Why do you give the behaviour of binary sort as 1 + log
2 N? Surely this is an implementation issue, and depends on exactly what you define as a step?
Sjn28
Binary search, you mean. It does depend on what one defines as a step. Usually we want a step to be something which can be performed in time that is bounded above by a constant which depends only on the implementation. In this case each lookup at a particular position in the array counts as a step, and we assume that any such lookup can be done in a guaranteed amount of time. But if we were able to look at many entries in a single step, that would change the efficiency.
Seb
Yes, I meant binary search. Sorry, slip of the brain. My point was that I don't think there's much point in saying any more than that the search is O(log N). Specifying something more specific like "3ceil(log_2(n)) + 2" or "2 log_2(n) - 1" is not terribly useful, because (a) you haven't defined what a step is; (b) it depends on exactly which datum you're looking for (you might be lucky); (c) other (unspecified) details of the implementation; but most importantly (d) it doesn't really matter what the constants are.
Sjn28
I've updated the page. Tell me if you think it's better.
Seb
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