What we saw Today in Class. Recall that we assumed that for some $h$, we had set cover instances with universe of size $u$, $s$ sets, each set of size $z$, where

- the Yes instances had a solution using at most $X$ sets, but
- the No instances had the property that even picking $h\cdot X$ sets would leave a $2^{-c \cdot h}$-fraction of the universe uncovered. (Let us call this the $h$-factor property.)

In the No instances, we proved a claim that if the total cost of edges bought in levels $1, 2, \ldots, j-1$ was at most $hX/2$, then the cost incurred at level $j$ itself would be at least $(u/4z)\cdot 2^{-ch}$. In particular, if $k$ were (at least) $(4zhX/u)\cdot 2^{ch}$, this claim implies that the total cost in the No instance is at least $hX/2$.

And hence we have a gap of $h/2$ between the Yes and the No instances.

Parameters (The Questions). So how do we set the parameters so that

- the set cover instance satisfies the $h$-factor property.
- The number of levels $k$ is at least $(4zhX/u) \cdot 2^{ch}$.
- the size of the construction $N \approx (2z)^k \cdot s$ is reasonable.

In turn, this gives us a set cover instance with universe size $u = n'd'2^{|K'|}$, number of sets $s = n'|L'|$, size of each set being $z = d' 2^{|K'| - 1}$, the optimal solution size is $X = 2n'$, and it satisfies the $h$-factor property above if $1/h^2 > O(\eta)$.

Parameters (The Answers). To answer the questions above:

- the set cover instance satisfies the $h$-factor property as long as $h^2 < O(1/\eta)$. This is satisfied if we set $\eta \approx h \approx \Theta(\log \log n)$.

- Note that the parameters derived above satisfy $zX/u = 4$, and hence $k = 16h 2^{ch} \approx \Theta(\log n)$ is sufficient.
- the size of the construction is $N \approx s (2z)^k = n^{- \log \eta} \cdot poly(1/\eta)^{(2^h)} \approx n^{\log \log \log n}$.

Questions and comments about today's lecture go here, as usual.

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