1) The document provides the proof of the Master Theorem for analyzing divide-and-conquer recurrences. 2) It introduces key lemmas used in the proof, including bounding the contribution of subproblem sizes. 3) The proof considers three cases depending on how the work at each level relates to the decrease in problem size, showing the recurrence has a runtime of O(nlogba), O(nklogn), or O(nk) respectively.

1. 1
Proof of Master Theorem
• The proof for the exact powers, n=bk
for k≥1.
• Lemma 4.2
– for T(n) = Θ(1) if n=1
– aT(n/b)+f(n) if n=bk
for k≥1
– where a ≥ 1, b>1, f(n) be a nonnegative function,
– Then
– T(n) = Θ(nlogba
)+ aj
f(n/bj
)
• Proof:
– By iterating the recurrence
– By recursion tree (See figure 4.3)
∑
j=0
logb
n-1

2. 2
Recursion tree for T(n)=aT(n/b)+f(n)

3. 3
Proof of Master Theorem (cont.)
• Lemma 4.3:
– Let a ≥ 1, b>1, f(n) be a nonnegative function defined on
exact power of b, then
– g(n)= aj
f(n/bj
) can be bounded for exact power of b as:
1. If f(n)=O(nlogb
a-ε) for some ε>0, then g(n)= O(nlogb
a
).
2. If f(n)= Θ(nlogb
a
), then g(n)= Θ(nlogb
a
lg n).
3. If f(n)= Ω(nlogb
a+ε) for some ε>0 and if af(n/b) ≤cf(n) for some
c<1 and all sufficiently large n ≥b, then g(n)= Θ(f(n)).
∑
j=0
logb
n-1

4. 4
Proof of Lemma 4.3
• For case 1: f(n)=O(nlogb
a-ε) implies f(n/bj
)=O((n /bj
)logb
a-ε), so
• g(n)= aj
f(n/bj
) =O( aj
(n /bj
)logb
a-ε)
• = O(nlogb
a-ε aj
/(blogb
a-ε)j
) = O(nlogb
a-ε aj
/(aj
(b-ε)j
))
• = O(nlogb
a-ε (bε)j
) = O(nlogb
a-ε(((bε)logbn
-1)/(bε
-1) )
• = O(nlogb
a-ε(((blogbn
)ε
-1)/(bε
-1)))=O(nlogb
a
n-ε
(nε
-1)/(bε
-1))
• = O(nlogb
a
)
∑
j=0
logb
n-1
∑
j=0
logb
n-1
∑
j=0
logb
n-1
∑
j=0
logb
n-1
∑
j=0
logb
n-1

5. 5
Proof of Lemma 4.3(cont.)
• For case 2: f(n)= Θ(nlogb
a
) implies f(n/bj
)= Θ((n /bj
)logb
a
), so
• g(n)= aj
f(n/bj
) = Θ( aj
(n /bj
)logb
a
)
• = Θ(nlogb
a
aj
/(blogb
a
)j
) = Θ(nlogb
a
1)
• = Θ(nlogb
a
logb
n) = Θ(nlogb
a
lg n)
∑
j=0
logb
n-1
∑
j=0
logb
n-1
∑
j=0
logb
n-1
∑
j=0
logb
n-1

6. 6
Proof of Lemma 4.3(cont.)
• For case 3:
– Since g(n) contains f(n), g(n) = Ω(f(n))
– Since af(n/b) ≤cf(n), aj
f(n/bj
) ≤cj
f(n) , why???
– g(n)= aj
f(n/bj
) ≤ cj
f(n) ≤ f(n) cj
– =f(n)(1/(1-c)) =O(f(n))
– Thus, g(n)=Θ(f(n))
∑
j=0
logb
n-1
∑
j=0
logb
n-1
∑
j=0
∞

7. 7
Proof of Master Theorem (cont.)
• Lemma 4.4:
– for T(n) = Θ(1) if n=1
– aT(n/b)+f(n) if n=bk
for k≥1
– where a ≥ 1, b>1, f(n) be a nonnegative function,
1. If f(n)=O(nlogb
a-ε) for some ε>0, then T(n)= Θ(nlogb
a
).
2. If f(n)= Θ(nlogb
a
), then T(n)= Θ(nlogb
a
lg n).
3. If f(n)=Ω(nlogb
a+ε) for some ε>0, and if af(n/b) ≤cf(n) for
some c<1 and all sufficiently large n, then T(n)=
Θ(f(n)).

8. 8
Proof of Lemma 4.4 (cont.)
• Combine Lemma 4.2 and 4.3,
– For case 1:
• T(n)= Θ(nlogb
a
)+O(nlogb
a
)=Θ(nlogb
a
).
– For case 2:
• T(n)= Θ(nlogb
a
)+Θ(nlogb
a
lg n)=Θ(nlogb
a
lg n).
– For case 3:
• T(n)= Θ(nlogb
a
)+Θ(f(n))=Θ(f(n)) because f(n)= Ω(nlogb
a+ε).

9. 9
Floors and Ceilings
• T(n)=aT(n/b)+f(n) and T(n)=aT(n/b)+f(n)
• Want to prove both equal to T(n)=aT(n/b)+f(n)
• Two results:
– Master theorem applied to all integers n.
– Floors and ceilings do not change the result.
• (Note: we proved this by domain transformation too).
• Since n/b≤n/b, and n/b≥ n/b, upper bound for
floors and lower bound for ceiling is held.
• So prove upper bound for ceilings (similar for lower
bound for floors).

10. 10
Upper bound of proof for T(n)=aT(n/b)+f(n)
• consider sequence n, n/b,  n/b/b,   n/b /b/b,
…
• Let us define nj as follows:
• nj= n if j=0
• = nj-1/b if j>0
• The sequence will be n0, n1, …, nlogb
n
• Draw recursion tree:

11. 11
Recursion tree of T(n)=aT(n/b)+f(n)

12. 12
The proof of upper bound for ceiling
– T(n) = Θ(nlogba
)+ aj
f(nj)
– Thus similar to Lemma 4.3 and 4.4, the upper
bound is proven.
∑
j=0
logb
n -1

13. 13
The simple format of master theorem
• T(n)=aT(n/b)+cnk
, with a, b, c, k are positive
constants, and a≥1 and b≥2,
O(nlogba
), if a>bk
.
• T(n) = O(nk
logn), if a=bk
.
O(nk
), if a<bk
.