Java——(高斯消去法/列主元消去法/LU消去)求解方程

心已赠人 2022-12-19 13:23 190阅读 0赞

## Java 高斯消去 / 列主元消去 / LU消去 ##

*  首先输入方程未知数个数
 *  输入系数矩阵，按顺序(↖为头)，每个数字输入后回车
 *  输入常量向量，按顺序(↑为头)，每个数字输入后回车
 *  选择计算方法/输入0退出
 *  源代码只包含高斯&列主元方法  
    [(点击此处)下载源代码][Link 1]

import java.util.Scanner;

package one;
    
    import java.util.Arrays;
    import java.util.List;
    import java.util.Scanner;
    
    public class Q2 { 
    	static double A[][];
    	static double b[];
    	static double x[];
    
    	static int n; // n表示未知数的个数
    	static int n_2; // 记录换行的次数
    
    	public static void main(String[] args) { 
    
    		while (true) { 
    			System.out.println("请输入数字选择计算方法：\n1.Gauss 2.LineGuess 3.LU 0.退出");
    			@SuppressWarnings("resource")
    			Scanner sn = new Scanner(System.in);
    			int c = sn.nextInt();
    			System.out.println("--------------输入方程组未知数的个数---------------");
    			@SuppressWarnings("resource")
    			Scanner sc = new Scanner(System.in);
    			n = sc.nextInt();
    
    			A = new double[n][n];
    			b = new double[n];
    			x = new double[n];
    
    			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] = sc.nextDouble();
    				}
    			}
    
    			System.out.println("--------------输入方程组的常量向量b:---------------");
    			for (int i = 0; i < n; i++) { 
    				b[i] = sc.nextDouble();
    			}
    			if (1 == c) { 
    				Gauss();
    			} else if (2 == c) { 
    				LineGuess();
    			} else if (3 == c) { 
    				LU();
    			} else if (0 == c) { 
    				System.out.println("Bye!");
    				System.exit(0);
    			}
    			
    		}
    	}
    
    	// 1
    	public static void Gauss() { 
    		Elimination();
    		BackSubstitution();
    		PrintRoot();
    	}
    
    	// 2
    	public static void LineGuess() { 
    		Elimination2();
    		BackSubstitution();
    		PrintRoot();
    	}
    
    	// 3
    	public static void LU() { 
    		double[] x = solve(A, b);
    		System.out.println("结果为：");
    		for (int z = 0; z < x.length; z++) { 
    			System.out.println("X" + z + " = " + x[z]);
    		}
    	}
    
    	// 高斯
    	public static void Elimination() { 
    		PrintA();
    		for (int k = 0; k < n;) { 
    			for (int i = k + 1; i < n; i++) { 
    				double l = A[i][k] / A[k][k];
    				A[i][k] = 0;
    
    				for (int j = k + 1; j < n; j++) { 
    					A[i][j] = A[i][j] - l * A[k][j];
    				}
    				b[i] = b[i] - l * b[k];
    			}
    			System.out.println("第" + ++k + "次消元后:");
    			PrintA();
    		}
    	}
    
    	// 列主元
    	public static void Elimination2() { 
    		PrintA();
    		for (int k = 0; k < n;) { 
    			WrapRow(k);
    			for (int i = k + 1; i < n; i++) { 
    				double l = A[i][k] / A[k][k];
    				A[i][k] = 0;
    
    				for (int j = k + 1; j < n; j++) { 
    					A[i][j] = A[i][j] - l * A[k][j];
    				}
    				b[i] = b[i] - l * b[k];
    			}
    			System.out.println("第" + ++k + "次消元后:");
    			PrintA();
    		}
    	}
    
    	// 交换矩阵的行
    	public static void WrapRow(int k) {  // k表示第k+1轮消元
    		double max = Math.abs(A[k][k]);
    		int Line = k; // 记住要交换的行
    		for (int i = k + 1; i < n; i++) { 
    			if (Math.abs(A[i][k]) > max) { 
    				Line = i;
    				max = A[i][k];
    				break;
    			}
    		}
    		if (Line != k) {  // 交换求得最大主元
    			n_2 += 1;
    			System.out.println("k = " + k + "时，" + "要交换的行为" + k + "和" + Line);
    			// 先交换A
    			for (int j = k; j < n; j++) { 
    				double[] arr = {  A[k][j], A[Line][j] };
    				Swap(arr, 0, 1);
    				A[k][j] = arr[0];
    				A[Line][j] = arr[1];
    
    			}
    			// 再交换b
    			double[] arr = {  b[k], b[Line] };
    			Swap(arr, 0, 1);
    			b[k] = arr[0];
    			b[Line] = arr[1];
    			System.out.println("--------------交换后---------------");
    			PrintA();
    		}
    	}
    
