昨天我使用fork / join框架here在Java 7中询问了有关并行矩阵乘法的问题。在axtavt的帮助下，我得到了我的示例程序。现在我只使用Java 6功能实现一个等效的程序。我得到了和昨天一样的问题，应用反馈axtavt给了我（我想）。我忽略了什么吗？ 代码：
package algorithms;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.TimeUnit;
public class Java6MatrixMultiply implements Algorithm {
private static final int SIZE = 1024;
private static final int THRESHOLD = 64;
private static final int MAX_THREADS = Runtime.getRuntime().availableProcessors();
private final ExecutorService executor = Executors.newFixedThreadPool(MAX_THREADS);
private float[][] a = new float[SIZE][SIZE];
private float[][] b = new float[SIZE][SIZE];
private float[][] c = new float[SIZE][SIZE];
@Override
public void initialize() {
init(a, b, SIZE);
}
@Override
public void execute() {
MatrixMultiplyTask task = new MatrixMultiplyTask(a, 0, 0, b, 0, 0, c, 0, 0, SIZE);
task.split();
executor.shutdown();
try {
executor.awaitTermination(Integer.MAX_VALUE, TimeUnit.DAYS);
} catch (InterruptedException e) {
System.out.println("Error: " + e.getMessage());
}
}
@Override
public void printResult() {
check(c, SIZE);
for (int i = 0; i < SIZE && i <= 10; i++) {
for (int j = 0; j < SIZE && j <= 10; j++) {
if(j == 10) {
System.out.print("...");
}
else {
System.out.print(c[i][j] + " ");
}
}
if(i == 10) {
System.out.println();
for(int k = 0; k < 10; k++) System.out.print(" ... ");
}
System.out.println();
}
System.out.println();
}
// To simplify checking, fill with all 1's. Answer should be all n's.
static void init(float[][] a, float[][] b, int n) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
a[i][j] = 1.0F;
b[i][j] = 1.0F;
}
}
}
static void check(float[][] c, int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (c[i][j] != n) {
throw new Error("Check Failed at [" + i + "][" + j + "]: " + c[i][j]);
//System.out.println("Check Failed at [" + i + "][" + j + "]: " + c[i][j]);
}
}
}
}
public class Seq implements Runnable {
private final MatrixMultiplyTask a;
private final MatrixMultiplyTask b;
public Seq(MatrixMultiplyTask a, MatrixMultiplyTask b, int size) {
this.a = a;
this.b = b;
if (size <= THRESHOLD) {
executor.submit(this);
} else {
a.split();
b.split();
}
}
public void run() {
a.multiplyStride2();
b.multiplyStride2();
}
}
private class MatrixMultiplyTask {
private final float[][] A; // Matrix A
private final int aRow; // first row of current quadrant of A
private final int aCol; // first column of current quadrant of A
private final float[][] B; // Similarly for B
private final int bRow;
private final int bCol;
private final float[][] C; // Similarly for result matrix C
private final int cRow;
private final int cCol;
private final int size;
MatrixMultiplyTask(float[][] A, int aRow, int aCol, float[][] B,
int bRow, int bCol, float[][] C, int cRow, int cCol, int size) {
this.A = A;
this.aRow = aRow;
this.aCol = aCol;
this.B = B;
this.bRow = bRow;
this.bCol = bCol;
this.C = C;
this.cRow = cRow;
this.cCol = cCol;
this.size = size;
}
public void split() {
int h = size / 2;
new Seq(new MatrixMultiplyTask(A,
aRow, aCol, // A11
B, bRow, bCol, // B11
C, cRow, cCol, // C11
h),
new MatrixMultiplyTask(A, aRow, aCol + h, // A12
B, bRow + h, bCol, // B21
C, cRow, cCol, // C11
h), h);
new Seq(new MatrixMultiplyTask(A,
aRow, aCol, // A11
B, bRow, bCol + h, // B12
C, cRow, cCol + h, // C12
h),
new MatrixMultiplyTask(A, aRow, aCol + h, // A12
