GaussianElimination 高斯消去法java(求逆,求解)
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java代码
import java.util.Scanner; public class GaussianElimination{ public static void Input(int n, double[][] A, double[] b, int length){ Scanner in = new Scanner(System.in); System.out.println("输入方程组的系数矩阵A;"); for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { A[i][j] = in.nextDouble(); } } if(length == 1){ System.out.println("输入方程组矩阵b:"); for (int i = 0; i < n; i++) { b[i] = in.nextDouble(); } } } public static void Merge1(int n, double[][] A, double[] b){ for (int i = 0; i < n; i++) { A[i][n] = b[i]; } Print(n, A, 1); return; } public static void Merge2(int n, double[][] A){ for (int i = 0; i < n; i++) { for (int j = n; j < 2*n; j++) { A[i][j] = 0; } A[i][n+i] = 1; } Print(n, A, n); return; }; public static void Print(int n, double[][] A, int length){ System.out.println("##################手 动 分 割 线#####################"); System.out.println("增广矩阵:"); for (int i = 0; i < n; i++) { for (int j = 0; j < n + length; j++) { System.out.printf("%-11.4f", A[i][j]); } System.out.print("\\n"); } } //length为1,求逆为n public static void Swap(int n, double[][]A, int i, int pivotrow, int length){ double temp = 0; for (int k = 0; k < n+length; k++) { temp = A[i][k]; A[i][k] = A[pivotrow][k]; A[pivotrow][k] = temp; } } //交换行 public static int SelectPivotrow(int n, double[][]A, int i) { int pivotrow = i; for (int j = i + 1; j < n; j++) { if (Math.abs(A[j][i]) > Math.abs(A[pivotrow][i])) pivotrow = j; } if (A[pivotrow][i] == 0) return -1; else return pivotrow; } //选择主元 public static void ForwardElimination(int n, double[][]A, int length){ for (int i = 0; i < n-1; i++) { int pivotrow = SelectPivotrow(n, A, i); if(pivotrow == -1){ System.out.println("输出结果不唯一"); break; } Swap(n, A, i, pivotrow, length); for (int j = i+1; j < n; j++) { double temp = A[j][i]/A[i][i]; for (int k = i; k < n+length; k++) { A[j][k] = A[j][k] - A[i][k] * temp; } } Print(n, A, length); } } //前向消去,求length为1,求逆为n public static void BackElimination(int n, double[][]A, int length){ for (int i = n-1; i > 0; i--) { for (int j = i-1; j >= 0; j--) { double temp = A[j][i]/A[i][i]; for (int k = n+length-1; k > j; k--) { A[j][k] = A[j][k] - A[i][k] * temp; } } Print(n, A, length); } } //反向替换,求length为1,求逆为n public static void Normalization(int n, double[][]A, int length){ for (int i = 0; i < n; i++) { for (int j = n; j < 2*n; j++) { A[i][j] = A[i][j]/A[i][i]; } A[i][i] = 1; } Print(n, A, length); } //归一化,length为1,求逆为n public static void PrintResult(int n, double[][] A, int length){ System.out.println("##################手 动 分 割 线#####################"); System.out.println("最终结果为:"); if(length == 1){ for (int i = 0; i < n; i++) { System.out.println("x"+(i+1)+" = "+A[i][n+length-1]); } } else if(n == length){ System.out.println("矩阵的逆:"); for (int i = 0; i < n; i++) { for (int j = n; j < 2*n; j++) { System.out.printf("%-11.2f", A[i][j]); } System.out.print("\\n"); } } return ; } public static void Do(int n, double[][]A, int length){ ForwardElimination(n, A, length); BackElimination(n, A, length); Normalization(n, A, length); PrintResult(n, A, length); } public static void main(String[] args) { Scanner in = new Scanner(System.in); while(true){ System.out.println("-----------------高 斯 消 元 法------------------"); System.out.println("求逆矩阵请输入1,求解请输入2,退出请输入0。"); int flag = in.nextInt(); if(flag != 0){ System.out.println("输入方程组的元数:"); int n = in.nextInt(); double[][] A = new double[n][2*n]; double[] b = new double[n]; double[] x = new double[n]; double[][] A_I = new double[n][n]; if(flag ==1){ Input(n, A, b, n); Merge2(n, A); Do(n, A, n); }else{ Input(n, A, b ,1); Merge1(n, A, b); Do(n, A,1); } }else break; } System.out.println("-----------------程 序 结 束 !!------------------"); } }
结果:
求解
题目
求逆
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