算法设计与分析作业答案

CS330:Introduction To Algorithm Spring2014

Lecture2:Asymptotic Analysis

TF:Xianrui Meng Scribes:Raman Bahdanouski,Ellie Vial

Exercise1

Suppose you have algorithms with thefive running times listed below.How much slower do each of these algorithms get when you(a)double the input size,or(b)increase the input size by one?

1.n2;

2.n3;

3.100n2;

4.nlog(n);

5.2n;

Solution

For doubling the input:

1.f(n)=n2;f (2n)=4n

2.;So the running time of the algorithm gets4times slower.

2.f(n)=n3;f (2n)=8n

3.;So the running time of the algorithm gets8times slower.

3.f(n)=100n2;f (2n)=400n2.;So the running time of the algorithm gets4times slower.

4.f(n)=nlog(n);f (2n)=2n∗log(2n)=2n∗(log(n)+log(2))=2n∗log(n)+2n∗log(2).;So the running

time of the algorithm increases by2n∗log(n)+2n∗log(2).

5.f(n)=2n;f (2n)=22n;f(2n)−f(n)=22n−2n=(2n)∗(2n−1);So the running time of the algorithm

gets slower by the factor of(2n−1).

For increasing the input by one:

1.f(n)=n2;f (n+1)=n2+2n+1.;So the running time of the algorithm increases by2n+1.

2.f(n)=n3;f (n+1)=n3+3n2+3n+1.;So the running time of the algorithm increases by3n2+3n+1.

3.f(n)=100n2;f (n+1)=100(n2+2n+1).;So the running time of the algorithm increases by

100(2n+1).

4.f(n)=nlog(n);f(n+1)=(n+1)∗log(n+1);f(n+1)−f(n)=(n+1)∗log(n+1)−nlog(n)=

n∗log(n+1)+log(n+1)−nlog(n)=log(n+1)+n∗(log(n+1)−log(n));So the running time increases by log(n+1)+n∗(log(n+1)−log(n)).

5.f(n)=2n;f (n+1)=2n+1=2∗2n;So the running time of the algorithm doubles.

2-1

2-2Lecture 2:Asymptotic Analysis Exercise 3

Take the following list of functions and arrange them in ascending order of growth rate.That is,if function g(n)immediately follows function f(n)in your list,it should be the case that f(n)is O(g(n)).

•f 1(n )=n 2.5

•f 2(n )=√2n

•f 3(n )=n +10

•f 4(n )=10n

•f 5(n )=100n

•f 6(n )=n 2log (n )

Solution

We can start approaching this problem by putting f 4and f 5at end of the list,because these functions are exponential and will grow the fastest.f 4