I am trying to implement bilinear interpolation as described in the paper Spatial Tranformer Networks by Jaderberg et. al (see link to paper). They describe bilinear interpolation in Equation 5 as:

$$ V_i^c = \sum_{n}^{H}\sum_{m}^{W} U_{nm}^c \max(0,1-|x_i^s - m|)\cdot\max(0,1-|y_i^s - n|), $$ where:

  • $V_i^c$ is the resulting pixel value in the new image
  • $H$ and $W$ are the height and width of the original image (or feature map) in pixels
  • $c$ refers to the channel (e.g. RGB)
  • $(x_i^s, y_i^s)$ are the coordinates where the original image is sampled (where the image is normalized such that $-1 \le x_i^s, y_i^s\le 1$)
  • $U_{nm}^c$ is defined as the pixel value at location $(n,m)$ in channel $c$.

I am having trouble interpreting the variables $n$ and $m$. Are these

  • coordinates in the normalized image (i.e. $-1 \le n, m\le 1$, where you would sum $n$ from $n=-1$ to $H=1$ in steps of the normalized resolution, e.g. steps of $1/100$ for an image that is 100 px in height)
  • or are these row and column values (e.g. you sum $n$ from $n=0$ to $n=100$ for an image that is 100px in height)?

I have tried out both to do downsampling of an image, but don't get consistent results.

If someone can help me out interpreting this, I would appreciate it very much.

Below I have included what I understand of bilinear interpolation. Maybe that someone can help me out based on this.

In the below figure, a single channel feature map (or image) with one channel is displayed that consists of four pixels with values $ U_{nm} $, where $ n $ and $ m $ are the coordinates of the center of the pixels, i.e. $ m,n \in \{-0.5, 0.5\} $. If we index $ m $ and $ n $ as $ m_k, n_k $, with $ k \in [1,4] $, we can also index the pixel values as $ U_{n_km_k} $. The values of all four pixels can be reduced to a single value $ V $ at position $ (x_i^s, y_i^s) $ by applying bilinear interpolation. Bilinear Interpolation

The procedure can be divided into three linear interpolations. First the value $ U_1' $ at position $ (x_{U_1'}, y_{U_1'}) $ can be computed by interpolating the values $ U_{n_1m_1} $ and $ U_{n_2m_2} $: \begin{equation} U_1' = \Delta x_2\ U_{n_1m_1} + \Delta x_1\ U_{n_2m_2}. \end{equation} As the sum of $ \Delta x_1 $ and $ \Delta x_2 $ is equal to one, due to normalization of the axes, the above equation can be rewritten as: \begin{equation} U_1' = (1-\Delta x_1) U_{n_1m_1} + (1-\Delta x_2) U_{n_2m_2}. \end{equation} The terms $ \Delta x_1 $ and $ \Delta x_2 $ can be expressed as: \begin{align} \Delta x_1 = |x_i^s - {m_1}|\\ \Delta x_2 = |x_i^s - {m_2}|, \end{align} which, substituted into the equation for $U_1'$ yields: \begin{equation} U_1' = U_{n_1m_1}(1-|x_i^s - {m_1}|) + U_{n_2m_2}(1-|x_i^s - {m_2}|). \end{equation}

Similarly the value for $ U_2' $ can be computed: \begin{equation} U_2' = U_{n_3m_3}(1-|x_i^s - {m_3}|) + U_{n_4m_4}(1-|x_i^s - {m_4}|). \end{equation}

Once $ U_1' $ and $ U_2' $ have been computed, $ V $ can be determined by linearly interpolating $ U_1' $ and $ U_2' $: \begin{equation} V = U_1'(1-\Delta y_1) + U_2'(1-\Delta y_2) . \end{equation} The values for $ \Delta y_1 $ and $ \Delta y_2 $ can be expressed as follows: \begin{align} \Delta y_1 = |y_i^s - y_{U_1'}| = |y_i^s - {n_1}| = |y_i^s - {n_2}|\\ \Delta y_2 = |y_i^s - y_{U_2'}| = |y_i^s - {n_3}| = |y_i^s - {n_4}| . \end{align}

Substituting the above equations and those of $\Delta x_1$ and $\Delta x_2$ into the equation for $V$ yields: \begin{equation} \begin{split} V &= U_{n_1m_1}\cdot (1-|x_i^s - {m_1}|) \cdot (1-|y_i^s - {n_1}|) \\ &+ U_{n_2m_2}\cdot (1-|x_i^s - {m_2}|) \cdot (1-|y_i^s - {n_2}|) \\ &+ U_{n_3m_3}\cdot (1-|x_i^s - {m_3}|) \cdot (1-|y_i^s - {n_3}|) \\ &+ U_{n_4m_4}\cdot (1-|x_i^s - {m_4}|) \cdot (1-|y_i^s - {n_4}|), \end{split} \end{equation} which can be written more compactly as:

\begin{equation} \begin{split} V &= \sum_{k=1}^{4} U_{n_km_k} \cdot (1-|x_i^s - {m_k}|) \cdot (1-|y_i^s - {n_k}|)\\ &=\sum_{n}^{H}\sum_{m}^{W} U_{nm} \cdot (1-|x_i^s - {m}|) \cdot (1-|y_i^s - {n}|). \end{split} \end{equation}

Edit to clarify my comment to @Dan Hulme

Initially I also thought that $n$ and $m$ are row and column indices as you normally do a summation over integer values. Also the summation is up to $H$ and $W$, respectively, which are the # of rows and # of columns. So it seems logical to think that $\sum_{n=0}^{H = \#rows-1}\sum_{m=0}^{W = \#columns-1}$, with $n=0,1,2,...,H-1$ and $m=0,1,2,...,W-1$.

