# How to jump directly from level 0 to level 5 in a Gaussian pyramid without having the levels 1, 2, 3, and 4?

I am current studying the Laplacian Pyramid as a compact image code by Peter J Burt and Edawrd Hudson.

I understood all the concepts but I am having trouble with the equivalent weighting function. I understand how it works but i can't find the relation between $$h(n,m)$$ and $$w(n,m)$$.

Here is the link for the paper. The equations are at the end of page 2.

• Could you copy relevant content to make your post self-contained?
– Evil
Commented Jun 24, 2019 at 2:10
• Convolutions are associative. Applying two consecutive convolutions has the same effect as convolving one kernel with the other, and convolving the image with the result. Then decimate.
– user16034
Commented Mar 11, 2022 at 13:23

$$w$$ is effectively defined by $$g_l(i,j) = \sum_{m=-2}^2 \sum_{n=-2}^2 w(m,n) g_{l-1}(2i+m, 2j+n)$$ and $$h_l$$ by $$g_l(i,j) = \sum_{m=-M_l}^{M_l} \sum_{n=-M_l}^{M_l} h_l(m,n) g_0(i2^l+m, j2^l + n)$$ Some straightforward substitution of the second into the first gives $$\sum_{m_l=-M_l}^{M_l} \sum_{n_l=-M_l}^{M_l} h_l(m_l,n_l) g_0(2^l i+m_l, 2^l j + n_l) = \\ \sum_{m=-2}^2 \sum_{n=-2}^2 w(m,n) \sum_{m'=-M_{l-1}}^{M_{l-1}} \sum_{n'=-M_{l-1}}^{M_{l-1}} h_{l-1}(m',n') g_0(2^l i+2^{l-1}m+m', 2^l j+2^{l-1}n + n')$$ We conclude firstly that $$M_l = 2^l + M_{l-1}$$, and since $$M_1 = 2$$ (and $$h_1 = w$$), we have $$M_l = 2^{l+1}-2$$. Secondly, that
$$h_l(m_l, n_l) = \sum_{\substack{-2 \le m \le 2 \\ |m_l - 2^{l-1}m| \le M_{l-1}}} \sum_{\substack{-2 \le n \le 2 \\ |n_l - 2^{l-1}n| \le M_{l-1}}} w(m,n) h_{l-1}(m_l - 2^{l-1}m, n_l - 2^{l-1}n)$$ by identification of weights of the same pixel of $$g_0$$.