# Is It Possible to Reconstruct the Original Image from Laplacian Pyramid?

In our computer vision course, my professor said we can reconstruct the original image from the Laplacian pyramid. It just feels to me that information is lost after doing subsampling and upsampling.

Is it possible? If so, why? Are the reconstructed image exactly the same as the original pixel by pixel? And is this answer correct?

Generate Laplacian pyramid $$(f_2, h_1, h_0)$$ from image $$f_0$$: Reconstruct $$f_0$$ from Laplacian pyramid: source

This depends on the order of upsampling and downsampling. If the order is correct, then you won't throw away anything and thus you should in principle be able to reconstruct the image.

In general:

$$\left(\uparrow_n\downarrow_n f\right) \neq \left(\downarrow_n\uparrow_n f\right)$$

Similar things are used when using the Wavelet decomposition on a signal, which again uses up- and downsampling combined with the Wavelet filters.

The important message is: "it can indeed be reconstructed", meaning under certain conditions, and not "always".

An image pyramid is hierarchical representation of an image with a collection of derived images at different resolutions (thus, sizes). In a Gaussian pyramid, derived images are smoothed at level $$l$$ by an operator $$S_l$$ (eg by a Gaussian filter) and downsampled by an operator $$D_l$$. As downsampled versions look a lot alike, Laplacian pyramids aim at storing only what differs, which requires to compare them at the same resolution with an upsampling operator $$U_l$$.

So, starting from image $$I_0$$, you get $$I_1$$ at lower resolution as $$I_1 = D_1 S_1 I_0$$, and the $$0$$-resolution version by $$\hat{I}_0 =U_1 I_1 = U_1 D_1 S_1 I_0$$, which is a form of prediction of $$I_0$$ from a filtered and downsampled version.

So at the first stage, you keep the difference $$L_0=I_0-\hat{I}_0$$ and $$I_1$$. Obviously, you can recover $$I_0$$ as $$L_0+U_1 I_1$$. This process can be repeated over any other level $$l$$: So once you have perfect reconstruction at each level, the whole process is perfect.

As a reminder, the term Laplacian stems from the observation that in $$L_l=I_l-\hat{I}_l = (\delta - U_l D_l S_l )I_l$$ where $$\delta$$ is the identity operator, the difference operator $$\delta - U_l D_l S_l$$ looks like an approximate Laplacian when $$S_l$$ is a smoothing operator: 