图像双立方插值——C实现

原理：详细参见https://en.wikipedia.org/wiki/Bicubic_interpolation
在数学上，双立方插值是2个变量函数在方形网格上立方插值的一个扩展，在图像处理中，双立方插值考虑插值像素点周围16个像素，当不在乎算法执行速度时，一般选择双立方插值，相对最近邻域插值和二次线性插值，双立方插值具有较少的插值畸变（artifacts ，不知道翻译成什么好）。
双立方插值与最近邻域插值和二次线性插值的比较如下图：

假设在(0,0),(1,0),(0,1),(1,1)单位正方形四个顶点函数值f 及其偏导数$f_x,f_y{f}_y和f_{xy}{f}_{xy}$已知，那么，单位正方形内部任意点的插值可以表示为：
$$p(x,y)=\sum _{i=0}^3\sum _{j=0}^3{a}_{ij}{x}^i{y}^j$$
因此插值问题变成求解16个参数的问题，四个点$(0,0),(1,0),(0,1),和(1,1)$代入上式可得：
$$\begin{gathered} f(0,0)=p(0,0)=a_{00}, \\ f(1,0)=p(1,0)=a_{00}+a_{10}+a_{20}+a_{30}, \\ f(0,1)=p(0,1)=a_{00}+a_{01}+a_{02}+a_{03}, \\ f(1,1)=p(1,1)=\sum _{i=0}^3\sum _{j=0}^3{a}_{ij}. \end{gathered}$$
根据4个点x和y方向上的偏导数，可得8个方程：
$$\begin{gathered} f_x(0,0)=p_x(0,0)=a_{10}, \\ {f}_x(1,0)=p_x(1,0)=a_{10}+2a_{20}+3a_{30}, \\ {f}_x(0,1)=p_x(0,1)=a_{10}+a_{11}+a_{12}+a_{13}, \\ {f}_x(1,1)=p_x(1,1)=\sum _{i=1}^3\sum _{j=0}^3{a}_{ij}i, \\ {f}_y(0,0)=p_y(0,0)=a_{01}, \\ {f}_y(1,0)=p_y(1,0)=a_{01}+a_{11}+a_{21}+a_{31}, \\ {f}_y(0,1)=p_y(0,1)=a_{01}+2a_{02}+3a_{03}, \\ {f}_y(1,1)=p_y(1,1)=\sum _{i=0}^3\sum _{j=1}^3{a}_{ij}j. \end{gathered}$$
四个点上的二阶混合偏导数（mixed partial derivative）可得四个方程：
$$\begin{gathered} f_{xy}(0,0)=p_{xy}(0,0)=a_{11}, \\ {f}_{xy}(1,0)=p_{xy}(1,0)=a_{11}+2a_{21}+3a_{31}, \\ {f}_{xy}(0,1)=p_{xy}(0,1)=a_{11}+2a_{12}+3a_{13}, \\ {f}_{xy}(1,1)=p_{xy}(1,1)=\sum _{i=1}^3\sum _{j=1}^3{a}_{ij}ij. \end{gathered}$$
上面的四个混合偏导数还可以表示为：
$$\begin{gathered} p_x(x,y)=\sum _{i=1}^3\sum _{j=0}^3{a}_{ij}i{x}^{i-1}{y}^j, \\ {p}_y(x,y)=\sum _{i=0}^3\sum _{j=1}^3{a}_{ij}{x}^{i}j{y}^{j-1}, \\ {p}_{xy}(x,y)=\sum _{i=1}^3\sum _{j=1}^3{a}_{ij}i{x}^{i-1}j{y}^{j-1}. \end{gathered}$$
将$a_{ij}$组合成一个向量：
$$\alpha ={\left[\begin{array}{cccccccccccccccc}{a}_{00}& a_{10}& a_{20}& a_{30}& a_{01}& a_{11}& a_{21}& a_{31}& a_{02}& a_{12}& a_{22}& a_{32}& a_{03}& a_{13}& a_{23}& a_{33}\end{array}\right]}^T$$令：
$$x={\left[\begin{array}{cccccccccccccccc}f(0,0)& f(1,0)& f(0,1)& f(1,1)& f_x(0,0)& f_x(1,0)& f_x(0,1)& f_x(1,1)& f_y(0,0)& f_y(1,0)& f_y(0,1)& f_y(1,1)& f_{xy}(0,0)& f_{xy}(1,0)& f_{xy}(0,1)& f_{xy}(1,1)\end{array}\right]}^T,$$
上面的16个式子可以写成$A\alpha =x$,求解这个线性方程组可得$a_{ij}$。
$A$的逆矩阵如下：
