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} $<$ 3/4.

With that idea noted, we come to:


Formal endgame of proof:

Define an edge $(u,v)$ to be "frugally suggested" if $\mid Sugg(u,v)\mid \le 10/\epsilon .$ 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$\left[va{l}_{}\right(f\left)\right]\ge \eta =:{\epsilon }^3/2000$, as desired. Supposing $(u,v)$ is good, it has consistent suggestions and hence there exist $a\in Sugg\left(u\right)$ and $\alpha \in Sugg\left(v\right)$ such that the labeling $(a,\alpha )$ satisfies the constraint on $(u,v)$ (that is, ${\pi }_{v\to u}\left(\alpha \right)=a$). Further, since $(u,v)$ is good, it is frugally suggested, so both $\mid Sugg\left(u\right)\mid$ and $\mid Sugg\left(v\right)\mid$ 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 $f\left(u\right)=a$ and $f\left(v\right)=\alpha$, 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$\left[\right(u,v)$ consistent suggestions$]$

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

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

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

$=\epsilon /4,$

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 $2/(1-\epsilon /4)\le 2+\epsilon$, 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

$1-2^{-(2+\epsilon )}=1-\frac{1}{4}{e}^{-\left(\ln 2\right)\epsilon }\le 1-\frac{1}{4}(1-(\ln 2\left)\epsilon \right)=3/4+.17\epsilon <3/4+(3/4)\epsilon$

a contradiction.