1. Week 12 - Monday

2.  What did we talk about last time?  Defining classes  Class practice  Lab 11

7.  We can extend the idea of moving balls further  Let’s make some balls hunters and some balls prey  Hunters search for the nearest prey  If they reach a particular prey, they eat it  Prey balls always head away from the nearest hunter

8.  The Prey class has:  x Location: double  y Location: double  Aliveness: boolean  Speed: double  It also has accessors for x, y, and aliveness and a method we can use to kill the prey  Finally, it has an update method where it decides what it will do next

9.  The Hunter class has:  x Location: double  y Location: double  Speed: double  It also has accessors for x and y  Finally, it also has an update method where it decides what it will do next

11.  Members are the data inside objects  But, sometimes, wouldn’t it be nice if some data were linked to the class as a whole, rather than just the object?  What if you wanted to keep track of the total number of a particular kind of object you create?

12.  Static members are stored with the class, not with the object public class Item { private static int count = 0;// one copy total private String name;// one copy per object public Item( String s ) { name = s; count++;// updates global counter } public String getName() { return name; } public static int getItemsInUniverse() { return count; } public class Item { private static int count = 0;// one copy total private String name;// one copy per object public Item( String s ) { name = s; count++;// updates global counter } public String getName() { return name; } public static int getItemsInUniverse() { return count; }

13.  Static members are also called class variables  Static members can be accessed by either static methods or regular methods (unlike normal members which cannot be accessed by static methods)  Static members can be either public or private  In general, static variables should not be used

15.  Sometimes a value will not change after an object has been created:  Example: A ball has a single color after it is created  You can enforce the fact that the value will not change with the final keyword  A member declared final can only be assigned a value once  Afterwards, it will never change

16.  Using final, we can fix the value of a member  It will only be set once, usually when the constructor is called  It is customary to use all caps for constant names public class Animal { private String name; private final boolean MAMMAL; // never changes public Animal( String s, boolean mammal ) { name = s; MALE = mammal; } public void evolve() { MAMMAL = !MAMMAL;// compile-time error! } public class Animal { private String name; private final boolean MAMMAL; // never changes public Animal( String s, boolean mammal ) { name = s; MALE = mammal; } public void evolve() { MAMMAL = !MAMMAL;// compile-time error! }

17.  It is possible to set a static member to be constant using the final keyword  Usually, this is used for global constants that will never ever change  Making these values public is reasonable  Since they never change, we never have to worry about a user corrupting the data inside of the object

18.  The number of sides of a pentagon is always 5  Other code can access this information by using the value Pentagon.SIDES  Exactly like Math.PI or Math.E public class Pentagon { private double x; private double y; public static final int SIDES = 5; // never changes public Pentagon( double newX, double newY ) { x = newX; y = newY; } public double getX() { return x; } public double getY() { return y; } } public class Pentagon { private double x; private double y; public static final int SIDES = 5; // never changes public Pentagon( double newX, double newY ) { x = newX; y = newY; } public double getX() { return x; } public double getY() { return y; } }

20.  We want to compare the running time of one program to another  We want a mathematical description with the following characteristics:  Worst case We care mostly about how bad things could be  Asymptotic We focus on the behavior as the input size gets larger and larger

21.  First, let’s compute an expression that gives us running time for a program  Assume that any single line of code that doesn’t call a method or execute a loop takes 1 unit of time  Then, we sum up the number of lines that are executed in a method or in a loop

22.  Add up the operations done by the following code:  Initialization: 1 operation  Loop: 1 initialization + n checks + n increments + n additions to sum = 3n + 1  Output: 1 operation  Total: 3n + 3 int sum = 0; for( int i = 0; i < n; i++ ) sum += i; System.out.println("Sum: " + sum); int sum = 0; for( int i = 0; i < n; i++ ) sum += i; System.out.println("Sum: " + sum);

23.  We could express the time taken by the code on the previous slide as a function of n: f(n) = 3n + 3  This approach has a number of problems:  We assumed that each line takes 1 time unit to accomplish, but the output line takes much longer than an integer operation  This program is 4 lines long, a longer program is going to have a very messy running time function  We can get nit picky: Does sum += i; take one operation or two if we count the addition and the store separately?

24.  In short, this way of getting a running time function is almost useless because:  It cannot be used to give us an idea of how long the program really runs in seconds  It is complex and unwieldy  The most important thing about the analysis of the code that we did is learning that the growth of the function should be linear  A general description of how the running time grows as the input grows would be useful

25.  Enter Big Oh notation  Big Oh simplifies a complicated running time function into a simple statement about its worst case growth rate  All constant coefficients are ignored  All low order terms are ignored  3n + 3 is O(n)  Big Oh is a statement that a particular running time is no worse than some function, with appropriate constants

26.  147n 3 + 2n 2 + 5n + 12083 is O(n3)O(n3)  n 1000 + 2 n is O(2 n )  15n 2 + 6n + 7log n + 145 is O(n2)O(n2)  659n + nlog n + 87829 is O(n log n)  Note: In CS, we use log 2 unless stated otherwise

28.  More on Big Oh notation  Searching  Sorting

29.  Keep working on Project 4  Due this Friday