I've got a vector of samples that form a curve. Let's imagine there are 1000 points in it. If I want to stretch it to fill 1500 points, what is the simplest algorithm that gives decent results? I'm looking for something that is just a few lines of C/C++.
I'll always want to increase the size of the vector, and the new vector can be anywhere from 1.1x to 50x the size of the current vector.
Thanks!
Here's C++ for linear and quadratic interpolation.
interp1( 5.3, a, n ) is a[5] + .3 * (a[6] - a[5]), .3 of the way from a[5] to a[6];
interp1array( a, 1000, b, 1500 ) would stretch a to b.
interp2( 5.3, a, n ) draws a parabola through the 3 nearest points a[4] a[5] a[6]: smoother than interp1 but still fast.
(Splines use 4 nearest points, smoother yet; if you read python, see
basic-spline-interpolation-in-a-few-lines-of-numpy.
// linear, quadratic interpolation in arrays
// from interpol.py denis 2010-07-23 July
#include <stdio.h>
#include <stdlib.h>
// linear interpolate x in an array
// inline
float interp1( float x, float a[], int n )
{
if( x <= 0 ) return a[0];
if( x >= n - 1 ) return a[n-1];
int j = int(x);
return a[j] + (x - j) * ( a[j+1] - a[j] );
}
// linear interpolate array a[] -> array b[]
void interp1array( float a[], int n, float b[], int m )
{
float step = float( n - 1 ) / (m - 1);
for( int j = 0; j < m; j ++ )
{
b[j] = interp1( j*step, a, n );
}
}
//..................................................................
// parabola through 3 points, -1 < x < 1
float parabola( float x, float f_1, float f0, float f1 )
{
if( x <= -1 ) return f_1;
if( x >= 1 ) return f1;
float l = f0 - x * (f_1 - f0);
float r = f0 + x * (f1 - f0);
return (l + r + x * (r - l)) / 2;
}
// quadratic interpolate x in an array
float interp2( float x, float a[], int n )
{
if( x <= .5 || x >= n - 1.5 )
return interp1( x, a, n );
int j = int( x + .5 );
float t = 2 * (x - j); // -1 .. 1
return parabola( t, (a[j-1] + a[j]) / 2, a[j], (a[j] + a[j+1]) / 2 );
}
// quadratic interpolate array a[] -> array b[]
void interp2array( float a[], int n, float b[], int m )
{
float step = float( n - 1 ) / (m - 1);
for( int j = 0; j < m; j ++ ){
b[j] = interp2( j*step, a, n );
}
}
int main( int argc, char* argv[] )
{
// a.out [n m] --
int n = 10, m = 100;
int *ns[] = { &n, &m, 0 },
**np = ns;
char* arg;
for( argv ++; (arg = *argv) && *np; argv ++, np ++ )
**np = atoi( arg );
printf( "n: %d m: %d\n", n, m );
float a[n], b[m];
for( int j = 0; j < n; j ++ ){
a[j] = j * j;
}
interp2array( a, n, b, m ); // a[] -> b[]
for( int j = 0; j < m; j ++ ){
printf( "%.1f ", b[j] );
}
printf( "\n" );
}
what is the simplest algorithm that gives decent results?
Catmull-Rom splines. (if you want a smooth curve)
http://www.mvps.org/directx/articles/catmull/
http://en.wikipedia.org/wiki/Cubic_Hermite_spline
For each new item calculate fractional position in old array, use use fractional part (f - floor(f)) as interpolation factor, and "integer" (i.e. floor(f)) part to find nearest elements.
That is assuming that you're operating on data that can be mathematically interpolated (floats). If data cannot be interpolated (strings), then the only solution is to use nearest available element of old array.
You'll need some tweaking if points in array aren't evenly distributed.
Simplest option I can think of is just a fn that expands the array based on mean averages, so:
x,y,z
becomes
x, avg(x,y), y, avg (y,z), z
If you need more data points, just run it multiple times on the vector.
Related
A non-empty zero-indexed array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P < Q < N, is called a slice of array A
(notice that the slice contains at least two elements). The average of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the
length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).
Write a function:
int solution(int A[], int N);
that, given a non-empty zero-indexed array A consisting of N integers, returns the starting position of the slice with the minimal average.
