Question
Transcribed Text
Q1 Order the following functions by asymptotic growth rate.
Anlogn + 2n 210 2logn
3n + 100logn 4n 27
n2+10n n° nlogn
Q2. Consider the following code fragment, taken from some package :
public class Maryland extends State {
Mayland() { /* null constructor */
}
public void printMe0) ( System.out.println("Read. } public static void main(String[]
largs) {
Region mid = new State(); State md = new Maryland(); Object obj = new Place(); Place
usa = new Region(); md.printMc();
mid.printMc();
((Place) obj).printMe();
obj=md;
((Maryland) obj).printMe();obj = usa;
((Place) obj).printMe();
usa=md;
((Place) usa).printMe();
11
class State extends Region {
State() { /* null constructor */ )
public void printMe() ( System.out.println("Ship it.");)
)
class Region extends Place {
Region() { /* null constructor */}
public void printMe0) ( System.out.println("Box it."); )
}
class Place extends Object {
Place() { null constructor */ }
public void printMc() { System.out.println("Buy it.");}
}
What is the output from calling the main() method of the Maryland class?
Solution Preview
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The order of growth for each function is given by the dominant term (the term of the highest degree). In order to be easier for ordering the functions by their asymptotic growth rate it is necessary to process some of them.2^logn = n =O(n) (according the basic properties of Math operations).
2^10=1,024 therefore O(1) (since it is a contant)
4nlogn +2n is O(nlogn)
3n+100logn is O(n)...
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