使用javascript实现矩阵LU分解

Love The Way You Lie 2022-05-27 05:54 276阅读 0赞

在线性代数中，LU分解是将一个矩阵分解为 L（单位下三角矩阵）和 U（上三角矩阵），可用于求解线性方程组、反矩阵和计算行列式。本文结合LU分解，用javascript实现线性方程组的求解；

假设存在线性系统Ax = b，其中A为n\*n矩阵， b为1\*n向量，先进行A=LU分解，得到 LUx = b, 再令Y = Ux，得到LY = b，求出Y，再由Y = Ux求解x，具体算法实现如下：

/**
     * 用于实现1-6元线性方程组求解
     * @author: Wujinhua
     * */
    
    function BuildArray () {
        this.Arr = [];
    }
    BuildArray.prototype.matrix = function(n) {
        for (let i = 0; i < n; i++) {
            let nRows = [];
            for (let j = 0; j < n; j++) {
                nRows.push(0);
            }
            this.Arr.push(nRows);
        }
    
        return this.Arr;
    };
    BuildArray.prototype.vector = function (n) {
        for (let i = 0; i < n; i++) {
            this.Arr.push(0);
        }
        return this.Arr;
    };
    
    function _argumentsToArray( args ) {
        return [].concat.apply( [], Array.prototype.slice.apply(args) );
    }
    
    /**
     * lu分解求解线性方程组
     * @param: n*n矩阵arraysA 和 n维向量arraysB
     * @example:
     * let A = [1,2,3,2,5,2,3,1,5];
     * let b = [4,18,20];
     * let x = lu(A, b);
     */
    
    function lu(arraysA, arraysB) {
    
        let n = arraysA.length;
        let result1 = _argumentsToArray(arraysA);
        let A = undefined;
    
        if(n === 2) A = mat2(result1);//2*2矩阵
        if(n === 3) A = mat3(result1);//3*3
        if(n === 4) A = mat4(result1);//4*4
        if(n === 5) A = mat5(result1);//5*5
        if(n === 6) A = mat6(result1);//6*6
    
        let ll = new BuildArray();//初始化矩阵L
        let l = ll.matrix(n);
        let uu = new BuildArray();//初始化U矩阵
        let u = uu.matrix(n);
    
    
        //进行U矩阵的第一行赋值
        for (let i = 0; i < n; i++) {
            u[0][i] = A[0][i];
        }
    
        //进行L矩阵的第一列赋值
        for (let i = 0; i < n; i++) {
            l[i][0] = A[i][0] / u[0][0];
        }
    
    
        //
        for (let i = 1; i < n; i++) {
            for (let j = i; j < n; j++) {
                let sum1 = 0.0;
                for (let k = 0; k < i; k++) {
                    sum1 += l[i][k] * u[k][j];
                }
                u[i][j] = A[i][j] - sum1;
            }
    
            if(i !== n) {
                for (let r = i+1; r < n; r++) {
                    let sum2 = 0.0;
                    for (let k = 0; k < i; k++) {
                        sum2 += l[r][k] * u[k][i];
                    }
    
                    l[r][i] = (A[r][i] - sum2) / u[i][i];
                }
            }
        }
    
        let y = new BuildArray();
        y.vector(n);
        y[0] = arraysB[0];
    
        for (let m = 0; m < n; m++) {
            let sum3 = 0.0;
            for (let k = 0; k < m; k++) {
                sum3 += l[m][k] * y[k];
                y[m] = arraysB[m] - sum3;
            }
        }
    
        let xx = new BuildArray();
        let x = xx.vector(n);
        x[n-1] = y[n-1] / u[n-1][n-1];
        for (let j = n - 2; j >= 0; j--) {
            let sum4 = 0.0;
            for (let k = j + 1; k < n; k++) {
                sum4 += u[j][k] * x[k];
                x[j] = (y[j] - sum4) / u[j][j];
            }
        }
    
        return x;
    }

其中用到构建n\*n矩阵的构造函数matn()，其代码如下：

向量的构造函数vecn()代码如下：

用一个五元一次方程组测试如下：

let A1 = mat5(
            2, 3, -5, 6, -7,
            1, 2, -3, 3, -2,
            3, -7, 2, 3, -4,
            4, 3, -5, 6, -7,
            5, 3, -3, 1, -2
        );
    
        let b = [10, 15, 13, 22, 33];
    
        let x = lu(A1, b);

结果：

![20180412192008752][]

结果与用MATLAB计算结果一致，证明程序可实现线性方程组求解。

如果需要下载本项目代码，请前往我的Github上下载：[https://github.com/Brucewu998/LU-Decomposition][https_github.com_Brucewu998_LU-Decomposition]

[20180412192008752]: /images/20220527/b61a581cad184cf498e872cc4b38cd4b.png
[https_github.com_Brucewu998_LU-Decomposition]: https://github.com/Brucewu998/LU-Decomposition