If there is more than one slice with a minimal average, you should return the smallest starting position of such a slice.
Assume that:
N is an integer within the range [2..100,000];
each element of array A is an integer within the range [−10,000..10,000].
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Can you post only solutions with order N only?
If A had only positive numbers, you could get away with this:
pos = 0
min_avg = A[0] + A[1]
for (i=2; i<N; i++)
m = A[i-1] + A[i]
if (m < min_avg)
min_avg = m
pos = i-1
return pos
This is only taking an average of a slice of two numbers, because a larger slice cannot have a smaller average than the minimum of a smaller slice.
If A has negative numbers, you could adjust all values upwards first:
offset = min(A)
for (i=0; i<N; i++)
A[i] -= offset
Combined with the previous algorithm:
offset = min(A) * 2 (because we're adding two numbers below)
pos = 0
min_avg = A[0] + A[1] - offset
for (i=2; i<N; i++)
m = A[i-1] + A[i] - offset
if (m < min_avg)
min_avg = m
pos = i-1
return pos
I think you're right, the best I can do is an O(N2) solution (this is in Python):
from random import randint
N = 1000
A = [randint(-10000, 10000) for _ in xrange(N)]
def solution(A, N):
min_avg = 10001
for p in xrange(N):
s = A[p]
for q in xrange(1,N-p):
s += A[p+q]
a = s / (q+1.)
if a < min_avg:
min_avg = a
pos = (p, q+1)
return pos
print solution(A, N)
However, averages of larger slices tend towards the mean (middle) value of the original range. In this case, the average is zero, halfway between -10000 and 10000. Most of the time, the smallest average is of a slice of two values, but sometimes it can be a slice of three values and rarely it can be even more values. So I think my previous answer works in most (>90%) of the cases. It really depends on the data values.
#include <assert.h>
struct Slice { unsigned P, Q; };
struct Slice MinSlice( int A[], unsigned N ) {
assert( N>=2 );
// find min slice of length 2
unsigned P = 0;
double min_sum = A[P] + A[P+1];
for (unsigned i = 1; i < N-1; ++i)
if ( min_sum > A[i] +A[i+1] ) {
P = i;
min_sum = A[P] + A[P+1];
}
unsigned Q = P+1;
double min_avg = min_sum / 2;
//extend the min slice if the avg can be reduced.
//(in the direction that most reduces the avg)
for (;;) {
if ( P > 0 && ( Q >= N-1 || A[P-1] <= A[Q+1] ) ) {
//reducing P might give the best reduction in avg
double new_sum = A[P-1] + min_sum;
double new_avg = new_sum / (Q - P + 2);
if ( min_avg < new_avg )
break;
min_sum = new_sum;
min_avg = new_avg;
--P;
} else if ( Q < N-1 && ( P <= 0 || A[P-1] >= A[Q+1] ) ) {
//increasing Q might give the best reduction in avg
double new_sum = min_sum + A[Q+1];
double new_avg = new_sum / (Q - P + 2);
if ( min_avg < new_avg )
break;
min_sum = new_sum;
min_avg = new_avg;
++Q;
} else
break;
}
struct Slice slice = { .P = P, .Q= Q };
return slice;
}
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 8 years ago.
Improve this question
How can I find an algorithm to solve this problem using C++: given an integer z<=10^100, find the smallest row of Pascal's triangle that contains the number z.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
For example if z=6 => result is on the 4th row.
Another way to describe the problem: given integer z<=10^100, find the smallest integer n: exist integer k so that C(k,n) = z.
C(k,n) is combination of n things taken k at a time without repetition
EDIT This solution needs Logarithmic time, it's O(Log z). Or maybe O( (Log z)^2 ).
Say you are looking for n,k where Binomial(n,k)==z for a given z.
Each row has its largest value in the middle, so starting from n=0 you increase the row number, n, as long as the middle value is smaller than the given number. Actually, 10^100 isn't that big, so before row 340 you find a position n0,k0=n0/2 where the value from the triangle is larger than or equal to z: Binomial(n0,k0)>=z
You walk to the left, i.e. you decrease the column number k, until eventually you find a value smaller than z. If there was a matching value in that row you would have hit it by now. k is not very large, less than 170, so this step won't be executed more often than that and does not present a performance problem.
