Constructing lossless conductors using zigzag product - a doubt

Reference - this survey: https://www.cs.huji.ac.il/~nati/PAPERS/expander_survey.pdf

I am reading the section on constructing lossless conductors using a bipartite variant of zigzag product (section 10, proof of theorem 10.2.2). I am having trouble following a line in the proof- in page 521 eq (20), the authors write:

Therefore, $$k+a = H_{\infty}(X_1, X_2, R_2) = H_{\infty}(X_1, Y_2, Z_2) = H_{\infty} (Y_1, Z_1, Z_2)$$. To conclude the argument, consider any $$y_1 \in \text{supp}(Y_1)$$ and note that by the bound on the total entropy and the lower bound proved above for $$H_{\infty} (Y_1)$$ ($$H_{\infty}(Y_1) \geq k-14a$$), we must have $$H_{\infty}(Z_1, Z_2|Y_1=y_1) \leq 15a$$

(For reference, the support of $$(Z_1, Z_2)$$ is $$\{0,1\}^{35a}$$.) I don't see why this follows. For example why can't the distribution of $$Z_1,Z_2$$ conditioned on $$Y_1=y_1$$ be uniform for certain values of $$y_1$$ (and for those values of $$y_1$$, the third step does not transfer all the entropy)?

One fix I can think of it is this: we can assume the source distribution is the uniform distribution on a set of size $$2^k$$ (since any distribution with min entropy $$k$$ can be written as a convex combination of distributions of this form). In this case $$p(y_1,z_1,z_2) = 1/2^{k+a}$$ for all $$(y_1,z_1,z_2)$$ in the support - and the upper bound on $$H_{\infty}(Z_1,Z_2|Y_1=y_1)$$ holds. But I don't see how this follows just from the assumption that $$H_{\infty}(Y_1,Z_1,Z_2)=k+a$$. Am I missing something?

• Would it make sense to contact the author(s) rather than hope that somebody who's very familiar with the paper chimes on?
– Kai
Aug 30, 2023 at 16:12