1. Recursion-tree method
A recursion tree models the costs (time) of a recursive execution of an algorithm.
The recursion-tree method can be unreliable.
The recursion-tree method promotes intuition, however.

2. Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2:

3. Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2:
T(n)

4. Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2
T(n/4)
T(n/2)

5. Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2
(n/4)2
(n/2)2
T(n/16)
T(n/8)
T(n/8)
T(n/4)

6. Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2
(n/4)2
(n/2)2
(n/16)2
(n/8)2
(n/8)2
(n/4)2 …
Q(1)

7. Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2
(n/4)2
(n/2)2
(n/16)2
(n/8)2
(n/8)2
(n/4)2 …
Q(1)

8. Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2
(n/4)2
(n/2)2
(n/16)2
(n/8)2
(n/8)2
(n/4)2 …
Q(1)

9. Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2
(n/4)2
(n/2)2
(n/16)2
(n/8)2
(n/8)2
(n/4)2 … …
Q(1)

10. Example of recursion tree
Solve T(n) = T(n/4) + T(n/2) + n2: n2
(n/4)2
(n/2)2
(n/16)2
(n/8)2
(n/8)2
(n/4)2 … …
Q(1)
Total =
= Q(n2)
geometric series

11. Appendix: geometric series
for x ¹ 1
for |x| < 1

12. The master method The master method applies to recurrences of the form
T(n) = a T(n/b) + f (n) ,
where a ³ 1, b > 1, and
f(n)  (nd), d  0

13. Idea of master theorem Recursion tree: f (n) f (n) a f (n/b)…
h = logbn a …
f (n/b)
f (n/b)
f (n/b) a …
f (n/b2)
f (n/b2)
f (n/b2) …
#leaves = ah
= alogbn
= nlogba
nlogbaT (1)
T (1)

14. Three common cases Compare a with bd: If a > bd, logba > d
nlogba > nd
We also know that f(n)  (nd)
f (n) grows polynomially slower than nlogba.
Solution: T(n) = Q(nlogba) .

15. Idea of master theorem Recursion tree: f (n) f (n) a … f (n/b) f (n/b)
h = logbn …
a2 f (n/b2)
f (n/b2)
f (n/b2)
f (n/b2) … …
CASE 1: The weight increases geometrically from the root to the leaves. The leaves hold a constant fraction of the total weight.
nlogbaT (1)
T (1)
Q(nlogba)

16. Three common cases Compare a with bd: If a = bd, logba = d Nlogba = nd
We also know that f(n)  (nd)
f (n) and nlogba grow at similar rates.
Solution: T(n) = Q( f(n) log n) .

17. Idea of master theorem Recursion tree: f (n) f (n) a … f (n/b) f (n/b)
h = logbn …
a2 f (n/b2)
f (n/b2)
f (n/b2)
f (n/b2) … …
CASE 2: The weight is approximately the same on each of the logbn levels.
nlogbaT (1)
T (1)
Q(nlogbalg n)

18. Three common cases (cont.)
Compare a with bd:
If a < bd,
logba < d
nlogba < nd
We also know that f(n)  (nd)
f (n) grows polynomially faster than nlogba.
Solution: T(n) = Q( f (n) ) = Q(nd) .

19. Idea of master theorem Recursion tree: f (n) f (n) a … f (n/b) f (n/b)
h = logbn …
a2 f (n/b2)
f (n/b2)
f (n/b2)
f (n/b2) … …
CASE 3: The weight decreases geometrically from the root to the leaves. The root holds a constant fraction of the total weight.
nlogbaT (1)
T (1)
Q( f (n))

20. Examples Ex. T(n) = 4T(n/2) + n
a = 4, b = 2, d = 1  nlogba = n2; f (n) = n.
(a=4) >( bd =21)
CASE 1:  T(n) = Q(n2).
Ex. T(n) = 4T(n/2) + n2
a = 4, b = 2, d=2  nlogba = n2; f (n) = n2.
(a =4) = (bd=22)
CASE 2:  T(n) = Q(n2log n).

21. Examples Ex. T(n) = 4T(n/2) + n3
a = 4, b = 2, d=3  nlogba = n2; f (n) = n3.
(a=4) < (bd = 23)
CASE 3:
 T(n) = Q(nd) = Q(n3).