"And a billion differs from 1 only by a constant - so that's O(1)"

You obviously don't understand what O(1) means.

Say we have an array of 1 billion elements. Let's look at two different search algorithms:

  1. Search from beginning to end, going thru each element one by one, until hit what you are searching for. In the worst case (the element is at the end of the array), you have to hit 1 billion elements, but according to you, that's O(1). I say it is O(n). We never put a restriction saying that an array can at most contain 1 billion elements (so the size of an array in general is not a constant, although it is a constant for a given array at one given observation point.)
  2. Do a binay search, in the worst case, you have to hit log2(1 billion) ~ 30 times. I call this O(log2(n)), according to you it is also O(1).

As everyone knows, the performance of those two approaches are so different, but according to your theory, they are both O(1)! The math here is so off! Well... I certainly don't mind if you insist your idea, but please don't confuse the general public.

What you said would be right, if we put a restriction saying that a hash can contain at most 1 billion elements. As O(1 billion) has the same complexity as O(1), even though 1 billion is much bigger than 1.

However O(n) is more complex than O(1 billion), even comparing with O(1 billion ** 1 billion), O(n) is still more complex. Why? because n is a variable, which can go to unlimit. 1 billion ** 1billion is huge, but n is going to unlimit, and evetually it will pass 1 billion ** 1 billion. In our context, please remember that, the size of a hash is a variable (that potentially goes to unlimit), and your analysis has to reflect this fact. Don't confuse it with the size of a given hash at a given time.

In reply to Re: Re: A short meditation about hash search performance by pg
in thread A short meditation about hash search performance by pg

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