Question

Solve the following recurrence relation T(1) = c1 T(n) = 2*T(n/2) + c2

Solve the following recurrence relation

T(1) = c1

T(n) = 2*T(n/2) + c2

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Homework Answers

Answer #1

T(1) = c1

T(n) = 2*T(n/2) + c2

Solution:

T(n) = 2*T(n/2) + c2 T(n/2) = 2*T(n/4) + c2

= 2*(2*T(n/4) + c2) + c2

= 4*T(n/4) + 3c2    T(n/4) = 2*T(n/8) + c2

= 4*(2*T(n/8) + c2) + 3c2

= 8*T(n/8) + 7c2 T(n/8) = 2*T(n/16) + c2

= 8*(2*T(n/16) + c2) + 7c2

= 16*T(n/16) + 15c2 T(n/16) = 2*T(n/32) + c2

= 16*(2*T(n/32) + c2 ) + 15c2

= 32*T(n/32) + 31c2

= .......

= 2k*T(n/2k) + (2k - 1)c2

T(1) = c1

T(n) = 2*T(n/2) + c2

T(n) = 2k*T(n/2k) + (2k - 1)c2

We need to get off T(n/2k) using T(1)

n/2k = 1

n = 2k

lg (n) = k

T(n) = 2lg (n) *T(n/2lg (n) ) + (2lg (n) - 1)c2

= nT(n/n) + (n - 1)c2

= nT(1) + (n - 1)c2

= nc1 + (n - 1)c2

  
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