To solve the recurrence relation $T(n) = 2T(n/2) + n$, we can use the Master Theorem, which provides a way to determine the time complexity of recurrence relations of this form.

The given recurrence relation is:

$T(n) = 2T(n/2) + n$

Step 1: Identify the components of the Master Theorem.

The general form of the Master Theorem is:

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

Where:

• $a$ is the number of subproblems at each level (in this case, $a = 2$).
• $b$ is the factor by which the problem size is reduced at each step (in this case, $b = 2$).
• $f(n)$ is the cost of the work done outside the recursive calls (in this case, $f(n) = n$).

Step 2: Calculate $n^{\log_b a}$.

We need to find $n^{\log_b a}$. In this case:

$a = 2, \quad b = 2$

$\log_b a = \log_2 2 = 1$

So:

$n^{\log_b a} = n^1 = n$

Step 3: Compare $f(n)$ with $n^{\log_b a}$.

Here, $f(n) = n$ and $n^{\log_b a} = n$.

Since $f(n) = \Theta(n^{\log_b a})$, we can use Case 2 of the Master Theorem, which states:

• If $f(n) = \Theta(n^{\log_b a} \log^k n)$ for some $k \geq 0$, then $T(n) = \Theta(n^{\log_b a} \log^{k+1} n)$.

In this problem, $k = 0$, so:

$T(n) = \Theta(n^{\log_b a} \log n) = \Theta(n \log n)$

Conclusion:

The time complexity of the given recurrence relation is $\Theta(n \log n)$.

Would you like more details or have any questions?

Follow-up Questions

1. What is the significance of the Master Theorem in analyzing algorithms?
2. Can you explain Case 1 and Case 3 of the Master Theorem?
3. How would the time complexity change if $f(n)$ were $\Theta(n^2)$?
4. What is the meaning of $\Theta$ notation?
5. How do we derive the logarithmic factor in $n \log n$?
6. Can you provide an example of a problem that has a similar time complexity?
7. What is the difference between $\Theta$, $O$, and $\Omega$ notations?
8. How does the divide-and-conquer method relate to the Master Theorem?

Tip

When analyzing algorithms, always pay attention to the recursive part and the non-recursive work done at each level; this helps determine the overall complexity efficiently.