From here you walk down, increasing n. Here you will find a steadily increasing value of Binomial[n,k]. Continue with 3 until the value gets bigger than or equal to z, then goto 2.
EDIT: This step 3 can loop for a very long time when the row number n is large, so instead of checking each n linearly you can do a binary search for n with Binomial(n,k) >= z > Binomial(n-1,k), then it only needs Log(n) time.
A Python implementation looks like this, C++ is similar but somewhat more cumbersome because you need to use an additional library for arbitrary precision integers:
# Calculate (n-k+1)* ... *n
def getnk( n, k ):
a = n
for u in range( n-k+1, n ):
a = a * u
return a
# Find n such that Binomial(n,k) >= z and Binomial(n-1,k) < z
def find_n( z, k, n0 ):
kfactorial = k
for u in range(2, k):
kfactorial *= u
xk = z * kfactorial
nk0 = getnk( n0, k )
n1=n0*2
nk1 = getnk( n1, k )
# duplicate n while the value is too small
while nk1 < xk:
nk0=nk1
n0=n1
n1*=2
nk1 = getnk( n1, k )
# do a binary search
while n1 > n0 + 1:
n2 = (n0+n1) // 2
nk2 = getnk( n2, k )
if nk2 < xk:
n0 = n2
nk0 = nk2
else:
n1 = n2
nk1 = nk2
return n1, nk1 // kfactorial
def find_pos( z ):
n=0
k=0
nk=1
# start by finding a row where the middle value is bigger than z
while nk < z:
# increase n
n = n + 1
nk = nk * n // (n-k)
if nk >= z:
break
# increase both n and k
n = n + 1
k = k + 1
nk = nk * n // k
# check all subsequent rows for a matching value
while nk != z:
if nk > z:
# decrease k
k = k - 1
nk = nk * (k+1) // (n-k)
else:
# increase n
# either linearly
# n = n + 1
# nk = nk * n // (n-k)
# or using binary search:
n, nk = find_n( z, k, n )
return n, k
z = 56476362530291763837811509925185051642180136064700011445902684545741089307844616509330834616
print( find_pos(z) )
It should print
(5864079763474581, 6)
Stirling estimation for n! can be used to find first row in triangle with binomial coefficient bigger or equal to a given x. Using this estimation we can derive lower and upper bound for
and then by observation that this is the maximum coefficient in row that expands 2n:
P( 2n, 0), P( 2n, 1), P( 2n, 2), ..., P( 2n, 2n -1), P( 2n, 2n)
we can find first row with maximum binomial coefficient bigger or equal to a given x. This is the first row in which x can be looking for, this is not possible that x can be found in the row smaller than this. Note: this may be right hint and give an answer immediately in some cases. At the moment I cannot see other way than to start a brute force search from this row.
template <class T>
T binomial_coefficient(unsigned long n, unsigned long k) {
unsigned long i;
T b;
if (0 == k || n == k) {
return 1;
}
if (k > n) {
return 0;
}
if (k > (n - k)) {
k = n - k;
}
if (1 == k) {
return n;
}
b = 1;
for (i = 1; i <= k; ++i) {
b *= (n - (k - i));
if (b < 0) return -1; /* Overflow */
b /= i;
}
return b;
}
Stirling:
double stirling_lower_bound( int n) {
double n_ = n / 2.0;
double res = pow( 2.0, 2 * n_);
res /= sqrt( n_ * M_PI);
return res * exp( ( -1.0) / ( 6 * n_));
}
double stirling_upper_bound( int n) {
double n_ = n / 2.0;
double res = pow( 2.0, 2 * n_) ;
res /= sqrt( n_ * M_PI);
return res * exp( 1.0 / ( 24 * n_));
}
int stirling_estimate( double x) {
int n = 1;
while ( stirling_lower_bound( n) <= x) {
if ( stirling_upper_bound( n) > x) return n;
++n;
}
return n;
}
usage:
long int search_coefficient( unsigned long int &n, unsigned long int x) {
unsigned long int k = n / 2;
long long middle_coefficient = binomial_coefficient<long long>( n, k);
if( middle_coefficient == x) return k;
unsigned long int right = binomial_coefficient<unsigned long>( n, ++k);
while ( x != right) {
while( x < right || x < ( right * ( n + 1) / ( k + 1))) {
right = right * ( n + 1) / ( ++k) - right;
}
if ( right == x) return k;
right = right * ( ++n) / ( ++k);
if( right > x) return -1;
}
return k;
}
/*
*
*/
int main(int argc, char** argv) {
long long x2 = 1365;
unsigned long int n = stirling_estimate( x2);
long int k = search_coefficient( n, x2);
std::cout << "row:" << n <<", column: " << k;
return 0;
}
output:
row:15, column: 11
I have GCD(n, i) where i=1 is increasing in loop by 1 up to n. Is there any algorithm which calculate all GCD's faster than naive increasing and compute GCD using Euclidean algorithm?
