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+2 votes
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asked in Algorithms by (111 points)   | 356 views
plz give me ans by back substitution.....

Here is an explaination for the solution http://techieme.in/solving-recurrences-master-method/

2 Answers

+1 vote

T(n) = 7T(n/2) + n^2

comparing with the equation (MASTER THEOREM)

T(n) = aT(n/b) + ⊖(n^k $\log_{n}p$)

we get ,a=7,b=2,k=2,p=0

now it satisfies a>b^k,

so case first of master theorem

T(n) = ⊖(n^$\log_{b}a$)

T(n)=⊖(n^3)

answered by Boss (8.1k points)  
T(n) = n^2.801
we know by master theoram

T(n) = aT(n/b) + n^klogp(base n)

where a>=1 , b>1 , k>=0 and p is a real number

In above question a =7, b=2 , k=2 and p=0

if a>b^k

then Tc = O(n^loga(base b))

            = O(n^log7(base 2))

            =O(n^3)
0 votes

Compare with master theorem ,a=7,b=2 and f(n)=n2

T(n)=nlog27 =n2.81 i.e f(n) is polynomially smaller than t(n), therefore it is first case

 therefore, ans is (nlog7) or (n2.81)

answered by Junior (523 points)  

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