Can Master Theorem be applied on any of these?

Solution 1:

For problem three, lets solve the recurrence $$T(n) = 9 T(\lfloor n/3\rfloor) + n \times (1+\lfloor\log_3 n\rfloor)^3$$ where we set $T(0)=0.$

Let the base three representation of $n$ be $$n = \sum_{k=0}^{\lfloor\log_3 n\rfloor} d_k 3^k.$$

Then we get the exact formula $$T(n) = \sum_{j=0}^{\lfloor\log_3 n\rfloor} 9^j \times (1+\lfloor\log_3 n\rfloor- j)^3 \times \sum_{k=j}^{\lfloor\log_3 n\rfloor} d_k 3^{k-j} \\ = \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times (1+\lfloor\log_3 n\rfloor - j)^3 \times \sum_{k=j}^{\lfloor\log_3 n\rfloor} d_k 3^k .$$

Now for an upper bound consider a string of two digits, which gives $$T(n)\le \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times (1+\lfloor\log_3 n\rfloor - j)^3 \times 2 \times \sum_{k=j}^{\lfloor\log_3 n\rfloor} 3^k \\ = \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times (1+\lfloor\log_3 n\rfloor - j)^3 \left(3^{\lfloor\log_3 n\rfloor + 1}-3^j\right) \\ = \sum_{j=1}^{\lfloor\log_3 n\rfloor+1} 3^{\lfloor\log_3 n\rfloor+1-j} \times j^3 \times \left(3^{\lfloor\log_3 n\rfloor + 1}-3^{\lfloor\log_3 n\rfloor + 1- j}\right) \\ = 3^{2(\lfloor\log_3 n\rfloor + 1)} \sum_{j=1}^{\lfloor\log_3 n\rfloor+1} 3^{-j} j^3 (1- 3^{-j}).$$ Now the sum term converges to a constant and we get the upper bound $$\frac{7917}{2048} \times 3^{2(\lfloor\log_3 n\rfloor + 1)}.$$ For the lower bound take a one followed by a string of zeros, which gives $$T(n)\ge \sum_{j=0}^{\lfloor\log_3 n\rfloor} 3^j \times (1+\lfloor\log_3 n\rfloor - j)^3 \times 3^{\lfloor\log_3 n\rfloor} \\ = 3^{\lfloor\log_3 n\rfloor} \sum_{j=1}^{\lfloor\log_3 n\rfloor+1} 3^{\lfloor\log_3 n\rfloor +1 -j} \times j^3 = 3^{2\lfloor\log_3 n\rfloor + 1} \sum_{j=1}^{\lfloor\log_3 n\rfloor+1} j^3 3^{-j}.$$ The sum term again converges to a constant and we get for the lower bound asymptotics the formula $$\frac{33}{8} 3^{2\lfloor\log_3 n\rfloor + 1} .$$ Note that for the upper bound we have $\lfloor\log_3 n\rfloor+1\to\log_3 n,$ so that it is in fact $$\frac{7917}{2048} n^2 \approx 3.86572265625\times n^2$$ but for the lower bound $\lfloor\log_3 n\rfloor = \log_3 n$, so that it is $$\frac{33}{8} \times 3 \times n^2 = \frac{99}{8} n^2 \approx 12.375 \times n^2.$$

Joining the two bounds we may conclude that $$T(n)\in \Theta\left(3^{2\lfloor\log_3 n\rfloor}\right) = \Theta(n^2).$$

A similar computation which is a bit simpler can be used to analyse the cost of Strassen matrix multiplication.

Solution 2:

You may be interested to know that there are exact solutions to your three problems. We will do problem one and provide references for the other two.

Suppose we have $T(0)$ and for $n\ge 1$ we have the recurrence $$T(n) = 6 T(\lfloor n/2 \rfloor) + 2^{3\lfloor \log_2 n \rfloor}.$$ We can unroll this recurrence to obtain the exact result that $$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor} 6^j \times 2^{3(\lfloor \log_2 n \rfloor - j)}.$$ This simplifies to $$T(n) = 2^{3\lfloor \log_2 n \rfloor} \sum_{j=0}^{\lfloor \log_2 n \rfloor} \left(\frac{6}{8}\right)^j = 2^{3\lfloor \log_2 n \rfloor} \frac{1-(3/4)^{\lfloor \log_2 n \rfloor +1}}{1-3/4} \\ =4 \times 2^{3\lfloor \log_2 n \rfloor} \left(1-(3/4)^{\lfloor \log_2 n \rfloor +1}\right).$$ The conclusion is that $$T(n) \in \Theta\left(2^{3\lfloor \log_2 n \rfloor}\right) = \Theta(n^3).$$

This MSE link points to a series of similar calculations.

Solution 3:

For your second problem, lets solve the recurrence $$T(n) = 8 T(\lfloor n/2 \rfloor) + \frac{n^3}{(1+\lfloor\log_2 n\rfloor)^4}$$ where $T(0) = 0.$

Let $$n = \sum_{k=0}^{\lfloor\log_2 n\rfloor} d_k 2^k$$ be the binary representation of $n.$ Unrolling the recursion we find the exact formula $$T(n) = \sum_{j=0}^{\lfloor\log_2 n\rfloor} \frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4} \left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} d_k 2^{k-j}\right)^3.$$

Now to get an upper bound on this consider the case of $n$ being a string of one digits, which gives $$T(n)\le \sum_{j=0}^{\lfloor\log_2 n\rfloor} \frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4} \left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} 2^{k-j}\right)^3 \\ = \sum_{j=0}^{\lfloor\log_2 n\rfloor} \frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4} \left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} 2^k\right)^3 = \sum_{j=0}^{\lfloor\log_2 n\rfloor} \frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4} (2^{\lfloor\log_2 n\rfloor+1} - 2^j)^3 \\ = \sum_{j=1}^{\lfloor\log_2 n\rfloor+1} \frac{1}{j^4} (2^{\lfloor\log_2 n\rfloor+1} - 2^{\lfloor\log_2 n\rfloor+1-j})^3 = 2^{3(\lfloor\log_2 n\rfloor+1)} \sum_{j=1}^{\lfloor\log_2 n\rfloor+1} \frac{1}{j^4} (1 - 2^{-j})^3.$$ The sum term converges rapidly to a constant and hence the asymptotics of the upper bound are $$\frac{1}{90} \left(\pi^4 -270 \mathrm{Li}_4(1/2) + 270 \mathrm{Li}_4(1/4) - 90 \mathrm{Li}_4(1/8)\right) \times 2^{3(\lfloor\log_2 n\rfloor+1)}.$$

For the a lower bound consider the case of a one digit followed by zeros, which gives $$T(n)\ge \sum_{j=0}^{\lfloor\log_2 n\rfloor} \frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4} \left(2^{\lfloor\log_2 n\rfloor-j}\right)^3 = 2^{3\lfloor\log_2 n\rfloor}\sum_{j=0}^{\lfloor\log_2 n\rfloor} \frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4} \\ = 2^{3\lfloor\log_2 n\rfloor}\sum_{j=1}^{\lfloor\log_2 n\rfloor+1} \frac{1}{j^4} .$$ The sum term once again converges to a constant and the asymptotics of the lower bound are $$\frac{\pi^4}{90} 2^{3\lfloor\log_2 n\rfloor}.$$

Joining the two bounds we see that $$T(n) \in \Theta\left(2^{3\lfloor\log_2 n\rfloor}\right) = \Theta(n^3).$$