i was saying that most often it happens that its base is 2 ... as it is
there on  binary search and all .... as your size is always reducing to half
... and like others ......

now by saying that it to not matter i want to say following proof  :-

let us have log with base x represented by logxn and log with base y
represented by logyn .....

then if we take logxn = p ........ (1)
and logyn = q ....... (2)

then n= x^p
and n = y^q

that means x^p = y^q

or in other taking log both sides

p * logx= q*logy ..... (3)

by 1 and 2 above 3 becomes

logxn*logx = logyn*logy

also logx and logy are constant ......

hence say logx=c1 and logy = c2
then c1*logxn = c2 * logyn

or simply saying logxn = clogyn
hence constant multiple of each other ......

its a very common thing ..... and for more detail see any basic data
structure book like tanenbaum


On 5/15/08, Dave <[EMAIL PROTECTED]> wrote:
>
>
> amitabh started his response fine, but strayed on the last line.
> Because of the change of base formula for logarithms, log_a(x) =
> log_b(x)/log_b(a), all logarithms are proportional.
>
> Now let's look at the definition of big O notation:
>
> We say that f(x) = O(g(x)) as x --> oo if and only if there exist
> constants x0 > 0 and M > 0 such that
>
> |f(x)| <= M|g(x)| whenever x > x0.
>
> Thus, any constants in f(x) or g(x), including the constant of
> proportionality between logarithms, just change the value of the
> constant M, and so are swallowed up in the big O notation. Thus,
> O(log2 x) = O(ln x) = O(log10 x).
>
> It makes no difference what logarithm you use in big O notation.
>
> Dave
>
> On May 15, 6:52 am, "amitabh chauhan" <[EMAIL PROTECTED]>
> wrote:
> > actually base do not matters base 2 and base 10 are constant multiple of
> > each other so complexity remains same ( ya constant multiple do change
> )...
> > but its base 2 most often...
> >
>


-- 
Amitabh S Chauhan
life is like a box of chocolates. You never know what you're gonna get....

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