Use iteration to find explicit formulas for the following recurrences. Assume T (1) = 1.
i. T (n) = 3T  n
3
 + n, n = 3k
ii. T (n) = 4T  n
4
 + n, n = 4k
3

To find the explicit formulas for the given recurrences using iteration, let's solve them step by step.

### i. \( T(n) = 3T\left(\frac{n}{3}\right) + n \), with \( n = 3^k \) and \( T(1) = 1 \)

Let's iterate the recurrence:

1. **First iteration:**
   \[
   T(n) = 3T\left(\frac{n}{3}\right) + n
   \]

2. **Second iteration:**
   \[
   T\left(\frac{n}{3}\right) = 3T\left(\frac{n}{9}\right) + \frac{n}{3}
   \]
   Substituting into the first equation:
   \[
   T(n) = 3\left(3T\left(\frac{n}{9}\right) + \frac{n}{3}\right) + n = 9T\left(\frac{n}{9}\right) + 3n + n = 9T\left(\frac{n}{9}\right) + 4n
   \]

3. **Third iteration:**
   \[
   T\left(\frac{n}{9}\right) = 3T\left(\frac{n}{27}\right) + \frac{n}{9}
   \]
   Substituting into the previous expression:
   \[
   T(n) = 9\left(3T\left(\frac{n}{27}\right) + \frac{n}{9}\right) + 4n = 27T\left(\frac{n}{27}\right) + 9n + 4n = 27T\left(\frac{n}{27}\right) + 13n
   \]

Continuing this pattern, we observe that after \( m \) iterations:

\[
T(n) = 3^m T\left(\frac{n}{3^m}\right) + n \left(\sum_{i=0}^{m-1} 3^i\right)
\]

Since \( n = 3^k \) and \( T(1) = 1 \), we have \( \frac{n}{3^k} = 1 \) when \( m = k \):

\[
T(n) = 3^k T(1) + n \left(\sum_{i=0}^{k-1} 3^i\right)
\]

\[
T(n) = 3^k \cdot 1 + n \left(\frac{3^k - 1}{2}\right)
\]

\[
T(n) = 3^k + \frac{n(3^k - 1)}{2}
\]

Since \( n = 3^k \), \( 3^k = n \):

\[
T(n) = n + \frac{n(n - 1)}{2} = \frac{n^2 + n}{2}
\]

Thus, the explicit formula is:

\[
\boxed{T(n) = \frac{n^2 + n}{2}}
\]

### ii. \( T(n) = 4T\left(\frac{n}{4}\right) + n \), with \( n = 4^k \) and \( T(1) = 1 \)

Let's iterate the recurrence:

1. **First iteration:**
   \[
   T(n) = 4T\left(\frac{n}{4}\right) + n
   \]

2. **Second iteration:**
   \[
   T\left(\frac{n}{4}\right) = 4T\left(\frac{n}{16}\right) + \frac{n}{4}
   \]
   Substituting into the first equation:
   \[
   T(n) = 4\left(4T\left(\frac{n}{16}\right) + \frac{n}{4}\right) + n = 16T\left(\frac{n}{16}\right) + 4n + n = 16T\left(\frac{n}{16}\right) + 5n
   \]

3. **Third iteration:**
   \[
   T\left(\frac{n}{16}\right) = 4T\left(\frac{n}{64}\right) + \frac{n}{16}
   \]
   Substituting into the previous expression:
   \[
   T(n) = 16\left(4T\left(\frac{n}{64}\right) + \frac{n}{16}\right) + 5n = 64T\left(\frac{n}{64}\right) + 16n + 5n = 64T\left(\frac{n}{64}\right) + 21n
   \]

Continuing this pattern, after \( m \) iterations:

\[
T(n) = 4^m T\left(\frac{n}{4^m}\right) + n \left(\sum_{i=0}^{m-1} 4^i\right)
\]

Since \( n = 4^k \) and \( T(1) = 1 \), we have \( \frac{n}{4^k} = 1 \) when \( m = k \):

\[
T(n) = 4^k T(1) + n \left(\sum_{i=0}^{k-1} 4^i\right)
\]

\[
T(n) = 4^k \cdot 1 + n \left(\frac{4^k - 1}{3}\right)
\]

\[
T(n) = 4^k + \frac{n(4^k - 1)}{3}
\]

Since \( n = 4^k \), \( 4^k = n \):

\[
T(n) = n + \frac{n(n - 1)}{3} = \frac{n^2 + 2n}{3}
\]

Thus, the explicit formula is:

\[
\boxed{T(n) = \frac{n^2 + 2n}{3}}
\]

Would you like more details on any of the steps? Here are some related questions you might find interesting:

1. How do these formulas relate to the Master Theorem for divide-and-conquer recurrences?
2. What are other common methods for solving recurrence relations?
3. How does the choice of base (3 in the first problem, 4 in the second) affect the solution?
4. What would the explicit formulas be for different initial conditions (e.g., \( T(1) = 2 \))?
5. How do these problems illustrate the growth rates of recursive functions?

**Tip:** Understanding the iterative method for solving recurrences is crucial for analyzing the time complexity of algorithms, especially in divide-and-conquer approaches.

Explore iterative methods to find explicit formulas for recurrence relations such as T(n) = 3T(n/3) + n and T(n) = 4T(n/4) + n. Understand the step-by-step process and mathematical concepts involved, including divide and conquer algorithms. Suitable for advanced undergraduate students and those interested in algorithm analysis.

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