B, bRow + h, bCol + h, // B22
C, cRow, cCol + h, // C12
h), h);
new Seq(new MatrixMultiplyTask(A, aRow
+ h, aCol, // A21
B, bRow, bCol, // B11
C, cRow + h, cCol, // C21
h),
new MatrixMultiplyTask(A, aRow + h, aCol + h, // A22
B, bRow + h, bCol, // B21
C, cRow + h, cCol, // C21
h), h);
new Seq(new MatrixMultiplyTask(A, aRow
+ h, aCol, // A21
B, bRow, bCol + h, // B12
C, cRow + h, cCol + h, // C22
h),
new MatrixMultiplyTask(A, aRow + h, aCol + h, // A22
B, bRow + h, bCol + h, // B22
C, cRow + h, cCol + h, // C22
h), h);
}
public void multiplyStride2() {
for (int j = 0; j < size; j += 2) {
for (int i = 0; i < size; i += 2) {
float[] a0 = A[aRow + i];
float[] a1 = A[aRow + i + 1];
float s00 = 0.0F;
float s01 = 0.0F;
float s10 = 0.0F;
float s11 = 0.0F;
for (int k = 0; k < size; k += 2) {
float[] b0 = B[bRow + k];
s00 += a0[aCol + k] * b0[bCol + j];
s10 += a1[aCol + k] * b0[bCol + j];
s01 += a0[aCol + k] * b0[bCol + j + 1];
s11 += a1[aCol + k] * b0[bCol + j + 1];
float[] b1 = B[bRow + k + 1];
s00 += a0[aCol + k + 1] * b1[bCol + j];
s10 += a1[aCol + k + 1] * b1[bCol + j];
s01 += a0[aCol + k + 1] * b1[bCol + j + 1];
s11 += a1[aCol + k + 1] * b1[bCol + j + 1];
}
C[cRow + i][cCol + j] += s00;
C[cRow + i][cCol + j + 1] += s01;
C[cRow + i + 1][cCol + j] += s10;
C[cRow + i + 1][cCol + j + 1] += s11;
}
}
}
}
}
答案 0 :(得分：3)
我尝试按照我的建议添加synchronized，这解决了问题。 ;）
我试过
-XX:-UseBiasedLocking 207 ms double代替float 237 ms。正如您所看到的，第五个选项比没有同步慢不到10％。如果你想进一步增加，我建议你改变访问数据的方式，使它们更加缓存友好。
总结我建议
private final ExecutorService executor = Executors.newFixedThreadPool(MAX_THREADS*2);
public void multiplyStride2() {
for (int i = 0; i < size; i += 2) {
for (int j = 0; j < size; j += 2) {
// code as is......
synchronized (C[cRow + i]) {
C[cRow + i][cCol + j] += s00;
C[cRow + i][cCol + j + 1] += s01;
C[cRow + i + 1][cCol + j] += s10;
C[cRow + i + 1][cCol + j + 1] += s11;
}
有趣的是，如果我计算2x4 2x2的块，则平均时间下降到172 ms。 （比没有同步的先前结果快）;）
答案 1 :(得分：2)
在阅读了这个this问题后，我决定调整我的程序。我的新程序运行良好，没有同步。谢谢你的想法彼得。
新代码：
package algorithms;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
import java.util.concurrent.FutureTask;
public class Java6MatrixMultiply implements Algorithm {
private static final int SIZE = 2048;
private static final int THRESHOLD = 64;
private static final int MAX_THREADS = Runtime.getRuntime().availableProcessors();
private final ExecutorService executor = Executors.newFixedThreadPool(MAX_THREADS);
private float[][] a = new float[SIZE][SIZE];
private float[][] b = new float[SIZE][SIZE];
private float[][] c = new float[SIZE][SIZE];
@Override
public void initialize() {
init(a, b, SIZE);
}
@Override
public void execute() {
MatrixMultiplyTask mainTask = new MatrixMultiplyTask(a, 0, 0, b, 0, 0, c, 0, 0, SIZE);
Future future = executor.submit(mainTask);
try {
future.get();
} catch (Exception e) {
System.out.println("Error: " + e.getMessage());
}
}
@Override
public void printResult() {
check(c, SIZE);
for (int i = 0; i < SIZE && i <= 10; i++) {
for (int j = 0; j < SIZE && j <= 10; j++) {
if(j == 10) {
System.out.print("...");
}
else {
System.out.print(c[i][j] + " ");
}
}
if(i == 10) {
System.out.println();
for(int k = 0; k < 10; k++) System.out.print(" ... ");
}
System.out.println();
}
System.out.println();
}
// To simplify checking, fill with all 1's. Answer should be all n's.
