CSC 205 – Java Programming II

Lecture 22March 1, 2002

User’s v.s. Developer’s Views

• User’s view– A better name could be client or external view– A user or client of an object o shouldn’t be

bothered with the implementation details of o • How does o store its state?

• How does o perform an action?

• Developer’s view– A better name could be internal view

Client of Server Roles

Program that uses collection

addremovesize

contains

Client object(e.g. a MyMusic object)

Server object(e.g. the LinkedCollection object)

Entities

Why Do We Need Abstraction

Using Collection Objects

cd1 cd2 null

cassette1 null

0

1

musics:Linked

Collection

MyMusics

aTitle1

cassette2

musics[0]:LinkedCollection

musics[1]:LinkedCollection

m:Music

Add a Music object m musics[m.getType()].add(m);

Remove a Music object m musics[m.getType()].remove(m);

Internal View

YESNO

MAYBE

head

newEntry

null

null

Add a new entry newEntry newEntry.next = head; head = newEntry;

The equals Method

prublic boolean equals (Object o)

{

if ( !(o instance of Date) )

return false;

else

Date d = (Date) o;

return (year==d.getYear() &&

month==d.getMonth() &&

day==d.getDay());

}

Big O Notations

• A rough estimation of how fast a function f(n) increase with the growth of n

• The hierarchy of orders O(1)O(log2n)O(n)O(n log2n)O(n 2 )… O(2 n )

• When n = 1,000,000– log2n ~ 20– n = 1,000,000– n log2n ~ 20,000,000– n 2 = 1,000,000,000,000

The Smallest Upper Bound

• Question 4(a)– Expand the expression

f(n) = (2+n)(3+log2n) = 6+2log2n+3n+n log2n– Find the highest order term

n log2n O(n log2n)– Use definition

f(n) <= C n log2n, when n >= K In this case K = 8, C = 4.

Show Equivalency

• Question 4(b)– Use definition f(n) is O(n+7) means f(n) <= C(n +7), when n >= K– Substitute C(n +7) = Cn +7C <= (C+7) n, when n >= C– Proved f(n) <= C’n , when n >= K’ f(n) is O(n) where C’ = C+7, K’ = C

Recursion

private static int puzzle (int n)

{

if ( (n % 3) == 2 )

return 1; //base case

else if ( (n % 3) == 1 )

return ( puzzle (n + 1) + 2 );

else

return ( puzzle (n / 3) + 1 );

}

Tracing Recursive Processespuzzle(9)LV: n=9RA: returnpuzzle(3)+1

puzzle(3)LV: n=3RA: returnpuzzle(2)+1

puzzle(2)LV: n=2RA: returnpuzzle(1)+1

puzzle(1)

return 1

Push execution state in stack: Local Variables Return Address

Pop execution state from stack: Local Variables Return Address

Towers of Hanoi

Algorithm

move (n, orig, dest, temp) { if (n == 1) Move disk from orig to dest; else { move (n - 1, orig, temp, dest); move (1, orig, dest, temp); move (n - 1, temp, dest, orig) ; } // else}

Tracing Recursive Processes