Question

Solve the following recurrence equations, expressing the answer in Big-Oh notation. Assume that T(n) is constant for sufficiently small n.

a.) T(n) = T(n - 1) + logn

b.) T(n) = T(n - 3) + n

Answer 1

a. T(n) = T(n-1) + log n

= T(n-2) + log (n-1) + log n

.

.

= T(1) + log 2 + log 3 +.....+ log n

Assuming T(1) = c (constant)

And by additive property of log, we get

T(n) = log (n(n-1)...2.1) + c = log (n!) + c

Since n! O((n/2)^n/2)

Hence T(n) =O( log (n/2)^n/2) + c = O(n log n)

b. Without loss of generality, assume n = 3k

T(n) = T(n-3) + n

= T(n-6) + (n-3) + n

= T(n-9) + (n-6) + (n-3) + n

.

.

= 3 + 6 + 9+ ...... + (3k -3) + 3k

= 3(1+2+....+(k-1) + k) = 3k(k+1)/2

=n(n/3 + 1)/2 = O(n²)

Hence T(n) = O(n²)

Please comment for any clarification .