PS I've noticed if n is prime I can assume that number from 1 to n-1 would give 1, because prime number would be co-prime to them. Any ideas for other numbers than prime?
C++ implementation, works in O(n * log log n) (assuming size of integers are O(1)):
#include <cstdio>
#include <cstring>
using namespace std;
void find_gcd(int n, int *gcd) {
// divisor[x] - any prime divisor of x
// or 0 if x == 1 or x is prime
int *divisor = new int[n + 1];
memset(divisor, 0, (n + 1) * sizeof(int));
// This is almost copypaste of sieve of Eratosthenes, but instead of
// just marking number as 'non-prime' we remeber its divisor.
// O(n * log log n)
for (int x = 2; x * x <= n; ++x) {
if (divisor[x] == 0) {
for (int y = x * x; y <= n; y += x) {
divisor[y] = x;
}
}
}
for (int x = 1; x <= n; ++x) {
if (n % x == 0) gcd[x] = x;
else if (divisor[x] == 0) gcd[x] = 1; // x is prime, and does not divide n (previous line)
else {
int a = x / divisor[x], p = divisor[x]; // x == a * p
// gcd(a * p, n) = gcd(a, n) * gcd(p, n / gcd(a, n))
// gcd(p, n / gcd(a, n)) == 1 or p
gcd[x] = gcd[a];
if ((n / gcd[a]) % p == 0) gcd[x] *= p;
}
}
}
int main() {
int n;
scanf("%d", &n);
int *gcd = new int[n + 1];
find_gcd(n, gcd);
for (int x = 1; x <= n; ++x) {
printf("%d:\t%d\n", x, gcd[x]);
}
return 0;
}
SUMMARY
The possible answers for the gcd consist of the factors of n.
You can compute these efficiently as follows.
ALGORITHM
First factorise n into a product of prime factors, i.e. n=p1^n1*p2^n2*..*pk^nk.
Then you can loop over all factors of n and for each factor of n set the contents of the GCD array at that position to the factor.
If you make sure that the factors are done in a sensible order (e.g. sorted) you should find that the array entries that are written multiple times will end up being written with the highest value (which will be the gcd).
CODE
Here is some Python code to do this for the number 1400=2^3*5^2*7:
prime_factors=[2,5,7]
prime_counts=[3,2,1]
N=1
for prime,count in zip(prime_factors,prime_counts):
N *= prime**count
GCD = [0]*(N+1)
GCD[0] = N
def go(i,n):
"""Try all counts for prime[i]"""
if i==len(prime_factors):
for x in xrange(n,N+1,n):
GCD[x]=n
return
n2=n
for c in xrange(prime_counts[i]+1):
go(i+1,n2)
n2*=prime_factors[i]
go(0,1)
print N,GCD
Binary GCD algorithm:
https://en.wikipedia.org/wiki/Binary_GCD_algorithm
is faster than Euclidean algorithm:
https://en.wikipedia.org/wiki/Euclidean_algorithm
I implemented "gcd()" in C for type "__uint128_t" (with gcc on Intel i7 Ubuntu), based on iterative Rust version:
https://en.wikipedia.org/wiki/Binary_GCD_algorithm#Iterative_version_in_Rust
Determining number of trailing 0s was done efficiently with "__builtin_ctzll()". I did benchmark 1 million loops of two biggest 128bit Fibonacci numbers (they result in maximal number of iterations) against gmplib "mpz_gcd()" and saw 10% slowdown. Utilizing the fact that u/v values only decrease, I switched to 64bit special case "_gcd()" when "<=UINT64_max" and now see speedup of 1.31 over gmplib, for details see:
https://www.raspberrypi.org/forums/viewtopic.php?f=33&t=311893&p=1873552#p1873552
inline int ctz(__uint128_t u)
{
unsigned long long h = u;
return (h!=0) ? __builtin_ctzll( h )
: 64 + __builtin_ctzll( u>>64 );
}
unsigned long long _gcd(unsigned long long u, unsigned long long v)
