I would like to prove that we need at least five comparisons to find median amoung five numbers.
And my proposition:
Let's consider tree of comparisons. There are at least $5 \cdot {4\choose 2} = 30$ possiblities, so in our tree we have at least $30$ leaves. So what is minimal height of this tree when we have at least $30$ leaves. (height is number of comparisons).
So $$height \ge \log_2(30) > 4$$
Hence, $height \ge 5$.
Is it correct proof ?
Your proof is correct. But I don't think 5 comparisons is achievable. Here is the algorithm that finds the median in 6 comparisons.
Sort the first two pairs. [ 2 comparisons]
Order the pairs w.r.t. their respective larger element. [ 1 comparison]
Call the result $[a,b,c,d,e]$; we know $a<b<d$ and $c<d$. Now there are 3 elements less than $d$, hence $d$ can't be the median( i.e. $3^{rd}$element of the sorted array)
WLOG say $c<e$ and $a<b$ (known already). Order the pairs w.r.t. their respective larger element. WLOG $a<b<e$ and $c<e$. Compare $c$ & $b$, greater one is the median. [ 3 comparisons - 1 for $c$ & $e$, 1 for ordering the pairs and 1 for $b$ & $c$ ]
Hence, total number of comparisons is equal to 6.
Idea taken from here .
The proof is correct as you are calculating lower bound for median of 5 elements. But this is data dependent, so it is not number of compares algorithm needs to find median. This is one of instances.
Unless you are happy to know that there are such cases...
But to derive algorithm that works always you have to compute upper bound on this, to have algorithm (that always works). Minimal number independent of order of elements to find median of five elements is six.
Try it yourself. Take 5 numbers, let it be 1,2,3,4,5, permute them and draw trees of comparisons. In fact there are instances with five compares, but you will find instances of six, and this upper bound is interesting.
You might be also interested in this: http://www.cs.hut.fi/~cessu/selection/