However, when you try to apply this to an image which is larger than 2x2 pixels, it doesn't hold.

To clarify this, imagine the original image is an 8x8 image and we wish to downsample the image to a 6x6 image (I cannot add a new image to expose this due to reputation constraints, I am new here). If we want to compute the value of the upper left pixel of the downsampled image with coordinates $(x_1^s, y_1^s) = (-0.833, 0.833)$, we would have: \begin{equation} \begin{split} V_{1} &= \sum_{n}^{H}\sum_{m}^{W} U_{nm}\cdot \max(0, 1-|x_1^s-m|)\cdot \max(0, 1-|y_1^s-n|) \\ &= U_{00}\cdot \max(0, 1-|-0.833-0|)\cdot \max(0, 1-|0.833-0|)\\ &+ U_{01}\cdot \max(0, 1-|-0.833-1|)\cdot \max(0, 1-|0.833-0|)\\ &+ U_{02}\cdot \max(0, 1-|-0.833-2|)\cdot \max(0, 1-|0.833-0|)\\ &+ ...\\ &+ U_{10}\cdot \max(0, 1-|-0.833-0|)\cdot \max(0, 1-|0.833-1|)\\ &+ U_{11}\cdot \max(0, 1-|-0.833-1|)\cdot \max(0, 1-|0.833-1|)\\ &+ ...\\ &+ U_{77}\cdot \max(0, 1-|-0.833-7|)\cdot \max(0, 1-|0.833-7|)\\ &= U_{00}\cdot 0.167^2 + U_{10}\cdot 0.167\cdot 0.833, \end{split} \end{equation} which is only a function of $U_{00}$ and $U_{10}$ and not of $U_{00}$, $U_{01}$, $U_{10}$ and $U_{11}$ as one would reason.

If we reason abouth the lower right pixel of the downsampled image with coordinates $(x_{49}^s, y_{49}^s) = (0.833, -0.833)$ and apply the same equation, we have: \begin{equation} \begin{split} V_{49} &= \sum_{n}^{H}\sum_{m}^{W} U_{nm}\cdot \max(0, 1-|x_{49}^s-m|)\cdot \max(0, 1-|y_{49}^s-n|) \\ &= U_{00}\cdot \max(0, 1-|0.833-0|)\cdot \max(0, 1-|-0.833-0|)\\ &+ U_{01}\cdot \max(0, 1-|0.833-1|)\cdot \max(0, 1-|-0.833-0|)\\ &+ U_{02}\cdot \max(0, 1-|0.833-2|)\cdot \max(0, 1-|-0.833-0|)\\ &+ ...\\ &+ U_{10}\cdot \max(0, 1-|0.833-0|)\cdot \max(0, 1-|-0.833-1|)\\ &+ U_{11}\cdot \max(0, 1-|0.833-1|)\cdot \max(0, 1-|-0.833-1|)\\ &+ ...\\ &+ U_{77}\cdot \max(0, 1-|0.833-7|)\cdot \max(0, 1-|-0.833-7|)\\ &= U_{00}\cdot 0.167^2 + U_{01}\cdot 0.833\cdot 0.167, \end{split} \end{equation} which again is only function of $U_{00}$ and $U_{01}$ and not of $U_{66}$, $U_{67}$, $U_{76}$ and $U_{77}$ as one would expect.

I have also tried normalizing $n$ and $m$, such that $n =-1, -1+ 2/8, -1 +4/8, ..., 1$ (and similarly for $m$, but I end up with similar problems.

  • $\begingroup$ Where does this -0.833 come from in your example? Obviously if you sample outside the original image only the two border pixels will be closest. You have to express the sample points in the same co-ordinate system you're using for the pixels in the old image. $\endgroup$
    – Dan Hulme
    Commented Oct 1, 2017 at 10:42
  • $\begingroup$ I am sorry for the incomplete addition to my question. Given that I am new, I cannot add another image to explain the values. You can view the image I was planning on putting in the following link dropbox.com/s/4jmcangptz1lftu/bilint_derivation11.png?dl=0. I will add this image to the question when my reputation is sufficient to include it. $\endgroup$ Commented Oct 2, 2017 at 7:14
  • $\begingroup$ The $x_s$ and $y_s$ in that image are clearly in [-1,1] co-ordinates, not [0,W) co-ordinates. Obviously to make the subtraction work they both have to be in the same co-ordinate system. $\endgroup$
    – Dan Hulme
    Commented Oct 2, 2017 at 9:21
  • $\begingroup$ In both cases the equation doesn't add up. $\endgroup$ Commented Oct 2, 2017 at 15:10

1 Answer 1


I think the formula at the top only makes sense if $n$ and $m$ are pixel indices, i.e. row and column numbers from 0 to $H - 1$ and 0 to $W - 1$. This way, for the four surrounding pixels, the weight varies from 0 to 1.

As a representation of the algorithm it's unnecessarily opaque, because you don't actually want to look at every pixel of the image and multiply most of them by zero.

To implement it, you want to do something closer to your diagram, where you only look at the four surrounding pixel values, and sum just those four, multiplying each by $(1 - |x - m|)(1 - |y - n|)$. This is exactly how your four-line equation for $V$ (near the bottom of your question) expresses it, and the code should look most like this.

  • $\begingroup$ Initially, I thought the same. However, for images larger than 2x2 this doesn't hold. This is because of the condition −1≤xsi,ysi≤1−1≤xis,yis≤1. I have added an example at the end of my post to clarify this. $\endgroup$ Commented Sep 30, 2017 at 11:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.