Bilinear interpolation

We first do linear interpolation in the x-direction. This yields

${\displaystyle {\begin{aligned}f(x,y_{1})={\frac {x_{2}-x}{x_{2}-x_{1}}}f(Q_{11})+{\frac {x-x_{1}}{x_{2}-x_{1}}}f(Q_{21}),\\f(x,y_{2})={\frac {x_{2}-x}{x_{2}-x_{1}}}f(Q_{12})+{\frac {x-x_{1}}{x_{2}-x_{1}}}f(Q_{22}).\end{aligned}}}$

We proceed by interpolating in the y-direction to obtain the desired estimate:

${\displaystyle {\begin{aligned}f(x,y)&={\frac {y_{2}-y}{y_{2}-y_{1}}}f(x,y_{1})+{\frac {y-y_{1}}{y_{2}-y_{1}}}f(x,y_{2})\\&={\frac {y_{2}-y}{y_{2}-y_{1}}}\left({\frac {x_{2}-x}{x_{2}-x_{1}}}f(Q_{11})+{\frac {x-x_{1}}{x_{2}-x_{1}}}f(Q_{21})\right)+{\frac {y-y_{1}}{y_{2}-y_{1}}}\left({\frac {x_{2}-x}{x_{2}-x_{1}}}f(Q_{12})+{\frac {x-x_{1}}{x_{2}-x_{1}}}f(Q_{22})\right)\\&={\frac {1}{(x_{2}-x_{1})(y_{2}-y_{1})}}\left(f(Q_{11})(x_{2}-x)(y_{2}-y)+f(Q_{21})(x-x_{1})(y_{2}-y)+f(Q_{12})(x_{2}-x)(y-y_{1})+f(Q_{22})(x-x_{1})(y-y_{1})\right)\\&={\frac {1}{(x_{2}-x_{1})(y_{2}-y_{1})}}{\begin{bmatrix}x_{2}-x&x-x_{1}\end{bmatrix}}{\begin{bmatrix}f(Q_{11})&f(Q_{12})\\f(Q_{21})&f(Q_{22})\end{bmatrix}}{\begin{bmatrix}y_{2}-y\\y-y_{1}\end{bmatrix}}.\end{aligned}}}$

Note that we will arrive at the same result if the interpolation is done first along the y direction and then along the x direction.[1]

An alternative way is to write the solution to the interpolation problem as a multilinear polynomial

${\displaystyle f(x,y)\approx a_{00}+a_{10}x+a_{01}y+a_{11}xy,}$

where the coefficients are found by solving the linear system

${\displaystyle {\begin{aligned}{\begin{bmatrix}1&x_{1}&y_{1}&x_{1}y_{1}\\1&x_{1}&y_{2}&x_{1}y_{2}\\1&x_{2}&y_{1}&x_{2}y_{1}\\1&x_{2}&y_{2}&x_{2}y_{2}\end{bmatrix}}{\begin{bmatrix}a_{00}\\a_{10}\\a_{01}\\a_{11}\end{bmatrix}}={\begin{bmatrix}f(Q_{11})\\f(Q_{12})\\f(Q_{21})\\f(Q_{22})\end{bmatrix}},\end{aligned}}}$

yielding the result

${\displaystyle {\begin{aligned}{\begin{bmatrix}a_{00}\\a_{10}\\a_{01}\\a_{11}\end{bmatrix}}={\frac {1}{(x_{2}-x_{1})(y_{2}-y_{1})}}{\begin{bmatrix}x_{2}y_{2}&-x_{2}y_{1}&-x_{1}y_{2}&x_{1}y_{1}\\-y_{2}&y_{1}&y_{2}&-y_{1}\\-x_{2}&x_{2}&x_{1}&-x_{1}\\1&-1&-1&1\end{bmatrix}}{\begin{bmatrix}f(Q_{11})\\f(Q_{12})\\f(Q_{21})\\f(Q_{22})\end{bmatrix}}.\end{aligned}}}$

A geometric visualisation of bilinear interpolation. The product of the value at the desired point (black) and the entire area is equal to the sum of the products of the value at each corner and the partial area diagonally opposite the corner (corresponding colours).

