O(log N) == O(1) - Why not?

Question

Whenever I consider algorithms/data structures I tend to replace the log(N) parts by constants. Oh, I know log(N) diverges - but does it matter in real world applications?

Answer 1

What do you mean by whether or not it "matters"?

If you're faced with the choice of an O(1) algorithm and a O(lg n) one, then you should not assume they're equal. You should choose the constant-time one. Why wouldn't you?

And if no constant-time algorithm exists, then the logarithmic-time one is usually the best you can get. Again, does it then matter? You just have to take the fastest you can find.

Can you give me a situation where you'd gain anything by defining the two as equal? At best, it'd make no difference, and at worst, you'd hide some real scalability characteristics. Because usually, a constant-time algorithm will be faster than a logarithmic one.

Even if, as you say, lg(n) < 100 for all practical purposes, that's still a factor 100 on top of your other overhead. If I call your function, N times, then it starts to matter whether your function runs logarithmic time or constant, because the total complexity is then O(n lg n) or O(n).

So rather than asking if "it matters" that you assume logarithmic complexity to be constant in "the real world", I'd ask if there's any point in doing that.

Often you can assume that logarithmic algorithms are fast enough, but what do you gain by considering them constant?