Solving a Fibonacci like recurrence in log n time

Question

Finding the nth term in Fibonacci series f(n) = f(n-1) + f(n-2) can be solved in O(n) time by memoization.

A more efficient way would be to find the nth power of matrix [ [1,1] , [1,0] ] using divide and conquer to solve the Fibonacci in log n time.

Is there similar approach which can be followed for f(n) = f(n-1) + f(n-x) + f(n-x+1) [ x is some constant ].

Just be storing previous x elements, this can solved in O(n) time.

Is there a better way to solve this recursion.

Answer 1

As you are already suspecting, this will work very similar. Use the n-th power of the x * x matrix

|1 0 0 0  .... 1 1|
|1 
|  1
|    1
|      1
|        1
...................
...................
|          ... 1 0|

This is easy to understand if you multiply this matrix with the vector

f(n-1), f(n-2), ... , f(n-x+1), f(n-x)

which results in

f(n), f(n-1), ... , f(n-x+1)

Matrix exponentiation can be done in O(log(n)) time (when x is considered to be constant).

For the Fibonacci recurrence, there is also a closed formula solution, see here http://en.wikipedia.org/wiki/Fibonacci_number, look for Binet's or Moivre's formula.

Answer 2

You may want to look into Tribonacci numbers (and other generalizations of Fiboniacci numbers.) They have been studied quite extensively. See e.g. Generalizations of Fibonacci numbers