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

Question

So given the following program:

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

I thought answering this in a separate question

Answer 1

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.