In mathematics, bicubic interpolation is an extension of cubic spline interpolation (a method of applying cubic interpolation to a data set) for interpolating data points on a two-dimensional regular grid. The interpolated surface (meaning the kernel shape, not the image) is smoother than corresponding surfaces obtained by bilinear interpolation or nearest-neighbor interpolation. Bicubic interpolation can be accomplished using either Lagrange polynomials, cubic splines, or cubic convolution algorithm.

In image processing, bicubic interpolation is often chosen over bilinear or nearest-neighbor interpolation in image resampling, when speed is not an issue. In contrast to bilinear interpolation, which only takes 4 pixels (2×2) into account, bicubic interpolation considers 16 pixels (4×4). Images resampled with bicubic interpolation can have different interpolation artifacts, depending on the b and c values chosen.

Computation

Suppose the function values ${\displaystyle f}$ and the derivatives ${\displaystyle f_{x))$, ${\displaystyle f_{y))$ and ${\displaystyle f_{xy))$ are known at the four corners ${\displaystyle (0,0)}$, ${\displaystyle (1,0)}$, ${\displaystyle (0,1)}$, and ${\displaystyle (1,1)}$ of the unit square. The interpolated surface can then be written as

$${\displaystyle p(x,y)=\sum \limits _{i=0}^{3}\sum _{j=0}^{3}a_{ij}x^{i}y^{j}.}$$

The interpolation problem consists of determining the 16 coefficients ${\displaystyle a_{ij))$. Matching ${\displaystyle p(x,y)}$ with the function values yields four equations:

1. ${\displaystyle f(0,0)=p(0,0)=a_{00},}$
2. ${\displaystyle f(1,0)=p(1,0)=a_{00}+a_{10}+a_{20}+a_{30},}$
3. ${\displaystyle f(0,1)=p(0,1)=a_{00}+a_{01}+a_{02}+a_{03},}$
4. ${\displaystyle f(1,1)=p(1,1)=\textstyle \sum \limits _{i=0}^{3}\sum \limits _{j=0}^{3}a_{ij}.}$

Likewise, eight equations for the derivatives in the ${\displaystyle x}$ and the ${\displaystyle y}$ directions:

1. ${\displaystyle f_{x}(0,0)=p_{x}(0,0)=a_{10},}$
2. ${\displaystyle f_{x}(1,0)=p_{x}(1,0)=a_{10}+2a_{20}+3a_{30},}$
3. ${\displaystyle f_{x}(0,1)=p_{x}(0,1)=a_{10}+a_{11}+a_{12}+a_{13},}$
4. ${\displaystyle f_{x}(1,1)=p_{x}(1,1)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=0}^{3}a_{ij}i,}$
5. ${\displaystyle f_{y}(0,0)=p_{y}(0,0)=a_{01},}$
6. ${\displaystyle f_{y}(1,0)=p_{y}(1,0)=a_{01}+a_{11}+a_{21}+a_{31},}$
7. ${\displaystyle f_{y}(0,1)=p_{y}(0,1)=a_{01}+2a_{02}+3a_{03},}$
8. ${\displaystyle f_{y}(1,1)=p_{y}(1,1)=\textstyle \sum \limits _{i=0}^{3}\sum \limits _{j=1}^{3}a_{ij}j.}$

And four equations for the ${\displaystyle xy}$ mixed partial derivative:

1. ${\displaystyle f_{xy}(0,0)=p_{xy}(0,0)=a_{11},}$
2. ${\displaystyle f_{xy}(1,0)=p_{xy}(1,0)=a_{11}+2a_{21}+3a_{31},}$
3. ${\displaystyle f_{xy}(0,1)=p_{xy}(0,1)=a_{11}+2a_{12}+3a_{13},}$
4. ${\displaystyle f_{xy}(1,1)=p_{xy}(1,1)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=1}^{3}a_{ij}ij.}$

The expressions above have used the following identities:

$${\displaystyle p_{x}(x,y)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=0}^{3}a_{ij}ix^{i-1}y^{j},}$$

$${\displaystyle p_{y}(x,y)=\textstyle \sum \limits _{i=0}^{3}\sum \limits _{j=1}^{3}a_{ij}x^{i}jy^{j-1},}$$

$${\displaystyle p_{xy}(x,y)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=1}^{3}a_{ij}ix^{i-1}jy^{j-1}.}$$

This procedure yields a surface ${\displaystyle p(x,y)}$ on the unit square ${\displaystyle [0,1]\times [0,1]}$ that is continuous and has continuous derivatives. Bicubic interpolation on an arbitrarily sized regular grid can then be accomplished by patching together such bicubic surfaces, ensuring that the derivatives match on the boundaries.