static void init(float[][] a, float[][] b, int n) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
a[i][j] = 1.0F;
b[i][j] = 1.0F;
}
}
}
static void check(float[][] c, int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (c[i][j] != n) {
throw new Error("Check Failed at [" + i + "][" + j + "]: " + c[i][j]);
//System.out.println("Check Failed at [" + i + "][" + j + "]: " + c[i][j]);
}
}
}
}
public class Seq implements Runnable {
private final MatrixMultiplyTask a;
private final MatrixMultiplyTask b;
public Seq(MatrixMultiplyTask a, MatrixMultiplyTask b) {
this.a = a;
this.b = b;
}
public void run() {
a.run();
b.run();
}
}
private class MatrixMultiplyTask implements Runnable {
private final float[][] A; // Matrix A
private final int aRow; // first row of current quadrant of A
private final int aCol; // first column of current quadrant of A
private final float[][] B; // Similarly for B
private final int bRow;
private final int bCol;
private final float[][] C; // Similarly for result matrix C
private final int cRow;
private final int cCol;
private final int size;
public MatrixMultiplyTask(float[][] A, int aRow, int aCol, float[][] B,
int bRow, int bCol, float[][] C, int cRow, int cCol, int size) {
this.A = A;
this.aRow = aRow;
this.aCol = aCol;
this.B = B;
this.bRow = bRow;
this.bCol = bCol;
this.C = C;
this.cRow = cRow;
this.cCol = cCol;
this.size = size;
}
public void run() {
//System.out.println("Thread: " + Thread.currentThread().getName());
if (size <= THRESHOLD) {
multiplyStride2();
} else {
int h = size / 2;
Seq seq1 = new Seq(new MatrixMultiplyTask(A,
aRow, aCol, // A11
B, bRow, bCol, // B11
C, cRow, cCol, // C11
h),
new MatrixMultiplyTask(A, aRow, aCol + h, // A12
B, bRow + h, bCol, // B21
C, cRow, cCol, // C11
h));
Seq seq2 = new Seq(new MatrixMultiplyTask(A,
aRow, aCol, // A11
B, bRow, bCol + h, // B12
C, cRow, cCol + h, // C12
h),
new MatrixMultiplyTask(A, aRow, aCol + h, // A12
B, bRow + h, bCol + h, // B22
C, cRow, cCol + h, // C12
h));
Seq seq3 = new Seq(new MatrixMultiplyTask(A, aRow
+ h, aCol, // A21
B, bRow, bCol, // B11
C, cRow + h, cCol, // C21
h),
new MatrixMultiplyTask(A, aRow + h, aCol + h, // A22
B, bRow + h, bCol, // B21
C, cRow + h, cCol, // C21
h));
Seq seq4 = new Seq(new MatrixMultiplyTask(A, aRow
+ h, aCol, // A21
B, bRow, bCol + h, // B12
C, cRow + h, cCol + h, // C22
h),
new MatrixMultiplyTask(A, aRow + h, aCol + h, // A22
B, bRow + h, bCol + h, // B22
C, cRow + h, cCol + h, // C22
h));
final FutureTask s1Task = new FutureTask(seq2, null);
final FutureTask s2Task = new FutureTask(seq3, null);
final FutureTask s3Task = new FutureTask(seq4, null);
executor.execute(s1Task);
executor.execute(s2Task);
executor.execute(s3Task);
seq1.run();
s1Task.run();
s2Task.run();
s3Task.run();
try {
s1Task.get();
s2Task.get();
s3Task.get();
} catch (Exception e) {
System.out.println("Error: " + e.getMessage());
executor.shutdownNow();
}
}
}
public void multiplyStride2() {
for (int j = 0; j < size; j += 2) {
for (int i = 0; i < size; i += 2) {
float[] a0 = A[aRow + i];
float[] a1 = A[aRow + i + 1];
float s00 = 0.0F;
float s01 = 0.0F;
float s10 = 0.0F;
float s11 = 0.0F;
for (int k = 0; k < size; k += 2) {
float[] b0 = B[bRow + k];
s00 += a0[aCol + k] * b0[bCol + j];
s10 += a1[aCol + k] * b0[bCol + j];
s01 += a0[aCol + k] * b0[bCol + j + 1];
s11 += a1[aCol + k] * b0[bCol + j + 1];
float[] b1 = B[bRow + k + 1];
s00 += a0[aCol + k + 1] * b1[bCol + j];
s10 += a1[aCol + k + 1] * b1[bCol + j];
s01 += a0[aCol + k + 1] * b1[bCol + j + 1];
s11 += a1[aCol + k + 1] * b1[bCol + j + 1];
}
C[cRow + i][cCol + j] += s00;
C[cRow + i][cCol + j + 1] += s01;
C[cRow + i + 1][cCol + j] += s10;
C[cRow + i + 1][cCol + j + 1] += s11;
}
}
}
}
}