我目前正在编写一个程序，该程序首先依次由Eratosthenes筛选生成素数，然后同时生成素数。该算法的并发版本应该比顺序版本更快，但在我的情况下，并发版本约为。慢10倍。我想知道在顺序解决方案中与主线程相比，我将更多工作放在我的线程上。这是我的程序（准备阅读一下！）：
Primes.java ：
public abstract class Primes {
byte[] bitArr;
int maxNum;
final int[] BITMASK = { 1, 2, 4, 8, 16, 32, 64 };
final int[] BITMASK2 = { 255 - 1, 255 - 2, 255 - 4, 255 - 8,
255 - 16, 255 - 32, 255 - 64 };
void setAllPrime() {
for (int i = 0; i < bitArr.length; i++) {
bitArr[i] = (byte) 127;
}
}
void crossOut(int i) {
bitArr[i/14] = (byte) (bitArr[i/14] - BITMASK[((i/2)%7)]);
}
boolean isPrime(int i) {
if(i == 2){
return true;
}
if((i%2) == 0){
return false;
}
return (bitArr[i/14] & BITMASK[(i%14)>>1]) != 0;
}
int nextPrime(int i) {
int k;
if ((i%2) == 0){
k =i+1;
}
else {
k = i+2;
}
while (!isPrime(k) && k < maxNum){
k+=2;
}
return k;
}
void printAllPrimes() {
for (int i = 2; i <= maxNum; i++){
if (isPrime(i)){
System.out.println("Prime: " + i);
}
}
}
}
PrimesSeq.java ：
import java.util.ArrayList;
public class PrimesSeq extends Primes{
PrimesSeq(int maxNum) {
this.maxNum = maxNum;
bitArr = new byte[(maxNum / 14) + 1];
setAllPrime();
generatePrimesByEratosthenes();
}
void generatePrimesByEratosthenes() {
crossOut(1); // 1 is not a prime
int curr = 3;
while(curr < Math.sqrt(maxNum)){
for(int i = curr*curr; i < maxNum; i+=2*curr){
if(isPrime(i)){ // 2*curr because odd*2 = even!
crossOut(i);
}
}
curr = nextPrime(curr);
}
}
}
PrimesPara.java ：
import java.util.ArrayList;
public class PrimesPara extends Primes {
PrimeThread[] threads;
int processors;
int currentState = 0;
//0 = Init
//1 = Generate primes after thread #0 finish
//2 = Factorize
public PrimesPara(int maxNum){
this.maxNum = maxNum;
this.processors = Runtime.getRuntime().availableProcessors();
bitArr = new byte[(maxNum / 14) + 1];
setAllPrime();
this.threads = new PrimeThread[processors*2];
generateErastothenesConcurrently();
//printAllPrimes();
}
public void generateErastothenesConcurrently(){
int[] starts = generateThreadIndexes();
for(int i = 0; i < threads.length; i++){
if(i != threads.length-1){
threads[i] = new PrimeThread(starts[i], starts[i+1]-1, i);
} else {
threads[i] = new PrimeThread(starts[i], maxNum, i);
}
}
//Start generating the first primes
crossOut(1);
Thread th = new Thread(threads[0]);
th.start();
try {
th.join();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
currentState = 1;
//Start generating the rest of the primes
Thread[] thrs = new Thread[threads.length];
for(int i = 0; i < thrs.length; i++){
thrs[i] = new Thread(threads[i]);
thrs[i].start();
}
for(int i = 0; i < thrs.length; i++){
try {
thrs[i].join();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
currentState = 2;
}
private int[] generateThreadIndexes(){
int[] indexes = new int[processors*2];
for(int i = 0; i < indexes.length; i++){
indexes[i] = (i*((maxNum/(processors*2))));
}
indexes[indexes.length-1]++;
return indexes;
}
public class PrimeThread implements Runnable {
int start;
int end;
int thridx;
public PrimeThread(int start, int end, int thridx){
this.start = start;
this.end = end;
this.thridx = thridx;
}
public void run() {
switch(currentState){
case 0:
generateSqrtPrimes();
break;
case 1:
generateMyPrimes();
break;
case 2:
break;
}
}
private void generateSqrtPrimes(){
int curr = 3;
while(curr < Math.sqrt(maxNum)+1){
for(int i = curr*curr; i < Math.sqrt(maxNum)+1; i+=2*curr){
if(isPrime(i)){ // 2*curr because odd*2 = even!