$$A^{-1}=\left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ -3& 3& 0& 0& -2& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 2& -2& 0& 0& 1& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& -3& 3& 0& 0& -2& -1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 2& -2& 0& 0& 1& 1& 0& 0\\ -3& 0& 3& 0& 0& 0& 0& 0& -2& 0& -1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& -3& 0& 3& 0& 0& 0& 0& 0& -2& 0& -1& 0\\ 9& -9& -9& 9& 6& 3& -6& -3& 6& -6& 3& -3& 4& 2& 2& 1\\ -6& 6& 6& -6& -3& -3& 3& 3& -4& 4& -2& 2& -2& -2& -1& -1\\ 2& 0& -2& 0& 0& 0& 0& 0& 1& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 2& 0& -2& 0& 0& 0& 0& 0& 1& 0& 1& 0\\ -6& 6& 6& -6& -4& -2& 4& 2& -3& 3& -3& 3& -2& -1& -2& -1\\ 4& -4& -4& 4& 2& 2& -2& -2& 2& -2& 2& -2& 1& 1& 1& 1\end{array}\right]$$
另外，上面的16个方程的简明表示：
$$\left[\begin{array}{cccc}f(0,0)& f(0,1)& f_y(0,0)& f_y(0,1)\\ f(1,0)& f(1,1)& f_y(1,0)& f_y(1,1)\\ f_x(0,0)& f_x(0,1)& f_{xy}(0,0)& f_{xy}(0,1)\\ f_x(1,0)& f_x(1,1)& f_{xy}(1,0)& f_{xy}(1,1)\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ 1& 1& 1& 1\\ 0& 1& 0& 0\\ 0& 1& 2& 3\end{array}\right]\left[\begin{array}{cccc}{a}_{00}& a_{01}& a_{02}& a_{03}\\ a_{10}& a_{11}& a_{12}& a_{13}\\ a_{20}& a_{21}& a_{22}& a_{23}\\ a_{30}& a_{31}& a_{32}& a_{33}\end{array}\right]\left[\begin{array}{cccc}1& 1& 0& 0\\ 0& 1& 1& 1\\ 0& 1& 0& 2\\ 0& 1& 0& 3\end{array}\right],$$
等式右边两边左乘$\left[\begin{array}{cccc}1& 0& 0& 0\\ 1& 1& 1& 1\\ 0& 1& 0& 0\\ 0& 1& 2& 3\end{array}\right]$的逆矩阵，然后右乘$\left[\begin{array}{cccc}1& 1& 0& 0\\ 0& 1& 1& 1\\ 0& 1& 0& 2\\ 0& 1& 0& 3\end{array}\right]$的逆矩阵，可得：
$$\left[\begin{array}{cccc}{a}_{00}& a_{01}& a_{02}& a_{03}\\ a_{10}& a_{11}& a_{12}& a_{13}\\ a_{20}& a_{21}& a_{22}& a_{23}\\ a_{30}& a_{31}& a_{32}& a_{33}\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 1& 0\\ -3& 3& -2& -1\\ 2& -2& 1& 1\end{array}\right]\left[\begin{array}{cccc}f(0,0)& f(0,1)& f_y(0,0)& f_y(0,1)\\ f(1,0)& f(1,1)& f_y(1,0)& f_y(1,1)\\ f_x(0,0)& f_x(0,1)& f_{xy}(0,0)& f_{xy}(0,1)\\ f_x(1,0)& f_x(1,1)& f_{xy}(1,0)& f_{xy}(1,1)\end{array}\right]\left[\begin{array}{cccc}1& 0& -3& 2\\ 0& 0& 3& -2\\ 0& 1& -2& 1\\ 0& 0& -1& 1\end{array}\right],$$

