What if we create a base 3 number from the given number N and then check if
there is only 1 bit with value 1 and all values should be 0. For example, if
lets say the number is 27. Its base 3 number will be 1 0 0 0, now since there
is only 1 single 1 present in this representation, it is a power of 3.
Gaurav
On Dec 7, 2011, at 6:03 PM, saurabh singh wrote:
Originaly problem rules out the use of log.Moreover log (or any floating
point operations) take lot of hardware time as they are emulated on the
floating point environment on most machines.Thirdly precision problem for
higher values of n may cause your solution to give wron g answers,,,
On Wed, Dec 7, 2011 at 11:54 PM, tech coder techcoderonw...@gmail.com wrote:
what about this log(base 3 n) is of integral type then n is a power of 3
On Mon, Dec 5, 2011 at 11:36 PM, Dave dave_and_da...@juno.com wrote:
@Carl: You can generate the constants the first time the procedure is
called. There is no need to do them every time. So the first call
would be O(wordsize) but subsequent calls are O(1).
Dave
On Dec 5, 10:28 am, Carl Barton odysseus.ulys...@gmail.com wrote:
Sorry, I was being a bit vague. I meant what would the time complexity be
then?
As I understand your Constant time is Dependant on the constant pre
computation you do, which is no longer the case when you generalise
On 5 December 2011 16:14, Dave dave_and_da...@juno.com wrote:
@Carl: Of course. For any given word size, extend the tables of powers
of 2 and of 3 and change the for loop limit.
Dave
On Dec 5, 9:36 am, Carl Barton odysseus.ulys...@gmail.com wrote:
Ah I see, in which case could you not generalise your solution for all
integers?
By taking into account the size of words on the computer for example?
On 5 December 2011 15:09, Dave dave_and_da...@juno.com wrote:
@Carl: Yes, as coded, my algorithm is for 32-bit integers. But the
original poster asked for a solution using bit manipulation, and
modulus and division are arithmetic operations, not bit operations.
Dave
On Dec 5, 8:56 am, Carl Barton odysseus.ulys...@gmail.com wrote:
@Dave Yours only works for a certain subset of all possible powers
or 3
doesn't it? So WgpShashank's would be more general?
On 5 December 2011 14:30, Dave dave_and_da...@juno.com wrote:
@WgpShashank: Yours is an O(log n) solution. Mine is O(1).
Dave
On Dec 5, 6:21 am, WgpShashank shashank7andr...@gmail.com wrote:
@SAMMM have a look
* *solution is to keep dividing the number by 3, i.e, do n = n/3
iteratively. In any iteration, if n%3 becomes non-zero and n is
not 1
then
n is not a power of 3, otherwise n is a power of 3
check it out ?http://codepad.org/863ptoBE
Thanks
Shashank
Computer Science
BIT Mesrahttp://
www.facebook.com/wgpshashankhttp://shashank7s.blogspot.com/
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successfully u play
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