The solution can also be written as a weighted mean of the f(Q):

${\displaystyle f(x,y)\approx w_{11}f(Q_{11})+w_{12}f(Q_{12})+w_{21}f(Q_{21})+w_{22}f(Q_{22}),}$

where the weights sum to 1 and satisfy the transposed linear system

${\displaystyle {\begin{bmatrix}1&1&1&1\\x_{1}&x_{1}&x_{2}&x_{2}\\y_{1}&y_{2}&y_{1}&y_{2}\\x_{1}y_{1}&x_{1}y_{2}&x_{2}y_{1}&x_{2}y_{2}\end{bmatrix}}{\begin{bmatrix}w_{11}\\w_{12}\\w_{21}\\w_{22}\end{bmatrix}}={\begin{bmatrix}1\\x\\y\\xy\end{bmatrix}},}$

yielding the result

${\displaystyle {\begin{aligned}{\begin{bmatrix}w_{11}\\w_{21}\\w_{12}\\w_{22}\end{bmatrix}}={\frac {1}{(x_{2}-x_{1})(y_{2}-y_{1})}}{\begin{bmatrix}x_{2}y_{2}&-y_{2}&-x_{2}&1\\-x_{2}y_{1}&y_{1}&x_{2}&-1\\-x_{1}y_{2}&y_{2}&x_{1}&-1\\x_{1}y_{1}&-y_{1}&-x_{1}&1\end{bmatrix}}{\begin{bmatrix}1\\x\\y\\xy\end{bmatrix}},\end{aligned}}}$

which simplifies to

${\displaystyle {\begin{aligned}w_{11}&=(x_{2}-x)(y_{2}-y)/((x_{2}-x_{1})(y_{2}-y_{1})),\\w_{12}&=(x_{2}-x)(y-y_{1})/((x_{2}-x_{1})(y_{2}-y_{1})),\\w_{21}&=(x-x_{1})(y_{2}-y)/((x_{2}-x_{1})(y_{2}-y_{1})),\\w_{22}&=(x-x_{1})(y-y_{1})/((x_{2}-x_{1})(y_{2}-y_{1})),\end{aligned}}}$

in agreement with the result obtained by repeated linear interpolation. The set of weights can also be interpreted as a set of generalized barycentric coordinates for a rectangle.

Combining the above, we have

${\displaystyle {\begin{aligned}f(x,y)\approx {\frac {1}{(x_{2}-x_{1})(y_{2}-y_{1})}}{\begin{bmatrix}f(Q_{11})&f(Q_{12})&f(Q_{21})&f(Q_{22})\end{bmatrix}}{\begin{bmatrix}x_{2}y_{2}&-y_{2}&-x_{2}&1\\-x_{2}y_{1}&y_{1}&x_{2}&-1\\-x_{1}y_{2}&y_{2}&x_{1}&-1\\x_{1}y_{1}&-y_{1}&-x_{1}&1\end{bmatrix}}{\begin{bmatrix}1\\x\\y\\xy\end{bmatrix}}.\end{aligned}}}$

If we choose a coordinate system in which the four points where f is known are (0, 0), (0, 1), (1, 0), and (1, 1), then the interpolation formula simplifies to

${\displaystyle f(x,y)\approx f(0,0)(1-x)(1-y)+f(0,1)(1-x)y+f(1,0)x(1-y)+f(1,1)xy,}$

or equivalently, in matrix operations:

${\displaystyle f(x,y)\approx {\begin{bmatrix}1-x&x\end{bmatrix}}{\begin{bmatrix}f(0,0)&f(0,1)\\f(1,0)&f(1,1)\end{bmatrix}}{\begin{bmatrix}1-y\\y\end{bmatrix}}.}$

Here we also recognize the weights:

${\displaystyle {\begin{aligned}w_{11}&=(1-x)(1-y),\\w_{12}&=(1-x)y,\\w_{21}&=x(1-y),\\w_{22}&=xy.\end{aligned}}}$

Alternatively, the interpolant on the unit square can be written as

${\displaystyle f(x,y)=a_{00}+a_{10}x+a_{01}y+a_{11}xy,}$

where

${\displaystyle {\begin{aligned}a_{00}&=f(0,0),\\a_{10}&=f(1,0)-f(0,0),\\a_{01}&=f(0,1)-f(0,0),\\a_{11}&=f(1,1)-f(1,0)-f(0,1)+f(0,0).\end{aligned}}}$

In both cases, the number of constants (four) correspond to the number of data points where f is given.

As seen in the example on the right, the intensity value at the pixel computed to be at row 20.2, column 14.5 can be calculated by first linearly interpolating between the values at column 14 and 15 on each rows 20 and 21, giving

${\displaystyle {\begin{aligned}I_{20,14.5}&={\frac {15-14.5}{15-14}}\cdot 91+{\frac {14.5-14}{15-14}}\cdot 210=150.5,\\I_{21,14.5}&={\frac {15-14.5}{15-14}}\cdot 162+{\frac {14.5-14}{15-14}}\cdot 95=128.5,\end{aligned}}}$

and then interpolating linearly between these values, giving

${\displaystyle I_{20.2,14.5}={\frac {21-20.2}{21-20}}\cdot 150.5+{\frac {20.2-20}{21-20}}\cdot 128.5=146.1.}$

This algorithm reduces some of the visual distortion caused by resizing an image to a non-integral zoom factor, as opposed to nearest-neighbor interpolation, which will make some pixels appear larger than others in the resized image.