To solve the recurrence relation $T(n) = T\left(\frac{n}{10}\right) + T\left(\frac{9n}{10}\right) + n$ using the Master Theorem, we first need to check if it fits into the standard form of the Master Theorem for divide-and-conquer recurrences:

$T(n) = a T\left(\frac{n}{b}\right) + f(n)$

where $a \geq 1$ and $b > 1$ are constants, and $f(n)$ is an asymptotically positive function.

For the given recurrence:

• $a = 2$ (since there are two recursive calls)
• $b$ is not explicitly stated in a direct form, but we have two different parts, $\frac{n}{10}$ and $\frac{9n}{10}$.

The given recurrence does not fit the standard form directly because $b$ should be a single value for both recursive calls. However, we can analyze it by considering the overall contribution of both terms.

To proceed with the analysis, we can use an alternate approach that combines the contributions of both terms. Let's consider the weighted average of the subproblem sizes:

Let $n_1 = \frac{n}{10}$ and $n_2 = \frac{9n}{10}$.

The combined size of subproblems is: $\frac{n}{10} + \frac{9n}{10} = n$

For a more precise method, we analyze the recurrence using the Akra-Bazzi method, which generalizes the Master Theorem for recurrences of this form:

$T(n) = g(n) + \sum_{i=1}^{k} T(b_i n)$

where $b_1, b_2, \ldots, b_k$ are constants. In our case:

• $g(n) = n$
• $b_1 = \frac{1}{10}$ and $b_2 = \frac{9}{10}$

The Akra-Bazzi theorem states that if $T(n) = \sum_{i=1}^{k} T(b_i n) + g(n)$, then the solution is given by:

$T(n) = \Theta \left( n^p \left( 1 + \int_1^n \frac{g(u)}{u^{p+1}} du \right) \right)$

where $p$ is a solution to the equation:

$\sum_{i=1}^{k} b_i^p = 1$

For our recurrence: $b_1^p + b_2^p = \left(\frac{1}{10}\right)^p + \left(\frac{9}{10}\right)^p = 1$

To find $p$, we solve the equation: $\left(\frac{1}{10}\right)^p + \left(\frac{9}{10}\right)^p = 1$

This is a transcendental equation and is typically solved numerically. However, we can approximate that for large values of $n$, $p$ is close to 1 because:

$\left(\frac{1}{10}\right)^1 + \left(\frac{9}{10}\right)^1 = 0.1 + 0.9 = 1$

Thus, $p \approx 1$.

Now, integrating $g(n)$ with respect to $n$: $\int_1^n \frac{u}{u^2} du = \int_1^n \frac{1}{u} du = \ln(n)$

Hence, $T(n) = \Theta \left( n^1 \left( 1 + \ln(n) \right) \right)$

Therefore, $T(n) = \Theta(n \log n)$

So, the solution to the recurrence $T(n) = T\left(\frac{n}{10}\right) + T\left(\frac{9n}{10}\right) + n$ is $T(n) = \Theta(n \log n)$.

Would you like further details or have any questions?

Here are 8 relative questions for further exploration:

1. How can we numerically solve the equation $\left(\frac{1}{10}\right)^p + \left(\frac{9}{10}\right)^p = 1$ for more precision?
2. What are some applications of the Akra-Bazzi method in solving complex recurrences?
3. Can you provide an example of another recurrence relation that can be solved using the Master Theorem?
4. How does the weighted average approach help in analyzing complex recurrences?
5. What are the limitations of the Master Theorem in solving recurrence relations?
6. How does the change of variables method simplify solving recurrences?
7. Can you explain the steps to derive $T(n) = \Theta(n \log n)$ using another method?
8. What are the key differences between the Master Theorem and the Akra-Bazzi method?

Tip: When dealing with complex recurrences, it's often helpful to break down the problem and consider the contributions of each recursive term separately before applying a general method.