Sequences and Series
Master theorem
R. Inkulu
http://www.iitg.ac.in/rinkulu/
(Asymptotic growth of functions)
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Recurrence relations
• useful in the analysis of recursive algorithms, primarily, in
divide-and-conquer algorithms
• R(n) = aR(n/b) + f (n): R(n) signifies the resources required to solve
the problem when n objects are involved; f (n) signifies the resources
involved in dividing n elements into n/b sized a numbered subproblems,
and combining the solutions of these subproblems into one.
(Asymptotic growth of functions)
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Recurrence tree corresponding to a recurrence relation
a
f(n)
f(n/b)
f(n/b)
f(n/b)
a
a
a
lg b n
Θ (1) Θ (1)
Θ (1) Θ (1)
lg a
n b
T(n) =
(Asymptotic growth of functions)
(
Θ(1) if n = 1
aT(n/b) + f (n)
if n > 1
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Master theorem
Let a ≥ 1 and b > 1 be constants, let f (n) be a function, and let T(n) be
defined on the nonnegative integers by the recurrence T(n) = aT(n/b) + f (n)
(with T(n) = Θ(1) when n = 1) where we interpret n/b to mean either ⌊n/b⌋
or ⌈n/b⌉. Then T(n) has the following asymptotic bounds:
(i) If f (n) = O(nlgb a−ǫ ) for some constant ǫ > 0, then T(n) = Θ(nlgb a ).
(ii) If f (n) = Θ(nlgb a ), then T(n) = Θ(nlgb a lg n).
(iii) If f (n) = Ω(nlgb a+ǫ ) for some constant ǫ > 0, and if af (n/b) ≤ cf (n)
for some constnat c < 1 and all sufficiently large n, then
T(n) = Θ(f (n)).
Do note that not every recurrence of the form T(n) = aT(n/b) + f (n) can be bounded using Master
Theorem.
(Asymptotic growth of functions)
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Examples
• T(n) = 9T(n/3) + n — case (i) leads to T(n) = Θ(n2 )
• T(n) = T(2n/3) + 1 — case (ii) leads to T(n) = Θ(lg (n))
• T(n) = 3T(n/4) + n lg n — case (iii) leads to T(n) = Θ(n lg n)
• T(n) = 2T(n/2) + n lg n — Master thm not applicable
(Asymptotic growth of functions)
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Proof of the master theorem1
From the recurrence tree, T(n) = Θ(nlgb a ) +
Plgb n−1
j=0
aj f ( bnj )
Then the second term of the RHS becomes
Plgb n−1 j n lg a−ǫ
• in case (i),
O( j=0
a ( bj ) b
) = O(nlgb a )
• for the case (ii),
Θ(
• for the case (iii),
(
further, (
Plgb n−1
j=0
Plgb n−1
j=0
Plgb n−1
j=0
aj ( bnj )lgb a ) = Θ(nlgb a lg n)
cj f (n)) + O(1) = O(f (n))
aj f ( bnj )) ∈ Ω(f (n)
1
we prove for the case wherein n is an exact power of b > 1; proof of the general case is
more or less the same
(Asymptotic growth of functions)
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Asymptotic bounds of typically occurring recurrences
T(n) ≤ qT(n/2) + cn when n > 2, and T(2) ≤ c
• If q = 1 then T(n) = Θ(n)
• If q = 2 then T(n) = Θ(n lg n)
• If q > 2 then T(n) = Θ(nlg2 q )
T(n) ≤ qT(n/2) + cn2 when n > 2, and T(2) ≤ c
• If q < 4 then T(n) = Θ(n2 )
• If q = 4 then T(n) = Θ(n2 lg n)
• If q > 4 then T(n) = Θ(nlg2 q )
(Asymptotic growth of functions)
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Akra-Bazzi method
— not presented
(Asymptotic growth of functions)
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