Grouping the unknown parameters ${\displaystyle a_{ij))$ in a vector

$${\displaystyle \alpha =\left[{\begin{smallmatrix}a_{00}&a_{10}&a_{20}&a_{30}&a_{01}&a_{11}&a_{21}&a_{31}&a_{02}&a_{12}&a_{22}&a_{32}&a_{03}&a_{13}&a_{23}&a_{33}\end{smallmatrix))\right]^{T))$$

$${\displaystyle x=\left[{\begin{smallmatrix}f(0,0)&f(1,0)&f(0,1)&f(1,1)&f_{x}(0,0)&f_{x}(1,0)&f_{x}(0,1)&f_{x}(1,1)&f_{y}(0,0)&f_{y}(1,0)&f_{y}(0,1)&f_{y}(1,1)&f_{xy}(0,0)&f_{xy}(1,0)&f_{xy}(0,1)&f_{xy}(1,1)\end{smallmatrix))\right]^{T},}$$

${\displaystyle A\alpha =x}$

Inverting the matrix gives the more useful linear equation ${\displaystyle A^{-1}x=\alpha }$, where

$${\displaystyle A^{-1}=\left[{\begin{smallmatrix}{\begin{array}{rrrrrrrrrrrrrrrr}1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\-3&3&0&0&-2&-1&0&0&0&0&0&0&0&0&0&0\\2&-2&0&0&1&1&0&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\0&0&0&0&0&0&0&0&-3&3&0&0&-2&-1&0&0\\0&0&0&0&0&0&0&0&2&-2&0&0&1&1&0&0\\-3&0&3&0&0&0&0&0&-2&0&-1&0&0&0&0&0\\0&0&0&0&-3&0&3&0&0&0&0&0&-2&0&-1&0\\9&-9&-9&9&6&3&-6&-3&6&-6&3&-3&4&2&2&1\\-6&6&6&-6&-3&-3&3&3&-4&4&-2&2&-2&-2&-1&-1\\2&0&-2&0&0&0&0&0&1&0&1&0&0&0&0&0\\0&0&0&0&2&0&-2&0&0&0&0&0&1&0&1&0\\-6&6&6&-6&-4&-2&4&2&-3&3&-3&3&-2&-1&-2&-1\\4&-4&-4&4&2&2&-2&-2&2&-2&2&-2&1&1&1&1\end{array))\end{smallmatrix))\right],}$$

${\displaystyle \alpha }$

There can be another concise matrix form for 16 coefficients:

$${\displaystyle {\begin{bmatrix}f(0,0)&f(0,1)&f_{y}(0,0)&f_{y}(0,1)\\f(1,0)&f(1,1)&f_{y}(1,0)&f_{y}(1,1)\\f_{x}(0,0)&f_{x}(0,1)&f_{xy}(0,0)&f_{xy}(0,1)\\f_{x}(1,0)&f_{x}(1,1)&f_{xy}(1,0)&f_{xy}(1,1)\end{bmatrix))={\begin{bmatrix}1&0&0&0\\1&1&1&1\\0&1&0&0\\0&1&2&3\end{bmatrix)){\begin{bmatrix}a_{00}&a_{01}&a_{02}&a_{03}\\a_{10}&a_{11}&a_{12}&a_{13}\\a_{20}&a_{21}&a_{22}&a_{23}\\a_{30}&a_{31}&a_{32}&a_{33}\end{bmatrix)){\begin{bmatrix}1&1&0&0\\0&1&1&1\\0&1&0&2\\0&1&0&3\end{bmatrix)),}$$

$${\displaystyle {\begin{bmatrix}a_{00}&a_{01}&a_{02}&a_{03}\\a_{10}&a_{11}&a_{12}&a_{13}\\a_{20}&a_{21}&a_{22}&a_{23}\\a_{30}&a_{31}&a_{32}&a_{33}\end{bmatrix))={\begin{bmatrix}1&0&0&0\\0&0&1&0\\-3&3&-2&-1\\2&-2&1&1\end{bmatrix)){\begin{bmatrix}f(0,0)&f(0,1)&f_{y}(0,0)&f_{y}(0,1)\\f(1,0)&f(1,1)&f_{y}(1,0)&f_{y}(1,1)\\f_{x}(0,0)&f_{x}(0,1)&f_{xy}(0,0)&f_{xy}(0,1)\\f_{x}(1,0)&f_{x}(1,1)&f_{xy}(1,0)&f_{xy}(1,1)\end{bmatrix)){\begin{bmatrix}1&0&-3&2\\0&0&3&-2\\0&1&-2&1\\0&0&-1&1\end{bmatrix)),}$$

$${\displaystyle p(x,y)={\begin{bmatrix}1&x&x^{2}&x^{3}\end{bmatrix)){\begin{bmatrix}a_{00}&a_{01}&a_{02}&a_{03}\\a_{10}&a_{11}&a_{12}&a_{13}\\a_{20}&a_{21}&a_{22}&a_{23}\\a_{30}&a_{31}&a_{32}&a_{33}\end{bmatrix)){\begin{bmatrix}1\\y\\y^{2}\\y^{3}\end{bmatrix)).}$$

