Bicubic interpolation

Extension to rectilinear grids

Often, applications call for bicubic interpolation using data on a rectilinear grid, rather than the unit square. In this case, the identities for ${\displaystyle p_x,p_y,}$ and ${\displaystyle p_{xy}}$ become $${\displaystyle p_x(x,y)=\sum _{i=1}^3\sum _{j=0}^3\frac{a_{ij}i{x}^{i-1}{y}^j}{\mathrm{\Delta }x},}$$ $${\displaystyle p_y(x,y)=\sum _{i=0}^3\sum _{j=1}^3\frac{a_{ij}{x}^{i}j{y}^{j-1}}{\mathrm{\Delta }y},}$$ $${\displaystyle p_{xy}(x,y)=\sum _{i=1}^3\sum _{j=1}^3\frac{a_{ij}i{x}^{i-1}j{y}^{j-1}}{\mathrm{\Delta }x\mathrm{\Delta }y},}$$ where ${\displaystyle \mathrm{\Delta }x}$ is the ${\displaystyle x}$ spacing of the cell containing the point ${\displaystyle (x,y)}$ and similar for ${\displaystyle \mathrm{\Delta }y}$. In this case, the most practical approach to computing the coefficients ${\displaystyle \alpha }$ is to let $${\displaystyle x={\left[\begin{array}{cccccccccccccccc}f(0,0)& f(1,0)& f(0,1)& f(1,1)& \mathrm{\Delta }x{f}_x(0,0)& \mathrm{\Delta }x{f}_x(1,0)& \mathrm{\Delta }x{f}_x(0,1)& \mathrm{\Delta }x{f}_x(1,1)& \mathrm{\Delta }y{f}_y(0,0)& \mathrm{\Delta }y{f}_y(1,0)& \mathrm{\Delta }y{f}_y(0,1)& \mathrm{\Delta }y{f}_y(1,1)& \mathrm{\Delta }x\mathrm{\Delta }y{f}_{xy}(0,0)& \mathrm{\Delta }x\mathrm{\Delta }y{f}_{xy}(1,0)& \mathrm{\Delta }x\mathrm{\Delta }y{f}_{xy}(0,1)& \mathrm{\Delta }x\mathrm{\Delta }y{f}_{xy}(1,1)\end{array}\right]}^T,}$$ then to solve ${\displaystyle \alpha =A^{-1}x}$ with ${\displaystyle A}$ as before. Next, the normalized interpolating variables are computed as $${\displaystyle \begin{array}{cc}\overline{x}& =\frac{x-x_0}{x_1-x_0},\\ \overline{y}& =\frac{y-y_0}{y_1-y_0}\end{array}}$$ where ${\displaystyle x_0,x_1,y_0,}$ and ${\displaystyle y_1}$ are the ${\displaystyle x}$ and ${\displaystyle y}$ coordinates of the grid points surrounding the point ${\displaystyle (x,y)}$. Then, the interpolating surface becomes $${\displaystyle p(x,y)=\sum _{i=0}^3\sum _{j=0}^3{a}_{ij}{\overline{x}}^i{\overline{y}}^j.}$$

Finding derivatives from function values

If the derivatives are unknown, they are typically approximated from the function values at points neighbouring the corners of the unit square, e.g. using finite differences.

To find either of the single derivatives, ${\displaystyle f_x}$ or ${\displaystyle f_y}$, using that method, find the slope between the two surrounding points in the appropriate axis. For example, to calculate ${\displaystyle f_x}$ for one of the points, find ${\displaystyle f(x,y)}$ for the points to the left and right of the target point and calculate the slope of the straight line between them, and similarly for ${\displaystyle f_y}$.

To find the cross derivative ${\displaystyle f_{xy}}$, take the derivative in both axes, one at a time. For example, one can first use the ${\displaystyle f_x}$ procedure to find the ${\displaystyle x}$ derivatives of the points above and below the target point, then use the ${\displaystyle f_y}$ procedure on those values (rather than, as usual, the values of ${\displaystyle f}$ for those points) to obtain the value of ${\displaystyle f_{xy}(x,y)}$ for the target point. (Or one can do it in the opposite direction, first calculating ${\displaystyle f_y}$ and then ${\displaystyle f_x}$ from those. The two give equivalent results.)

At the edges of the dataset, when one is missing some of the surrounding points, the missing points can be approximated by a number of methods. A simple and common method is to assume that the slope from the existing point to the target point continues without further change, and using this to calculate a hypothetical value for the missing point.