crossOut(i);
}
}
curr = nextPrime(curr);
}
}
private void generateMyPrimes(){
int curr = start>(int)Math.sqrt(maxNum)?start:(int)Math.sqrt(maxNum);
while(curr < end){
for(int i = 3; i < Math.sqrt(maxNum)+1; i = nextPrime(i)){
if((curr%i) == 0){
if(isPrime(curr)){
crossOut(curr);
}
}
}
curr = nextPrime(curr);
}
}
}
}
如果有人能告诉我并发计划的瓶颈在哪里，我会非常高兴。提前谢谢！
答案 0 :(得分：3)
我不是JAVA编码器所以我坚持使用C ++。这也不是你问题的直接答案（对不起，但是我无法调试JAVA）把它作为一些指针去哪个方向去检查......
Eratosthenes的筛子
可以实现并行化，但速度增益不够大。相反，我使用更多的筛选标签，每个筛选器都有自己的子分区，每个表大小是所有子除数的共同乘法。这样您只需启动一次表，然后只需在O(1)
<强>并行化
检查完所有的筛子后，我会使用线程对所有未使用的除数进行明显的除法测试
<强> memoize的
如果你有所有找到的素数的有效表，那么除以素数并添加所有找到的新素数
我正在使用非并行素数搜索，这对我来说足够快......
[Edit1]更新代码
//---------------------------------------------------------------------------
int bits(DWORD p)
{
DWORD m=0x80000000; int b=32;
for (;m;m>>=1,b--)
if (p>=m) break;
return b;
}
//---------------------------------------------------------------------------
DWORD sqrt(const DWORD &x)
{
DWORD m,a;
m=(bits(x)>>1);
if (m) m=1<<m; else m=1;
for (a=0;m;m>>=1) { a|=m; if (a*a>x) a^=m; }
return a;
}
//---------------------------------------------------------------------------
List<int> primes_i32; // list of precomputed primes
const int primes_map_sz=4106301; // max size of map for speedup search for primes max(LCM(used primes per bit)) (not >>1 because SOE is periodic at double LCM size and only odd values are stored 2/2=1)
const int primes_map_N[8]={ 4106301,3905765,3585337,4026077,3386981,3460469,3340219,3974653 };
const int primes_map_i0=33; // first index of prime not used in mask
const int primes_map_p0=137; // biggest prime used in mask
BYTE primes_map[primes_map_sz]; // factors map for first i0-1 primes
bool primes_i32_alloc=false;
int isprime(int p)
{
int i,j,a,b,an,im[8]; BYTE u;
an=0;
if (!primes_i32.num) // init primes vars
{
primes_i32.allocate(1024*1024);
primes_i32.add( 2); for (i=1;i<primes_map_sz;i++) primes_map[i]=255; primes_map[0]=254;
primes_i32.add( 3); for (u=255- 1,j= 3,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 5); for (u=255- 2,j= 5,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 7); for (u=255- 4,j= 7,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 11); for (u=255- 8,j= 11,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 13); for (u=255- 16,j= 13,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 17); for (u=255- 32,j= 17,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 19); for (u=255- 64,j= 19,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 23); for (u=255-128,j= 23,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 29); for (u=255- 1,j=137,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 31); for (u=255- 2,j=131,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 37); for (u=255- 4,j=127,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 41); for (u=255- 8,j=113,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 43); for (u=255- 16,j= 83,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 47); for (u=255- 32,j= 61,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 53); for (u=255- 64,j=107,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 59); for (u=255-128,j=101,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 61); for (u=255- 1,j=103,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 67); for (u=255- 2,j= 67,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 71); for (u=255- 4,j= 37,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 73); for (u=255- 8,j= 41,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 