Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > When n = 22kfor some k ≥ 0, the recurrence...
Start Learning for Free
When n = 22k for some k ≥ 0, the recurrence relation
T(n) = √(2) T(n/2) + √n, T(1) = 1
evaluates to :
T(n) = √(2) T(n/2) + √n, T(1) = 1
evaluates to :
- a)√(n) (log n + 1)
- b)√(n) (log n )
- c)√(n) log √(n)
- d)n log √(n)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2...
Please note that the question is asking about exact solution. Master theorem provides results in the form of asymptotic notations. So we can't apply Master theorem here. We can solve this recurrence using simple expansion or recurrence tree method.
T(n) = √(2) T(n/2) + √n = √(2) [√(2) T(n/4) + √(n/2) ] + √n = 2 T(n/4) + √2 √(n/2) +√n = 2[ √2 T(n/8) + √(n/4) ]+√2 √(n/2)+√n = √(2^3) T(n/8)+ 2 √(n/4) + √2 √(n/2) +√n = √(2^3) T(n/8)+√n +√n +√n = √(2^3) T(n/(2^3))+3√n ............................................. = √(2^k) T(n/(2^k))+k√n = √(2^logn) + logn √n = √n + logn √n = √n(logn +1)
Alternate Solution : This question can be easily done by substitution method look: T(1)= 1; GIVEN. Now use n=2 in the given recurrence relation which gives 2*(1.414) (since value of root over 2 is 1.414) now by looking at the options use n=2 which satisfies option A.
Most Upvoted Answer
When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2...
Please note that the question is asking about exact solution. Master theorem provides results in the form of asymptotic notations. So we can't apply Master theorem here. We can solve this recurrence using simple expansion or recurrence tree method.
T(n) = √(2) T(n/2) + √n = √(2) [√(2) T(n/4) + √(n/2) ] + √n = 2 T(n/4) + √2 √(n/2) +√n = 2[ √2 T(n/8) + √(n/4) ]+√2 √(n/2)+√n = √(2^3) T(n/8)+ 2 √(n/4) + √2 √(n/2) +√n = √(2^3) T(n/8)+√n +√n +√n = √(2^3) T(n/(2^3))+3√n ............................................. = √(2^k) T(n/(2^k))+k√n = √(2^logn) + logn √n = √n + logn √n = √n(logn +1)
Alternate Solution : This question can be easily done by substitution method look: T(1)= 1; GIVEN. Now use n=2 in the given recurrence relation which gives 2*(1.414) (since value of root over 2 is 1.414) now by looking at the options use n=2 which satisfies option A.
Free Test
FREE
| Start Free Test |
Community Answer
When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2...
Please note that the question is asking about exact solution. Master theorem provides results in the form of asymptotic notations. So we can't apply Master theorem here. We can solve this recurrence using simple expansion or recurrence tree method.
T(n) = √(2) T(n/2) + √n = √(2) [√(2) T(n/4) + √(n/2) ] + √n = 2 T(n/4) + √2 √(n/2) +√n = 2[ √2 T(n/8) + √(n/4) ]+√2 √(n/2)+√n = √(2^3) T(n/8)+ 2 √(n/4) + √2 √(n/2) +√n = √(2^3) T(n/8)+√n +√n +√n = √(2^3) T(n/(2^3))+3√n ............................................. = √(2^k) T(n/(2^k))+k√n = √(2^logn) + logn √n = √n + logn √n = √n(logn +1)
Alternate Solution : This question can be easily done by substitution method look: T(1)= 1; GIVEN. Now use n=2 in the given recurrence relation which gives 2*(1.414) (since value of root over 2 is 1.414) now by looking at the options use n=2 which satisfies option A.
|
Explore Courses for Computer Science Engineering (CSE) exam
|
|
Top Courses for Computer Science Engineering (CSE)View all
Question Description
When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer?.
When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer?.
Solutions for When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.
Here you can find the meaning of When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer?, a detailed solution for When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice When n = 22kfor some k ≥ 0, the recurrence relationT(n) = √(2) T(n/2) + √n, T(1) = 1evaluates to :a)√(n) (log n + 1)b)√(n) (log n )c)√(n) log √(n)d)n log √(n)Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
|
|
Explore Courses for Computer Science Engineering (CSE) exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup