big o - solving recurrence T(n) = T(n/2) + T(n/2 - 1) + n/2 + 2 -


need on solving runtime recurrence, using big-oh:

t(n) = t(n/2) + t(n/2 - 1) + n/2 + 2

i don't quite how use master theorem here

for n big enough can assume t(n/2 - 1) == t(n/2), can change

  t(n) = t(n/2) + t(n/2 - 1) + n/2 + 2

into 

  t(n) = 2*t(n/2) + n/2 + 2

and use master theorem (http://en.wikipedia.org/wiki/master_theorem) for

  t(n) = a*t(n/b) + f(n)    = 2   b = 2   f(n) = n/2 + 2   c = 1    k = 0    log(a, b) = 1 = c

and have (case 2, since log(a, b) = c)

  t(n) = o(n**c * log(n)**(k + 1))   t(n) = o(n * log(n))

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