You find this so useful .... http://zobayer.blogspot.com/2010/11/matrix-exponentiation.html
On Sun, Jan 9, 2011 at 6:10 PM, shady <[email protected]> wrote: > I am not familiar with such techniques, can you please refer to some article > where such techniques are explained, ... > Regards, > Shady > > On Sun, Jan 9, 2011 at 5:44 PM, juver++ <[email protected]> wrote: >> >> There is very useful technique for this kind of problem. >> A(n) = C1 * A(n-1) - C2 * A(n-2) + C3 * A(n - 3). >> Let's introduce matrix B: >> C1 -C2 C3 >> 1 0 0 >> 0 1 0. >> A(n-1) A(n) >> B x A(n-2) = A(n-1) >> A(n-3) A(n-2) >> and so on.. >> So to find A(n) you should muultiply matrix B^n by initial vector (A2, A1, >> A0). >> B^n can be calculated using fast exponention algorithm. >> Complexity is O(k^3 * log(n)), where k - size of the matrix B, so k=3. >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
