@jammy: Sorry for wrong answer
thats O(n^3 *lg(K))

On Thu, Jan 13, 2011 at 12:28 AM, Jammy <xujiayiy...@gmail.com> wrote:
> I think you meant n^3*log(K).
> Multiplication on two matrices would take O(n) per entry. And total
> #entries is n*n, thus multiplying 2 matrices takes O(n^3)
> To get a^(K), use simple divide-and-conquer technique. Since once one
> gets a^(K/2), to get a^(K) only needs to square it.
> Now we get T(K) = T(K/2)+O(multiply)--->T(K) = logK*O(multiply) =
> n^3*logK.
>
> On Jan 12, 8:04 am, radha krishnan <radhakrishnance...@gmail.com>
> wrote:
>> Using Matrix Exponentiation technique
>> Complexity is n^3*(lg n) for matrix n x n
>> we need two n x n matrix to raise the matrix to power K ..
>> Correct me if am wronf
>>
>> On Wed, Jan 12, 2011 at 6:28 PM, bittu <shashank7andr...@gmail.com> wrote:
>> > How will you raise a n x n matrix to the power k efficiently. What's
>> > the time complexity and space complexity of you algorithm? How will
>> > you optimize this?
>>
>> > Thanks &Regards
>> > Shashank "Don't b Evil U Can Earn While U Learn"
>>
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