高斯消去 Java
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Gaussian Elimination Java
提问by user2458768
I have this example matrix:
我有这个示例矩阵：
[4,1,3]
[2,1,3]
[4,-1,6]
and i want to solve exuotions:
我想解决exuotions：
4x1+1x2+3x3=v
2x1+1x2+2x3=v
4x1-1x2+6x3=v
x1+x2+x3=1
it will be: 4x1+1x2+3x3 = 2x1+1x2+2x3 = 4x1-1x2+6x3
它将是：4x1+1x2+3x3 = 2x1+1x2+2x3 = 4x1-1x2+6x3
-2x1+x2-5x3 =0
-2x1+x2-5x3 =0
and I use the code:
我使用代码：
import java.util.*;
public class GaussianElimination {
// This is the problem we solved in class
private static double[][] problem1 = {
// x = 1, y = 2, z = 3
{ 1, 2, 3, 14 }, // 1x + 2y + 3z = 14
{ 1, -1, 1, 2 }, // 1x - 1y + 1z = 2
{ 4, -2, 1, 3 } // 4x - 2y + 1z = 3
};
public static void solve(double[][] c, int row) {
int rows = c.length;
int cols = rows + 1;
// 1. set c[row][row] equal to 1
double factor = c[row][row];
for (int col=0; col<cols; col++)
c[row][col] /= factor;
// 2. set c[row][row2] equal to 0
for (int row2=0; row2<rows; row2++)
if (row2 != row) {
factor = -c[row2][row];
for (int col=0; col<cols; col++)
c[row2][col] += factor * c[row][col];
}
}
public static void solve(double[][] c) {
int rows = c.length;
for (int row=0; row<rows; row++)
solve(c,row);
}
public static void print(double[][] c) {
int rows = c.length;
int cols = rows + 1;
for (int row=0; row<rows; row++) {
for (int col=0; col<cols; col++)
System.out.printf("%5.1f ",c[row][col]);
System.out.println();
}
System.out.println();
}
public static void printSolution(double[][] c) {
int rows = c.length, cols = rows + 1;
char variable = (char)((rows > 3) ? ('z' - (rows-1)) : 'x');
System.out.println("Solution:\n");
for (int row=0; row<rows; row++)
System.out.printf(" %c = %1.1f\n",(char)variable++,c[row][cols-1]);
System.out.println();
}
public static void doProblem(double[][] problem, String description) {
System.out.printf("******* %s ********\n",description);
System.out.println("Original Equations:");
print(problem);
solve(problem);
System.out.println("Solved (reduced row echelon form):");
print(problem);
printSolution(problem);
}
public static void main(String[] args) {
doProblem(problem1,"Problem 1 (from class)");
}
}
How do I set the matrix in private static double[][] problem1so that I get x1,x2,x3?
如何设置矩阵private static double[][] problem1以便得到 x1、x2、x3？
回答by Martijn Courteaux
I don't really understand your question or problem. However I see some bugs in the row reduction echelon form solving method. I recently wrote this method as well. Mine works. Since I don't suspect this to be a Java homework assignment but rather an interest in programming mathematical algorithms, I will just throw in my code. I recommend taking a look at how the rref method is actually defined in the world of maths.
我真的不明白你的问题或问题。但是我在行减少梯形求解方法中看到了一些错误。我最近也写了这个方法。我的作品。由于我不怀疑这是一项 Java 作业，而是对编程数学算法的兴趣，因此我将直接提交我的代码。我建议看看 rref 方法是如何在数学世界中实际定义的。
The bug I spotted is that the factoryou use is wrong. Take a look at my code (note that it doesn't put zero rows to the bottom of the matrix):
我发现的错误是factor您使用的错误。看看我的代码（请注意，它不会将零行放在矩阵的底部）：
public static double[][] rref(double[][] mat)
{
double[][] rref = new double[mat.length][mat[0].length];
/* Copy matrix */
for (int r = 0; r < rref.length; ++r)
{
for (int c = 0; c < rref[r].length; ++c)
{
rref[r][c] = mat[r][c];
}
}
for (int p = 0; p < rref.length; ++p)
{
/* Make this pivot 1 */
double pv = rref[p][p];
if (pv != 0)
{
double pvInv = 1.0 / pv;
for (int i = 0; i < rref[p].length; ++i)
{
rref[p][i] *= pvInv;
}
}
/* Make other rows zero */
for (int r = 0; r < rref.length; ++r)
{
if (r != p)
{
double f = rref[r][p];
for (int i = 0; i < rref[r].length; ++i)
{
rref[r][i] -= f * rref[p][i];
}
}
}
}
return rref;
}
回答by Pixel
The following code adapted from Rosettacode.orgtakes into account moving rows up/down as well:
以下改编自 Rosettacode.org 的代码也考虑了向上/向下移动行：
static public void rref(double [][] m)
{
int lead = 0;
int rowCount = m.length;
int colCount = m[0].length;
int i;
boolean quit = false;
for(int row = 0; row < rowCount && !quit; row++)
{
print(m);
println();
if(colCount <= lead)
{
quit = true;
break;
}
i=row;
while(!quit && m[i][lead] == 0)
{
i++;
if(rowCount == i)
{
i=row;
lead++;
if(colCount == lead)
{
quit = true;
break;
}
}
}
if(!quit)
{
swapRows(m, i, row);
if(m[row][lead] != 0)
multiplyRow(m, row, 1.0f / m[row][lead]);
for(i = 0; i < rowCount; i++)
{
if(i != row)
subtractRows(m, m[i][lead], row, i);
}
}
}
}
// swaps two rows
static void swapRows(double [][] m, int row1, int row2)
{
double [] swap = new double[m[0].length];
for(int c1 = 0; c1 < m[0].length; c1++)
swap[c1] = m[row1][c1];
for(int c1 = 0; c1 < m[0].length; c1++)
{
m[row1][c1] = m[row2][c1];
m[row2][c1] = swap[c1];
}
}
static void multiplyRow(double [][] m, int row, double scalar)
{
for(int c1 = 0; c1 < m[0].length; c1++)
m[row][c1] *= scalar;
}
static void subtractRows(double [][] m, double scalar, int subtract_scalar_times_this_row, int from_this_row)
{
for(int c1 = 0; c1 < m[0].length; c1++)
m[from_this_row][c1] -= scalar * m[subtract_scalar_times_this_row][c1];
}
static public void print(double [][] matrix)
{
for(int c1 = 0; c1 < matrix.length; c1++)
{
System.out.print("[ ");
for(int c2 = 0; c2 < matrix[0].length-1; c2++)
System.out.print(matrix[c1][c2] + ", ");
System.out.println(matrix[c1][matrix[c1].length-1] + " ]");
}
}
static public void println()
{
System.out.println();
}
