1. 1 Iteration Chapter 4 pt. 2 Trey Kirk

2. 2 Java looping  Options while do-while for  Allow programs to control how many times a statement list is executed

3. 3 Averaging values

4. 4 Averaging  Problem Extract a list of positive numbers from standard input and produce their average  Numbers are one per line  A negative number acts as a sentinel to indicate that there are no more numbers to process  Observations Cannot supply sufficient code using just assignments and conditional constructs to solve the problem  Don’t how big of a list to process Need ability to repeat code as needed

5. 5 Averaging  Algorithm Prepare for processing Get first input While there is an input to process do {  Process current input  Get the next input } Perform final processing

6. 6 Averaging  Problem Extract a list of positive numbers from standard input and produce their average  Numbers are one per line  A negative number acts as a sentinel to indicate that there are no more numbers to process  Sample run Enter positive numbers one per line. Indicate end of list with a negative number. 4.5 0.5 1.3 Average 2.1

7. public class NumberAverage { // main(): application entry point public static void main(String[] args) { // set up the input // prompt user for values // get first value // process values one-by-one while (value >= 0) { // add value to running total // processed another value // prepare next iteration - get next value } // display result if (valuesProcessed > 0) // compute and display average else // indicate no average to display }

8. int valuesProcessed = 0; double valueSum = 0; // set up the input Scanner stdin = new Scanner (System.in); // prompt user for values System.out.println("Enter positive numbers 1 per line.\n" + "Indicate end of the list with a negative number."); // get first value double value = stdin.nextDouble(); // process values one-by-one while (value >= 0) { valueSum += value; ++valuesProcessed; value = stdin.nextDouble(); } // display result if (valuesProcessed > 0) { double average = valueSum / valuesProcessed; System.out.println("Average: " + average); } else { System.out.println("No list to average"); }

9. 9 Program Demo NumberAverage.java NumberAverage.java

10. 10 While syntax and semantics Logical expression that determines whether Action is to be executed while ( Expression ) Action Action is either a single statement or a statement list within braces

11. 11 While semantics for averaging problem // process values one-by-one while ( value >= 0 ) { // add value to running total valueSum += value; // we processed another value ++valueProcessed; // prepare to iterate – get the next input value = stdin.nextDouble(); } Test expression is evaluated at the start of each iteration of the loop. If test expression is true, these statements are executed. Afterward, the test expression is reevaluated and the process repeats

12. 12 While Semantics Expression Action true false Expression is evaluated at the start of each iteration of the loop If Expression is true, Action is executed If Expression is false, program execution continues with next statement

13. 13 int valuesProcessed = 0; double valueSum = 0; double value = stdin.nextDouble(); while (value >= 0) { valueSum += value; ++valuesProcessed; value = stdin.nextDouble(); } if (valuesProcessed > 0) { double average = valueSum / valuesProcessed; System.out.println("Average: " + average); } else { System.out.println("No list to average"); } int valuesProcessed = 0; double valueSum = 0; double value = stdin.nextDouble(); while (value >= 0) { valueSum += value; ++valuesProcessed; value = stdin.nextDouble(); if (valuesProcessed > 0) { double average = valueSum / valuesProcessed; System.out.println("Average: " + average); Execution Trace Suppose input contains: 4.5 0.5 1.3 -1 0 valuesProcessed valueSum 0 value 4.5 Suppose input contains: 4.5 0.5 1.3 -1 4.5 1 Suppose input contains: 4.5 0.5 1.3 -1 0.5 5.0 2 1.3 6.3 Suppose input contains: 4.5 0.5 1.3 -1 3 Suppose input contains: 4.5 0.5 1.3 -1 average 2.1

14. 14 What do these pictures mean? Light beer Light beer Dandy lions Dandy lions Assaulted peanut Assaulted peanut Eggplant Eggplant Dr. Pepper Dr. Pepper Pool table Pool table Tap dancers Tap dancers Card shark Card shark King of pop King of pop I Pod I Pod Gator aide Gator aide Knight mare Knight mare Hole milk Hole milk

15. 15 Converting text to lower case

16. 16 Converting text to strictly lowercase public static void main(String[] args) { Scanner stdin = new Scanner (System.in); System.out.println("Enter input to be converted:"); String converted = ""; while (stdin.hasNext()) { String currentLine = stdin.nextLine(); String currentConversion = currentLine.toLowerCase(); converted += (currentConversion + "\n"); } System.out.println("\nConversion is:\n" + converted); }

17. 17 Sample run A Ctrl+z was entered. It is the Windows escape sequence for indicating end-of-file An empty line was entered

18. 18 Program Demo LowerCaseDisplay.java LowerCaseDisplay.java

19. 19 Program trace public static void main(String[] args) { Scanner stdin = new Scanner (System.in); System.out.println("Enter input to be converted:"); String converted = ""; while (stdin.hasNext()) { String currentLine = stdin.nextLine(); String currentConversion = currentLine.toLowerCase(); converted += (currentConversion + "\n"); } System.out.println("\nConversion is:\n" + converted); } public static void main(String[] args) { Scanner stdin = new Scanner (System.in); System.out.println("Enter input to be converted:"); String converted = ""; while (stdin.hasNext()) { String currentLine = stdin.nextLine(); String currentConversion = currentLine.toLowerCase(); converted += (currentConversion + "\n"); } System.out.println("\nConversion is:\n" + converted); }