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

$${\displaystyle p_{x}(x,y)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=0}^{3}{\frac {a_{ij}ix^{i-1}y^{j)){\Delta x)),}$$

$${\displaystyle p_{y}(x,y)=\textstyle \sum \limits _{i=0}^{3}\sum \limits _{j=1}^{3}{\frac {a_{ij}x^{i}jy^{j-1)){\Delta y)),}$$

$${\displaystyle p_{xy}(x,y)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=1}^{3}{\frac {a_{ij}ix^{i-1}jy^{j-1)){\Delta x\Delta y)),}$$

${\displaystyle \Delta x}$

${\displaystyle x}$

${\displaystyle (x,y)}$

${\displaystyle \Delta y}$

${\displaystyle \alpha }$

$${\displaystyle x=\left[{\begin{smallmatrix}f(0,0)&f(1,0)&f(0,1)&f(1,1)&\Delta xf_{x}(0,0)&\Delta xf_{x}(1,0)&\Delta xf_{x}(0,1)&\Delta xf_{x}(1,1)&\Delta yf_{y}(0,0)&\Delta yf_{y}(1,0)&\Delta yf_{y}(0,1)&\Delta yf_{y}(1,1)&\Delta x\Delta yf_{xy}(0,0)&\Delta x\Delta yf_{xy}(1,0)&\Delta x\Delta yf_{xy}(0,1)&\Delta x\Delta yf_{xy}(1,1)\end{smallmatrix))\right]^{T},}$$

${\displaystyle \alpha =A^{-1}x}$

${\displaystyle A}$

$${\displaystyle {\begin{aligned}{\overline {x))&={\frac {x-x_{0)){x_{1}-x_{0))},\\{\overline {y))&={\frac {y-y_{0)){y_{1}-y_{0))}\end{aligned))}$$

${\displaystyle x_{0},x_{1},y_{0},}$

${\displaystyle y_{1))$

${\displaystyle x}$

${\displaystyle y}$

${\displaystyle (x,y)}$

$${\displaystyle p(x,y)=\sum \limits _{i=0}^{3}\sum _{j=0}^{3}a_{ij}{\overline {x))^{i}{\overline {y))^{j}.}$$

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 their slope, 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:

$${\displaystyle W(x)={\begin{cases}(a+2)|x|^{3}-(a+3)|x|^{2}+1&{\text{for ))|x|\leq 1,\\a|x|^{3}-5a|x|^{2}+8a|x|-4a&{\text{for ))1<|x|<2,\\0&{\text{otherwise)),\end{cases))}$$

${\displaystyle a}$

${\displaystyle W(0)=1}$

${\displaystyle W(n)=0}$

${\displaystyle n}$

This approach was proposed by Keys, who showed that ${\displaystyle a=-0.5}$ produces third-order convergence with respect to the sampling interval of the original function.^[1]

If we use the matrix notation for the common case ${\displaystyle a=-0.5}$, we can express the equation in a more friendly manner:

$${\displaystyle p(t)={\tfrac {1}{2)){\begin{bmatrix}1&t&t^{2}&t^{3}\end{bmatrix)){\begin{bmatrix}0&2&0&0\\-1&0&1&0\\2&-5&4&-1\\-1&3&-3&1\end{bmatrix)){\begin{bmatrix}f_{-1}\\f_{0}\\f_{1}\\f_{2}\end{bmatrix))}$$

${\displaystyle t}$

${\displaystyle t}$

For two dimensions first applied once in ${\displaystyle x}$ and again in ${\displaystyle y}$:

$${\displaystyle {\begin{aligned}b_{-1}&=p(t_{x},f_{(-1,-1)},f_{(0,-1)},f_{(1,-1)},f_{(2,-1)}),\\[1ex]b_{0}&=p(t_{x},f_{(-1,0)},f_{(0,0)},f_{(1,0)},f_{(2,0)}),\\[1ex]b_{1}&=p(t_{x},f_{(-1,1)},f_{(0,1)},f_{(1,1)},f_{(2,1)}),\\[1ex]b_{2}&=p(t_{x},f_{(-1,2)},f_{(0,2)},f_{(1,2)},f_{(2,2)}),\end{aligned))}$$

$${\displaystyle p(x,y)=p(t_{y},b_{-1},b_{0},b_{1},b_{2}).}$$

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.

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.