1. Using The Master Method Case 1
T(n) = 9T(n/3) + n
a=9, b=3, f(n) = n
nlogb a = nlog3 9 = (n2)
Since f(n) = O(nlog3 9 - ) = O(n2-0.5) = O(n1.5)
where =0.5
case 1 applies:
Thus the solution is
T(n) = (n2)

2. Using The Master Method Case 2
T(n) = T(2n/3) + 1
a=1, b=3/2, f(n) = 1
nlogba = nlog3/21 = n0 = 1
Since f(n) = (nlogba) = (1)
case 2 applies:
Thus the solution is
T(n) = (lg n)

3. Using The Master Method Case 3
T(n) = 3T(n/4) + n lg n
a=3, b=4, f(n) = n lg n
nlogba = nlog43 = n0.793 = n0.8
Since f(n) = W(nlog43+) = W(n )= W(n)
where   0.2, and for sufficiently large n,
a . f(n/b) = 3(n/4) lg(n/4) < (3/4) n lg n for c = 3/4
case 3 applies:
Thus the solution is
T(n) = (n lg n)

4. When the Master Method does not apply to recurrence
T(n) = 2T(n/2) + n lg n
a = 2, b = 2, f(n) = n lg n, and nlogba = n
Note that f(n) = n lg n is asymptotically larger than nlogba = n …. It might seem that case 3 should apply.
The problem is that it is not polynomially larger.
The ratio f(n) / nlogba = (n lg n) / n = lg n is asymptotically less than nε for any positive constant ε.
The recurrence falls into the gap between case 2 and case 3.

5. Recursion tree view f(n) f(n/b) f(n) a … af(n/b) f(n/b2) a2f(n/b2)
Θ(1)
f(n)
f(n/b)
f(n/b2) a …
f(n)
af(n/b)
a2f(n/b2)

6. Examples Ex. T(n) = 4T(n/2) + n Ex. T(n) = 4T(n/2) + n2
a = 4, b = 2, f (n) = n
nlogba = n2 f
Since f (n) = O(n2 – ) for  = 1
Case 1 applies
T(n) = Θ(n2).
Ex. T(n) = 4T(n/2) + n2
a = 4, b = 2, f(n) = n2
nlogba = n2
Since f (n) = Θ(n2).
Case 2 applies
T(n) = Θ(n2lgn).

7. Examples Ex. T(n) = 4T(n/2) + n3 Ex. T(n) = 4T(n/2) + n2/lgn
a = 4, b = 2, f (n) = n3
nlogba = n2
case 3: f (n) = Ω(n2 + ) for  = 1
and 4(n/2)3 ≤ cn3 for c = 1/2.
T(n) = Θ(n3).
Ex. T(n) = 4T(n/2) + n2/lgn
a = 4, b = 2, f (n) = n2/lgn
Master method does not apply.

8. Master Method – Examples
T(n) = 16T(n/4)+n
a = 16, b = 4, nlogba = nlog416 = n2.
f(n) = n = O(nlogba-) = O(n2- ), where  = 1  Case 1.
Hence, T(n) = (nlogba ) = (n2).
T(n) = T(3n/7) + 1
a = 1, b=7/3, and nlogba = nlog 7/3 1 = n0 = 1
f(n) = 1 = (nlogba)  Case 2.
Therefore, T(n) = (nlogba lg n) = (lg n)

9. Simplified Master Theorem
Let a  1 and b > 1 be constants and let T(n) be the recurrence
T(n) = a T(n/b) + c nk
defined for n  0.
1. If a > bk, then T(n) = ( nlogba ).
2. If a = bk, then T(n) = ( nk lg n ).
3. If a < bk, then T(n) = ( nk ).

10. Examples T(n) = 16T(n/4) + n T(n) = T(3n/7) + 1 a = 16, b = 4, k=1
T(n) = ( nlogba ) = ( nlog416 )
T(n) = (n2)
T(n) = T(3n/7) + 1
a = 1, b=7/3, k = 0
1 = (7/3)0  1 = 1  a = bk
T(n) = ( nk lg n ) = ( n0 lg n )
T(n) = ( lg n ).