I`m studying Master Theorem, and I got stuck in the case 3. The example is :
T(n) = 3T(n/4) + nlogn.
I have no idea how my teacher got the final value, c = 3/4, based on the equation below :
3*[n/4 * log(n/4)] <= c * nlogn.
On the internet, I found the same example but with no real explanation on the resolution : (slide 26) http://www.cs.bgu.ac.il/~ds142/wiki.files/Presentation02[2].pdf
Can someone explain me step by step ?
Thanks
There is another closely related recurrence that admits an exact solution. Suppose we have $T(0)=0$ and $T(1)=T(2)=T(3)=1$ and for $n\ge 4$ $$T(n) = 3 T(\lfloor n/4 \rfloor) + n \lfloor \log_4 n \rfloor.$$
It seems reasonable to use integer values here as the running time of an algorithm is a function of discrete quantities.
Furthermore let the base four representation of $n$ be $$n = \sum_{k=0}^{\lfloor \log_4 n \rfloor} d_k 4^k.$$
Then we can unroll the recurrence to obtain the following exact formula for $n\ge 4$ $$T(n) = 3^{\lfloor \log_4 n \rfloor} + \sum_{j=0}^{\lfloor \log_4 n \rfloor -1} 3^j \times (\lfloor \log_4 n \rfloor - j) \times \sum_{k=j}^{\lfloor \log_4 n \rfloor} d_k 4^{k-j}.$$
Now to get an upper bound consider a string of value three digits to obtain $$T(n) \le 3^{\lfloor \log_4 n \rfloor} + \sum_{j=0}^{\lfloor \log_4 n \rfloor -1} 3^j \times (\lfloor \log_4 n \rfloor - j) \times \sum_{k=j}^{\lfloor \log_4 n \rfloor} 3 \times 4^{k-j}.$$ This simplifies to $$(16 \lfloor \log_4 n \rfloor - 48) \times 4^{\lfloor \log_4 n \rfloor} + \frac{193}{4} \times 3^{\lfloor \log_4 n \rfloor} + \frac{1}{2} \lfloor \log_4 n \rfloor + \frac{3}{4}.$$
This bound is actually attained and cannot be improved upon, just like the lower bound, which occurs with a one digit followed by zeroes to give $$T(n) \ge 3^{\lfloor \log_4 n \rfloor} + \sum_{j=0}^{\lfloor \log_4 n \rfloor -1} 3^j \times (\lfloor \log_4 n \rfloor - j) \times 4^{\lfloor \log_4 n \rfloor-j}.$$ This simplifies to $$ (4 \lfloor \log_4 n \rfloor - 12) \times 4^{\lfloor \log_4 n \rfloor} + 13 \times 3^{\lfloor \log_4 n \rfloor}.$$
Joining the dominant terms of the upper and the lower bound we obtain the asymptotics $$\lfloor \log_4 n \rfloor \times 4^{\lfloor \log_4 n \rfloor} \in \Theta\left(\log_4 n \times 4^{\log_4 n}\right) = \Theta\left(\log n \times n\right).$$
Observe that there is a lower order term $$3^{\lfloor \log_4 n \rfloor} \in \Theta\left(4^{\log_4 3 \times \log_4 n}\right) = \Theta\left(n^{\log_4 3}\right).$$
This term represents the asymptotics being generated by the recursive term in the recurrence.
These are both in agreement with what the Master theorem would produce.
This MSE link I may be relevant as may be this MSE link II.
