1. CSC 205 – Java Programming II Lecture 22 March 1, 2002

2. 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

3. 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

4. Why Do We Need Abstraction

5. 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);

6. Internal View YESNO MAYBE head newEntry null Add a new entry newEntry newEntry.next = head; head = newEntry;

7. 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()); }

8. 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

9. 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.

10. 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

11. 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 ); }

12. 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

13. Towers of Hanoi

14. 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 }

15. Tracing Recursive Processes