Homework 1 Solution

Problem 2-1, Page 39
a. Insertion sort runs in θ(n2) worst case time on an input of size n, so on each sublist of k
length it runs in θ(k2) time. We have n/k sublists, so n/k θ(k2) = θ(nk) time overall.
b. At each level we have log(n/k) merges and each merge takes θ(n) time, so merging n/k
sublists takes θ(nlog(n/k)) worst-case time.
c. Total time θ(nk + nlg(n/k)) = θ(nlg). Both nk and nlg(n/k) must be no more than
O(nlgn). Therefor the largest value of k is (lgn).
d. We should pick a value of k where insertion sort is faster than merge sort. It would be
best if k is the smallest size.

Exercise 4.1-5, Page 75
Maximum-Subarray(A,j)
left = 0
right = 0
temp_max = A[0]
total_max = 0
for i = 1 to j do
temp_max = MAX(A[i], temp_max + A[i])
if temp_max > total_max
total_max= temp_max
right = i
if temp_max == A[i]
left = i
return (left, right, tatol_max)

Exercise 4.4-9, Page 93
𝑛
Assume 1- α > α. The tree is full for 𝑙𝑜𝑔1/(1−α)
levels, each contributing cn, we guess
𝑛
𝑛
Ω(n𝑙𝑜𝑔1/(1−α)) = Ω(nlogn). It has 𝑙𝑜𝑔1/α levels, each contributing ≤ cn, so we guess
𝑛
O(𝑛𝑙𝑜𝑔1/α
) = O(nlogn). We show T(n)=θ(nlogn) by substitution.
(let 1-α=β)
T(n) ≤ dnlogn
T(n)=T(αn) + T(βn) + cn
≤dαnlg(αn)+cβnlg(βn)+cn
≤dαnlgn+dβnlgn+dαnlgα+dβnlgβ+cn
≤dnlgn+(d(αlgα+βlgβ)+c)n
≤dnlgn
it is true for the value of d where (d(αlgα+βlgβ)+c≤0), therefor:
d ≤
d ≥
−c
αlgα + (1 − α)lg(1 − α)
c
−αlgα − (1 − α)lg(1 − α)
In order to show the lower bound we should do the same manipulation to show that
T(n) ≥ dnlogn. Therefor T(n) = θ(nlogn).

Problem 4-1, Page 107
a. T(n) = 2T(n/2) + n4
a=2, b=2, f(n) = n4
nlog22= n
Case 3 of master’s method
F(n) = n4 = Ω(nlog22 + 2)
a/bk = 2/24 = 1/8 < 1
Θ(n4)
b. T(n) = T(7n/10) + n
a=1, b=10/7, f(n) = n
n^(log10/71)= 1
Case 3 of the master’s method
F(n) = n = Ω(n^(log10/71+1)
a/bk = 1/(10/7)1
= 7/10 < 1
Θ(n)
c. T(n) = 16T(n/4) + n2
a=16, b=4, f(n) = n2
n^(log416) = n2= f(n) = n2
Case 2 of the master method where k = 0
T(n) = Θ(n^(log416) log 0+1 n)
= Θ(n2 log 1 n) = Θ(n2 log n)
d. F. T(n) = 2T(n/4) + sqrt(n)
a=2, b=4, f(n) = sqrt(n)
n^(logb a) = n^(log4 2) = n1/2 = sqrt(n)
Case 2 of the master method where k =0
T(n) = Θ(n^(log42) log 0+1 n)
= Θ(n1/2 log 1 n) = Θ(sqrt(n) log n)

Problem 5
Missing_item(A)
n=length(A)
item = (n*(n+1))/2
for i=1 to n do
item = item – A[i]
return item