我正在尝试计算矩阵(任意大小)的行列式，用于 self 编码/面试练习。我的第一次尝试是使用递归，这使我实现了以下实现:
import java.util.Scanner.*;
public class Determinant {
double A[][];
double m[][];
int N;
int start;
int last;
public Determinant (double A[][], int N, int start, int last){
this.A = A;
this.N = N;
this.start = start;
this.last = last;
}
public double[][] generateSubArray (double A[][], int N, int j1){
m = new double[N-1][];
for (int k=0; k<(N-1); k++)
m[k] = new double[N-1];
for (int i=1; i<N; i++){
int j2=0;
for (int j=0; j<N; j++){
if(j == j1)
continue;
m[i-1][j2] = A[i][j];
j2++;
}
}
return m;
}
/*
* Calculate determinant recursively
*/
public double determinant(double A[][], int N){
double res;
// Trivial 1x1 matrix
if (N == 1) res = A[0][0];
// Trivial 2x2 matrix
else if (N == 2) res = A[0][0]*A[1][1] - A[1][0]*A[0][1];
// NxN matrix
else{
res=0;
for (int j1=0; j1<N; j1++){
m = generateSubArray (A, N, j1);
res += Math.pow(-1.0, 1.0+j1+1.0) * A[0][j1] * determinant(m, N-1);
}
}
return res;
}
}
到目前为止一切都很好，它给了我一个正确的结果。现在我想通过使用多个线程来计算这个行列式值来优化我的代码。 我尝试使用 Java Fork/Join 模型对其进行并行化。这是我的方法:
@Override
protected Double compute() {
if (N < THRESHOLD) {
result = computeDeterminant(A, N);
return result;
}
for (int j1 = 0; j1 < N; j1++){
m = generateSubArray (A, N, j1);
ParallelDeterminants d = new ParallelDeterminants (m, N-1);
d.fork();
result += Math.pow(-1.0, 1.0+j1+1.0) * A[0][j1] * d.join();
}
return result;
}
public double computeDeterminant(double A[][], int N){
double res;
// Trivial 1x1 matrix
if (N == 1) res = A[0][0];
// Trivial 2x2 matrix
else if (N == 2) res = A[0][0]*A[1][1] - A[1][0]*A[0][1];
// NxN matrix
else{
res=0;
for (int j1=0; j1<N; j1++){
m = generateSubArray (A, N, j1);
res += Math.pow(-1.0, 1.0+j1+1.0) * A[0][j1] * computeDeterminant(m, N-1);
}
}
return res;
}
/*
* Main function
*/
public static void main(String args[]){
double res;
ForkJoinPool pool = new ForkJoinPool();
ParallelDeterminants d = new ParallelDeterminants();
d.inputData();
long starttime=System.nanoTime();
res = pool.invoke (d);
long EndTime=System.nanoTime();
System.out.println("Seq Run = "+ (EndTime-starttime)/100000);
System.out.println("the determinant valaue is " + res);
}
但是对比性能后发现，Fork/Join方式的性能很差，矩阵维数越高越慢(相比第一种方式)。开销在哪里？谁能阐明如何改进这一点？
最佳答案
使用此类，您可以计算具有任何维度的矩阵的行列式
此类使用许多不同的方法使矩阵成为三角形，然后计算它的行列式。它可用于 500 x 500 甚至更大的高维矩阵。此类的优点是您可以获得BigDecimal 中的结果，因此没有无穷大并且您将始终获得准确的答案。顺便说一句，使用许多不同的方法并避免递归导致更快的方法和更高的答案性能。希望对您有所帮助。
import java.math.BigDecimal;
public class DeterminantCalc {
private double[][] matrix;
private int sign = 1;
DeterminantCalc(double[][] matrix) {
this.matrix = matrix;
}
public int getSign() {
return sign;
}
public BigDecimal determinant() {
BigDecimal deter;
if (isUpperTriangular() || isLowerTriangular())
deter = multiplyDiameter().multiply(BigDecimal.valueOf(sign));
else {
makeTriangular();
deter = multiplyDiameter().multiply(BigDecimal.valueOf(sign));
}
return deter;
}
/* receives a matrix and makes it triangular using allowed operations
on columns and rows
*/
public void makeTriangular() {
for (int j = 0; j < matrix.length; j++) {
sortCol(j);
for (int i = matrix.length - 1; i > j; i--) {
if (matrix[i][j] == 0)
continue;
double x = matrix[i][j];
double y = matrix[i - 1][j];
multiplyRow(i, (-y / x));
addRow(i, i - 1);
multiplyRow(i, (-x / y));
}
}
}
public boolean isUpperTriangular() {
if (matrix.length < 2)
return false;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < i; j++) {
if (matrix[i][j] != 0)
return false;
}
}
return true;
}
public boolean isLowerTriangular() {
if (matrix.length < 2)
return false;
for (int j = 0; j < matrix.length; j++) {
for (int i = 0; j > i; i++) {
if (matrix[i][j] != 0)
return false;
}
}
return true;
}
public BigDecimal multiplyDiameter() {
BigDecimal result = BigDecimal.ONE;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix.length; j++) {
if (i == j)
result = result.multiply(BigDecimal.valueOf(matrix[i][j]));
}
}
return result;
}
// when matrix[i][j] = 0 it makes it's value non-zero
