Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$\begin{align*} &T(n) = 2T(n/2) + \log_2(n)\\ &T(1) = 0 \end{align*}$$

$n$ is a power of $2$

solve the recurrence relation

my work so far:

unrolling this, we have

$$\begin{align*} T(n) &= 4T(n/4) + \log_2(n) -1\\ &= 8T(n/8) + 2\log_2(n) -2\\ &=\log_2(n-1) \log_2(n) - \log_2(n) + 1 \end{align*}$$

after substituting for base case.

where is my mistake?

share|cite|improve this question

You didn’t perform the unrolling correctly.

I will write $\lg n$ for $\log_2 n$. Suppose that $n=2^m$; then

$$\begin{align*} T(n)&=T(2^m)\\ &=2T(2^{m-1})+\lg 2^m\\ &=2T(2^{m-1})+m\\ &=2\Big(2T(2^{m-2})+\lg 2^{m-1}\Big)+m\\ &=2^2T(2^{m-2}+2(m-1)+m\\ &=2^2\Big(2T(2^{m-3})+\lg 2^{m-2}\Big)+2(m-1)+m\\ &=2^3T(2^{m-3})+2^2(m-2)+2(m-1)+m\\ &\;\vdots\\ &=2^kT(2^{m-k})+\sum_{i=0}^{k-1}2^i(m-i)\\ &\;\vdots\\ &=2^{m-1}T(1)+\sum_{i=0}^{m-2}2^i(m-i)\\ &=\sum_{i=0}^{m-2}2^i(m-i)\\ &=m\sum_{i=0}^{m-2}2^i-\sum_{i=0}^{m-2}i2^i\;. \end{align*}$$

Can you finish it by evaluating those two summations to get a closed form?

share|cite|improve this answer

Substituting $n=2^k$ we have: $$\begin{align*}T(n)=T(2^k)&=2T(2^{k-1})+k=2(2T(2^{k-2})+k-1)+k=4T(2^{k-2})+3k-1\\ &=4(2T(2^{k-3})+k-3)+3k-1=8T(2^{k-3})+7k-13=...=\\ &= 2^mT(2^{k-m})+\sum_{t=1}^m2^{t-1}(k-t+1)=...=2^kT(1)+\sum_{t=0}^{k-1}2^t(k-t)\\ &=2^k\cdot0+k\sum_{t=0}^{k-1}2^t-\sum_{t=0}^{k-1}t2^t=k\frac{2^k-1}{2-1}-2\frac{(k-1)2^k-k2^{k-1}+1}{2-1}\\ &=k2^k-k-2(k-1)2^k+k2^k-2=2k2^k-2k2^k+2\cdot2^k-k-2=\\ =&2\cdot2^k-k-2=2n-\log_2n-2\end{align*}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.