Is the time complexity of the empty algorithm O(0)?

Question

So given the following program:

---

Is the time complexity of this program O(0)? In other words, is 0 O(0)?

I thought answering this in a separate question would shed some light on this question.

EDIT: Lots of good answers here! We all agree that 0 is O(1). The question is, is 0 O(0) as well?

Answer 1

From Wikipedia:

A description of a function in terms of big O notation usually only provides an upper bound on the growth rate of the function.

From this description, since the empty algorithm requires 0 time to execute, it has an upper bound performance of O(0). This means, it's also O(1), which happens to be a larger upper bound.

Edit:

More formally from CLR (1ed, pg 26):

For a given function g(n), we denote O(g(n)) the set of functions

O(g(n)) = { f(n): there exist positive constants c and n₀ such that 0 ≤ f(n) ≤ cg(n) for all n ≥ n₀ }

The asymptotic time performance of the empty algorithm, executing in 0 time regardless of the input, is therefore a member of O(0).

Edit 2:

We all agree that 0 is O(1). The question is, is 0 O(0) as well?

Based on the definitions, I say yes.

Furthermore, I think there's a bit more significance to the question than many answers indicate. By itself the empty algorithm is probably meaningless. However, whenever a non-trivial algorithm is specified, the empty algorithm could be thought of as lying between consecutive steps of the algorithm being specified as well as before and after the algorithm steps. It's nice to know that "nothingness" does not impact the algorithm's asymptotic time performance.

Edit 3:

Adam Crume makes the following claim:

For any function f(x), f(x) is in O(f(x)).

Proof: let S be a subset of R and T be a subset of R^* (the non-negative real numbers) and let f(x):S ->T and c ≥ 1. Then 0 ≤ f(x) ≤ f(x) which leads to 0 ≤ f(x) ≤ cf(x) for all x∈S. Therefore f(x) ∈ O(f(x)).

Specifically, if f(x) = 0 then f(x) ∈ O(0).