{
for(;;) {
if (u > v) { unsigned long long a=u; u=v; v=a; }
v -= u;
if (v == 0) return u;
v >>= __builtin_ctzll(v);
}
}
__uint128_t gcd(__uint128_t u, __uint128_t v)
{
if (u == 0) { return v; }
else if (v == 0) { return u; }
int i = ctz(u); u >>= i;
int j = ctz(v); v >>= j;
int k = (i < j) ? i : j;
for(;;) {
if (u > v) { __uint128_t a=u; u=v; v=a; }
if (v <= UINT64_MAX) return _gcd(u, v) << k;
v -= u;
if (v == 0) return u << k;
v >>= ctz(v);
}
}
I'm trying to use the repeated squaring algorithm (using recursion) to perform matrix exponentiation. I've included header files from the NEWMAT library instead of using arrays. The original matrix has elements in the range (-5,5), all numbers being of type float.
# include "C:\User\newmat10\newmat.h"
# include "C:\User\newmat10\newmatio.h"
# include "C:\User\newmat10\newmatap.h"
# include <iostream>
# include <time.h>
# include <ctime>
# include <cstdlib>
# include <iomanip>
using namespace std;
Matrix repeated_squaring(Matrix A, int exponent, int n) //Recursive function
{
A(n,n);
IdentityMatrix I(n);
if (exponent == 0) //Matrix raised to zero returns an Identity Matrix
return I;
else
{
if ( exponent%2 == 1 ) // if exponent is odd
return (A * repeated_squaring (A*A, (exponent-1)/2, n));
else //if exponent is even
return (A * repeated_squaring( A*A, exponent/2, n));
}
}
Matrix direct_squaring(Matrix B, int k, int no) //Brute Force Multiplication
{
B(no,no);
Matrix C = B;
for (int i = 1; i <= k; i++)
C = B*C;
return C;
}
//----Creating a matrix with elements b/w (-5,5)----
float unifRandom()
{
int a = -5;
int b = 5;
float temp = (float)((b-a)*( rand()/RAND_MAX) + a);
return temp;
}
Matrix initialize_mat(Matrix H, int ord)
{
H(ord,ord);
for (int y = 1; y <= ord; y++)
for(int z = 1; z<= ord; z++)
H(y,z) = unifRandom();
return(H);
}
//---------------------------------------------------
void main()
{
int exponent, dimension;
cout<<"Insert exponent:"<<endl;
cin>>exponent;
cout<< "Insert dimension:"<<endl;
cin>>dimension;
cout<<"The number of rows/columns in the square matrix is: "<<dimension<<endl;
cout<<"The exponent is: "<<exponent<<endl;
Matrix A(dimension,dimension),B(dimension,dimension);
Matrix C(dimension,dimension),D(dimension,dimension);
B= initialize_mat(A,dimension);
cout<<"Initial Matrix: "<<endl;
cout<<setw(5)<<setprecision(2)<<B<<endl;
//-----------------------------------------------------------------------------
cout<<"Repeated Squaring Result: "<<endl;
clock_t time_before1 = clock();
C = repeated_squaring (B, exponent , dimension);
cout<< setw(5) <<setprecision(2) <<C;
clock_t time_after1 = clock();
float diff1 = ((float) time_after1 - (float) time_before1);
cout << "It took " << diff1/CLOCKS_PER_SEC << " seconds to complete" << endl<<endl;
//---------------------------------------------------------------------------------
cout<<"Direct Squaring Result:"<<endl;
clock_t time_before2 = clock();
D = direct_squaring (B, exponent , dimension);
cout<<setw(5)<<setprecision(2)<<D;
clock_t time_after2 = clock();
float diff2 = ((float) time_after2 - (float) time_before2);
cout << "It took " << diff2/CLOCKS_PER_SEC << " seconds to complete" << endl<<endl;
}
I face the following problems:
The random number generator returns only "-5" as each element in the output.
