1.9k views

Let $T(n)$ be a function defined by the recurrence

$T(n) = 2T(n/2) + \sqrt n$ for $n \geq 2$ and
$T(1) = 1$

Which of the following statements is TRUE?

1. $T(n) = \Theta(\log n)$
2. $T(n) = \Theta(\sqrt n)$
3. $T(n) = \Theta(n)$
4. $T(n) = \Theta(n \log n)$

edited | 1.9k views
+3
Is the question ok?
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I am getting D, after subsituting values for different numbers.
+2
can someone solve this qus by using recursion TREE!!!?

Option $C$ is the answer. It can be done by Master's theorem.

$n^{\log_b a} = n^{\log_2 2} = n$.

$f(n) = \sqrt n = n^{\frac{1}{2}}$.

So, $f(n) = O\left(n^{\log_b a -\epsilon}\right)$ is true for any real $\epsilon$, $0 < \epsilon < \frac{1}{2}$. Hence Master theorem Case 1 satisfied,
$$T(n) = \Theta\left(n^{\log_b a}\right) = \Theta (n).$$

by Boss (14.4k points)
edited
+1
this is not in aT(n/b) +n power k log n .

How can this be one in ny Master's?
+1
master theorem /...how!!1!
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How?
+1
There was a typo in question. Corrected now.
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Thanks!
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How to do it using back substitution?  let me know if i am doing any mistake.

by (131 points)
0
Shouldn't it be Math.root(n/2) instead of Math.root(n)/2 ?
Answer is C it can also be solved by master theorem. by case 1(a>=b^k)

T(n)= aT(n/b)+n^k

Here a = 2 b=2 and k =1/2

so a>= b^k

T(n)=Θ (n^logba)

hence T(n)=Θ(n)
by (403 points) option c is right