Quiz
CS3381 Design and Analysis of Algorithms
17 Oct, 2001
You have 45 minutes to finish this quiz.
Do not spend too much time on any problem.
Write your answers neatly.
Student ID: ___________________________
Section A. Fill in the blanks.
6 marks
Optimal Substructure
State two key ingredients of dynamic programming: ________________________________ and
Overlapping Subproblems
__________________________________.
6 marks
Optimal Substructure
State two key ingredients of greedy algorithms: ________________________________ and
Greedy-Choice Property
__________________________________.
6 marks
Divide
Combine
Conquer
State the steps in the divide-and-conquer paradigm: ________________________________________________
14 marks
Section B. True or False
1. Memoization techniques store solutions of sub-problems in tables.
2. Memoization is a bottom-up approach.
True
False
3. Memoization solves problems recursively.
True
4. In some cases when some subproblems do not overlap with one another, memoization is faster than basic
dynamic programming approaches.
False
5. Usually, if a problem can be solved by greedy strategy, then it can be solved by dynamic programming. True
6. Greedy algorithms are usually simpler but slower than basic dynamic programming solutions.
7. Greedy algorithms solve problems in a top-down manner.
Section C. Short questions
1. Give asymptotic tight bound for T(n) = 8T(n/2) + n2. Show your work.
14 marks
T(n) = a T(n/b) + f(n)
a = 8,
b=2
f(n) = n2
nlogba = nlog28 = n3
f(n) = n2 = nlogba-e, where e=1 > 0
=> case 1 => T(n) = (n3)
False
True
Quiz
CS3381 Design and Analysis of Algorithms
17 Oct, 2001
2
2
To show that there exists positive c and no s.t. 0<=(n+1)2 <= c * n2 for all n > n0.
Take c as 2, n0 as 10, n >=n0
To prove (n+1)2>=0: (n+1)2 = n2+2n+1 >= 0 (proved)
To prove (n+1)2 <= cn2 :
we first show that (n+1)2 /n2 <= cn2 /n2,
ie. 1+2/n+1/n2 <= c.
Since 1+2/n+1/n2 <= 2 is true for all n>=10,
We have 1+2/n+1/n2 <= c.
ie. (n+1)2 /n2 <= cn2 /n2
=> (n+1)2 <= cn2. (because n2 is positive)
Since we can find +ve c and n0 s.t. 0<=(n+1)2 <= c * n2 for all n > n0. Therefore
(n+1)2 = O(n2)
2.
Prove that (n+1) = O(n )
3.
2
2
)
a.) Given an array A of n integers:So
a1,a(n+1)
theO(n
algorithm
Longest(A) is designed to determine the length
2,..an, =
of the longest non-decreasing consecutive subsequence of integers starting with a1 (LN1 of A).
For example, one such subsequence (LN1 of A) for A=<34,57,53,54,78>, is <34,57>. Hence
Longest(A) should return 2.
Prove the correctness of this algorithm by using a loop invariant.
12 marks
Longest(A)
1 j=1
2 while j<n and A[j+1]>=A[j]
3
j=j+1
4 return j
Loop Invariant: At the start of each iteration of the while loop, a1, a2, a3, .. aj are non-decreasing consecutive subsequence of A starting with a1.
18 marks
Initialization: Before the first iteration, j=1, then there is only one integer a1, in a1, a2, a3, .. aj, which is non-decreasing consecutive subsequence.
Maintenance: In each iteration, the cycle is executed only if aj+1>=aj, then j is incremented by one. So aj-1, aj is non-decreasing. Hence, if the loop
invariant is true before an iteration, it remains true before next iteration.
Termination: The while loop continues whenever aj+1>=aj and j<n. It ends with j=n, or aj+1<aj. In former case a1, a2, a3, .. aj, is the longest possible
LN1 of A. In latter case, aj+1 is not involved in LN1 of A. Then a1, a2, a3, .. aj is LN1 of A.
b.) What are the worst and best cases of the algorithm? Describe the running time of the algorithm in
terms of asymptotic tight, upper, and lower bounds.
12 marks
Best case: LN1 of A only has one element. (1)
Worse case: LN1 of A only has n element. (n)
Algorithm: O(n), (1)
12 marks
4.
A training center offers a series of courses in various levels of difficulties. A student ‘graduates’ from a
program by finishing a program-compulsory course, eg. CS0005. But before he takes CS0005, he needs to
finish either CS0022 or CS0017 first. CS0022 is elementary and has no prerequisite. However CS0017
would need the student to take either CS0009 or CS0020 first.
Let S=s1,s2,s3,..sn be the courses provided.
And k1,k2,..kn are the durations of the courses in number of weeks.
Each course si may have two prerequisites: sp(i) and sq(i) or otherwise it has no prerequisites: p(i)=q(i)=0.
Formulate a recursive function to compute the shortest time that a new student can graduate from a
program. (Assume that the courses can be started anytime whenever there is a request, and each student
cannot take more than one course at the same time.)
Let d(si) be the shortest time that a new student can finish a course:
d(si) =
ki
if p(i)=q(i)=0
min (ki+d(sp(i)), ki+d(sq(i))) otherwise