$p(x,y)$插值可由下式给出：
$$p(x,y)=\left[\begin{array}{cccc}1& x& x^2& x^3\end{array}\right]\left[\begin{array}{cccc}{a}_{00}& a_{01}& a_{02}& a_{03}\\ a_{10}& a_{11}& a_{12}& a_{13}\\ a_{20}& a_{21}& a_{22}& a_{23}\\ a_{30}& a_{31}& a_{32}& a_{33}\end{array}\right]\left[\begin{array}{c}1\\ y\\ y^2\\ y^3\end{array}\right].$$

差分计算可以采用下面公式，参考：https://en.wikipedia.org/wiki/Finite_difference
$$\begin{array}{rl}{f}_x(x,y)& \approx \frac{f(x+h,y)-f(x-h,y)}{2h}\\ f_y(x,y)& \approx \frac{f(x,y+k)-f(x,y-k)}{2k}\\ f_{xx}(x,y)& \approx \frac{f(x+h,y)-2f(x,y)+f(x-h,y)}{h^2}\\ f_{yy}(x,y)& \approx \frac{f(x,y+k)-2f(x,y)+f(x,y-k)}{k^2}\\ f_{xy}(x,y)& \approx \frac{f(x+h,y+k)-f(x+h,y-k)-f(x-h,y+k)+f(x-h,y-k)}{4hk}\end{array}$$
$h=1$

另外一种实现如下：
上面的实现对每个$p(x,y)$都要求解一个线性方程，可以采用卷积方式达到同样的目的，卷积核如下：
$$W(x)=\{\begin{array}{ll}(a+2)|x{|}^3-(a+3)|x{|}^2+1& \text{for }|x|\le 1,\\ a|x{|}^3-5a|x{|}^2+8a|x|-4a& \text{for }1<|x|<2,\\ 0& \text{otherwise},\end{array}$$
$a=-0.5$。
上式可以表示成矩阵形式：
$$p(t)=\frac{1}{2}\left[\begin{array}{cccc}1& t& t^2& t^3\end{array}\right]\left[\begin{array}{cccc}0& 2& 0& 0\\ -1& 0& 1& 0\\ 2& -5& 4& -1\\ -1& 3& -3& 1\end{array}\right]\left[\begin{array}{c}{f}_{-1}\\ f_0\\ f_1\\ f_2\end{array}\right]$$
对图像进行插值时，可以先进行x方向插值，然后再进行y方向插值。
$$\begin{gathered} b_{-1}=p(t_x,f_{(-1,-1)},f_{(0,-1)},f_{(1,-1)},f_{(2,-1)}), \\ {b}_0=p(t_x,f_{(-1,0)},f_{(0,0)},f_{(1,0)},f_{(2,0)}), \\ {b}_1=p(t_x,f_{(-1,1)},f_{(0,1)},f_{(1,1)},f_{(2,1)}), \\ {b}_2=p(t_x,f_{(-1,2)},f_{(0,2)},f_{(1,2)},f_{(2,2)}), \\ p(x,y)=p(t_y,b_{-1},b_0,b_1,b_2). \end{gathered}$$

算法实现
方法一：

static inline void matrix_mul(float c[4][4], float a[4][4], float b[4][4])
{
	int n, k; 
	for (n = 0; n < 4; n++)
	{
		for (k = 0; k < 4; k++)
		{
			c[n][k] = a[n][0] * b[0][k] + a[n][1] * b[1][k] + a[n][2] * b[2][k] + a[n][3] * b[3][k]; 
		}
	}
}

void img_resize_using_bicubic(uint8_2d* src, uint8_2d* dst)
{
	int src_rows, src_cols;
	int dst_rows, dst_cols;
	int i, j;
	if (!src || !dst)return;
	uint8_t** src_arr;
	uint8_t** dst_arr;
	float xratio;
	float yratio;
	float x, y;
	int   xi, yi;
	int   xminus1, yminus1;
	int   xadd1, yadd1;
	int   xadd2, yadd2;

	float u, v;
	float u2, u3;
	float v2, v3;

