计算矩阵行列式

12

我正在尝试计算矩阵(任意大小)的行列式,以进行编码/面试练习。我的第一次尝试是使用递归,并导致了以下实现:

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;
    }
}

目前为止一切都很好,我得到了正确的结果。现在我想通过使用多线程来优化我的代码来计算这个行列式的值。 我尝试使用Java Fork/Join模型并行化它。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);
}
然而,经过比较性能后,我发现Fork/Join方法的性能非常差,矩阵维度越高,速度变得越慢(与第一种方法相比)。开销在哪里?有人可以指点如何改进吗?
然而,比较性能后,我发现 Fork/Join 方法的性能非常差,矩阵维度越高,速度变得越慢(与第一种方法相比)。开销在哪里?有人可以指点如何改进吗?

1
在引入线程之前,我会停止在循环中进行分配。一个选项可能是使用两个数组参数来确定要计算的列和行,而不是使用N。 - Stefan Haustein
我建议您查看一些旨在并行化设计的算法。虽然我没有详细审查过您的算法,但是在我的经验中,通过搜索常见问题,可以找到许多聪明的优化方法。 - Osama Javed
5个回答

2
使用这个类,你可以计算任意维度的矩阵的行列式。
这个类使用了许多不同的方法来使矩阵三角化,然后计算它的行列式。它可以用于高维矩阵,如500 x 500甚至更大。这个类的优点是你可以得到BigDecimal类型的结果,所以没有无穷大,而且你总是能得到精确的答案。顺便说一下,使用许多不同的方法和避免递归导致了更快速、性能更高的答案。希望这对你有所帮助。
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];
    }
} }

这个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。它还显示解决方案和最终的上三角矩阵。

 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;
}}

1
他并没有要求算法/解决方案来找到行列式!他在问题中明确提到了多线程! - Yahya
1
这个算法解决了性能问题,对于非常高的维度来说结果非常快,我相信值得测试。 - Seyyed Mohsen Mousavi
1
@SeyyedMohsenMousavi 哥们,这个算法超级快!在互联网上有没有它的描述? - Václav
谢谢,您说的描述是指算法本身还是速度? - Seyyed Mohsen Mousavi
1
@SeyyedMohsenMousavi 算法本身。也许是一章书或一篇文章。这些三角形-我以前从未见过。 - Václav
1
@Václav 如果你正在寻找一本书,可以看看这本来自Adams的,其中有矩阵部分。而如果你只是想理解这个算法的基础,可以在youtube上观看这个视频 - Seyyed Mohsen Mousavi

1
ForkJoin代码较慢的主要原因是它实际上是序列化的,并带有一些线程开销。要从fork/join中受益,您需要先fork所有实例,然后等待结果。将“compute”中的循环拆分为两个循环:一个用于fork(将ParallelDeterminants的实例存储在数组中),另一个用于收集结果。
另外,我建议仅在最外层级别进行fork,而不是任何内部级别。您不希望创建O(N^2)个线程。

1

有一种新的计算矩阵行列式的方法,你可以从这里了解更多。

我已经用简单的Java实现了一个简单版本,没有使用任何高级优化技术或库,并且我已经测试过以前描述的方法,平均速度比之前的快了10倍。

public class Test {
public static double[][] reduce(int row , int column , double[][] mat){
    int n=mat.length;
    double[][] res = new double[n- 1][n- 1];
    int r=0,c=0;
    for (int i = 0; i < n; i++) {
        c=0;
        if(i==row)
            continue;
        for (int j = 0; j < n; j++) {
            if(j==column)
                continue;
            res[r][c] = mat[i][j];

            c++;
        }
        r++;
    }
    return res;
}

public static double det(double mat[][]){
    int n = mat.length;
    if(n==1)
        return mat[0][0];
    if(n==2)
        return mat[0][0]*mat[1][1] - (mat[0][1]*mat[1][0]);
    //TODO : do reduce more efficiently
    double[][] m11 = reduce(0,0,mat);
    double[][] m1n = reduce(0,n-1,mat);
    double[][] mn1 = reduce(n-1 , 0 , mat);
    double[][] mnn = reduce(n-1,n-1,mat);
    double[][] m11nn = reduce(0,0,reduce(n-1,n-1,mat));
    return (det(m11)*det(mnn) - det(m1n)*det(mn1))/det(m11nn);
}

public static double[][] randomMatrix(int n , int range){
    double[][] mat = new double[n][n];
    for (int i=0; i<mat.length; i++) {
        for (int j=0; j<mat[i].length; j++) {
            mat[i][j] = (Math.random()*range);
        }
    }
    return mat;
}

public static void main(String[] args) {
    double[][] mat = randomMatrix(10,100);
    System.out.println(det(mat));
}
}

