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I think it's usefull to have a look at a more formal way of saying what the Big O means after having read this informal orientation. In the end, it's not that hard to understand.

You can express the complexity of an algorithm as a complete function that includes constants (startup cost) and constant factors (for example if you iterate over a list thrice), based on the assumption that all primitve actions (+,-,/,*,assignation,...) cost 1 "step" of time. Let's call this function f(n,m,...) where n,m,... are all the variable factors that influence the problem. How can we now find a function g(n,m,...) so that O(g(n,m,...)) = f(n,m,...)? Simple

f(n,m,...) = O(g(n,m,...)) means that there exists a constant "c", for which c * g(n,m,...) >= f(n,m,...) is true for large enough n,m,...

If you think about this relation it's clear that you can forget all constant factors or constants in your program, because a large enough c will always suffice to make c*g() grow faster. Because O(n^2) = O(n^3) you should always find the smallest g() for which f() = O(g()) is true in order to have a representative function g().

f(n,m,...) = &Omega;(g(n,m,...) means that there is .... c * g(n,m,...) <= f(n,m,...)

This is the lower bound. And finally Θ marks an average "middle":

f(n,m,...) = &Theta;(g(n,m,...)) means that f(n,m,...) = O( g(n,m,...) ) and f(n,m,...) = &Omega;( g(n,m,...) )
--
http://fruiture.de

In reply to Re: An informal introduction to O(N) notation by fruiture
in thread An informal introduction to O(N) notation by dws

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