Complexity of the recursion: T(n) = T(n-1) + T(n-2) + C
问题 I want to understand how to arrive at the complexity of the below recurrence relation. T(n) = T(n-1) + T(n-2) + C Given T(1) = C and T(2) = 2C; Generally for equations like T(n) = 2T(n/2) + C (Given T(1) = C), I use the following method. T(n) = 2T(n/2) + C => T(n) = 4T(n/4) + 3C => T(n) = 8T(n/8) + 7C => ... => T(n) = 2^k T (n/2^k) + (2^k - 1) c Now when n/2^k = 1 => K = log (n) (to the base 2) T(n) = n T(1) + (n-1)C = (2n -1) C = O(n) But, I'm not able to come up with similar approach for