public void makeNonZero(int rowPos, int colPos) {
int len = matrix.length;
outer:
for (int i = 0; i < len; i++) {
for (int j = 0; j < len; j++) {
if (matrix[i][j] != 0) {
if (i == rowPos) { // found "!= 0" in it's own row, so cols must be added
addCol(colPos, j);
break outer;
}
if (j == colPos) { // found "!= 0" in it's own col, so rows must be added
addRow(rowPos, i);
break outer;
}
}
}
}
}
//add row1 to row2 and store in row1
public void addRow(int row1, int row2) {
for (int j = 0; j < matrix.length; j++)
matrix[row1][j] += matrix[row2][j];
}
//add col1 to col2 and store in col1
public void addCol(int col1, int col2) {
for (int i = 0; i < matrix.length; i++)
matrix[i][col1] += matrix[i][col2];
}
//multiply the whole row by num
public void multiplyRow(int row, double num) {
if (num < 0)
sign *= -1;
for (int j = 0; j < matrix.length; j++) {
matrix[row][j] *= num;
}
}
//multiply the whole column by num
public void multiplyCol(int col, double num) {
if (num < 0)
sign *= -1;
for (int i = 0; i < matrix.length; i++)
matrix[i][col] *= num;
}
// sort the cols from the biggest to the lowest value
public void sortCol(int col) {
for (int i = matrix.length - 1; i >= col; i--) {
for (int k = matrix.length - 1; k >= col; k--) {
double tmp1 = matrix[i][col];
double tmp2 = matrix[k][col];
if (Math.abs(tmp1) < Math.abs(tmp2))
replaceRow(i, k);
}
}
}
//replace row1 with row2
public void replaceRow(int row1, int row2) {
if (row1 != row2)
sign *= -1;
double[] tempRow = new double[matrix.length];
for (int j = 0; j < matrix.length; j++) {
tempRow[j] = matrix[row1][j];
matrix[row1][j] = matrix[row2][j];
matrix[row2][j] = tempRow[j];
}
}
//replace col1 with col2
public void replaceCol(int col1, int col2) {
if (col1 != col2)
sign *= -1;
System.out.printf("replace col%d with col%d, sign = %d%n", col1, col2, sign);
double[][] tempCol = new double[matrix.length][1];
for (int i = 0; i < matrix.length; i++) {
tempCol[i][0] = matrix[i][col1];
matrix[i][col1] = matrix[i][col2];
matrix[i][col2] = tempCol[i][0];
}
} }
此类从用户那里接收一个 n x n 的矩阵，然后计算它的行列式。它还显示了解决方案和最终的三角矩阵。
import java.math.BigDecimal;
import java.text.NumberFormat;
import java.util.Scanner;
public class DeterminantTest {
public static void main(String[] args) {
String determinant;
//generating random numbers
/*int len = 300;
SecureRandom random = new SecureRandom();
double[][] matrix = new double[len][len];
for (int i = 0; i < len; i++) {
for (int j = 0; j < len; j++) {
matrix[i][j] = random.nextInt(500);
System.out.printf("%15.2f", matrix[i][j]);
}
}
System.out.println();*/
/*double[][] matrix = {
{1, 5, 2, -2, 3, 2, 5, 1, 0, 5},
{4, 6, 0, -2, -2, 0, 1, 1, -2, 1},
{0, 5, 1, 0, 1, -5, -9, 0, 4, 1},
{2, 3, 5, -1, 2, 2, 0, 4, 5, -1},
{1, 0, 3, -1, 5, 1, 0, 2, 0, 2},
{1, 1, 0, -2, 5, 1, 2, 1, 1, 6},
{1, 0, 1, -1, 1, 1, 0, 1, 1, 1},
{1, 5, 5, 0, 3, 5, 5, 0, 0, 6},
{1, -5, 2, -2, 3, 2, 5, 1, 1, 5},
{1, 5, -2, -2, 3, 1, 5, 0, 0, 1}
};
*/
double[][] matrix = menu();
DeterminantCalc deter = new DeterminantCalc(matrix);
BigDecimal det = deter.determinant();
determinant = NumberFormat.getInstance().format(det);
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix.length; j++) {
System.out.printf("%15.2f", matrix[i][j]);
}
System.out.println();
}
System.out.println();
System.out.printf("%s%s%n", "Determinant: ", determinant);
System.out.printf("%s%d", "sign: ", deter.getSign());
}
public static double[][] menu() {
Scanner scanner = new Scanner(System.in);
System.out.print("Matrix Dimension: ");
int dim = scanner.nextInt();
double[][] inputMatrix = new double[dim][dim];
System.out.println("Set the Matrix: ");
for (int i = 0; i < dim; i++) {
System.out.printf("%5s%d%n", "row", i + 1);
for (int j = 0; j < dim; j++) {
System.out.printf("M[%d][%d] = ", i + 1, j + 1);
inputMatrix[i][j] = scanner.nextDouble();
}
System.out.println();
}
scanner.close();
return inputMatrix;
}}
关于java - 计算矩阵行列式，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/16602350/