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我正在用 Java 制作一个项目,我必须使用 BigInteger 类来实现加密方法。

我有 nxn 方阵,其中 n 可以是 200,我需要计算行列式。我使用子矩阵的行列式做了这个方法,但它需要永远计算。

public BigInteger determinant(Matrix matrix){
    if (matrix.getColumns()!=matrix.getRows()){
        System.out.println("The matrix is not square");
        return BigInteger.valueOf(-1);
    }
    if (matrix.getColumns() == 1) {
    return matrix.getMatrix()[0][0];
    }
    if (matrix.getRows()==2) {
        return ((matrix.getValueAt(0, 0).multiply(matrix.getValueAt(1, 1)))).subtract(( matrix.getValueAt(0, 1).multiply(matrix.getValueAt(1, 0))));
    }
    BigInteger sum = BigInteger.valueOf(0);
    for (int i=0; i<matrix.getColumns(); i++) {
        sum = sum.add(this.changeSign(BigInteger.valueOf(i)).multiply(matrix.getValueAt(0, i)).multiply(determinant(createSubMatrix(matrix, 0, i))));// * determinant(createSubMatrix(matrix, 0, i));
    }
    return sum;
    } 

有没有一种非递归的方法来计算行列式?

提前致谢。

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4 回答 4

1

我已将此作为评论发布,但我认为这实际上可以解决您的问题,因此我也将其发布为答案。你可以使用这个包: http: //math.nist.gov/javanumerics/jama/

于 2013-12-01T18:21:38.300 回答
0

计算大矩阵的确定性的一种常见做法是使用 LUP 分解。在这种情况下,行列式可以用以下思路计算:

{L, U, P} = LUP(A)
sign = -1 ^ 'number of permutations in P'
det(A) = diagonalProduct(U) * sign

这就是大型数学包如何做到这一点。您可能应该自己实现 LU。

于 2013-12-02T08:30:22.157 回答
0

我相信这正是您所需要的。使用此类,您可以计算具有任何维度的矩阵的行列式

这个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。它可以用于高维矩阵,如 500 x 500 甚至更大。这门课的好处是你可以得到BigDecimal 的结果,所以没有无穷大,你总能得到准确的答案。顺便说一句,使用许多不同的方法并避免递归会导致更快的方法和更高的答案性能。希望它会有所帮助。

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];
    }
}

}

然后这个类从用户那里接收一个 nxn 的矩阵,或者可以生成一个 nxn 的随机矩阵,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.security.SecureRandom;
import java.text.NumberFormat;
import java.util.Scanner;


public class DeterminantTest {

public static void main(String[] args) {

    String determinant;

    //generating random numbers
int len = 500;
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;
}

}

于 2019-05-21T06:00:44.840 回答
0

递归方法需要很长时间才能找到尺寸超过 10x10 的矩阵的行列式。您将需要进行 LU 分解和高斯归约。我用它来找到 1000x1000 矩阵的行列式,它在一秒钟内产生了正确的结果。您可以在 Numerical Recipes Book(仅使用第 3 版)中获得此代码:第 52 行。它是用 C++ 编写的,但您可以轻松地将其转换为 Java

或者在这个 https://www.cc.gatech.edu/gvu/people/Phd/warren/matrix.c中检查ludcmp()

于 2020-08-20T07:28:06.030 回答