79); for (u=255- 16,j= 43,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 83); for (u=255- 32,j= 47,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 89); for (u=255- 64,j= 53,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add( 97); for (u=255-128,j= 59,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add(101); for (u=255- 1,j= 97,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add(103); for (u=255- 2,j= 89,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add(107); for (u=255- 4,j=109,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add(109); for (u=255- 8,j= 79,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add(113); for (u=255- 16,j= 73,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add(127); for (u=255- 32,j= 71,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add(131); for (u=255- 64,j= 31,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
primes_i32.add(137); for (u=255-128,j= 29,i=j>>1;i<primes_map_sz;i+=j) primes_map[i]&=u;
}
if (!primes_i32_alloc)
{
if (p <=1) return 0; // ignore too small walues
if (p&1==0) return 0; // not prime if even
if (p>primes_map_p0)
{
j=p>>1;
i=j; if (i>=primes_map_N[0]) i%=primes_map_N[0]; if (!(primes_map[i]& 1)) return 0;
i=j; if (i>=primes_map_N[1]) i%=primes_map_N[1]; if (!(primes_map[i]& 2)) return 0;
i=j; if (i>=primes_map_N[2]) i%=primes_map_N[2]; if (!(primes_map[i]& 4)) return 0;
i=j; if (i>=primes_map_N[3]) i%=primes_map_N[3]; if (!(primes_map[i]& 8)) return 0;
i=j; if (i>=primes_map_N[4]) i%=primes_map_N[4]; if (!(primes_map[i]& 16)) return 0;
i=j; if (i>=primes_map_N[5]) i%=primes_map_N[5]; if (!(primes_map[i]& 32)) return 0;
i=j; if (i>=primes_map_N[6]) i%=primes_map_N[6]; if (!(primes_map[i]& 64)) return 0;
i=j; if (i>=primes_map_N[7]) i%=primes_map_N[7]; if (!(primes_map[i]&128)) return 0;
}
}
an=primes_i32[primes_i32.num-1];
if (an>=p)
{
// linear table search
if (p<127) // 31st prime
{
if (an>=p) for (i=0;i<primes_i32.num;i++)
{
a=primes_i32[i];
if (a==p) return 1;
if (a> p) return 0;
}
}
// approximation table search
else{
for (j=1,i=primes_i32.num-1;j<i;j<<=1); j>>=1; if (!j) j=1;
for (i=0;j;j>>=1)
{
i|=j;
if (i>=primes_i32.num) { i-=j; continue; }
a=primes_i32[i];
if (a==p) return 1;
if (a> p) i-=j;
}
return 0;
}
}
a=an; a+=2;
for (j=a>>1,i=0;i<8;i++) im[i]=j%primes_map_N[i];
an=(1<<((bits(p)>>1)+1))-1; if (an<=0) an=1;
an=an+an;
for (;a<=p;a+=2)
{
for (j=1,i=0;i<8;i++,j<<=1) // check if map is set
if (!(primes_map[im[i]]&j)) { j=-1; break; } // if not dont bother with division
for (i=0;i<8;i++) { im[i]++; if (im[i]>=primes_map_N[i]) im[i]-=primes_map_N[i]; }
if (j<0) continue;
for (i=primes_map_i0;i<primes_i32.num;i++)
{
b=primes_i32[i];
if (b>an) break;
if ((a%b)==0) { i=-1; break; }
}
if (i<0) continue;
primes_i32.add(a);
if (a==p) return 1;
if (a> p) return 0;
}
return 0;
}
//---------------------------------------------------------------------------
void getprimes(int p) // compute and allocate primes up to p
{
if (!primes_i32.num) isprime(3);
int p0=primes_i32[primes_i32.num-1]; // biggest prime computed yet
if (p>p0+10000) // if too big difference use sieves to fast precompute
{
// T((0.3516+0.5756*log10(n))*n) -> O(n.log(n))
// sieves N/16 bytes p=100 000 000 t=1903.031 ms
// ------------------------------
// 0 1 2 3 4 5 6 7 bit
// ------------------------------
// 1 3 5 7 9 11 13 15 +-> +2
// 17 19 21 23 25 27 29 31 |
// 33 35 37 39 41 43 45 47 V +16
// ------------------------------
int N=(p|15),M=(N>>4); // store only odd values 1,3,5,7,... each bit ...
