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

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?

From http://en.wikipedia.org/wiki/Big_O_notation">Wikipedia</a>;:

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:

https://stackoverflow.com/users/25498/adam-crume">Adam Crume makes the following https://stackoverflow.com/questions/3209139/is-the-time-complexity-of-the-empty-algorithm-o0/3215301#3215301">claim</a>:</p>

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).

It takes the same amount of time to run regardless of the input, therefore it is O(1) by definition.

Several answers say that the complexity is O(1) because the time is a constant and the time is bounded by the product of some coefficient and 1. Well, it is true that the time is a constant and it is bounded that way, but that doesn't mean that the best answer is O(1).

Consider an algorithm that runs in linear time. It is ordinarily designated as O(n) but let's play devil's advocate. The time is bounded by the product of some coefficient and n^2. If we consider O(n^2) to be a set, the set of all algorithms whose complexity is small enough, then linear algorithms are in that set. But it doesn't mean that the best answer is O(n^2).

The empty algorithm is in O(n^2) and in O(n) and in O(1) and in O(0). I vote for O(0).

I have a very simple argument for the empty algorithm being O(0): For any function f(x), f(x) is in O(f(x)). Simply let f(x)=0, and we have that 0 (the runtime of the empty algorithm) is in O(0).

On a side note, I hate it when people write f(x) = O(g(x)), when it should be f(x) ∈ O(g(x)).

Big O is asymptotic notation. To use big O, you need a function - in other words, the expression must be parametrized by n, even if n is not used. It makes no sense to say that the number 5 is O(n), it's the constant function f(n) = 5 that is O(n).

So, to analyze time complexity in terms of big O you need a function of n. Your algorithm always makes arguably 0 steps, but without a varying parameter talking about asymptotic behaviour makes no sense. Assume that your algorithm is parametrized by n. Only now you may use asymptotic notation. It makes no sense to say that it is O(n²), or even O(1), if you don't specify what is n (or the variable hidden in O(1))!

As soon as you settle on the number of steps, it's a matter of the definition of big O: the function f(n) = 0 is O(0).

Since this is a low-level question it depends on the model of computation. Under "idealistic" assumptions, it is possible you don't do anything. But in Python, you cannot say def f(x):, but only def f(x): pass. If you assume that every instruction, even pass (NOP), takes time, then the complexity is f(n) = c for some constant c, and unless c != 0 you can only say that f is O(1), not O(0).

It's worth noting big O by itself does not have anything to do with algorithms. For example, you may say sin x = x + O(x³) when discussing Taylor expansion. Also, O(1) does not mean constant, it means bounded by constant.

All of the answers so far address the question as if there is a right and a wrong answer. But there isn't. The question is a matter of definition. Usually in complexity theory the time cost is an integer --- although that too is just a definition. You're free to say that the empty algorithm that quits immediately takes 0 time steps or 1 time step. It's an abstract question because time complexity is an abstract definition. In the real world, you don't even have time steps, you have continuous physical time; it may be true that one CPU has clock cycles, but a parallel computer could easily have asynchronoous clocks and in any case a clock cycle is extremely small.

That said, I would say that it's more reasonable to say that the halt operation takes 1 time step rather than that it takes 0 time steps. It does seem more realistic. For many situations it's arguably very conservative, because the overhead of initialization is typically far greater than executing one arithmetic or logical operation. Giving the empty algorithm 0 time steps would only be reasonable to model, for example, a function call that is deleted by an optimizing compiler that knows that the function won't do anything.

It should be O(1). The coefficient is always 1.

Consider:

If something grows like 5n, you don't say O(5n), you say O(n) [in other words, O(1n)]

If something grows like 7n^2, you don't say O(7n^2), you say O(n^2) [in other words, O(1n^2)]

Likewise you should say O(1), not O(some other constant)

There is no such thing as O(0). Even an oracle machine or a hypercomputer require the time for one operation, i.e. solve(the_goldbach_conjecture), ergo:

All machines, theoretical or real, finite or infinite produce algorithms with a minimum time complexity of O(1).

But then again, this code right here is O(0):

// Hello world!