The Matrix multiplication yield different results with brute force multiplication and using the repeated squaring algorithm.
I'm timing the execution time of my code to compare the times taken by brute force multiplication and by repeated squaring.
Could someone please find out what's wrong with the recursion and with the matrix initialization?
NOTE: While compiling this program, make sure you've imported the NEWMAT library.
Thanks in advance!
rand() returns an int so rand()/RAND_MAX will truncate to an integer = 0. Try your
repeated square algorithm by hand with n = 1, 2 and 3 and you'll find a surplus A *
and a gross inefficiency.
Final Working code has the following improvements:
Matrix repeated_squaring(Matrix A, int exponent, int n) //Recursive function
{
A(n,n);
IdentityMatrix I(n);
if (exponent == 0) //Matrix raised to zero returns an Identity Matrix
return I;
if (exponent == 1)
return A;
{
if (exponent % 2 == 1) // if exponent is odd
return (A*repeated_squaring (A*A, (exponent-1)/2, n));
else //if exponent is even
return (repeated_squaring(A*A, exponent/2, n));
}
}
Matrix direct_squaring(Matrix B, int k, int no) //Brute Force Multiplication
{
B(no,no);
Matrix C(no,no);
C=B;
for (int i = 0; i < k-1; i++)
C = B*C;
return C;
}
//----Creating a matrix with elements b/w (-5,5)----
float unifRandom()
{
int a = -5;
int b = 5;
float temp = (float) ((b-a)*((float) rand()/RAND_MAX) + a);
return temp;
}
Is there any faster method of matrix exponentiation to calculate Mn (where M is a matrix and n is an integer) than the simple divide and conquer algorithm?
You could factor the matrix into eigenvalues and eigenvectors. Then you get
M = V * D * V^-1
Where V is the eigenvector matrix and D is a diagonal matrix. To raise this to the Nth power, you get something like:
M^n = (V * D * V^-1) * (V * D * V^-1) * ... * (V * D * V^-1)
= V * D^n * V^-1
Because all the V and V^-1 terms cancel.
Since D is diagonal, you just have to raise a bunch of (real) numbers to the nth power, rather than full matrices. You can do that in logarithmic time in n.
Calculating eigenvalues and eigenvectors is r^3 (where r is the number of rows/columns of M). Depending on the relative sizes of r and n, this might be faster or not.
It's quite simple to use Euler fast power algorith. Use next algorith.
#define SIZE 10
//It's simple E matrix
// 1 0 ... 0
// 0 1 ... 0
// ....
// 0 0 ... 1
void one(long a[SIZE][SIZE])
{
for (int i = 0; i < SIZE; i++)
for (int j = 0; j < SIZE; j++)
a[i][j] = (i == j);
}
//Multiply matrix a to matrix b and print result into a
void mul(long a[SIZE][SIZE], long b[SIZE][SIZE])
{
long res[SIZE][SIZE] = {{0}};
for (int i = 0; i < SIZE; i++)
for (int j = 0; j < SIZE; j++)
for (int k = 0; k < SIZE; k++)
{
res[i][j] += a[i][k] * b[k][j];
}
for (int i = 0; i < SIZE; i++)
for (int j = 0; j < SIZE; j++)
a[i][j] = res[i][j];
}
//Caluclate a^n and print result into matrix res
void pow(long a[SIZE][SIZE], long n, long res[SIZE][SIZE])
{
one(res);
while (n > 0) {
if (n % 2 == 0)
{
mul(a, a);
n /= 2;
}
else {
mul(res, a);
n--;
}
}
}
Below please find equivalent for numbers:
long power(long num, long pow)
{
if (pow == 0) return 1;
if (pow % 2 == 0)
return power(num*num, pow / 2);
else
return power(num, pow - 1) * num;
}
Exponentiation by squaring is frequently used to get high powers of matrices.
I would recommend approach used to calculate Fibbonacci sequence in matrix form. AFAIK, its efficiency is O(log(n)).