我有[-1,0.69]和[0,0.71]两个点,需要在中间插15个值,连成一条平滑的曲线,这样的曲线我要画15条,而且不能重合,该怎么写这个算法,我用的编程语言是js
时间: 2024-03-19 22:43:25 浏览: 16
你可以使用三次B样条插值算法来实现这个需求。以下是一个Javascript实现的例子:
```javascript
function cubic_spline_interpolation(points, num_interp_points) {
// 根据给定的点,创建一个三次B样条插值对象
var spline = new Spline(points);
// 计算插值点,注意要去掉端点
var step = 1 / (num_interp_points + 1);
var interp_points = [];
for (var i = 1; i <= num_interp_points; i++) {
var t = i * step;
interp_points.push(spline.interpolate(t));
}
return interp_points;
}
// 三次B样条插值算法
function Spline(points) {
// 创建对角线矩阵
function createDiagonalMatrix(n, value) {
var matrix = [];
for (var i = 0; i < n; i++) {
var row = [];
for (var j = 0; j < n; j++) {
if (i == j) {
row.push(value);
} else {
row.push(0);
}
}
matrix.push(row);
}
return matrix;
}
// 创建三次B样条插值对象
function createSpline(points) {
var n = points.length - 1;
var h = [];
var alpha = [];
var l = createDiagonalMatrix(n + 1, 1);
var u = createDiagonalMatrix(n + 1, 0);
var z = [];
for (var i = 1; i <= n; i++) {
h[i] = points[i][0] - points[i - 1][0];
alpha[i] =
3 / h[i] *
(points[i + 1][1] - points[i][1]) -
3 / h[i - 1] *
(points[i][1] - points[i - 1][1]);
}
for (var i = 1; i < n; i++) {
u[i][i - 1] = h[i];
u[i][i] = 2 * (h[i] + h[i + 1]);
u[i][i + 1] = h[i + 1];
z[i] =
3 *
(alpha[i + 1] - alpha[i]) /
(h[i] + h[i + 1]) -
u[i][i - 1] * z[i - 1];
l[i + 1][i] = h[i + 1] / u[i][i];
u[i + 1][i] = u[i][i + 1] - l[i + 1][i] * u[i][i];
z[i + 1] =
(z[i] -
h[i] * z[i - 1] -
(h[i] + h[i + 1]) * alpha[i] / 3) /
u[i + 1][i];
}
var c = [];
for (var i = n - 1; i >= 0; i--) {
c[i] =
z[i] -
l[i + 1][i] * c[i + 1] -
(h[i + 1] * alpha[i + 1]) / 3;
}
c[n] = 0;
var b = [];
var d = [];
for (var i = 0; i < n; i++) {
b[i] =
(points[i + 1][1] - points[i][1]) / h[i] -
h[i] * (c[i + 1] + 2 * c[i]) / 3;
d[i] = (c[i + 1] - c[i]) / (3 * h[i]);
}
b[n] = 0;
d[n] = 0;
return {
points: points,
b: b,
c: c,
d: d,
interpolate: function (t) {
var i;
for (i = 0; i < n; i++) {
if (t <= this.points[i + 1][0]) {
break;
}
}
var x = t - this.points[i][0];
return (
this.points[i][1] +
this.b[i] * x +
this.c[i] * x * x +
this.d[i] * x * x * x
);
},
};
}
return createSpline(points);
}
// 测试代码
var points = [
[-1, 0.69],
[0, 0.71],
];
var num_interp_points = 15;
var interp_points = cubic_spline_interpolation(points, num_interp_points);
console.log(interp_points);
```
这个算法的核心是 `Spline` 类,它根据给定的点,创建一个三次B样条插值对象,并提供一个 `interpolate` 方法,可以根据插值参数 `t`,返回对应的插值点的值。`cubic_spline_interpolation` 方法则是使用 `Spline` 类,计算出需要插值的点,并返回结果。