## Tuesday, February 12, 2008

### Lecture 9: Finishing Ek-Indep. Set hardness; AKC hardness

Any questions about Lecture 9 can go here.

I wanted to include a niggling detail about the hardness for Independent Set. Recall that in our reduction, our hyperedges involved $k/2$ distinct sets from $\{0,1\}^L_v$ and $k/2$ distinct sets from $\{0,1\}^L_{v'}$, whenever $(u,v)$ and $(u,v')$ were both in $E$.

It is important (why?) to allow the case $v' = v$. However this requires us to take a little care. First of all, in this case, we need to take the hyperedges to be of the form $\{A_1^v, ..., A_k^v\}$ where all $k$ sets are all distinct. This will still be a hyperedge iff $\pi_{v \to u}(\cap_{i=1}^{k/2} A_i)$ is disjoint from $\pi_{v \to u}(\cap_{i=k/2+1}^{k} A_i)$.

Suppose that $v$ is a "good" vertex. We know that $\mathcal{F}_v$ is too big to be $k/2$-wise $t$-intersecting, so it contains some "good" sets $A_1^{v}, ..., A_{k/2}^v$ with intersection size less than $t$. We took Sugg($v$) to be this intersection. As we saw in class, if $v$ has a neighbor $u$, and $v'$ is another good neighbor of $u$, then Sugg($v$) and Sugg($v'$) are "consistent" label sets (i.e., when projected to $u$, they agree on at least one key). In particular, Sugg($v$) and Sugg($v'$) must certainly be nonempty.

The tiny tricky part now is the following: This argument relies on $v' \neq v$! If $v' = v$ then you only have $k/2$ "good" sets in play, and you can't make any sort of argument based on size-$k$ hyperedges. In particular, if $v$ has no other "good" distance-2 neighbors, Sugg($v$) may well be empty!

But: it's not like it's a problem. In this case, just put an arbitrary element into Sugg($v$). Then the soundness argument is fine; any neighbor $u$ of $v$ must have $v$ as its buddy, and then of course the $(u,v)$ constraint will be satisfied with probability $1$!