Question

Be $T(1) = b$ and $T(n) = aT(n/c) + bn^d$ for constants $a, b, c, d > 0$, 1. Show that $T(n) = \sum_{i=0}^{k} (\frac{a}{c^d})^i \cdot bn^d$ for all $n = c^k$ With $k \in \mathbb{N}$ applies. 2. Prove that for $n = c^k$ applies: $\circ T(n) \leq O(n^d \log n)$, if $a = c^d$ $\circ T(n) \leq O(n^{\log_c a})$, if $a > c^d$

          Be $T(1) = b$ and $T(n) = aT(n/c) + bn^d$ for constants
$a, b, c, d > 0$,
1. Show that $T(n) = \sum_{i=0}^{k} (\frac{a}{c^d})^i \cdot bn^d$ for all $n = c^k$ With
$k \in \mathbb{N}$ applies.
2. Prove that for $n = c^k$ applies:
$\circ T(n) \leq O(n^d \log n)$, if $a = c^d$
$\circ T(n) \leq O(n^{\log_c a})$, if $a > c^d$

Be T(1) = b and T(n) = aT(n/c) + bn^d for constants
a, b, c, d > 0,
1. Show that T(n) = ∑i=0^k ((a)/(c^d))^i · bn^d for all n = c^k With
k ∈ℕ applies.
2. Prove that for n = c^k applies:
∘ T(n) ≤ O(n^d log n), if a = c^d
∘ T(n) ≤ O(n^ a), if a > c^d

Added by Tara H.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert Cora Marcet

Step 1

In this case, the recurrence relation becomes T(n) = cdT(n/c) + bnd. Using the Master Theorem, we can see that the runtime complexity can be expressed as T(n) = O(n^d log n). Now, let's prove that for n = ck, T(n) < O(nd log n). We can rewrite the recurrence Show more…

Show all steps

Thanks for your feedback!

Master theorem: BeT(1) = b and T(n) = aT(n/c) + bnd for constants a, b, c, d > 0, - bnd for all n = ck with k âˆˆ N applies. 2. Prove that for n = ck applies: oT(n) < O(ndlogn), if a = cd oT(n) < O(nlogca), if a > cd