Recurrence Relations T(n)=T(√n)+logn Using Master's Theorem || GATECSE || DAA

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In this video, we dive into the fascinating world of recurrence relations with a focus on the equation T(n) = T(√n) + logn. We explore the application of Master's Theorem to solve this intriguing recurrence relation, providing a comprehensive understanding of its mechanics and implications.
What You'll Learn:
Understanding Recurrence Relations: We'll start by demystifying recurrence relations and their significance in algorithm analysis and time complexity determination.
Master's Theorem Unveiled: We'll uncover the power and utility of Master's Theorem in solving recurrence relations like T(n) = T(√n) + logn, simplifying complex equations with its structured approach.
Step-by-Step Analysis: Through a detailed walkthrough, we'll break down the Master's Theorem application for T(n) = T(√n) + logn, demonstrating the step-by-step process to solve this specific type of recurrence relation.
Insights into Time Complexity: By the end of the video, you'll gain insights into the time complexity of algorithms represented by T(n) = T(√n) + logn, enhancing your ability to analyze and compare different algorithms efficiently.
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📚 Subject Wise Playlist 📚
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[@ramch7915]

U explain excellent

[@Pranav-mc6cp]

Sir on timeline of 4 min i am confuse there How you can substitute 2^(m/2) with m/2 please respond if the replacement is incorrect

[@priyamgupta2779]

haan bhaii sahii kahe rahe ho
S(m/2) nahii hoga waha par S(m^(1/2)) hoga

[@RahulKumar-qy7pz]

@@priyamgupta2779 aaiesa kr skte hai .... phele T ka function chal rha tha use S ke function se change kiya hai

[@factsandtales8329]

aisa t ke function mein hoga

[@shahinakhan1205]

Mashaallah

[@avneeshkaushik6474]

Theta log(logn)

[@polymers9987]

t(n)=t(n/2) + 2^n. By master theorem.

[@ankitmishra3185]

Sir substitution method se to ( logn.loglogn ) aa rha hai aapne hi karaya hai

[@THEGATEHUB]

That question is T(n)= 2T(√n)+logn..

[@__harshchaturvedi_3931]

Bina termination condition ke kaise nikal diye bhai

[@polymers9987]

​@@THEGATEHUB t(n)=t(n/2) + 2^n. Solution?

[@newsForEveryoneFast]

The Master Theorem is used to determine the asymptotic behavior of recurrences of the form:
\[ T(n) = aT\left(\frac{n}{b}
ight) + f(n) \]
For the Master Theorem to be applicable, the constants \(a\) and \(b\) must satisfy the following conditions:
- \(a \geq 1\)
- \(b > 1\)
Here’s a brief overview of the Master Theorem cases based on these conditions:
1. **Case 1:** If \( f(n) = O(n^c) \) where \( c < \log_b{a} \), then \( T(n) = \Theta(n^{\log_b{a}}) \).
2. **Case 2:** If \( f(n) = \Theta(n^c) \) where \( c = \log_b{a} \), then \( T(n) = \Theta(n^{\log_b{a}} \log{n}) \).
3. **Case 3:** If \( f(n) = \Omega(n^c) \) where \( c > \log_b{a} \) and \( af\left(\frac{n}{b}
ight) \leq kf(n) \) for some \( k < 1 \) and sufficiently large \( n \), then \( T(n) = \Theta(f(n)) \).
These cases help in determining the asymptotic bounds of the recurrence relation.

[@JaskaranSingh-hw5jf]

u did it wrong... last me log ki power p+1 krni thi

[@user-vs9dk9bw3v]

nhi 3rd case h ye, 2nd nhi

[@newsForEveryoneFast]

@@user-vs9dk9bw3v arey bhai... a>=1, and b>1 . since, condition is not satisfied , how can we ever solve this porblem using master method.. ?