Lecture 18: Last bits of the proof

Recall the low-diameter decomposition procedure: we picked a random radius $X$ and a random permutation $\pi$. We then constructed the random partition by considering all the vertices one by one, in the order given by $\pi$: when considering $w$, we assigned all the yet-unassigned vertices $v$ with $d(w,v)\le X$ to $w$'s partition.

Just for the analysis: suppose we renumber the vertices $w_1,w_2,\stackrel{.}{s},w_n$ in order of their distance from the edge $e=(u,v)$.

The crucial definitions (that avoid pain later) are the following:

1. At some time instant in this procedure, one (or both) of $u$ or $v$ gets assigned to some $w_i$. Say that $w_i$ settles the edge $(u,v)$.
   
   
2. At the moment the edge is settled, if only one endpoint of this edge gets assigned, then we say that $w_i$ cuts the edge $(u,v)$.
   

Note that each edge is settled at exactly one time instant in the procedure, and it may or may not be cut at that point in time. Of course, once the edge is settled (with or without being cut), it is never cut in the future.

Consider $w_j$ and let its distance $d(w_j,u)=a_j$ and $d(w_j,v)=b_j$. Assume , the other case is identical. If $w_j$ cuts $(u,v)$ when the random values are $X$ and $\pi$, the following two properties must hold:

1. The random variable $X$ must lie in the interval $[a_j,b_j]$ (else either none or both of $(u,v)$ would get marked).
   
   
2. The node $w_j$ must come before $w_1,...,w_{j-1}$ in the permutation $\pi$.
   
   Suppose not, and one of them came before $w_j$ in the permutation. Since all these vertices are closer to the edge than $w_j$ is, then for the current value of $X$, they would have settle the edge (either capture one or both of the endpoints) at some previous time point, and hence $w_j$ would not settle---and hence not cut---the edge $(u,v)$.
   

Now the rest of the argument is as in class: Pr[ edge $e$ is cut ] = $\sum _j$ Pr[ $w_j$ cuts the edge $e$]. Moreover, Pr[$w_j$ cuts $e$] $\le$ Pr[$X\in [a_j,b_j]$ and $w_j$ comes before $w_1,...w_{j-1}$ in the permutation $\pi$] $\le$ $(d_{uv}/(r/2\left)\right)\times (1/j)$. And summing this over all vertices gives us $2(d_{uv}/r)H_n$.