	int val;
	src_arr = src->arr;
	dst_arr = dst->arr;
	src_rows = src->rows;
	src_cols = src->cols;
	dst_rows = dst->rows;
	dst_cols = dst->cols;
	float m1[4][4] = { { 1, 0, 0, 0 },
					{ 0, 0, 1, 0},
					{-3, 3, -2,-1},
					{ 2,-2, 1, 1} };

	float m0[4][4] = { { 1, 0, -3, 2 },
					{ 0, 0, 3, -2 },
					{ 0, 1, -2, 1 },
					{ 0, 0, -1, 1 } };
	float d[4][4];
	float a[4][4];
	xratio = (float)src_cols / (float)dst_cols;
	yratio = (float)src_rows / (float)dst_rows;

	int v00, v01, v02, v03;
	int v10, v11, v12, v13;
	int v20, v21, v22, v23;
	int v30, v31, v32, v33;
	
	float tmp[4];
	for (i = 0; i < dst_rows; i++)
	{
		for (j = 0; j < dst_cols; j++)
		{
			x = xratio*j;
			y = yratio*i;
			xi = (int)x;
			yi = (int)y;

			u = x - xi;
			v = y - yi;

			xminus1 = xi - 1;
			yminus1 = yi - 1;
			xadd1 = xi + 1;
			yadd1 = yi + 1;
			xadd2 = xi + 2;
			yadd2 = yi + 2;

			if (xi >= src_cols)xi = src_cols - 1;
			if (yi >= src_rows)yi = src_rows - 1;
			if (xminus1 < 0)xminus1 = 0;
			if (yminus1 < 0)yminus1 = 0;
			if (xadd1 >= src_cols)xadd1 = src_cols - 1;
			if (yadd1 >= src_rows)yadd1 = src_rows - 1;
			if (xadd2 >= src_cols)xadd2 = src_cols - 1;
			if (yadd2 >= src_rows)yadd2 = src_rows - 1;

			v00 = src_arr[yminus1][xminus1];
			v01 = src_arr[yminus1][xi];
			v02 = src_arr[yminus1][xadd1];
			v03 = src_arr[yminus1][xadd2];

			v10 = src_arr[yi][xminus1];
			v11 = src_arr[yi][xi];
			v12 = src_arr[yi][xadd1];
			v13 = src_arr[yi][xadd2];

			v20 = src_arr[yadd1][xminus1];
			v21 = src_arr[yadd1][xi];
			v22 = src_arr[yadd1][xadd1];
			v23 = src_arr[yadd1][xadd2];

			v30 = src_arr[yadd2][xminus1];
			v31 = src_arr[yadd2][xi];
			v32 = src_arr[yadd2][xadd1];
			v33 = src_arr[yadd2][xadd2];

			d[0][0] = v11;//f00
			d[0][1] = v21;//f01
			d[1][0] = v12;//f10
			d[1][1] = v22;//f11

			//d[0][2] = (v21-v01)/2.0;//fy00
			//d[0][3] = (v31-v11)/2.0;//fy01
			//d[1][2] = (v22-v02)/2.0;//fy10;
			//d[1][3] = (v32-v12)/2.0;//fy11;

			//d[2][0] = (v12-v10)/2.0;//fx00;
			//d[2][1] = (v22-v20)/2.0;//fx01;
			//d[3][0] = (v13-v11)/2.0;//fx10;
			//d[3][1] = (v23-v21)/2.0;//fx11;
           //这里采用sobel算子求一阶偏导数，为了避免引入一些噪声
			d[0][2] = (v20 + 2*v21+v22 - v00 - 2*v01- v02) / 8.0;//fy00
			d[0][3] = (v30 + 2*v31+v32 - v10 - 2*v11- v12) / 8.0;//fy01
			d[1][2] = (v21 + 2*v22+v23 - v01 - 2*v02- v03) / 8.0;//fy10;
			d[1][3] = (v31 + 2*v32+v33 - v11 - 2*v12- v13) / 8.0;//fy11;

