A slight modification to statement of the theorem today.

All we care about is that we get one set A such that: n/3 <= |A| <= 2n/3. The original statement follows from this one but it allows us to focus on getting a small partition into A, S, and the rest of the graph.

We had a little review of the previous lecture material but today's focus is on partitioning the set between t_0 and t_2 (defined last time).

First, we add an artificial vertex "s" and put in edges to every vertex of L(t_0) and keep the BFS tree from before up until L(t_2). The depth of this new tree, T, is at most sqrt(n) because t_0 and t_2 were chosen so that t_2 – t_0 <= sqrt(n).

We'll triangulate T and start with an two paths starting at s to a leaf in L(t_2). The vertices along these two paths will be part of the separator set S. For each vertex in T, we can calculate the number of vertices interior and exterior to the separator set using some sort of dynamic programming. We shrink the interior until we have at between n/3 and 2n/3 vertices in it.

We'll probably get another lecture on the separator theorem and cover some applications of it. Please review the boards and the book for Monday's lecture.