Lecture 3: 1 vs. 3/4 + eps hardness for Max-Coverage: now with bonus endgame calculations

Any questions about the lecture can go here. I'll also show how to finish off the calculations.

First, there was one small extra idea that I didn't have time to mention: It uses the fact that the function $1 - 2^{-t}$ is a concave function of $t$. Recall that the overall average number of suggestions per edges is $2$. We also know that edges with t total suggestions but no consistent suggestions have coverage at most $1 - 2^{-t}$. We want that they have overall coverage at most $1 - 2^{-2}$. We need to check that the Suggester can't "cheat" by achieving awesome coverage on some edges by "spending" a lot of suggestions, overcoming bad coverage on edges where few suggestions were spent.

This indeed follows from concavity. As an example, suppose the Suggester is labeling 2 edges inconsistently, and it has 4 labels to spend (average of 2, as needed). It could spend 2 on each and achieve avg{$1 - 2^{-2}$, $1 - 2^{-2}$) $= 3/4$. Or it could try spending 3 on one and 1 on the other. But this only achieves avg{$1 - 2^{-3}$, $1 - 2^{-1}$} = avg{7/8, 1/2} $\lt$ 3/4.

With that idea noted, we come to:


Formal endgame of proof:

Define an edge $(u,v)$ to be "frugally suggested" if $|Sugg(u,v)| \leq 10/\epsilon.$ Next, define an edge to be "good" if it is both frugally and consistently suggested.

Suppose the fraction of good edges $(u,v)$ is at least $\epsilon/20$. We argue that this implies E$[val_{\mathcal{G}}(f)] \geq \eta = : \epsilon^3/2000$, as desired. Supposing $(u,v)$ is good, it has consistent suggestions and hence there exist $a \in Sugg(u)$ and $\alpha \in Sugg(v)$ such that the labeling $(a, \alpha)$ satisfies the constraint on $(u,v)$ (that is, $\pi_{v \rightarrow u}(\alpha) = a$). Further, since $(u,v)$ is good, it is frugally suggested, so both $|Sugg(u)|$ and $|Sugg(v)|$ are at most $10/\epsilon$. Thus when $f$ is randomly chosen, there is at least an $(\epsilon/10)^2$ chance that it will get both $f(u) = a$ and $f(v) = \alpha$, thus satisfying $(u,v)$. Hence in expectation, $f$ satisfies at least an $(\epsilon/20)(\epsilon/10)^2 = \eta$ fraction of edges $(u,v)$, as desired.

It remains to show that indeed the fraction of good edges cannot be less than $\epsilon/20$. If this were the case, then for a randomly chosen edge $(u,v)$,

P$[(u,v)$ consistent suggestions$]$

$\leq P$[(u,v)$ good$] +$ P$[(u,v)$ not frugally sugg'd$]$

$\lt \epsilon/20 +$ P$[(u,v)$ more than $10/\epsilon$ sugg'd labels$]$

$\leq \epsilon/20 + 2(\epsilon/10) $

$ = \epsilon/4,$

where the second-to-last step used Markov's inequality based on the fact that the average number of suggested labels per edge is exactly $2$.

Even though the consistently suggested edges have perfect coverage, when their fraction is only $\epsilon/4$, the fact that the overall coverage is assumed to be at least $3/4 + \epsilon$ implies that the inconsistently suggested edges have average coverage at least $3/4 + (3/4)\epsilon$. Our claim implied that an inconsistently suggested edge with $t$ suggested labels has coverage at most $1 - 2^{-t}$. The average number of labels per inconsistently suggested edge is at most $2/(1 - \epsilon/4) \leq 2 + \epsilon$, where we again used that the fraction of consistently suggested edges is at most $\epsilon/4$. Thus by concavity of the function $1 - 2^{-t}$, the average coverage of ground elements on inconsistently suggested edges is at most

$1 - 2^{-(2 + \epsilon)} = 1 - \frac{1}{4} e^{-(\ln 2) \epsilon} \leq 1 - \frac{1}{4} (1 - (\ln 2)\epsilon) = 3/4 + .17\epsilon \lt 3/4 + (3/4)\epsilon$

a contradiction.