gaps = new int [gs.size()];
for (int i = 0 ; i < gs.size() ; i++)
gaps[i] = gs.get(i); // Arrays are faster than lists, so moving the list of gaps to an array
l = gaps.length;
sievingPrimes = simpleSieve(sqrtx); //initializing the sieving primes
}
}
Currently, it produces 50847534 primes below 10^9 in about 1.6 seconds. This is very impressive, at least by my standards, but I am looking to make it faster, possibly break the 1 second barrier. Even then, I believe it can be made much faster still.
The whole program is based on wheel factorization: https://en.wikipedia.org/wiki/Wheel_factorization. I have noticed I am getting the fastest results using a wheel of all primes up to 19.
public static int [] wheels = new int [] {2,3,5,7,11,13,17,19}; // base wheel primes
This means that the multiples of those primes are skipped, resulting in a much smaller searching range. The gaps between numbers which we need to take are then calculated in the preCalc method. If we make those jumps between the the numbers in the searching range we skip the multiples of the base primes.
public static void preCalc ()
{
sqrtx = (int) Math.sqrt(x);
int prod = 1;
for (long p : wheels)
prod *= p; // primorial
nextp = BigInteger.valueOf(wheels[wheels.length-1]).nextProbablePrime().intValue(); //the first prime that comes after the wheel
int lim = prod + nextp; // circumference of the wheel
boolean [] marks = new boolean [lim + 1];
Arrays.fill(marks, true);
for (int j = 2 * 2 ;j <= lim ; j += 2)
marks[j] = false;
for (int i = 1 ; i < wheels.length ; i++)
{
int p = wheels[i];
for (int j = p * p ; j <= lim ; j += 2 * p)
marks[j]=false; // removing all integers that are NOT comprime with the base wheel primes
}
ArrayList <Integer> gs = new ArrayList <Integer>(); //list of the gaps between the integers that are coprime with the base wheel primes
int d = nextp;
for (int p = d + 2 ; p < marks.length ; p += 2)
{
if (marks[p]) //d is prime. if p is also prime, then a gap is identified, and is noted.
{
gs.add(p - d);
d = p;
}
}
gaps = new int [gs.size()];
for (int i = 0 ; i < gs.size() ; i++)
gaps[i] = gs.get(i); // Arrays are faster than lists, so moving the list of gaps to an array
l = gaps.length;
sievingPrimes = simpleSieve(sqrtx); //initializing the sieving primes
}
At the end of the preCalc method, the simpleSieve method is called, efficiently sieving all the sieving primes mentioned before, the primes <= sqrtx. This is a simple Eratosthenes sieve, rather than segmented, but it is still based on wheel factorization, perviously computed.
public static boolean [] simpleSieve (int l)
{
long sqrtl = (long)Math.sqrt(l);
boolean [] primes = new boolean [l/2+2];
Arrays.fill(primes, true);
int g = -1;
for (int i = nextp ; i <= sqrtl ; i += gaps[g])
{
if (primes[(i + 1) / 2])
for (int j = i * i ; j <= l ; j += i * 2)
primes[(j + 1) / 2]=false;
g++;
if (g == l)
g=0;
}
return primes;
}
Finally, we reach the heart of the algorithm. We start by enumerating all primes <= sqrtx, with the following call:
long pi = pisqrtx();`
which used the following method:
public static long pisqrtx ()
{
int pi = wheels.length;
if (x < wheels[wheels.length-1])
{
if (x < 2)
return 0;
int k = 0;
while (wheels[k] <= x)
k++;
return k;
}
int g = -1;
for (int i = nextp ; i <= sqrtx ; i += gaps[g])
{
if(sievingPrimes[( i + 1 ) / 2])
pi++;
g++;
if (g == l)
g=0;
}
return pi;
}
Then, after initializing the pi variable which keeps track of the enumeration of primes, we perform the mentioned segmentation, starting the enumeration from the first prime > sqrtx:
int segSize = Math.max(sqrtx, 32768*8); //size of each segment
long u = nextp; // 'u' is the running index of the program. will continue from one segment to the next
int wh = 0; // the will be the gap index, indicating by how much we increment 'u' each time, skipping the multiples of the wheel primes
long pi = pisqrtx(); // the primes count. initialize with the number of primes <= sqrtx
for (long low = 0 ; low < x ; low += segSize) //the heart of the code. enumerating the primes through segmentation. enumeration will begin at p > sqrtx
{
long high = Math.min(x, low + segSize);
boolean [] segment = new boolean [(int) (high - low + 1)];
int g = -1;
for (int i = nextp ; i <= sqrtx ; i += gaps[g])
{
if (sievingPrimes[(i + 1) / 2])
{
long firstMultiple = (long) (low / i * i);
if (firstMultiple < low)
firstMultiple += i;
if (firstMultiple % 2 == 0) //start with the first odd multiple of the current prime in the segment
firstMultiple += i;
for (long j = firstMultiple ; j < high ; j += i * 2)
segment[(int) (j - low)] = true;
}
g++;
//if (g == l) //due to segment size, the full list of gaps is never used **within just one segment** , and therefore this check is redundant.
//should be used with bigger segment sizes or smaller lists of gaps
//g = 0;
}
while (u <= high)
{
if (!segment[(int) (u - low)])
pi++;
u += gaps[wh];
wh++;
if (wh == l)
wh = 0;
}
}
I have also included it as a note, but will explain as well. Because the segment size is relatively small, we will not go through the entire list of gaps within just one segment, and checking it - is redundant. (Assuming we use a 19-wheel). But in a broader scope overview of the program, we will make use of the entire array of gaps, so the variable u has to follow it and not accidentally surpass it:
while (u <= high)
{
if (!segment[(int) (u - low)])
pi++;
u += gaps[wh];
wh++;
if (wh == l)
wh = 0;
}
Using higher limits will eventually render a bigger segment, which might result in a neccessity of checking we don't surpass the gaps list even within the segment. This, or tweaking the wheel primes base might have this effect on the program. Switching to bit-sieving can largely improve the segment limit though.
L1 & L2 cache-sizes into account. I get the
fastest results using a segment size of 32,768 * 8 = 262,144 = 2^18. I am not sure what the cache-size of my computer is, but I do
not think it can be that big, as I see most cache sizes <= 32,768.
Still, this produces the fastest run time on my computer, so this is
why it's the chosen segment size.4, using 4 threads (corresponding to 4
cores). The idea is that each thread will still use the idea of the
segmented sieve, but work on different portions. Divide the n
into 4 equal portions - threads, each in turn performing the
segmentation on the n/4 elements it is responsible for, using the
above program. My question is how do I do that? Reading about
multithreading and examples, unfortunately, did not bring to me any
insight on how to implement it in the case above efficiently. It
seems to me, as opposed to the logic behind it, that the threads were
running sequentially, rather than simultaneously. This is why I
excluded it from the code to make it more readable. I will really
appreciate a code sample on how to do it in this specific code, but a
good explanation and reference will maybe do the trick too. Additionally, I would like to hear about more ways of speeding-up this program even more, any ideas you have, I would love to hear! Really want to make it very fast and efficient. Thank you!
像这样的示例应该可以帮助您入门。
解决方案概要：
通过将结果连接到读取输出队列的单独任务中，或者甚至通过更新 下的可变共享输出结构，可能（或可能不会）实现额外的加速synchronized，具体取决于连接步骤涉及的工作量。
希望这可以帮助。