Question

1. Solve with Masters theorm, show all steps T(n) = 2T(n/4) + n

Answer 1

The master method works only for following type of recurrences or for recurrences that can be transformed to following type.

T(n) = aT(n/b) + f(n) where a >= 1 and b > 1
Following are the 3 cases possible
1. If f(n) = Θ(n^c) where c < Log_ba then T(n) = Θ(n^Log_ba)

2. If f(n) = Θ(n^clog^kn) where c = Log_ba then T(n) = Θ(n^cLog^k+1 n)

3.If f(n) = Θ(n^c) where c > Log_ba then T(n) = Θ(f(n))

Solution :

According to question :

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

As we see the formula we get the following values

a = 2

b = 4

f(n) = n

f(n) = Θ(n^c)

c = 1.

then , log_ba = log₄(2) =.log₄(4)^1/2

= 1/2

log_ba = 0.5 -------- (i)

as, log_ba < 1

or  c > log_ba

soit follows the Case (iii) of master's theorem.

T(n) = Θ(f(n)) = Θ(n)

Thus the given recurrence relation T(n) was in Θ(n) that complies with the f(n) of the original formula.

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