			d[2][0] = (v02 + 2*v12+v22 - v00 - 2*v10 - v20) / 8.0;//fx00;
			d[2][1] = (v12 + 2*v22+v32 - v10 - 2*v20 - v30) / 8.0;//fx01;
			d[3][0] = (v03 + 2*v13+v23 - v01 - 2*v11 - v21) / 8.0;//fx10;
			d[3][1] = (v13 + 2*v23+v33 - v11 - 2*v21 - v31) / 8.0;//fx11;

			d[2][2] = (v22+v00-v02-v20) / 4.0; //fxy00
			d[2][3] = (v32+v10-v30-v12) / 4.0; //fxy01;
			d[3][2] = (v23+v01-v03-v21) / 4.0; //fxy10;
			d[3][3] = (v33+v11-v13-v31) / 4.0; //fxy11

			matrix_mul(a, d, m0);
			matrix_mul(d, m1, a);

			u2 = u*u;
			u3 = u2*u;
			v2 = v*v;
			v3 = v2*v;
			tmp[0] = d[0][0] + d[0][1] * v + d[0][2] * v2 + d[0][3] * v3;
			tmp[1] = d[1][0] + d[1][1] * v + d[1][2] * v2 + d[1][3] * v3;
			tmp[2] = d[2][0] + d[2][1] * v + d[2][2] * v2 + d[2][3] * v3;
			tmp[3] = d[3][0] + d[3][1] * v + d[3][2] * v2 + d[3][3] * v3;

			val = tmp[0] + tmp[1] * u + tmp[2] * u2 + tmp[3] * u3;
			if (val < 0)val = 0;
			if (val > 255)val = 255;
			dst_arr[i][j] = val;
		}
	}
}

方法二：

static inline void matrix_mul1(float c[4], float a[4][4], float b[4])
{
	int n;
	for (n = 0; n < 4; n++)
	{
		c[n] = a[n][0] * b[0] + a[n][1] * b[1] + a[n][2] * b[2] + a[n][3] * b[3];
	}
}

void img_resize_using_bicubic1(uint8_2d* src, uint8_2d* dst)
{
	int src_rows, src_cols;
	int dst_rows, dst_cols;
	int i, j;
	if (!src || !dst)return;
	uint8_t** src_arr;
	uint8_t** dst_arr;
	float xratio;
	float yratio;
	float x, y;
	int   xi, yi;
	int   xminus1, yminus1;
	int   xadd1, yadd1;
	int   xadd2, yadd2;

	float u, v;
	float u2, u3;
	float v2, v3;

	int val;
	src_arr = src->arr;
	dst_arr = dst->arr;
	src_rows = src->rows;
	src_cols = src->cols;
	dst_rows = dst->rows;
	dst_cols = dst->cols;

	float m[4][4] = { { 0, 2, 0,   0},
					   { -1, 0, 1,  0},
					   { 2, -5, 4,  -1},
					   { -1, 3, -3, 1}};

	xratio = (float)src_cols / (float)dst_cols;
	yratio = (float)src_rows / (float)dst_rows;

	float tmp[4];
	float pix[4];
	float b[4];
	for (i = 0; i < dst_rows; i++)
	{
		for (j = 0; j < dst_cols; j++)
		{
			x = xratio*j;
			y = yratio*i;
			xi = (int)x;
			yi = (int)y;

			u = x - xi;
			v = y - yi;

			xminus1 = xi - 1;
			yminus1 = yi - 1;
			xadd1 = xi + 1;
			yadd1 = yi + 1;
			xadd2 = xi + 2;
			yadd2 = yi + 2;

			if (xi >= src_cols)xi = src_cols - 1;
			if (yi >= src_rows)yi = src_rows - 1;
			if (xminus1 < 0)xminus1 = 0;
			if (yminus1 < 0)yminus1 = 0;
			if (xadd1 >= src_cols)xadd1 = src_cols - 1;
			if (yadd1 >= src_rows)yadd1 = src_rows - 1;
			if (xadd2 >= src_cols)xadd2 = src_cols - 1;
			if (yadd2 >= src_rows)yadd2 = src_rows - 1;

			u2 = u*u;
			u3 = u2*u;
			v2 = v*v;
			v3 = v2*v;

