Introduction to Algorithms
Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Question
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Chapter 4, Problem 1P

(a)

Program Plan Intro

To determine the asymptotic bounds for the recurrence relation using master method.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given Information: The recurrence relation is T(n)=2T(n/2)+n4 .

Explanation:

For a divide and conquer recurrence of the form T(n)=aT(n/b)+f(n) where a1,b>1 and f(n)>0, the following three cases can happen:

Case 1: If f(n)=O(n logbaε) for some constant ε>0 then T(n)=Θ(n lgba) .

Case 2: If f(n)=Θ(n lgba) then T(n)=Θ(n lgbalogn) .

Case 3: If f(n)=Ω(n logba+ε) for some constant ε>0 and if af(n/b)cf(n) for some constants c>1 and sufficiently large n then T(n)=Θ(f(n)) .

The values of a,b and f(n) are 2, 2 and n4 respectively.

Therefore, nlogba=nlog22=n and f(n)=1 . Here, f(n)=Ω(n log22+ε) where ε=3 .

So, case 3 of the master method applies.

Hence, T(n)=Θ(f(n))=Θ(n4) .

(b)

Program Plan Intro

To determine the asymptotic bounds for the recurrence relation using master method.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given Information: The recurrence relation is T(n)=T(7n/10)+n .

Explanation:

The values of a,b and f(n) are 1, 10/7 and n respectively.

Therefore, nlogba=nlog10/71=n0=1 and f(n)=n . Here, f(n)=Ω(n log42+ε) where ε=1/2 .

So, case 3 of the master method applies.

Hence, T(n)=Θ(f(n))=Θ(n) .

(c)

Program Plan Intro

To determine the asymptotic bounds for the recurrence relation using master method.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given Information: The recurrence relation is T(n)=16T(n/4)+n2 .

Explanation:

The values of a,b and f(n) are 16, 4 and n2 respectively.

Therefore, nlogba=nlog416=n2 and f(n)=Θ(n logba)=Θ(n2) .

So, case 2 of the master method applies.

Hence, T(n)=Θ(nclgn)=Θ(n2lgn) .

(d)

Program Plan Intro

To determine the asymptotic bounds for the recurrence relation using master method.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given Information: The recurrence relation is T(n)=7T(n/3)+n2 .

Explanation:

The values of a,b and f(n) are 7, 3 and n2 respectively.

Therefore, nlogba=nlog37 and f(n)=n2 . Here, f(n)=Ω(n log42+ε) where ε=3/2 .

So, case 3 of the master method applies.

Hence, T(n)=Θ(f(n))=Θ(n2) .

(e)

Program Plan Intro

To determine the asymptotic bounds for the recurrence relation using master method.

(e)

Expert Solution
Check Mark

Explanation of Solution

Given Information: The recurrence relation is T(n)=7T(n/2)+n2 .

Explanation:

The values of a,b and f(n) are 7, 2 and n2 respectively.

Therefore, nlogba=nlog27 and f(n)=n2 . Here, f(n)=O(n log27ε) where ε>0 .

So, case 1 of the master method applies.

Hence, T(n)=Θ(n logba)=Θ(n log27) .

(f)

Program Plan Intro

To determine the asymptotic bounds for the recurrence relation using master method.

(f)

Expert Solution
Check Mark

Explanation of Solution

Given Information: The recurrence relation is T(n)=2T(n/4)+n .

Explanation:

The values of a,b and f(n) are 2, 4 and n respectively.

Therefore, nlogba=nlog42=n and f(n)=Θ(n logba)=Θ(n) .

So, case 2 of the master method applies.

Hence, T(n)=θ(n lg42lgn)=θ(nlgn) .

(g)

Program Plan Intro

To determine the asymptotic bounds for the recurrence relation using master method.

(g)

Expert Solution
Check Mark

Explanation of Solution

Given Information: The recurrence relation is T(n)=T(n2)+n2 .

Explanation:

The recurrence relation is not in the form of master theorem. Therefore, it cannot be solve by master theorem.

Solve the recurrence relation T(n)=T(n2)+n2

as follows:

  T(n)=T(n2)+n2=n2+(n2)2+T(n4)=n2i=0n/21+4i=0n/2i24ni=0n/2i=n2n2+413(2n3+6n2+4n)+4n12(n3+2n2)=2n33+n2+43n=Θ(n3)

Therefore, the asymptotic notation of the recurrence T(n)=T(n2)+n2 is Θ(n3) .

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