:)

I would say it's O(1) by definition, but O(0) if you want to get technical about it: since O(k₁g(n)) is equivalent to O(k₂g(n)) for any constants k₁ and k₂, it follows that O(1 * 1) is equivalent to O(0 * 1), and therefore O(0) is equivalent to O(1).

However, the empty algorithm is not like, for example, the identity function, whose definition is something like "return your input". The empty algorithm is more like an empty statement, or whatever happens between two statements. Its definition is "do absolutely nothing with your input", presumably without even the implied overhead of simply having input.

Consequently, the complexity of the empty algorithm is unique in that O(0) has a complexity of zero times whatever function strikes your fancy, or simply zero. It follows that since the whole business is so wacky, and since O(0) doesn't already mean something useful, and since it's slightly ridiculous to even discuss such things, a reasonable special case for O(0) is something like this:

The complexity of the empty algorithm is O(0) in time and space. An algorithm with time complexity O(0) is equivalent to the empty algorithm.

So there you go.

Given the formal definition of Big O:

Let f(x) and g(x) be two functions defined over the set of real numbers. Then, we write:

f(x) = O(g(x)) as x approaches infinity iff there exists a real M and a real x0 so that:

|f(x)| <= M * |g(x)| for every x > x0

As I see it, if we substitute g(x) = 0 (in order to have a program with complexity O(0)), we must have:

|f(x)| <= 0, for every x > x0 (the constraint of existence of a real M and x0 is practically lifted here)

which can only be true when f(x) = 0.

So I would say that not only the empty program is O(0), but it is the only one for which that holds. Intuitively, this should've been true since O(1) encompasses all algorithms that require a constant number of steps regardless of the size of its task, including 0. It's essentially useless to talk about O(0); it's already in O(1). I suspect it's purely out of simplicity of definition that we use O(1), where it could as well be O(c) or something similar.

0 = O(f) for all function f, since 0 <= |f|, so it is also O(0).

Not only is this a perfectly sensible question, but it is important in certain situations involving amortized analysis, especially when "cost" means something other than "time" (for example, "atomic instructions").

Let's say there is a datastructure featuring multiple operation types, for which an amortized analysis is being conducted. It could well happen that one type of operation can always be funded fully using "coins" deposited during previous operations.

There is a simple example of this: the "multipop queue" described in Cormen, Leiserson, Rivest, Stein [CLRS09, 17.2, p. 457], and also on Wikipedia. Each time an item is pushed, a coin is put on the item, for a total amortized cost of 2. When (multi) pops occur, they can be fully paid for by taking one coin from each item popped, so the amortized cost of MULTIPOP(k) is O(0). To wit:

> Note that the amortized cost of MULTIPOP is a constant (0) > ... > Moreover, we can also charge MULTIPOP operations nothing. To pop the > first plate, we take the dollar of credit off the plate and use it to > pay the actual cost of a POP operation. To pop a second plate, we > again have a dollar of credit on the plate to pay for the POP > operation, and so on. Thus, we have always charged enough up front to > pay for MULTIPOP operations. In other words, since each plate on the > stack has 1 dollar of credit on it, and the stack always has a > nonnegative number of plates, we have ensured that the amount of > credit is always nonnegative.

Thus O(0) is an important "complexity class" for certain amortized operations.

O(1) means the algorithm's time complexity is always constant.

Let's say we have this algorithm (in C):

void doSomething(int[] n)
{
  int x = n[0]; // This line is accessing an array position, so it is time consuming.
  int y = n[1]; // Same here.
  return x + y;
}

I am ignoring the fact that the array could have less than 2 positions, just to keep it simple.

If we count the 2 most expensive lines, we have a total time of 2.

2 = O(1), because:

2 <= c * 1, if c = 2, for every n > 1

If we have this code:

public void doNothing(){}

And we count it as having 0 expansive lines, there is no difference in saying it has O(0) O(1), or O(1000), because for every one of these functions, we can prove the same theorem.

Normally, if the algorithm takes a constant number of steps to complete, we say it has O(1) time complexity.

I guess this is just a convention, because you could use any constant number to represent the function inside the O().

No. It's O(c) by convention whenever you don't have dependence on input size, where c is any positive constant (typically 1 is used - O(1) = O(12.37)).