I am trying to understand the LidarBoost algorithm as explained in this paper (PDF warning).

I don't understand how they take the original depth-images $Y_k$ and transform them into the up-sampled images $D_k$. I get how optical flow is used to align the $Y_k$ into a chosen reference frame, but is the transform between $Y_k$ and $D_k$ just a standard image up-sampling with nearest neighbors for the "interpolation" step? If that's the case, can someone explain how they get the term $W_k$ in their data term of the energy function? They say this about it, but I don't understand how to construct $W_k$ for each $k$:

$W_k \in \mathbb{R}^{\beta_n \times \beta_m}$ is a banded matrix that encodes the positions of $D_k$ which one samples from during resampling on the high-resolution target grid

I'm trying (so far frustratingly ineffectively) to get enough of a sense of the algorithm to try to implement it.


1 Answer 1


It sounds like $W_k$ is just a matrix of 1's and 0's. When $W_k$ is a 1, then the corresponding range value in $D_k$ is used. If it's a 0, then that range value is not used.

Similarly, $T_k$ should be mostly 1, and 0 when the corresponding $D_k$ values are unreliable.

Does that tally with your understanding? If not, can you elaborate on your question with more details of what you suspect?

  • $\begingroup$ Two things confuse me with that. First of all, it says that $W_k$ is a banded matrix, and I don't see how treating it as an indicator matrix would give the banded structure; same thing for $T_k$, which the paper claims is diagonal. Secondly, how would I actually construct $W_k$ from a given $Y_k$ and $D_k$? I can't figure out when to decide to use a sample or not. $\endgroup$
    – anjruu
    Commented Apr 30, 2013 at 20:46
  • 1
    $\begingroup$ I suspect $W_k$ is "banded" because the top-right and bottom-left corners may be zero (depending on the re-sampling). I, too, do not understand why $W_k$ would be diagonal. I emailed Dr Schuon to see if he has any more detailed explanation of the algorithm; let's see if he responds. $\endgroup$
    – Peter K.
    Commented Apr 30, 2013 at 21:53

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