参考资料

• https://en.wikipedia.org/wiki/Bilinear_interpolation

前言

双线性插值，又称为双线性内插。在数学上，双线性插值是对线性插值在二维直角网格上的扩展，用于对双变量函数（例如 x 和 y）进行插值。其核心思想是在x，y两个方向分别进行一次线性插值。

线性插值可以查看 之前的博客文章。

推导

假如我们想得到未知函数 $f$ 在点 $P=(x,y)$ 的值，假设我们已知函数 $f$ 在 $Q_{11}=\left(x_1,y_1\right)，Q_{12}=\left(x_1,y_2\right)，Q_{21}=\left(x_2,y_1\right)$ ， $Q_{22}=\left(x_2,y_2\right)$ 四个点的值。

先x方向，后y方向

首先在 $x$ 方向进行线性插值 (即$x$变，$y$不变)，得到
$$\begin{array}{rl}& f\left(x,y_1\right)\approx \frac{x_2-x}{x_2-x_1}f\left(Q_{11}\right)+\frac{x-x_1}{x_2-x_1}f\left(Q_{21}\right),\\ & f\left(x,y_2\right)\approx \frac{x_2-x}{x_2-x_1}f\left(Q_{12}\right)+\frac{x-x_1}{x_2-x_1}f\left(Q_{22}\right).\end{array}$$
然后在 $y$ 方向进行线性插值，得到
$$\begin{array}{rl}f(x,y)& \approx \frac{y_2-y}{y_2-y_1}f\left(x,y_1\right)+\frac{y-y_1}{y_2-y_1}f\left(x,y_2\right)\\ & =\frac{y_2-y}{y_2-y_1}\left(\frac{x_2-x}{x_2-x_1}f\left(Q_{11}\right)+\frac{x-x_1}{x_2-x_1}f\left(Q_{21}\right)\right)+\frac{y-y_1}{y_2-y_1}\left(\frac{x_2-x}{x_2-x_1}f\left(Q_{12}\right)+\frac{x-x_1}{x_2-x_1}f\left(Q_{22}\right)\right)\\ & =\frac{1}{\left(x_2-x_1\right)\left(y_2-y_1\right)}\left(f\left(Q_{11}\right)\left(x_2-x\right)\left(y_2-y\right)+f\left(Q_{21}\right)\left(x-x_1\right)\left(y_2-y\right)+f\left(Q_{12}\right)\left(x_2-x\right)\left(y-y_1\right)+f\left(Q_{22}\right)\left(x-x_1\right)\left(y-y_1\right)\right)\\ & =\frac{1}{\left(x_2-x_1\right)\left(y_2-y_1\right)}\left[x_2-xx-x_1\right]\left[\begin{array}{ll}f\left(Q_{11}\right)& f\left(Q_{12}\right)\\ f\left(Q_{21}\right)& f\left(Q_{22}\right)\end{array}\right]\left[\begin{array}{l}{y}_2-y\\ y-y_1\end{array}\right].\end{array}$$

