Lecture 13: The Hardness Endgame

Here's the final steps to the hardness of Priority Steiner tree. The details are in the paper On the Approximability of Network Design Problems, J. Chuzhoy, A. Gupta, J. Naor and A. Sinha, SODA 2006.

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.)
  

The Yes instances mapped to instances of PST where the solution cost was at most $X$.

In the No instances, we proved a claim that if the total cost of edges bought in levels $1,2,\dots ,j-1$ 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

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

Our Friend the Set Cover Instance. The construction of the set cover instance is the usual thing as discussed in Lecture 11 (or in Homework 3): note that we can start from a SAT instance of size $n$, and get a label cover instance with $nʹ$ nodes in each partition, each node participating in $dʹ$ constraints, label set $Lʹ$, key set $Kʹ$, where $nʹ=n^{-\log \eta }$, and $dʹ,\mid Lʹ\mid ,\mid Kʹ\mid$ are all $poly(1/\eta )$.

In turn, this gives us a set cover instance with universe size $u=nʹdʹ2^{\mid Kʹ\mid }$, number of sets $s=nʹ\mid Lʹ\mid$, size of each set being $z=dʹ2^{\mid Kʹ\mid -1}$, the optimal solution size is $X=2nʹ$, and it satisfies the $h$-factor property above if $1/h^2>O\left(\eta \right)$.

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

1. the set cover instance satisfies the $h$-factor property as long as . This is satisfied if we set $\eta \approx h\approx \Theta \left(\log \log n\right)$.
   
   
2. Note that the parameters derived above satisfy $zX/u=4$, and hence $k=16h{2}^{ch}\approx \Theta \left(\log n\right)$ is sufficient.
   
   
3. the size of the construction is $N\approx s(2z{)}^k=n^{-\log \eta }\cdot poly(1/\eta {)}^{\left(2^h\right)}\approx n^{\log \log \log n}$.
   

Finally, the gap between the Yes and No instances is $h=\Omega \left(\log \log n\right)=\log \log N$ as well. This completes the proof.

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