1. Recurrence Examples

2. Recursion Tree for T(n)=3T(n/4)+(n2)
cn2
cn2
c(n/4)2
T(n/16)
(c)
T(n/4)
T(n/4)
T(n/4)
(a)
(b)
cn2
cn2
(3/16)cn2
c(n/4)2
c(n/4)2
c(n/4)2
log 4n
(3/16)2cn2
c(n/16)2
c(n/16)2
c(n/16)2
c(n/16)2
c(n/16)2
c(n/16)2
c(n/16)2
c(n/16)2
c(n/16)2
(nlog 43)
T(1)
T(1)
T(1)
T(1)
T(1)
T(1)
3log4n= nlog 43
Total O(n2)
(d)

3. Solution to T(n)=3T(n/4)+(n2)
The height is log 4n,
#leaf nodes = 3log 4n= nlog 43. Leaf node cost: T(1).
Total cost T(n)=cn2+(3/16) cn2+(3/16)2 cn2+
 +(3/16)log 4n-1 cn2+ (nlog 43)
=(1+3/16+(3/16)2+  +(3/16)log 4n-1) cn2 + (nlog 43)
<(1+3/16+(3/16)2+  +(3/16)m+ ) cn2 + (nlog 43)
=(1/(1-3/16)) cn2 + (nlog 43)
=16/13cn2 + (nlog 43)
=O(n2).

4. Recursion Tree of T(n)=T(n/3)+ T(2n/3)+O(n)

5. Recursion tree for T(n)=aT(n/b)+f(n)

6. General Divide-and-Conquer Recurrence
T(n) = aT(n/b) + f (n) where f(n)  (nd), d  0
Master Theorem: If a < bd, T(n)  (nd)
If a = bd, T(n)  (nd log n)
If a > bd, T(n)  (nlog b a )
Examples:
T(n) = T(n/2) + n  T(n)  (n )
Here a = 1, b = 2, d = 1, a < bd
T(n) = 2T(n/2) + 1  T(n)  (n log 2 2 ) = (n )
Here a = 2, b = 2, d = 0, a > bd

7. Examples T(n) = T(n/2) + 1  T(n)  (log(n) )
Here a = 1, b = 2, d = 0, a = bd
T(n) = 4T(n/2) + n  T(n)  (n log 2 4 ) = (n 2 )
Here a = 4, b = 2, d = 1, a > bd
T(n) = 4T(n/2) + n2  T(n)  (n2 log n)
Here a = 4, b = 2, d = 2, a = bd
T(n) = 4T(n/2) + n3  T(n)  (n3)
Here a = 4, b = 2, d = 3, a < bd