最小二乘法来求解矩阵
// 计算Homography矩阵
function calculateHomography(srcPoints, dstPoints) {if (srcPoints.length !== dstPoints.length || srcPoints.length < 4) {throw new Error('需要至少四个点进行计算');}// 设置矩阵 方程 Ax = blet A = [];let b = [];// 构建方程的A矩阵 和b向量for (let i = 0; i < srcPoints.length; i++) {let x1 = srcPoints[i][0];let y1 = srcPoints[i][1];let x2 = dstPoints[i][0];let y2 = dstPoints[i][1];// 方程:x2 = H11 * x1 + H12 * y1 + H13// 方程:y2 = H21 * x1 + H22 * y1 + H23A.push([-x1, -y1, -1, 0, 0, 0, x1 * x2, y1 * x2]);A.push([0, 0, 0, -x1, -y1, -1, x1 * y2, y1 * y2]);b.push(x2);b.push(y2);}// 使用高斯-约旦消元法求解A * h = blet H = gaussJordan(A, b);// 构造Homography矩阵 let homographyMatrix = [[H[0], H[1], H[2]],[H[3], H[4], H[5]],[H[6], H[7], 1]];return homographyMatrix;
}// 高斯-约旦消元法解线性方程组
function gaussJordan(A, b) {let n = A.length;let augmentedMatrix = A.map((row, i) => row.concat(b[i])); // A 与 b 合并为增广矩阵 for (let i = 0; i < n; i++) {// 找到最大值所在的行进行行交换let maxRow = i;for (let j = i + 1; j < n; j++) {if (Math.abs(augmentedMatrix[j][i]) > Math.abs(augmentedMatrix[maxRow][i])) {maxRow = j;}}[augmentedMatrix[i], augmentedMatrix[maxRow]] = [augmentedMatrix[maxRow], augmentedMatrix[i]];// 将主对角线上的元素变为1let scale = augmentedMatrix[i][i];for (let j = i; j < n + 1; j++) {augmentedMatrix[i][j] /= scale;}// 通过行操作消去其他列的元素for (let j = 0; j < n; j++) {if (j !== i) {scale = augmentedMatrix[j][i];for (let k = i; k < n + 1; k++) {augmentedMatrix[j][k] -= augmentedMatrix[i][k] * scale;}}}}// 提取解向量(即矩阵 的最后一列)let solution = augmentedMatrix.map(row => row[n]);return solution;
}// 示例:输入源点和目标点
let srcPoints = [[7.606727752, 13.23349844],[5.727826596, 13.23044123],[4.788626754, 13.23106254],[2.906026488, 13.23060659]
];let dstPoints = [[7.623758876, 13.24447489],[5.744538875, 13.24339615],[4.804678876, 13.24468678],[2.923728876, 13.24552804]
];// 计算Homography矩阵
let homographyMatrix = calculateHomography(srcPoints, dstPoints);
console.log("Homography Matrix:");
console.log(homographyMatrix);