Question

Recurrence time complexity and solution (Analysis of Algorithms- Haskell Language)

Answer 1

Solution:

The recurrence relation will be:

T(n)= T(n/2) + c

As per the master theorem,

here,

a = 1 and b = 2, f(n) = c

n^(log_b a) = n^(log₂ 1) = n^0 = c

comparing k^(log_b a) and f(k), we can see that this is case 2 of master theorem

which means

T(n) =
The recurrence relation will be: T(n) = T(n/2) + c As per the master theorem, here, a = 1 and b = 2, f(n) = c n^(log_b a) = n^(log_2 1) = n^0 = c comparing k^(log_b a) and f(k), we can see that this is case 2 of master theorem which means T(n) = theta (log_2 n) I hope this help if you find any problem.
(log₂ n)

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