    	// 交换Swap函数
    	public static void Swap(double[] ar, int x, int y) { 
    		Double tmp = ar[x];
    		ar[x] = ar[y];
    		ar[y] = tmp;
    	}
    
    	// 回代法
    	public static void BackSubstitution() { 
    		x[n - 1] = b[n - 1] / A[n - 1][n - 1];
    		for (int i = n - 2; i >= 0; i--) { 
    			x[i] = (b[i] - solve(i)) / A[i][i];
    		}
    	}
    
    	public static double solve(int i) { 
    		double result = 0.0;
    		for (int j = i; j < n; j++)
    			result += A[i][j] * x[j];
    		return result;
    	}
    
    	// 输出方程组的根
    	public static void PrintRoot() { 
    		System.out.println("--------------方程组的根为---------------");
    		for (int i = 0; i < n; i++) { 
    				System.out.println("X" + i + " = " + x[i]);
    		}
    	}
    
    	// 输出A的增广矩阵
    	public static void PrintA() { 
    		System.out.println("--------------增广矩阵---------------");
    		for (int i = 0; i < n; i++) { 
    			for (int j = 0; j < n; j++) { 
    				System.out.print(+A[i][j] + " ");
    			}
    			System.out.println(b[i]);
    		}
    	}
    
    	// LU
    	public static double[] solve(double[][] a, double[] b) { 
    		List<double[][]> LAndU = decomposition(a); // LU decomposition
    		double[][] L = LAndU.get(0);
    		double[][] U = LAndU.get(1);
    		double[] UMultiX = getUMultiX(a, b, L); // 前代
    		return getSolution(a, U, UMultiX); // 回代
    	}
    
    	private static double[] getSolution(double[][] a, double[][] U, double[] UMultiX) { 
    		double[] solutions = new double[a[0].length];
    		for (int i = U.length - 1; i >= 0; i--) { 
    			double right_hand = UMultiX[i];
    			for (int j = U.length - 1; j > i; j--) { 
    				right_hand -= U[i][j] * solutions[j];
    			}
    			solutions[i] = right_hand / U[i][i];
    		}
    		return solutions;
    	}
    
    	private static double[] getUMultiX(double[][] a, double[] b, double[][] L) { 
    		double[] UMultiX = new double[a.length];
    		for (int i = 0; i < a.length; i++) { 
    			double right_hand = b[i];
    			for (int j = 0; j < i; j++) { 
    				right_hand -= L[i][j] * UMultiX[j]; //
    			}
    			UMultiX[i] = right_hand / L[i][i];
    		}
    		return UMultiX;
    	}
    
    	private static List<double[][]> decomposition(double[][] a) { 
    		double[][] U = a; // a是要分解的矩阵
    		double[][] L = createIndentityMatrix(a.length);
    
    		for (int j = 0; j < a[0].length - 1; j++) { 
    			if (a[j][j] == 0) { 
    				throw new IllegalArgumentException("zero pivot encountered.");
    			}
    
    			for (int i = j + 1; i < a.length; i++) { 
    				double mult = a[i][j] / a[j][j];
    				for (int k = j; k < a[i].length; k++) { 
    					U[i][k] = a[i][k] - a[j][k] * mult;
    					// 得出上三角矩阵U,通过减去矩阵的第一行,第二行,第一行(第二行)得到上三角矩阵
    				}
    				L[i][j] = mult; // 得到下三角矩阵是得出上三角矩阵的乘积因子
    			}
    		}
    		System.out.println("--------------L矩阵---------------");
    		PrintLU(L);
    		System.out.println("--------------U矩阵---------------");
    		PrintLU(U);
    		return Arrays.asList(L, U);
    
    	}
    	
    	public static void PrintLU(double [][]z){ 
    		for (int i = 0; i < n; i++) { 
    			for (int j = 0; j < n; j++) { 
    				System.out.print(z[i][j] + " ");
    			}
    			System.out.println("\n");
    		}
    	}
    
    	private static double[][] createIndentityMatrix(int n) { 
    		double[][] identityMatrix = new double[n][n];
    		for (int i = 0; i < identityMatrix.length; i++) { 
    			for (int j = i; j < identityMatrix[i].length; j++) { 
    				if (j == i) { 
    					if (j == i) { 
    						identityMatrix[i][j] = 1;
    					} else { 
    						identityMatrix[i][j] = 0;
    					}
    				}
    			}
    		}
    		return identityMatrix;
    	}
    
    }

![高斯消去][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L0dvZE91Tw_size_16_color_FFFFFF_t_70_pic_center]  
![列主元消去][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L0dvZE91Tw_size_16_color_FFFFFF_t_70_pic_center 1]

[Link 1]: https://download.csdn.net/download/GodOuO/13096399
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