note fruiture <p>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.</p> <p>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</p> <code> 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,... </code> <p>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().</p> <code> f(n,m,...) = &Omega;(g(n,m,...) means that there is .... c * g(n,m,...) <= f(n,m,...) </code> <p>This is the lower bound. And finally &Theta; marks an average "middle":</p> <code> f(n,m,...) = &Theta;(g(n,m,...)) means that f(n,m,...) = O( g(n,m,...) ) and f(n,m,...) = &Omega;( g(n,m,...) ) </code> -- <br/> http://[fruiture].de 227909 227909