No, each computer after getting the median of medians, partitions the numbers it has and sends two values to the dedicated computer. Those two values of number of elements which are less than the median of medians and number of elements that are greater than the median of medians. Now the dedicated computer knows how many elements in totality are greater and less than the median of medians. So if more n^2/2 are less, then the median will on the left side else it'll be on the right side. Take the case of number of elements less than median of medians (denoted by L) being greater than n^2 /2. (note that if L = n^2-1/2 and G=n^2-1/2, then median of medians is the median of the whole set of n^2 elements). So L > n^2 / 2 and G < n^2 / 2, in this case the median n^2 elements is the (n^2-G)th element in the L set. The process is repeated again but this time, each computer ignores all the values that are greater than the median of medians and considers only those which are less.
The original problem asks for median but this can be applied to select any 'k'th element in the total n^2 set.
