Question:
Consider the following recurrence relation:
\[
T(n) =
\begin{cases}
\sqrt{n}T(\sqrt{n}) + n & \text{for } n \geq 1,
1 & \text{for } n = 1.
\end{cases}
\]
Which one of the following options is CORRECT?
Consider the following recurrence relation:
\[
T(n) =
\begin{cases}
\sqrt{n}T(\sqrt{n}) + n & \text{for } n \geq 1,
1 & \text{for } n = 1. \end{cases} \] Which one of the following options is CORRECT?
1 & \text{for } n = 1. \end{cases} \] Which one of the following options is CORRECT?
Show Hint
For recurrence relations involving nested square roots, use transformations like ( n = 2^m ) to simplify and determine the number of levels.
Updated On: Jan 22, 2025
- \( T(n) = \Theta(n \log \log n) \)
- \( T(n) = \Theta(n \log n) \)
- \( T(n) = \Theta(n^2 \log n) \)
- \( T(n) = \Theta(n^2 \log \log n) \)
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The Correct Option is A
Solution and Explanation
Step 1: Recurrence Analysis.
The given recurrence is: \[ T(n) = \sqrt{n}T(\sqrt{n}) + n. \] Let \( n = 2^m \). Then, \( \sqrt{n} = 2^{m/2} \), and the recurrence becomes: \[ T(2^m) = 2^{m/2} T(2^{m/2}) + 2^m. \] Step 2: Substitution.
Substitute \( T(2^m) = f(m) \). The recurrence now becomes: \[ f(m) = 2^{m/2} f(m/2) + 2^m. \] Step 3: Expanding the Recurrence.
Expand the recurrence: \[ f(m) = 2^{m/2} \left( 2^{m/4} f(m/4) + 2^{m/2} \right) + 2^m. \] \[ f(m) = 2^{m/2} \cdot 2^{m/4} f(m/4) + 2^m + 2^m. \] This continues for \( k \) levels, with each level contributing \( 2^m \). The number of levels is approximately \( \log \log n \). Step 4: Complexity.
The total complexity is: \[ T(n) = \Theta(n \log \log n). \] Final Answer: \[ \boxed{\Theta(n \log \log n)} \]
The given recurrence is: \[ T(n) = \sqrt{n}T(\sqrt{n}) + n. \] Let \( n = 2^m \). Then, \( \sqrt{n} = 2^{m/2} \), and the recurrence becomes: \[ T(2^m) = 2^{m/2} T(2^{m/2}) + 2^m. \] Step 2: Substitution.
Substitute \( T(2^m) = f(m) \). The recurrence now becomes: \[ f(m) = 2^{m/2} f(m/2) + 2^m. \] Step 3: Expanding the Recurrence.
Expand the recurrence: \[ f(m) = 2^{m/2} \left( 2^{m/4} f(m/4) + 2^{m/2} \right) + 2^m. \] \[ f(m) = 2^{m/2} \cdot 2^{m/4} f(m/4) + 2^m + 2^m. \] This continues for \( k \) levels, with each level contributing \( 2^m \). The number of levels is approximately \( \log \log n \). Step 4: Complexity.
The total complexity is: \[ T(n) = \Theta(n \log \log n). \] Final Answer: \[ \boxed{\Theta(n \log \log n)} \]
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