2. (a) Use the master theorem to find the exact solution of the following recurrence equation. Make sure you find the constants. Assume n is a power of 2. +n, n ≥ 2 4T 3, n = 1 T(n) = 2(b) Use repeated substitution to find the exact solution of the following recurrence. T(n) = { 1, n+T(n-1), n ≥2 n = 1
2. (a) Use the master theorem to find the exact solution of the following recurrence equation. Make sure you find the constants. Assume n is a power of 2. +n, n ≥ 2 4T 3, n = 1 T(n) = 2(b) Use repeated substitution to find the exact solution of the following recurrence. T(n) = { 1, n+T(n-1), n ≥2 n = 1
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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can you solve 2020 image one I also provided refrences how MY prof solved so based on that pleease solve this image 2020 one step by step formet please use from refrnecs
![**Problem Statement:**
2. Use the master theorem to find the exact solution of the following recurrence equation.
Find the constants. Assume \( n \) is an integer power of 2, \( n = 2^k \).
\[
T(n) =
\begin{cases}
27 \left(\frac{n}{2}\right) + n^2, & n \geq 2 \\
1, & n = 1
\end{cases}
\]
**Solution Using Master Theorem:**
1. **Master Theorem Constants:**
- \( a = 2 \), \( b = 2 \), \( \beta = 2 \)
- \( h = \log_b a = 1 \)
Since \( h \neq \beta \), we proceed with:
\[
T(n) = A n^h + B n^\beta
\]
Substituting the values for \( h \) and \( \beta \):
\[
T(n) = A n + B n^2
\]
2. **Finding Constants \( A \) and \( B \):**
- \( T(1) = 1 = A + B \)
- \( T(2) = 2T(1) + 2^2 = 2 + 4 = 6 = 2A + 4B \)
From the equations:
- \( A + B = 1 \)
- \( 2A + 4B = 6 \)
Solving the equations:
- Rearrange: \( 2B = 4 \) implies \( B = 2 \)
- Substitute: \( A + 2 = 1 \) implies \( A = -1 \)
3. **Final Solution:**
Substitute the constants back into the equation:
\[
T(n) = 2n^2 - n
\]
This gives the exact solution to the recurrence using the master theorem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bad4e48-dad8-4710-a64b-b24b80d1efcf%2F9edc7a7e-c6df-482e-b3b3-3a254fe442c9%2F6arjl3_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
2. Use the master theorem to find the exact solution of the following recurrence equation.
Find the constants. Assume \( n \) is an integer power of 2, \( n = 2^k \).
\[
T(n) =
\begin{cases}
27 \left(\frac{n}{2}\right) + n^2, & n \geq 2 \\
1, & n = 1
\end{cases}
\]
**Solution Using Master Theorem:**
1. **Master Theorem Constants:**
- \( a = 2 \), \( b = 2 \), \( \beta = 2 \)
- \( h = \log_b a = 1 \)
Since \( h \neq \beta \), we proceed with:
\[
T(n) = A n^h + B n^\beta
\]
Substituting the values for \( h \) and \( \beta \):
\[
T(n) = A n + B n^2
\]
2. **Finding Constants \( A \) and \( B \):**
- \( T(1) = 1 = A + B \)
- \( T(2) = 2T(1) + 2^2 = 2 + 4 = 6 = 2A + 4B \)
From the equations:
- \( A + B = 1 \)
- \( 2A + 4B = 6 \)
Solving the equations:
- Rearrange: \( 2B = 4 \) implies \( B = 2 \)
- Substitute: \( A + 2 = 1 \) implies \( A = -1 \)
3. **Final Solution:**
Substitute the constants back into the equation:
\[
T(n) = 2n^2 - n
\]
This gives the exact solution to the recurrence using the master theorem.
![***Educational Resource: Master Theorem and Repeated Substitution for Recurrence Relations***
**Problem 2: Solving Recurrence Relations**
**2(a) Using the Master Theorem**
To find the exact solution of the following recurrence equation, apply the Master Theorem. Ensure the constants are identified. Assume \( n \) is a power of 2.
\[
T(n) =
\begin{cases}
4T\left(\frac{n}{2}\right) + n, & n \geq 2 \\
3, & n = 1
\end{cases}
\]
**2(b) Using Repeated Substitution**
Employ repeated substitution to find the exact solution for the following recurrence.
\[
T(n) =
\begin{cases}
n + T(n-1), & n \geq 2 \\
1, & n = 1
\end{cases}
\]
---
Each part provides a distinct approach to solving recurrence relations, offering insight into how recursive algorithms' time complexities can be derived.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bad4e48-dad8-4710-a64b-b24b80d1efcf%2F9edc7a7e-c6df-482e-b3b3-3a254fe442c9%2Fugx4cq_processed.png&w=3840&q=75)
Transcribed Image Text:***Educational Resource: Master Theorem and Repeated Substitution for Recurrence Relations***
**Problem 2: Solving Recurrence Relations**
**2(a) Using the Master Theorem**
To find the exact solution of the following recurrence equation, apply the Master Theorem. Ensure the constants are identified. Assume \( n \) is a power of 2.
\[
T(n) =
\begin{cases}
4T\left(\frac{n}{2}\right) + n, & n \geq 2 \\
3, & n = 1
\end{cases}
\]
**2(b) Using Repeated Substitution**
Employ repeated substitution to find the exact solution for the following recurrence.
\[
T(n) =
\begin{cases}
n + T(n-1), & n \geq 2 \\
1, & n = 1
\end{cases}
\]
---
Each part provides a distinct approach to solving recurrence relations, offering insight into how recursive algorithms' time complexities can be derived.
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