Outline
•
•
•
•
Recurrence, Sloppiness
Recitation 2: Recurrence, sloppiness
1
Master Method
•
2.
3.
If f (n) = O(nlogba – ε) for ε > 0, then
T(n) = Θ(nlogba) .
If f (n) = Θ(nlogba lgkn) for some constant k ≥ 0, then
T(n) = Θ(nlogba lgk+1n) .
If f (n) = Ω(nlogba + ε) for some constant ε > 0 and
a f (n/b) ≤ c f (n) for some constant c < 1, then
T(n) = Θ( f (n) ) .
Recitation 2: Recurrence, sloppiness
Recitation 2: Recurrence, sloppiness
3
• For each of the following recurrences, give
the asymptotic solution using the Master
Theorem (state which case), or else state
that the Master Theorem does not apply
• T(n) = 4T(n/2) + n
Recitation 2: Recurrence, sloppiness
• T(n) = 4T(n/2) + n2
• T(n) = 4T(n/2) + n2 lg n
• T(n) = 4T(n/2) + n3
• T(n) = T(n-2) + n
• T(n) = 3T(n/4) + n lg n
• T(n) = 5T(n/2) + n2 lg n
Recitation 2: Recurrence, sloppiness
2
Solving Recurrence (Practice)
Let a ≥ 1, b > 1 and T(n) = aT(n/b) + f(n), where
n/b means either ⎣n/b⎦ or ⎡n/b⎤. Then
1.
Master Method review
Sloppiness
Solving recurrence (practice)
Solving recurrence (tricks)
5
Recitation 2: Recurrence, sloppiness
4
6
1
• T(n) = 4T(n/4) + n/lg n
• T(n) = 2T(n/4) + √n
• T(n) = T(n/4) + lg n
• T(n) = 8 T((n - √n)/4) + n2
• T(n) = 2T(n/4) + lg n
• T(n) = 2T(n/4) + n0.51
Recitation 2: Recurrence, sloppiness
7
Recitation 2: Recurrence, sloppiness
• T(n) = 4T(n/2) + n/lg n
• T(n) = T(n/2) + lg(n!)
• T(n) = 8T(n/3) + 2n
• T(n) = T(n/2) + n!
• T(n) = 7T(n/3) + n2
• T(n) =16T(n/4) + n2 log 2 n
Recitation 2: Recurrence, sloppiness
9
Change of Variable
10
Drop Lower Order Terms
• An unfamiliar looking recurrence can be
sometimes be made into a familiar one by a
change of variable
• Example: T(n) = T(√n) + lg n
• Let m = lg n, so n = 2m and √n = 2m/2
• So T(2m) = T(2m/2) + m
• Define S(m) = T(2m), then
• For recurrences like
T(n) = 2T(n/2 + √n) + Θ(1), note that n
dominates √n.
• Guess that solution is the same as
S(n) = 2S(n/2) + Θ(1) = Θ(n)
• Solution must be verified by the substitution
method
– S(m) = S(m/2) + m
• So S(m) = Θ(m)
• Substituting back: T(n) = Θ(lg n)
Recitation 2: Recurrence, sloppiness
Recitation 2: Recurrence, sloppiness
8
11
Recitation 2: Recurrence, sloppiness
12
2
Theorem (Sloppiness):
• Suppose that:
Sloppiness
1.
2.
3.
• What about the merge sort recurrence?
– T(n) = T(⎣n/2⎦ ) + T(⎡n/2⎤) + Θ(n)
– Not strictly covered by Master Theorem
– In the case where n is a power of 2, no problem:
• T(n) = 2T(n/2)+Θ(n)
• By case 2 of Master Theorem, T(n) = Θ(n log n)
– Under what conditions can we say that recurrences such
as the merge sort recurrence has the same solution as
the simplified version that works for powers of 2??
Recitation 2: Recurrence, sloppiness
13
T(n) and f(n) are monotonically increasing
T(bi) ≤ f(bi) for all i > 0
f(n) = O(f(n/b))
- polynomial growth
• Then T(n) = O(f(n))
Proof:
• T(n) ≤ T(b ⎡logbn⎤)
by (1)
• ≤ f(b ⎡logbn⎤) by (2)
• ≤ cf(b ⎡logbn⎤-1) by (3)
• ≤ c(f(n))
by (1)
Recitation 2: Recurrence, sloppiness
14
Summary
• Master Theorem
– Memorize - can do most recurrences in your head
• Sloppiness
– Recurrences in algorithms often not defined cleanly
– Only need to be defined cleanly for powers of b
• Tricks for solving recurrences
– Change of variable
– Drop lower order terms
• Akra-Bazzi
– Useful in some cases not covered by Master Theorem
• If these methods don’t work, use your ingenuity,
then verify with substitution method!
Recitation 2: Recurrence, sloppiness
18
3