Bicubic convolution algorithm

Bicubic spline interpolation requires the solution of the linear system described above for each grid cell. An interpolator with similar properties can be obtained by applying a convolution with the following kernel in both dimensions: $${\displaystyle W(x)=\{\begin{array}{ll}(a+2)|x{|}^3-(a+3)|x{|}^2+1& \text{for }|x|\le 1,\\ a|x{|}^3-5a|x{|}^2+8a|x|-4a& \text{for }1<|x|<2,\\ 0& \text{otherwise},\end{array}}$$ where ${\displaystyle a}$ is usually set to −0.5 or −0.75. Note that ${\displaystyle W(0)=1}$ and ${\displaystyle W(n)=0}$ for all nonzero integers ${\displaystyle n}$.

This approach was proposed by Rifman of TRW Systems Group, targeting image data from ERTS (Earth Resources Technology Satellite, later Landsat 1). Initially, ${\displaystyle a}$ was fixed at -1, but later it was parameterized by Simon of the same company, making -0.5, -0.75, and -1.0 meaningful values. However, these proposals did not sufficiently present the derivation process or formulas, so Keys later re-proposed it in a complete form, demonstrating that with ${\displaystyle a=-0.5}$, cubic convergence is achieved regarding the sampling interval of the original function.

The convolution kernel ${\displaystyle W(x)}$ is derived as follows: $${\displaystyle W(x)=\{\begin{array}{ll}{A}_1|x{|}^3+B_1|x{|}^2+C_1|x|+D_1& \text{for }|x|\le 1\\ A_2|x{|}^3+B_2|x{|}^2+C_2|x|+D_2& \text{for }1<|x|<2\\ 0& \text{otherwise}\end{array}}$$

With the conditions: ${\displaystyle W(0)=1}$, ${\displaystyle W(1)=W(2)=0}$, also for first derivative: ${\displaystyle W^{\prime }(0)=W^{\prime }(2)=0}$, ${\displaystyle W^{\prime }(1)=a}$, we get;

1. ${\displaystyle W(0)=D_1=1,}$
2. ${\displaystyle W(1^{-})=A_1+B_1+C_1+D_1=0,}$
3. ${\displaystyle W(1^{+})=A_2+B_2+C_2+D_2=0,}$
4. ${\displaystyle W(2^{-})=8A_2+4B_2+2C_2+D_2=0,}$
5. ${\displaystyle W^{\prime }(0)=C_1=0,}$
6. ${\displaystyle W^{\prime }(1^{-})=3A_1+2B_1+C_1=a,}$
7. ${\displaystyle W^{\prime }(1^{+})=3A_2+2B_2+C_2=a,}$
8. ${\displaystyle W^{\prime }(2^{-})=12A_2+4B_2+C_2=0.}$

Solving the above simultaneous equations gives us; $${\displaystyle A_1=a+2,B_1=-(a+3),C_1=0,D_1=1,}$$ $${\displaystyle A_2=a,B_2=-5a,C_2=8a,D_2=-4a,}$$ and the above equation.

Matrix notation

If we use the matrix notation, we can express the equation in a more friendly manner: $${\displaystyle p(t)=\left[\begin{array}{cccc}1& t& t^2& t^3\end{array}\right]\left[\begin{array}{cccc}0& 1& 0& 0\\ a& 0& -a& 0\\ -2a& -a-3& 2a+3& a\\ a& a+2& -a-2& -a\end{array}\right]\left[\begin{array}{c}{f}_{-1}\\ f_0\\ f_1\\ f_2\end{array}\right]}$$ for ${\displaystyle t}$ between 0 and 1 for one dimension. Note that for 1-dimensional cubic convolution interpolation 4 sample points are required. For each inquiry two samples are located on its left and two samples on the right. These points are indexed from −1 to 2 in this text. The distance from the point indexed with 0 to the inquiry point is denoted by ${\displaystyle t}$ here.

Since ${\displaystyle a=-0.5}$, it coincides with the Catmull-Rom spline. Therefore, bicubic interpolation with ${\displaystyle a}$ set to ${\displaystyle -0.5}$ is sometimes called "Catmull-Rom interpolation".

