Question

I came across places where floors and ceilings are neglected while solving recurrences.

Example from CLRS (chapter 4, pg.83) where floor is neglected:

Here (pg.2, exercise 4.1–1) is an example where ceiling is ignored:
(EDIT: I gather from public opinion that this is somewhat fishy.)

In fact in CLRS (pg.88) its mentioned that:

"Floors and ceilings USUALLY do not matter when solving recurrences"

My questions:

1. Here does "usually" means ALL cases ? If yes, i can just forget them all the time.
2. If not, then when do floors and ceilings really count while solving recurrences ?

Note: This ain’t a homework problem. I thought about it while I was refreshing my DS and algo concepts.

Answer 1

The floor and ceiling functions satisfy the following inequalities for all x:

• x−1 < ⌊x⌋ ≤ x

• x ≤ ⌈x⌉ < x+1

Thus, in the first example, we have ⌊n / 2⌋ ≤ n / 2. Also, since the logarithm is a monotone increasing function, we know that lg ⌊n / 2⌋ ≤ lg(n / 2). Putting these together, we get the first inequality 2(c ⌊n / 2⌋ lg ⌊n / 2⌋) + n ≤ cn lg(n / 2) + n.

The second example actually contains an error: c lg ⌈n / 2⌉ + 1 is never less than but can be equal to c lg(n / 2) + 1. However, it is true that c lg ⌈n / 2⌉ + 1 ≤ c lg(n / 2 + 1) + 1, which we can then bound from above by, say, c lg(n / 2) + 2 (assuming that n ≥ 2) and thus derive the desired conclusion that T(n) ∈ O(lg n).

In fact, the second example contains other errors too: even with the assumptions stated in the following paragraph (which you did not quote), the last = sign should be ≤.

(Ps. Phew, this was a real pain to type without LaTeX. That, if nothing else, is why questions like these are better asked on math.SE.)