			pix[0] = src_arr[yminus1][xminus1];
			pix[1] = src_arr[yminus1][xi];
			pix[2] = src_arr[yminus1][xadd1];
			pix[3] = src_arr[yminus1][xadd2];
			matrix_mul1(tmp, m, pix);
			b[0] = 0.5*(tmp[0] + tmp[1] * u + tmp[2] * u2 + tmp[3] * u3);

			pix[0] = src_arr[yi][xminus1];
			pix[1] = src_arr[yi][xi];
			pix[2] = src_arr[yi][xadd1];
			pix[3] = src_arr[yi][xadd2];
			matrix_mul1(tmp, m, pix);
			b[1] = 0.5*(tmp[0] + tmp[1] * u + tmp[2] * u2 + tmp[3] * u3);

			pix[0] = src_arr[yadd1][xminus1];
			pix[1] = src_arr[yadd1][xi];
			pix[2] = src_arr[yadd1][xadd1];
			pix[3] = src_arr[yadd1][xadd2];
			matrix_mul1(tmp, m, pix);
			b[2] = 0.5*(tmp[0] + tmp[1] * u + tmp[2] * u2 + tmp[3] * u3);

			pix[0] = src_arr[yadd2][xminus1];
			pix[1] = src_arr[yadd2][xi];
			pix[2] = src_arr[yadd2][xadd1];
			pix[3] = src_arr[yadd2][xadd2];
			matrix_mul1(tmp, m, pix);
			b[3] = 0.5*(tmp[0] + tmp[1] * u + tmp[2] * u2 + tmp[3] * u3);

			matrix_mul1(tmp, m, b);
			val = 0.5*(tmp[0] + tmp[1] * v + tmp[2] * v2 + tmp[3] * v3);
			if (val < 0)val = 0;
			if (val > 255)val = 255;
			dst_arr[i][j] = val;
		}
	}
}

测试代码：

void test_bicubic_resize(dai_image* img, float factor)
{
	uint8_2d* r = NULL;
	uint8_2d* g = NULL;
	uint8_2d* b = NULL;

	uint8_2d* r1 = NULL;
	uint8_2d* g1 = NULL;
	uint8_2d* b1 = NULL;

	dai_image* img1 = NULL;
	if (!img)return;
	split_img_data(img, &r, &g, &b);
	int w, h;
	int w1, h1;
	w = img->width;
	h = img->height;
	w1 = factor*w;
	h1 = factor*h;

	r1 = create_uint8_2d(h1, w1);
	g1 = create_uint8_2d(h1, w1);
	b1 = create_uint8_2d(h1, w1);

	if (factor < 1.0)
	{
		real_2d* kernel = create_gaussian_kernel(1.4);
		real_2d* r2 = uint8_to_real(r->data, r->cols, r->rows);
		gaussian_blur(r2, kernel, 0);
		real_2d* g2 = uint8_to_real(g->data, g->cols, g->rows);
		gaussian_blur(g2, kernel, 0);
		real_2d* b2 = uint8_to_real(b->data, b->cols, b->rows);
		gaussian_blur(b2, kernel, 0);
		destroy_uint8_2d(&r);
		destroy_uint8_2d(&g);
		destroy_uint8_2d(&b);
		r = real_to_uint8(r2);
		g = real_to_uint8(g2);
		b = real_to_uint8(b2);
	}
	img_resize_using_bicubic(r, r1);
	img_resize_using_bicubic(g, g1);
	img_resize_using_bicubic(b, b1);

	merge_img_data(r1, g1, b1, &img1);
	if (img1)
	{
		img1->type = EJPEG;
		dai_save_image("resize_bicubic.jpg",img1);
		dai_destroy_image(&img1);
	}
	destroy_uint8_2d(&r);
	destroy_uint8_2d(&g);
	destroy_uint8_2d(&b);
	destroy_uint8_2d(&r1);
	destroy_uint8_2d(&g1);
	destroy_uint8_2d(&b1);
}

对于scale小于1.0的情况，需要先进行低通滤波然后再进行亚采样插值，split_img_data和merge_img_data可以参看本人上一篇关于二次线性插值的博客。

效果图：
原图：

放大图：