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Bilinear interpolation

1   Algorithm

A digital image has the intensity function \(f(x,y)\) at a point \((x,y)\). Any point \((x, y)\) has four neighbor pixels \((x_1,y_1),\,(x_1,y_2),\,(x_2,y_1),\,\text{and}\,(x_2,y_2)\), and we already know their pixel values \(f(x_1,y_1),\,f(x_1,y_2),\,f(x_2,y_1),\,\text{and}\,f(x_2,y_2)\).

In the \(x\)-direction for \(y=y_1\), \begin{equation} \begin{split} f(x,y_1)&=f(x_1,y_1)+\frac{f(x_2,y_1)-f(x_1,y_1)}{x_2-x_1}(x-x_1)\\ &=\frac{f(x_1,y_1)(x_2-x_1)+f(x_2,y_1)(x-x_1)-f(x_1,y_1)(x-x_1)}{x_2-x_1}\\ &=\frac{f(x_1,y_1)(x_2-x_1-x+x_1)+f(x_2,y_1)(x-x_1)}{x_2-x_1}\\ &=\frac{x_2-x}{x_2-x_1}f(x_1,y_1)+\frac{x-x_1}{x_2-x_1}f(x_2,y_1) \end{split} \end{equation} or \begin{equation} \begin{split} f(x,y_1)&=\frac{f(x_2,y_1)-f(x_1,y_1)}{x_2-x_1}x+\frac{f(x_1,y_1)x_2-f(x_2,y_1)x_1}{x_2-x_1}\\ &=a_1x+b_1,\\ \\ a_1&=\frac{f(x_2,y_1)-f(x_1,y_1)}{x_2-x_1}\\ b_1&=\frac{f(x_1,y_1)x_2-f(x_2,y_1)x_1}{x_2-x_1}. \end{split} \end{equation}

Similarly for \(y=y_2\), \begin{equation} f(x,y_2)=\frac{x_2-x}{x_2-x_1}f(x_1,y_2)+\frac{x-x_1}{x_2-x_1}f(x_2,y_2) \end{equation} or \begin{equation} \begin{split} f(x,y_2)&=\frac{f(x_2,y_2)-f(x_1,y_2)}{x_2-x_1}x+\frac{f(x_1,y_2)x_2-f(x_2,y_2)x_1}{x_2-x_1}\\ &=a_2x+b_2,\\ \\ a_2&=\frac{f(x_2,y_2)-f(x_1,y_2)}{x_2-x_1}\\ b_2&=\frac{f(x_1,y_2)x_2-f(x_2,y_2)x_1}{x_2-x_1}. \end{split} \end{equation}

Now, in the \(y\)-direction, \begin{equation} \begin{split} f(x,y)&=f(x,y_1)+\frac{f(x,y_2)-f(x,y_1)}{y_2-y_1}(y-y_1)\\ &=\frac{f(x,y_1)(y_2-y_1)+f(x,y_2)(y-y_1)-f(x,y_1)(y-y_1)}{y_2-y_1}\\ &=\frac{f(x,y_1)(y_2-y_1-y+y_1)+f(x,y_2)(y-y_1)}{y_2-y_1}\\ &=\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}f(x,y_1)+\frac{y-y_1}{y_2-y_1}f(x,y_2) \end{split} \end{equation} or \begin{equation} \begin{split} f(x,y)&=\frac{f(x,y_2)-f(x,y_1)}{y_2-y_1}y+\frac{f(x,y_1)y_2-f(x,y_2)y_1}{y_2-y_1}\\ &=\frac{a_2x+b_2-a_1x-b_1}{y_2-y_1}y+\frac{(a_1x+b_1)y_2-(a_2x+b_2)y_1}{y_2-y_1}\\ &=\frac{(a_2-a_1)xy+(b_2-b_1)y+(a_1y_2-a_2y_1)x+b_1y_2-b_2y_1}{y_2-y_1}\\ &=\frac{(a_1y_2-a_2y_1)x+(b_2-b_1)y+(a_2-a_1)xy+b_1y_2-b_2y_1}{y_2-y_1}\\ &=ax+by+cxy+d,\\ \\ a&=\frac{a_1y_2-a_2y_1}{y_2-y_1}\\ b&=\frac{b_2-b_1}{y_2-y_1}\\ c&=\frac{a_2-a_1}{y_2-y_1}\\ d&=\frac{b_1y_2-b_2y_1}{y_2-y_1}. \end{split} \end{equation}

2   Matrix method

We know that bilinear interpolation has a functional form of \(f(x, y)=ax+by+cxy+d\) and has four neighbor pixels: \((x_1,y_1),\,(x_1,y_2),\,(x_2,y_1),\,\text{and}\,(x_2,y_2)\). At the same time, we already know the \(f(x,y)\) values of these neighbor pixels and there are four unknown variables: \(a,\,b,\,c,\,\text{and}\,d\). From the following system of equations: \begin{equation} \begin{split} f(x_1,y_1)&=ax_1+by_1+cx_1y_1+d\\ f(x_1,y_2)&=ax_1+by_2+cx_1y_2+d\\ f(x_2,y_1)&=ax_2+by_1+cx_2y_1+d\\ f(x_2,y_2)&=ax_2+by_2+cx_2y_2+d, \end{split} \end{equation} we can obtain this matrix equation: \begin{equation} \begin{bmatrix} x_1& y_1& x_1y_1& 1\\ x_1& y_2& x_1y_2& 1\\ x_2& y_1& x_2y_1& 1\\ x_2& y_2& x_2y_2& 1 \end{bmatrix} \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix} = \begin{bmatrix} f(x_1,y_1)\\ f(x_1,y_2)\\ f(x_2,y_1)\\ f(x_2,y_2) \end{bmatrix}. \end{equation} Now, the unknown variables can be obtained by \begin{equation} \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix} = \begin{bmatrix} x_1& y_1& x_1y_1& 1\\ x_1& y_2& x_1y_2& 1\\ x_2& y_1& x_2y_1& 1\\ x_2& y_2& x_2y_2& 1 \end{bmatrix}^{-1} \begin{bmatrix} f(x_1,y_1)\\ f(x_1,y_2)\\ f(x_2,y_1)\\ f(x_2,y_2) \end{bmatrix}. \end{equation} \(f(x,y)\) is calculated as \begin{equation} f(x,y)= \begin{bmatrix} x& y& xy& 1 \end{bmatrix} \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix} = \begin{bmatrix} x& y& xy& 1 \end{bmatrix} \begin{bmatrix} x_1& y_1& x_1y_1& 1\\ x_1& y_2& x_1y_2& 1\\ x_2& y_1& x_2y_1& 1\\ x_2& y_2& x_2y_2& 1 \end{bmatrix}^{-1} \begin{bmatrix} f(x_1,y_1)\\ f(x_1,y_2)\\ f(x_2,y_1)\\ f(x_2,y_2) \end{bmatrix}. \end{equation}