Step 1: For each group of 8, find the 4th smallest.                 A*n/8

Step 2: Find the median of those candidates.                        T(n/8)

Step 3: Partition the list around that pivot.                         n

Step 4: In the worst case, our recursion is on the larger 
        side, and that side has as many values as possible. 
        We need to determine how many values are required
        to partition into the smaller side using the "X"s 
        visualization technique.

           The "less than" side of the partitioning will 
           have (roughly) at least 1/2 of 4/8 of the values
           so (4/16)ths.

           The "greater than" side of the partitioning will 
           have (roughly) at least 1/2 of 5/8 of the values 
           so (5/16)ths.

           The remaining 7/16 of the values could go on
           either side.

           We will assume the adversary will construct things 
           so that the 7/16 goes with the 5/16 and that our 
           recursive call will be on that quantity of 12/16ths 
           of the input, or (3/4)n.
                                                                    T(3n/4)


So, the run-time would be:

	T(n) = An/8 + T(n/8) + n + T(3n/4)


* From here, you should be able to prove that this will be a linear run-time 
  using a recursion tree.

* Of course you can then use constructive induction to solve for   c   such that
  T(n)<=cn.