For two dimensions first applied once in ${\displaystyle x}$ and again in ${\displaystyle y}$: $${\displaystyle \begin{array}{cc}{b}_{-1}& =p(t_x,f_{(-1,-1)},f_{(0,-1)},f_{(1,-1)},f_{(2,-1)}),\\ b_0& =p(t_x,f_{(-1,0)},f_{(0,0)},f_{(1,0)},f_{(2,0)}),\\ b_1& =p(t_x,f_{(-1,1)},f_{(0,1)},f_{(1,1)},f_{(2,1)}),\\ b_2& =p(t_x,f_{(-1,2)},f_{(0,2)},f_{(1,2)},f_{(2,2)}),\end{array}}$$ $${\displaystyle p(x,y)=p(t_y,b_{-1},b_0,b_1,b_2).}$$

Differential Continuity

It is continuous in the first derivative by definition, and can be easily verified using the following matrix notation.

${\displaystyle p_i^{\prime }(1)=\left[\begin{array}{cccc}0& a& 0& -a\end{array}\right]\left[\begin{array}{c}{f}_{i-1}\\ f_i\\ f_{i+1}\\ f_{i+2}\end{array}\right]=p_{i+1}^{\prime }(0)=\left[\begin{array}{cccc}a& 0& -a& 0\end{array}\right]\left[\begin{array}{c}{f}_i\\ f_{i+1}\\ f_{i+2}\\ f_{i+3}\end{array}\right]}$

Hence, it is continuous in the first derivative.

${\displaystyle p_i^{″}(1)=\left[\begin{array}{cccc}2a& 4a+6& -2a-6& -4a\end{array}\right]\left[\begin{array}{c}{f}_{i-1}\\ f_i\\ f_{i+1}\\ f_{i+2}\end{array}\right]\ne p_{i+1}^{″}(0)=\left[\begin{array}{cccc}-4a& -2a-6& 4a+6& 2a\end{array}\right]\left[\begin{array}{c}{f}_i\\ f_{i+1}\\ f_{i+2}\\ f_{i+3}\end{array}\right]}$

Hence, it is not continuous in the second derivative.

Comparison with other methods

Below is a comparison of the kernel functions for each method:

Use in computer graphics

The bicubic algorithm is frequently used for scaling images and video for display (see bitmap resampling). It preserves fine detail better than the common bilinear algorithm.

Bicubic algorithm is also used for image downscaling, but because resampling during downscaling differs from the effect of zooming out in real optics, other methods may be more suitable.

Effect of parameter "a"

However, due to the negative lobes on the kernel, it causes overshoot (haloing). This can cause clipping, and is an artifact (see also ringing artifacts), but it increases acutance (apparent sharpness), and can be desirable. This effect is larger when the parameter ${\displaystyle a}$ is ${\displaystyle -0.75}$ or ${\displaystyle -1.0}$ than when it is ${\displaystyle -0.5}$.

When ${\displaystyle a=-0.5}$, the interpolation result matches the Taylor approximation up to the second derivative (excluding the remainder term) of the original image (the continuous function before sampling). A clear example of this effect is when the original image is a linear function (a simple gradient image), and the interpolated result is a linear function only when ${\displaystyle a=-0.5}$. For example, if the input is $${\displaystyle \left[\begin{array}{cccc}{f}_{i-1}& f_i& f_{i+1}& f_{i+2}\end{array}\right]=\left[\begin{array}{cccc}-1& 0& 1& 2\end{array}\right],}$$ the interpolated result is $${\displaystyle p(t)=-2(2a+1)t^3+3(2a+1)t^2-2at,}$$ and it is a linear function (quadratic and cubic terms are 0) only when ${\displaystyle a=-0.5}$.

Below are a simple gradient image enlarged 16x, and a portion of the standard test image "cameraman" enlarged 8x. The vertical stripes in the former and the artificial mottled pattern around the temples in the latter are artifacts caused by the above-mentioned effect.

Note that when ${\displaystyle a=-0.75}$, the second derivative of the convolution kernel is continuous at ${\displaystyle x=1}$, but the interpolation result does not have a continuous second derivative. Also, when ${\displaystyle a=-1.0}$, the derivative of the convolution kernel matches the derivative of the sinc function at ${\displaystyle x=1}$, but this has no significance other than for comparison with sinc interpolation.

Resampling Position

When enlarging an image by an integer multiple, it is common to set the resampling position between the input image pixels so that the center of gravity of the enlarged image does not shift relative to the input image (see the left image below).
On the other hand, there is also a method of setting the resampling position to overlap the input image pixels (see the right image below). This method ensures the reversibility of the transformation.

Extrapolation

Two pixels of the input image are required on the left and right (or top and bottom) for interpolation. Since the outer edges of the image are invalid, extrapolation is required. The extrapolation method varies depending on the implementation.
The following is an example of copying and extrapolating the outermost pixels.