Question

Let H(n) = H(n/2) + log(n). Give matching upper and lower bounds for H(n) by substitution.

Answer 1

H(n) = H(n/2) log(n)

Let log n = k so that 2^k = n

So our Recurrence becomes

M(2^k) = M(2^k-1) + k

Let P(k) = M(2^k). This gives us

P(k) = P(k - 1) + k

Further expanding P(k),

P(k) = P(k - 1) + k

= P(k - 2) + (k - 1) + k

= P(k - 3) + (k - 2) + (k - 1) + k

...

= P(0) + 1 + 2 + 3 + ... + k

= P(0) + Θ(k²)

Given P(0) is a constant, say 1 or 0

Recurrence solution becomes P(k) = Θ(k²) = Θ (log² n) or Θ((log n)²)