The given recurrence relation is:

$T(n) = 3T\left(\frac{n}{2}\right) + n + 1$

To find the Big O complexity, we can apply the Master Theorem. The recurrence is of the form:

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

Here, $a = 3$, $b = 2$, and $f(n) = n + 1$ (which behaves like $O(n)$).

Step 1: Determine $\log_b{a}$

We first compute $\log_b{a}$, where $a = 3$ and $b = 2$:

$\log_b{a} = \log_2{3} \approx 1.585$

Step 2: Compare $f(n)$ with $n^{\log_b{a}}$

Now, we compare the function $f(n) = n$ (which is $O(n)$) with $n^{\log_b{a}} = n^{1.585}$:

• If $f(n)$ grows slower than $n^{\log_b{a}}$, then the solution is $T(n) = O(n^{\log_b{a}})$.
• If $f(n)$ grows at the same rate as $n^{\log_b{a}}$, then the solution is $T(n) = O(n^{\log_b{a}} \log n)$.
• If $f(n)$ grows faster than $n^{\log_b{a}}$, then the solution is $T(n) = O(f(n))$.

In this case, $f(n) = O(n)$ grows slower than $n^{\log_b{a}} = n^{1.585}$.

Step 3: Apply the Master Theorem

Since $f(n) = O(n)$ and $n^{\log_b{a}} = n^{1.585}$, the first case of the Master Theorem applies, and the complexity is dominated by the term $n^{\log_b{a}}$.

Thus, the Big O for this recurrence is:

$T(n) = O(n^{\log_2{3}}) \approx O(n^{1.585})$

Summary:

The Big O complexity for $T(n) = 3T(n/2) + n + 1$ is $O(n^{1.585})$.

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Would you like further details or clarification on this method?

Here are some related questions you may find helpful:

1. What is the general form of the Master Theorem for divide-and-conquer recurrences?
2. How can we solve recurrence relations without using the Master Theorem?
3. What are the differences between the three cases of the Master Theorem?
4. How does the choice of $a$, $b$, and $f(n)$ affect the Big O complexity?
5. What happens if $f(n)$ is $O(n^k)$ for some constant $k$?

Tip: When solving recurrences using the Master Theorem, carefully identify the terms that correspond to $a$, $b$, and $f(n)$, and be sure to compare the growth rates properly to determine the correct case.