@Dave :but it will take more time. any other solution <O(n^2).?

On Sat, Dec 11, 2010 at 7:48 PM, Dave <dave_and_da...@juno.com> wrote:

> @Naresh: The sequence of numbers generated by this rule for any given
> starting number is called a Collatz Sequence. Try googling it.
>
> Here is a list of the number of iterations required for n between 1
> and 10,000: http://oeis.org/A006577/b006577.txt. Maybe that will help.
>
> Dave
>
> On Dec 11, 7:20 am, Naresh A <suryanar...@gmail.com> wrote:
> > Given range of numbers between A and B (A<= B)
> > Find the number within given range which has more number of iterations as
> > per the following
> >
> >      n { stop ; return iteration number }  if n=1;
> >      n = 3n+1                              if n is odd
> >      n =  n/2                              if n is even
> >
> > for eg :
> >
> > n=3 odd
> > ----
> > n=10;
> > n=5;
> > n=16;
> > n=8;
> > n=4;
> > n=2;
> > n=1;
> >
> > iterations : 7
> >
> > --
> > *
> > Time complexity= (<n^2)
> >
> > *************************************************************
> > *NARESH ,A*
> > **
>
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-- 
Shoban babu.B
M.E.,CSA,
IISc.

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