Book edition 7th

Author(s) Kenneth H. Rosen

Pages 808 pages

ISBN 9780073383095

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Short Answer

Suppose that the function $f$ satisfies the recurrence relation $f (n) = 2 f (\sqrt{n}) + l o g n$ whenever $n$ is a perfect square greater than $1$ and $f (2) = 1$ .
a) Find $f (16)$ .
b) Give a big - $O$ estimate for $f (n)$ . [Hint: Make the substitution $m = l o g n$ ].

(a) The required function $f (16) = 12$ .

(b) The required function $f (n) = O (\log n loglog n)$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

The function $f$ satisfies the recurrence relation $f (n) = 2 f (\sqrt{n}) + 1$ , Where $n$ is a perfect square greater than $1$ and $f (2) = 1$ .

Step 2: Explain the Master theorem

The master theorem is a formula for solving recurrence relations of the form: $T (n) = aT (n / b) + f (n)$ , where $n =$ the size of the input $a =$ number of sub-problems is in the recursion $n / b =$ size of each subproblem. All subproblems are assumed to have the same size.

Step 3: Calculate the functionf(16)

(a)

To find $f (16)$ :

$f (16) = 2 f (4) + \log (16)$

$f (16) = 2 (2 f (2) + \log (4)) + \log (16)$

$\begin{matrix} f (16) = 2 (2 + 2) + 4 \\ f (16) = 12 \end{matrix}$

Step 4: Calculate the value off(n)by the Master theorem

(b)

Use the suggested substitution and let $m = \log n$ .

So, $2^{m} = n$

Now, it will may be $f (n) = f (2^{m})$ .

$\begin{array}{l} f (n) = 2 f (\sqrt{2^{m}}) + \log (2^{m}) \\ f (n) = 2 f (2^{m / 2}) + m \end{array}$

Rewrite the function, for clarity, as $g (m) = f (2^{m})$ .

And so, $g (m) = 2 g (m / 2) + m$ .

Next, apply the Master theorem with $a = 2, b = 2, c = 1, d = 1$ :

g (m) = O (m \log_{2} m)

Hence, $f (n) = O (\log n loglog n)$

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Discrete Mathematics and its Applications

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Short Answer

Suppose that the function $f$ satisfies the recurrence relation $f (n) = 2 f (\sqrt{n}) + l o g n$ whenever $n$ is a perfect square greater than $1$ and $f (2) = 1$ .
a) Find $f (16)$ .
b) Give a big - $O$ estimate for $f (n)$ . [Hint: Make the substitution $m = l o g n$ ].

Step by Step Solution

Step 1: Given information

Step 2: Explain the Master theorem

Step 3: Calculate the functionf(16)

Step 4: Calculate the value off(n)by the Master theorem

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Discrete Mathematics and its Applications

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Short Answer

Suppose that the function f satisfies the recurrence relation f(n)=2f(n)+log n whenever n is a perfect square greater than 1and f(2) = 1.a) Find f(16).b) Give a big -Oestimate forf(n). [Hint: Make the substitution m=logn].

Step by Step Solution

Step 1: Given information

Step 2: Explain the Master theorem

Step 3: Calculate the functionf(16)

Step 4: Calculate the value off(n)by the Master theorem

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Suppose that the function $f$ satisfies the recurrence relation $f (n) = 2 f (\sqrt{n}) + l o g n$ whenever $n$ is a perfect square greater than $1$ and $f (2) = 1$ .
a) Find $f (16)$ .
b) Give a big - $O$ estimate for $f (n)$ . [Hint: Make the substitution $m = l o g n$ ].