char *m=new char[M+1]; // m[i] -> is 1+i+i prime? (factors map)
int i,j,k,n;
// init sieves
m[0]=254; for (i=1;i<=M;i++) m[i]=255;
for(n=sqrt(p),i=1;i<=n;)
{
int a=m[i>>4];
if (int(a& 1)!=0) for(k=i+i,j=i+k;j<=N;j+=k) m[j>>4]&=255-(1<<((j>>1)&7)); i+=2;
if (int(a& 2)!=0) for(k=i+i,j=i+k;j<=N;j+=k) m[j>>4]&=255-(1<<((j>>1)&7)); i+=2;
if (int(a& 4)!=0) for(k=i+i,j=i+k;j<=N;j+=k) m[j>>4]&=255-(1<<((j>>1)&7)); i+=2;
if (int(a& 8)!=0) for(k=i+i,j=i+k;j<=N;j+=k) m[j>>4]&=255-(1<<((j>>1)&7)); i+=2;
if (int(a& 16)!=0) for(k=i+i,j=i+k;j<=N;j+=k) m[j>>4]&=255-(1<<((j>>1)&7)); i+=2;
if (int(a& 32)!=0) for(k=i+i,j=i+k;j<=N;j+=k) m[j>>4]&=255-(1<<((j>>1)&7)); i+=2;
if (int(a& 64)!=0) for(k=i+i,j=i+k;j<=N;j+=k) m[j>>4]&=255-(1<<((j>>1)&7)); i+=2;
if (int(a&128)!=0) for(k=i+i,j=i+k;j<=N;j+=k) m[j>>4]&=255-(1<<((j>>1)&7)); i+=2;
}
// compute primes
i=p0&0xFFFFFFF1; k=m[i>>4]; // start after last found prime in list
if ((int(k& 1)!=0)&&(i>p0)) primes_i32.add(i); i+=2;
if ((int(k& 2)!=0)&&(i>p0)) primes_i32.add(i); i+=2;
if ((int(k& 4)!=0)&&(i>p0)) primes_i32.add(i); i+=2;
if ((int(k& 8)!=0)&&(i>p0)) primes_i32.add(i); i+=2;
if ((int(k& 16)!=0)&&(i>p0)) primes_i32.add(i); i+=2;
if ((int(k& 32)!=0)&&(i>p0)) primes_i32.add(i); i+=2;
if ((int(k& 64)!=0)&&(i>p0)) primes_i32.add(i); i+=2;
if ((int(k&128)!=0)&&(i>p0)) primes_i32.add(i); i+=2;
for(j=i>>4;j<M;i+=16,j++) // continue with 16-blocks
{
k=m[j];
if (!k) continue;
if (int(k& 1)!=0) primes_i32.add(i );
if (int(k& 2)!=0) primes_i32.add(i+ 2);
if (int(k& 4)!=0) primes_i32.add(i+ 4);
if (int(k& 8)!=0) primes_i32.add(i+ 6);
if (int(k& 16)!=0) primes_i32.add(i+ 8);
if (int(k& 32)!=0) primes_i32.add(i+10);
if (int(k& 64)!=0) primes_i32.add(i+12);
if (int(k&128)!=0) primes_i32.add(i+14);
}
k=m[j]; // do the last primes
if ((int(k& 1)!=0)&&(i<=p)) primes_i32.add(i); i+=2;
if ((int(k& 2)!=0)&&(i<=p)) primes_i32.add(i); i+=2;
if ((int(k& 4)!=0)&&(i<=p)) primes_i32.add(i); i+=2;
if ((int(k& 8)!=0)&&(i<=p)) primes_i32.add(i); i+=2;
if ((int(k& 16)!=0)&&(i<=p)) primes_i32.add(i); i+=2;
if ((int(k& 32)!=0)&&(i<=p)) primes_i32.add(i); i+=2;
if ((int(k& 64)!=0)&&(i<=p)) primes_i32.add(i); i+=2;
if ((int(k&128)!=0)&&(i<=p)) primes_i32.add(i); i+=2;
delete[] m;
}
else{
bool b0=primes_i32_alloc;
primes_i32_alloc=true;
isprime(p);
primes_i32_alloc=false;
primes_i32_alloc=b0;
}
}
//---------------------------------------------------------------------------
getprimes(p)功能，可以快速将所有primes<=p添加到列表中，如果它们还没有那么getprimes(15 485 863)在我设置的175.563毫秒内解决了这个问题 isprime的速度要慢一些
primes_i32是int s的动态列表
primes_i32.num是列表中int的数量primes_i32[i]是i - int i = <0,primes_i32.num-1> primes_i32.add(x)将x添加到列表末尾primes_i32.allocate(N)为列表中的N项预分配空间，以避免重新分配速度减慢<强> [注释]
我使用这个非并行版本的欧拉问题10（所有素数的总和低于2000000）
----------------------------------------------------------------------------------
Time ID Reference | My solution | Note
----------------------------------------------------------------------------------
[ 35.639 ms] Problem010. 142913828922 | 142913828922 | 64_bit
primes_map[]数组中的一个位片，只使用奇数值（甚至不需要记住筛子）。isprime(max value)并阅读primes_i32[]的内容希望我没有忘记在这里复制一些东西