基于圆弧点集推导最优拟合圆/椭圆的无矩阵算法问询

架构老林

2026-8-12

圆弧点序列的最优拟合圆（及椭圆）求解方案

需求说明

拥有构成圆弧的点序列，需要推导最优拟合的圆形（或椭圆形）曲线
分割圆弧的依据：圆弧长度范围内的角度变化相对稳定
偏好无矩阵的解决方案，熟练掌握C/Java/Kotlin语言，优先这类实现，也可接受伪代码或其他语言方案

最优拟合圆的Kotlin实现

感谢Renat提供的解决方案，效果极佳，以下是适配Kotlin的实现代码：

class Circle(val x: Int, val y: Int, val radius: Int) {
    companion object {
        /**
         * 为点集合拟合最优圆
         */
        fun fit(points: List<Point>): Circle {
            val sx = points.sumOf { it.x.toDouble() }
            val sy = points.sumOf { it.y.toDouble() }
            val sx2 = points.sumOf { it.x.toDouble() * it.x }
            val sy2 = points.sumOf { it.y.toDouble() * it.y }
            val sxy = points.sumOf { it.x.toDouble() * it.y }
            val sx3 = points.sumOf { it.x.toDouble() * it.x * it.x }
            val sy3 = points.sumOf { it.y.toDouble() * it.y * it.y }
            val sx2y = points.sumOf { it.x.toDouble() * it.x * it.y }
            val sxy2 = points.sumOf { it.x.toDouble() * it.y * it.y }

            val n = points.size
            val s11 = n * sxy - sx * sy
            val s20 = n * sx2 - sx * sx
            val s02 = n * sy2 - sy * sy
            val s30 = n * sx3 - sx2 * sx
            val s03 = n * sy3 - sy * sy2
            val s21 = n * sx2y - sx2 * sy
            val s12 = n * sxy2 - sx * sy2
            val a = ((s30 + s12)*s02 - (s03 + s21)*s11) / (2*( s20 * s02 - s11 * s11))
            val b = ((s03 + s21)*s20 - (s30 + s12)*s11) / (2*( s20 * s02 - s11 * s11))
            val c = (sx2 + sy2 - 2*a*sx - 2*b*sy) / n
            val radius = Math.sqrt(c + a*a + b*b)
            return Circle(a.roundToInt(), b.roundToInt(), radius.roundToInt())
        }
    }
}