先y方向，后x方向

首先在 $y$ 方向进行线性插值 (即$y$变，$x$不变)，得到
$$\begin{array}{rl}& f\left(x_1,y\right)\approx \frac{y_2-y}{y_2-y_1}f\left(Q_{11}\right)+\frac{y-y_1}{y_2-y_1}f\left(Q_{12}\right),\\ & f\left(x_2,y\right)\approx \frac{y_2-y}{y_2-y_1}f\left(Q_{21}\right)+\frac{y-y_1}{y_2-y_1}f\left(Q_{22}\right).\end{array}$$
然后在 $x$ 方向进行线性插值，得到
$$\begin{array}{rl}f(x,y)& \approx \frac{x_2-x}{x_2-x_1}f\left(x_1,y\right)+\frac{x-x_1}{x_2-x_1}f\left(x_2,y\right)\\ & =\frac{x_2-x}{x_2-x_1}\left(\frac{y_2-y}{y_2-y_1}f\left(Q_{11}\right)+\frac{y-y_1}{y_2-y_1}f\left(Q_{12}\right)\right)+\frac{x-x_1}{x_2-x_1}\left(\frac{y_2-y}{y_2-y_1}f\left(Q_{21}\right)+\frac{y-y_1}{y_2-y_1}f\left(Q_{22}\right)\right)\\ & =\frac{1}{\left(x_2-x_1\right)\left(y_2-y_1\right)}\left(f\left(Q_{11}\right)\left(x_2-x\right)\left(y_2-y\right)+f\left(Q_{21}\right)\left(x-x_1\right)\left(y_2-y\right)+f\left(Q_{12}\right)\left(x_2-x\right)\left(y-y_1\right)+f\left(Q_{22}\right)\left(x-x_1\right)\left(y-y_1\right)\right)\\ & =\frac{1}{\left(x_2-x_1\right)\left(y_2-y_1\right)}\left[x_2-xx-x_1\right]\left[\begin{array}{ll}f\left(Q_{11}\right)& f\left(Q_{12}\right)\\ f\left(Q_{21}\right)& f\left(Q_{22}\right)\end{array}\right]\left[\begin{array}{l}{y}_2-y\\ y-y_1\end{array}\right].\end{array}$$

可见，无论哪个方向先进行插值，双线性插值的结果是一样的。

简化后的双线性插值

如果选择一个坐标系统使得 $f$ 的四个已知点坐标分别为 $(x_1,y_1)=(0,0)、(x_1,y_2)=(0,1)、(x_2,y_1)=(1,0)$ 和 $(x_2,y_2)=(1,1)$ ，那么插值公式就可以化简为
$$\begin{array}{rl}f(x,y)& \approx f(0,0)(1-x)(1-y)+f(1,0)x(1-y)+f(0,1)(1-x)y+f(1,1)xy\\ & \approx \left[\begin{array}{ll}1-x& x\end{array}\right]\left[\begin{array}{ll}f(0,0)& f(0,1)\\ f(1,0)& f(1,1)\end{array}\right]\left[\begin{array}{c}1-y\\ y\end{array}\right]\end{array}$$

双线性插值的一阶导

在进行code reading 时，发现有时候会需要计算插值后的结果在x,y方向上的一阶导，因此，这里也给出双线性插值的一阶导。

$f(x,y)$在$x$方向上的一阶导：
$$\frac{\partial f}{\partial x}=\frac{1}{\left(x_2-x_1\right)\left(y_2-y_1\right)}\left[-11\right]\left[\begin{array}{ll}f\left(Q_{11}\right)& f\left(Q_{12}\right)\\ f\left(Q_{21}\right)& f\left(Q_{22}\right)\end{array}\right]\left[\begin{array}{l}{y}_2-y\\ y-y_1\end{array}\right].$$

$f(x,y)$在$y$方向上的一阶导：

$$\frac{\partial f}{\partial y}=\frac{1}{\left(x_2-x_1\right)\left(y_2-y_1\right)}\left[x_2-xx-x_1\right]\left[\begin{array}{ll}f\left(Q_{11}\right)& f\left(Q_{12}\right)\\ f\left(Q_{21}\right)& f\left(Q_{22}\right)\end{array}\right]\left[\begin{array}{l}-1\\ 1\end{array}\right].$$

简化后：

$$\frac{\partial f}{\partial x}=[-1,1]\left[\begin{array}{ll}f(0,0)& f(0,1)\\ f(1,0)& f(1,1)\end{array}\right]\left[\begin{array}{c}1-y\\ y\end{array}\right]$$

$$\frac{\partial f}{\partial y}=[1-x,x]\left[\begin{array}{ll}f(0,0)& f(0,1)\\ f(1,0)& f(1,1)\end{array}\right]\left[\begin{array}{c}-1\\ 1\end{array}\right]$$

c++实现参考如下：