获得了 90% 的 JavaScript 代码 - 无法弄清楚其余的

Question

所以我正在尝试为任何大小的N×N矩阵建模Gram-Schmidt，我已经正式遇到了一个我无法逾越的障碍。我知道这是一个正确循环的问题，但我不知道问题是什么。记住，我不想传入一个3×3矩阵，而是任何大小的N×N。

课程笔记QR分解与Gram-Schmidt解释了我想做什么。很简单的计算。在课程笔记中||u||表示它是元素的平方和，所以sqrt(x₁² + x_{22 + x3² + ....+ x <子> n <一口> 2> p>乘法符号实际上是点积。}

到目前为止，我编写的代码如下所示。有什么问题吗?

function qrProjection(arr) {
    var qProjected = [];
    var tempArray = [];
    var aTemp = arr;
    var uTemp = new Array(arr.length);
    var uSquareSqrt = new Array(arr.length);
    var eTemp = [];
    var sum = 0;
    var sumOfSquares = 0;
    var breakCondition = 0;
    var secondBreakCondition = 0;
    var iterationCounter = 0;
    //Build uTemp Array
    for (i = 0; i < arr.length; i++) {
        uTemp[i] = new Array(arr[i].length);
    }
    for (i = 0; i < arr.length; i++) {
        eTemp[i] = new Array(arr[i].length);
    }
    uTemp[0] = aTemp[0];
    for (j = 0; j <= arr.length; j++) {
        for (l = 0; l < arr[j].length; l++) {
            if (breakCondition == 1) break;
            sumOfSquares = Math.pow(uTemp[j][l], 2) + sumOfSquares;
        }
        if (breakCondition == 0) {
            uSquareSqrt[j] = Math.sqrt(sumOfSquares);
            sumOfSquares = 0;
        }
        for (i = 0; i < arr[j].length; i++) {
            if (breakCondition == 1) break;
            eTemp[j][i] = (1 / (uSquareSqrt[j])) * (uTemp[j][i]);
        }
        breakCondition = 1;
        if (iterationCounter == 0) {
            for (m = 0; m < arr[j].length; m++) {
                matrixDotProduct = aTemp[j + 1][m] * eTemp[j][m] + matrixDotProduct;
            }
        }
        else {
            for (m = 0; m < arr[j].length; m++) {
                for (s = 0; s <= iterationCounter; s++) {
                    matrixDotProduct = aTemp[j + 1][s] * eTemp[m][s] + matrixDotProduct;
                }
                for (t = 0; t < arr[j].length; t++) {
                    uTemp[j + 1][t] = aTemp[j + 1][t] - eTemp[j][t] * matrixDotProduct;
                }
            }
        }
        if (iterationCounter == 0) {
            for (m = 0; m < arr[j].length; m++) {
                uTemp[j + 1][m] = aTemp[j + 1][m] - eTemp[j][m] * matrixDotProduct;
            }
        }
        matrixDotProduct = 0;
        for (l = 0; l < arr[j].length; l++) {
            sumOfSquares = Math.pow(uTemp[j + 1][l], 2) + sumOfSquares;
        }
        uSquareSqrt[j + 1] = Math.sqrt(sumOfSquares);
        sumOfSquares = 0;
        for (i = 0; i < arr[j].length; i++) {
            eTemp[j + 1][i] = (1 / (uSquareSqrt[j + 1])) * (uTemp[j + 1][i]);
        }
        iterationCounter++;
    }
    qProjected = eTemp;
    return qProjected;
}

Answer 1

我必须道歉，我没有调整您的代码，而是从头开始编写自己的代码:

/* Main function of interest */
// Each entry of a matrix object represents a column
function gramSchmidt(matrixA, n) {        
    var totalVectors = matrixA.length;
    for (var i = 0; i < totalVectors; i++) {
        var tempVector = matrixA[i];
        for (var j = 0; j < i; j++) {
            var dotProd = dot(matrixA[i], matrixA[j], n);
            var toSubtract = multiply(dotProd, matrixA[j], n);
            tempVector = subtract(tempVector, toSubtract, n);
        }
        var nrm = norm(tempVector, n);
        matrixA[i] = multiply(1 / nrm, tempVector, n);
    }
}
/*
 * Example usage:
 * var myMatrix = [[1,0,0],[2,3,0],[5,4,7]];
 * gramSchmidt(myMatrix, 3);
 *   ==> myMatrix now equals [[1,0,0],[0,1,0],[0,0,1]]
 * 3 here equals the number of dimensions per vector
 */

/* Simple vector arithmetic */
function subtract(vectorX, vectorY, n) {
    var result = new Array(n);
    for (var i = 0; i < n; i++)
        result[i] = vectorX[i] - vectorY[i];
    return result;
}
function multiply(scalarC, vectorX, n) {
    var result = new Array(n);
    for (var i = 0; i < n; i++)
        result[i] = scalarC * vectorX[i];
    return result;
}
function dot(vectorX, vectorY, n) {
    var sum = 0;
    for (var i = 0; i < n; i++)
        sum += vectorX[i] * vectorY[i];
    return sum;
}
function norm(vectorX, n) {
    return Math.sqrt(dot(vectorX, vectorX, n));
}

注意，上述算法计算的是Gram-Schmidt正交化，即矩阵[e₁ | e₂ |…]| e_n]， 不 QR分解!