[Solved]: Counting elements that are greater than the median of medians

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Problem Detail: Short version: I want to know where the $-2$ comes from in the formula on p. 221 of CLRS 3rd edition. Long version: CLRS (3rd ed.) give an algorithm for $O(n)$ worst case arbitrary order statistic of $n$ distinct numbers. The algorithm is roughly:

Input: an array of $n$ elements and $i$, the number of the order statistic to return from the elements.

  1. Divide the $n$ elements into $lfloor n/5 rfloor$ groups of 5 elements each along with an optional group containing $nmod{5}$ elements (resulting in $lceil n/5 rceil$ groups.)
  2. Find the median of each of the groups by sorting.
  3. Recurse, using the $lceil n/5 rceil$ medians as the array and $lfloorlceil n/5 rceil/2rfloor$ as the order statistic, resulting in the median-of-medians.
  4. Partition the $n$ elements around the median-of-medians (using a quicksort-like $O(n)$ partitioning algorithm.
  5. Letting $k-1$ be the number of elements less than the median-of-medians, if $i = k$, return the median-of-medians. Otherwise recurse: if $i < k$ then recurse finding the $i$th order statistic of the $k-1$ elements less than the median-of-medians; if $i > k$, then recurse finding the $i-k$th order statistic of the $n-k$ elements greater than the median-of-medians.

Output: the $i$th order statistic of the $n$ numbers.

In the proof of the runtime, CLRS argue that the number of elements greater than the median-of-medians is at least: $$ 3 bigg(bigglceil frac{1}2 bigglceil{frac{n}5} biggrceil biggrceil – 2bigg) $$ The reasoning is that half of the medians are greater than the median-of-medians, and each of those medians’ groups has at least three elements greater than the median-of-medians (the median itself plus the two elements greater than the median.) That would result in $$ 3 bigg(bigglceil frac{1}2 bigglceil{frac{n}5} biggrceil biggrceilbigg) $$ for the lower bound on the number of elements greater than the median-of-medians. But we must account for two things: the group containing the median-of-medians (the median-of-medians is not greater than itself) and the group that contains the modulo leftovers. To account for the group containing the median-of-medians, we subtract 1, resulting in: $$ 3 bigg(bigglceil frac{1}2 bigglceil{frac{n}5} biggrceil biggrceilbigg) – 1 $$ and I think that for the modulo leftovers group, we should subtract 4, because the least number of elements in the group is 1. So that would give: $$ 3 bigg(bigglceil frac{1}2 bigglceil{frac{n}5} biggrceil biggrceilbigg) – 5 $$ which can be transformed into $$ 3 bigg(bigglceil frac{1}2 bigglceil{frac{n}5} biggrceil biggrceil – 2bigg) + 1 $$ Why does my analysis lead to a lower-bound 1 greater than that given in CLRS?

Asked By : Carl G

Answered By : Massimo Cafaro

In $$ 3 bigg(bigglceil frac{1}2 bigglceil{frac{n}5} biggrceil biggrceil – 2bigg) $$ we are subtracting 2 in order to discard the group containing the median of medians and the group of leftovers. So, 2 is the number of groups we are discarding. First of all, note that there may not be a group of leftover elements: if $n$ is an exact multiple of 5 there will be no leftover group. We are interested in bounding from below the number of elements greater than the median-of-medians in the worst case, so suppose that a leftover group exists. Therefore, it will contain at least an element and no more than 4 elements. If you want to reason in terms of elements to be discarded and not in terms of groups, then we must discard exactly 3 elements for the group containing the median of medians (including the median of the medians), and at most 2 element from the leftovers group (if this group contains 1 or 2 elements you do not discard any element; if this group contains 3 or 4 elements, then you discard respectively 1 or 2 elements which are greater than the group’s median). So, you discard in the worst case (leftover group with 4 elements) 3 + 2 elements: $$ 3 bigg(bigglceil frac{1}2 bigglceil{frac{n}5} biggrceil biggrceilbigg) – 5 $$
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