如果m11nn行列式为零,会出现一点小错误,此时程序会崩溃,需要进行检查。我对100个随机样本进行了测试,很少发生这种情况,但我认为值得一提。同时,使用更好的索引方案也可以提高效率。


0
这是我的Matrix类的一部分,它使用一个名为data的double[][]成员变量来存储矩阵数据。 _determinant_recursivetask_impl()函数使用一个带有ForkJoinPool的RecursiveTask对象,尝试使用多个线程进行计算。
与获取上/下三角矩阵的矩阵运算相比,这种方法速度非常慢。例如,尝试计算一个13x13矩阵的行列式。
public class Matrix
{
    // Dimensions
    private final int I,J;
    private final double[][] data;
    private Double determinant = null;
    static class MatrixEntry
    {
        public final int I,J;
        public final double value;
        private MatrixEntry(int i, int j, double value) {
            I = i;
            J = j;
            this.value = value;
        }
    }

    /**
     * Calculates determinant of this Matrix recursively and caches it for future use.
     * @return determinant
     */
    public double determinant()
    {
        if(I!=J)
            throw new IllegalStateException(String.format("Can't calculate determinant of (%d,%d) matrix, not a square matrix.", I,J));
        if(determinant==null)
            determinant = _determinant_recursivetask_impl(this);
        return determinant;
    }
    private static double _determinant_recursivetask_impl(Matrix m)
    {
        class determinant_recurse extends RecursiveTask<Double>
        {
            private final Matrix m;
            determinant_recurse(Matrix m) {
                this.m = m;
            }

            @Override
            protected Double compute() {
                // Base cases
                if(m.I==1 && m.J==1)
                    return m.data[0][0];
                else if(m.I==2 && m.J==2)
                    return m.data[0][0]*m.data[1][1] - m.data[0][1]*m.data[1][0];
                else
                {
                    determinant_recurse[] tasks = new determinant_recurse[m.I];
                    for (int i = 0; i <m.I ; i++) {
                        tasks[i] = new determinant_recurse(m.getSubmatrix(0, i));
                    }
                    for (int i = 1; i <m.I ; i++) {
                        tasks[i].fork();
                    }
                    double ret = m.data[0][0]*tasks[0].compute();
                    for (int i = 1; i < m.I; i++) {
                        if(i%2==0)
                            ret += m.data[0][i]*tasks[i].join();
                        else
                            ret -= m.data[0][i]*tasks[i].join();
                    }
                    return ret;
                }
            }
        }
        return ForkJoinPool.commonPool().invoke(new determinant_recurse(m));
    }

    private static void _map_impl(Matrix ret, Function<Matrix.MatrixEntry, Double> operator)
    {
        for (int i = 0; i <ret.I ; i++) {
            for (int j = 0; j <ret.J ; j++) {
                ret.data[i][j] = operator.apply(new Matrix.MatrixEntry(i,j,ret.data[i][j]));
            }
        }
    }
    /**
     * Returns a new Matrix that is sub-matrix without the given row and column.
     * @param removeI row to remove
     * @param removeJ col. to remove
     * @return new Matrix.
     */
    public Matrix getSubmatrix(int removeI, int removeJ)
    {
        if(removeI<0 || removeJ<0 || removeI>=this.I || removeJ>=this.J)
            throw new IllegalArgumentException(String.format("Invalid element position (%d,%d) for matrix(%d,%d).", removeI,removeJ,this.I,this.J));
        Matrix m = new Matrix(this.I-1, this.J-1);
        _map_impl(m, (e)->{
            int i = e.I, j = e.J;
            if(e.I >= removeI) ++i;
            if(e.J >= removeJ) ++j;
            return this.data[i][j];
        });
        return m;
    }
    // Constructors
    public Matrix(int i, int j) {
        if(i<1 || j<1)
            throw new IllegalArgumentException(String.format("Invalid array dimensions: (%d,%d)", i, j));
        I = i;
        J = j;
        data = new double[I][J];
    }
}

-1
int det(int[][] mat) {
    if (mat.length == 1)
        return mat[0][0];
    if (mat.length == 2)
        return mat[0][0] * mat[1][1] - mat[1][0] * mat[0][1];
    int sum = 0, sign = 1;
    int newN = mat.length - 1;
    int[][] temp = new int[newN][newN];
    for (int t = 0; t < newN; t++) {
        int q = 0;
        for (int i = 0; i < newN; i++) {
            for (int j = 0; j < newN; j++) {
                temp[i][j] = mat[1 + i][q + j];
            }
            if (q == i)
                q = 1;
        }
        sum += sign * mat[0][t] * det(temp);
        sign *= -1;
    }
    return sum;
}

4
你能否通过为代码添加注释并详细说明问题中提到的性能问题来改进你的答案?请尽可能地完善。 - Jindra Helcl

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