Median of distinct numbers What is the least number of comparisons we need to find the median of 6 distinct numbers?
I am able to find the answer to the median of 5 distinct numbers to be 6 comparisons, and it makes sense, however in the case of 6 numbers I can't find an answer.
The best I was able to do it in by hand was 9 comparisons.  Can that be minimized further?
Edit:  Median in this case, we are assuming to be the lower median.

It takes at least 8 and can be done in exactly 8.

EDIT 
As pointed out by hardmath, the lower bound below is actually an upper
  bound on the lower bound. JeffE's notes which I link at the bottom
  give a value of $7$, not $8$.
See this cstheory question here (again thanks to hardmath):
  https://cstheory.stackexchange.com/questions/12257/exact-number-of-comparisons-to-compute-the-median/12287#12287
which does indeed say the value is $8$.

Note: the below is an incomplete proof.
Lower Bound
A lower bound to find all the $k$ smallest elements is given here: Lower bounds for Selection.
Which says we need at least
$$ n - k + \sum_{n+1-k < j \le n} \lceil \log_{2} j \rceil $$
For $n=6, k=3$, this gives $9$.
$$ 6 - 3 + \lceil \log_{2} 5 \rceil + \lceil \log_{2} 6 \rceil = 9$$
Thus to find the smallest, second smallest and the third smallest, we need at least $9$ compares.
Now any algorithm which finds the $3^{rd}$ smallest element (say $z$) will have to find the set $S = \\{x, y \\}$ such that $x < z$ and $y < z$ (otherwise how does it know that $z$ is the third largest?).
An extra compare between $x$ and $y$ will now give us the smallest, second smallest and third smallest numbers.
Thus any algorithm which finds the third smallest must use at least 8 comparisons.

Upper Bound
As I.J. Kennedy notes, there is an algorithm to find the 3rd smallest element of 6 elements in 8 comparisons and can be found in Knuth's Art of Computer Programming Vol 3, Page 212.
According to Knuth, this is a result obtained by A. Hadian and M. Sobel [Univ of Minnesota, Dept of Statistics Report 121 (169)] and they show that the $k^{th}$ largest element can be found in
$$ n - k + (k-1) \lceil \log_{2} (n+2-k) \rceil $$
comparisons. For $n=6$ and $k=4$ we get the upper bound of $8$.

Algorithm
The way they do it (Knuth's book has all this information) is:
First create a knockout tournament on $n-k+2$ items, which takes $n-k+1$ compares.
The largest item (say $L$) of this cannot be a candidate. So of the remaining $k-2$ pick one and follow the path of $L$ up the tree, which gives the the new largest in at most $\lceil \log_{2} n+2-k \rceil$ compares. Remove this new largest, replacing it with one of the remaining and following up the path of this new largest.
Continue till the rest of the $k-2$ are exhausted. Finally replace the largest with $-\infty$ and determine the largest of the remaining $n+1-k$ which will be the $k^{th}$ largest as we have removed the largest $k-1$ so far.

Related: http://compgeom.cs.uiuc.edu/~jeffe/teaching/497/02-selection.pdf
