Problem Set 4
4.4-4 Use a recursion tree to determine a good asymptotic upper bound on the
recurrence T (n) = 2T (n − 1) + 1. Use the substitution method to verify your
answer.
Using a recursion tree or 'unwrapping' the recurrence relation allows one to
conclude that
T (n) = 2T (n − 1) + 1
= 2 (2T (n − 2) + 1) + 1 = 22 T (n − 2) + 1 + 2
..
.
= 2k T (n − k) + 1 + 2 + . . . 2k−1
This terminates at n − k = 1, i.e. k = n − 1.
Then
Pn=2
T (n) = 2n−1 T (1) + i=0 2i = c2n (noting 2n−1 = 12 2n ).
The asymptotic upper bound that this suggests is T (n) = Θ (2n ). You
will nd that guessing T (n) ≤ c2n doesn't quite work, though. Instead, guess
T (n) ≤ c2n − d.
For xed d, c can be made suciently large that the inequality is true for
n = 1.
The inductive hypothesis allows us to conclude
n−1
T (n) = 2T (n − 1) + 1 ≤ 2 c2
≤ c2n − 2d + 1.
or
−d +1
In order to conclude T (n) ≤ c2n − d , we need that c2n − 2d + 1 ≤ c2n − d,
−2d + 1 ≤ −d
1 ≤ d.
Setting d = 1 and allowing c to be as large as necessary to make the base
case true, completes the proof that T (n) ≤ c2n − 1, so T (n) = O (2n ).
4-1 Give asymptotic upper and lower bounds for T (n) in each of the following
recurrences. Assume T (n) is constant for n ≤ 2. Make your bounds as tight as
possible, and justify your answers.
a. T (n) = 2T (n/2) + n4 : T (n) = Θ n4 : in terms of the Master Theorem,
4
logb a+3
a = 2, b = 2, so logb a = log2 2 =
; this is looking like
1. f (n) = n = n
n
case 3. We need to check af b ≤ cf (n) for some c < 1 and n large enough.
4
In fact, af nb = 2 n2 = 18 n4 = 81 f (n), and case 3 applies.
Conclude T (n) = Θ n4 .
b. T (n) = T (7n/10) + n: T (n) = Θ (n) : in terms of the Master Theorem,
logb a+1
10
a = 1, b = 10
; this is looking like
7 , so logb a = log 7 1 = 0. f (n) = n = n
n
case 3. We need to check af b ≤ cf (n) for some c < 1 and n large enough.
1
7
7
In fact, af nb = 1 7n
= 10
n = 10
f (n), and case 3 applies.
10
1
Conclude T (n) = Θ (n).
c. T (n) = 16T (n/4) + n2 : T (n) = Θ n2 lg n : in terms of the Master
Theorem, a = 16, b = 4, so logb a = log4 16 = 2. f (n) = n2 = nlogb a ; this is
case 2.
Conclude T (n) = Θ n2 lg n .
d. T (n) = 7T (n/3) + n2 : T (n) = Θ n2 lg n : in terms of the Master
Theorem, a = 7, b = 3, so 1 < logb a = log3 7 < 2. f (n) = n2 = nlogb a+ε for
some ε > 0; this is looking like case 3. We need to check af nb ≤ cf (n) for
2
some c < 1 and n large enough. In fact, af nb = 7 n3 = 79 n2 = 79 f (n), and
case 3 applies.
Conclude T (n) = Θ n2 lg n .
e. T (n) = 7T (n/2)+n2 : T (n) = Θ nlog2 7 : in terms of the Master Theorem,
2
logb a−ε
a = 7, b = 2, so 2 < logb a = log
; this is case 1.
2 7 < 3. f (n) = n = n
log2 7
Conclude T (n) = Θ n
.
√
√
f. T (n) = 2T (n/4) + n: T (n) = Θ ( n lg n): in terms of the Master
1
Theorem, a = 2, b = 4, so logb a = log4 2 = 12 . f (n) = n 2 = nlogb a ; this is case
2.
√
Conclude T (n) = Θ ( n lg n).
g. T (n) = T (n−2)+n2 : T (n) = Θ n3 : the Master Theorem doesn't apply.
A recursion tree or 'unwrapping' the recurrence relation allows one to conclude
2
2
T (n) = T (n − 2k) + n2 + (n − 2) + . . . (n − 2 (k − 1)) .
n
For odd n, set k = b 2 c to get
Pb n2 c−1
Pn−1
Pn
2
2
T (n) = T (1)+ i=0
(n − 2i) ≤ T (1)+ i=0 (n − i) = T (1)+ i=1 n2 =
. This last expression is Θ n3 , allowing us to conclude that
T (1) + n(n+1)(2n+1)
6
T (n) = O n3 .
For a lower bound,n
Pb
c−1
2
T (n) = T (1) + i=0
(n − 2i)2
2
2
2
1 2
= T (1) + n + (n − 2) + (n − 4)2 + . . . 32 (n
− 4) + . . . 2 3
= T (1) +
1
2
2
Pn−2
i=0
(n − 2i)2 ≥ T (1) +
1
2
2
Pb n2 c−1
i=0
(n − 2i)2
= T (1)+ 21 n2 + 21 n2 + 12 (n − 2)2 + 21 (n − 2)2 + 12 (n − 4)2 + 21 (n − 4)2 +. . . 12 3+ 21 3.
≥ T (1) + 12 n2 + 12 (n − 1)2 + 21 (n − 2)2 + 12 (n − 3)2 + 12 (n − 4)2 + 12 (n − 5)2 +
1
. . . 2 3 + 21 2
Pn
≥ T (1) + 21 i=2 n2 = T (1) + n(n+1)(2n+1)
− 12 .
12
This last expression is Θ n3 , allowing us to conclude that T (n) = Ω n3 .
A similar argument, using T (0) as the base, gives the same asymptotic
bounds for n even.
Conclude T (n) = Θ n3 .
2