CSC 205 – Java Programming II Lecture 22 March 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 add remove size 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 cd2null cassette1null 0 1 musics: Linked Collection MyMusics aTitle 1 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 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(log 2 n)  O(n)  O(n log 2 n)  O(n 2 )  …  O(2 n ) When n = 1,000,000 – log 2 n ~ 20 – n = 1,000,000 –n log 2 n ~ 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+log 2 n) = 6+2log 2 n+3n+n log 2 n –Find the highest order term n log 2 n  O(n log 2 n) –Use definition f(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) = K –Substitute C(n +7) = Cn +7C = C –Proved f(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 Processes puzzle(9) LV: n=9 RA: return puzzle(3)+1 puzzle(3) LV: n=3 RA: return puzzle(2)+1 puzzle(2) LV: n=2 RA: return puzzle(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