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For parameters a and b, both of which are ω(1), T(n)=T(n1/a)+1, and T(b)=1. Then T(n) is
- a)Θ(logalogbn)
- b)Θ(logabn)
- c)Θ(logblogan)
- d)Θ(log2log2n)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
For parameters a and b, both of which are ω(1), T(n)=T(n1/a)+1, ...
Given,
T(n) = T(n1/a)+1, T(b) = 1
Now, using iterative method,
= T(n) = [T(n1/a2)+1] + 1 = [T(n1/a3)+1] + 2 = [T(n1/a4)+1] + 3 . . . = [T(n1/ak)+1] + (k-1) = T(n1/ak) + k
Let,
→ n1/ak = b → log(n1/ak) = log(b) → ak = log(n) / log (b) = logb(n) → k = logalogb(n)
Therefore,
= T(n1/ak) + k = T(b) + logalogb(n) = 1 + logalogb(n) = Θ(logalogb(n))
Option (A) is correct.
Most Upvoted Answer
For parameters a and b, both of which are ω(1), T(n)=T(n1/a)+1, ...
Explanation:
T(n) = T(n1/a) + 1 represents a recursive function where the input size decreases by a factor of a each time, and the function makes constant time operations at each step.
Using Master Theorem:
- We can apply the Master Theorem to analyze the time complexity of the given recursive function T(n).
- According to the Master Theorem, if T(n) = aT(n/b) + f(n), where a >= 1, b > 1, and f(n) is a polynomial function, then:
- If f(n) = Θ(nc) where c < />ba, then T(n) = Θ(nlogba).
- If f(n) = Θ(nc) where c = logba, then T(n) = Θ(nc * log n).
- If f(n) = Θ(nc) where c > logba, then T(n) = Θ(f(n)).
Applying Master Theorem:
- In this case, a = 1, b = a, and f(n) = 1.
- We have c = 0, which is less than logab = log11 = 0.
- Therefore, according to the Master Theorem, T(n) = Θ(logan) = Θ(log n).
Conclusion:
- The correct answer is option 'A' (Θ(log n)), as the time complexity of the given recursive function T(n) is logarithmic in nature.
T(n) = T(n1/a) + 1 represents a recursive function where the input size decreases by a factor of a each time, and the function makes constant time operations at each step.
Using Master Theorem:
- We can apply the Master Theorem to analyze the time complexity of the given recursive function T(n).
- According to the Master Theorem, if T(n) = aT(n/b) + f(n), where a >= 1, b > 1, and f(n) is a polynomial function, then:
- If f(n) = Θ(nc) where c < />ba, then T(n) = Θ(nlogba).
- If f(n) = Θ(nc) where c = logba, then T(n) = Θ(nc * log n).
- If f(n) = Θ(nc) where c > logba, then T(n) = Θ(f(n)).
Applying Master Theorem:
- In this case, a = 1, b = a, and f(n) = 1.
- We have c = 0, which is less than logab = log11 = 0.
- Therefore, according to the Master Theorem, T(n) = Θ(logan) = Θ(log n).
Conclusion:
- The correct answer is option 'A' (Θ(log n)), as the time complexity of the given recursive function T(n) is logarithmic in nature.
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Community Answer
For parameters a and b, both of which are ω(1), T(n)=T(n1/a)+1, ...
Given,
T(n) = T(n1/a)+1, T(b) = 1
Now, using iterative method,
= T(n) = [T(n1/a2)+1] + 1 = [T(n1/a3)+1] + 2 = [T(n1/a4)+1] + 3 . . . = [T(n1/ak)+1] + (k-1) = T(n1/ak) + k
Let,
→ n1/ak = b → log(n1/ak) = log(b) → ak = log(n) / log (b) = logb(n) → k = logalogb(n)
Therefore,
= T(n1/ak) + k = T(b) + logalogb(n) = 1 + logalogb(n) = Θ(logalogb(n))
Option (A) is correct.
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Solutions for For parameters a and b, both of which are ω(1), T(n)=T(n1/a)+1, and T(b)=1. Then T(n) isa)Θ(logalogbn)b)Θ(logabn)c)Θ(logblogan)d)Θ(log2log2n)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).
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