20. 20 Program trace Representation of lower case conversion of current input line converted += (currentConversion + "\n"); The append assignment operator updates the representation of converted to include the current input line Newline character is needed because method nextLine() "strips" them from the input

21. 21 Another optical illusion

22. 22 Loop Design & Reading From a File

23. 23 Loop design  Questions to consider in loop design and analysis What initialization is necessary for the loop’s test expression? What initialization is necessary for the loop’s processing? What causes the loop to terminate? What actions should the loop perform? What actions are necessary to prepare for the next iteration of the loop? What conditions are true and what conditions are false when the loop is terminated? When the loop completes what actions are need to prepare for subsequent program processing?

24. 24 Reading a file  Background Same Scanner class! Scanner fileIn = new Scanner (new File (filename) ); The File class allows access to files It’s in the java.io package filename is a String

25. 25 Reading a file  Class File Allows access to files (etc.) on a hard drive  Constructor File (String s) Opens the file with name s so that values can be extracted Name can be either an absolute pathname or a pathname relative to the current working folder

26. 26 Reading a file Scanner stdin = new Scanner (System.in); System.out.print("Filename: "); String filename = stdin.nextLine(); Scanner fileIn = new Scanner (new File (filename)); String currentLine = fileIn.nextLine(); while (currentLine != null) { System.out.println(currentLine); currentLine = fileIn.nextLine(); } Scanner stdin = new Scanner (System.in); System.out.print("Filename: "); String filename = stdin.nextLine(); Scanner fileIn = new Scanner (new File (filename)); String currentLine = fileIn.nextLine(); while (currentLine != null) { System.out.println(currentLine); currentLine = fileIn.nextLine(); } Set up standard input streamDetermine file nameSet up file streamProcess lines one by oneGet first lineMake sure got a line to processDisplay current lineGet next lineMake sure got a line to process If not, loop is done Close the file stream

27. 27 Today’s demotivators

28. 28 End of lecture on 23 February 2007

29. 29 The For statement

30. 30 The For Statement currentTerm = 1; for ( int i = 0; i < 5; ++i ) { System.out.println(currentTerm); currentTerm *= 2; } After each iteration of the body of the loop, the update expression is reevaluated The body of the loop iterates while the test expression is true int Initialization step is performed only once -- just prior to the first evaluation of the test expression The body of the loop displays the current term in the number series. It then determines what is to be the new current number in the series

31. ForExpr Action truefalse ForInit ForUpdate Evaluated once at the beginning of the for statements's execution The ForExpr is evaluated at the start of each iteration of the loop If ForExpr is true, Action is executed After the Action has completed, the PostExpression is evaluated If ForExpr is false, program execution continues with next statement After evaluating the PostExpression, the next iteration of the loop starts

32. 32 for statement syntax Logical test expression that determines whether the action and update step are executed for ( ForInit ; ForExpression ; ForUpdate ) Action Update step is performed after the execution of the loop body Initialization step prepares for the first evaluation of the test expression The body of the loop iterates whenever the test expression evaluates to true

33. 33 for vs. while  A for statement is almost like a while statement for ( ForInit; ForExpression; ForUpdate ) Action is ALMOST the same as: ForInit; while ( ForExpression ) { Action; ForUpdate; }  This is not an absolute equivalence! We’ll see when they are different in a bit

34. 34 Variable declaration  You can declare a variable in any block: while ( true ) { int n = 0; n++; System.out.println (n); } System.out.println (n); Variable n gets created (and initialized) each time Thus, println() always prints out 1 Variable n is not defined once while loop ends As n is not defined here, this causes an error

35. 35 Variable declaration  You can declare a variable in any block: if ( true ) { int n = 0; n++; System.out.println (n); } System.out.println (n); Only difference from last slide

36. 36 System.out.println("i is " + i); } System.out.println("all done"); System.out.println("i is " + i); } System.out.println("all done"); i is 0 i is 1 i is 2 all done Execution Trace i 0 int i = 0;i < 3;++ifor () {int i = 0;i < 3;++i 123 Variable i has gone out of scope – it is local to the loop

37. 37 for vs. while  An example when a for loop can be directly translated into a while loop: int count; for ( count = 0; count < 10; count++ ) { System.out.println (count); }  Translates to: int count; count = 0; while (count < 10) { System.out.println (count); count++; }

38. 38 for vs. while  An example when a for loop CANNOT be directly translated into a while loop: for ( int count = 0; count < 10; count++ ) { System.out.println (count); }  Would (mostly) translate as: int count = 0; while (count < 10) { System.out.println (count); count++; } count IS defined here count is NOT defined here only difference

39. 39 for loop indexing  Java (and C and C++) indexes everything from zero  Thus, a for loop like this: for ( int i = 0; i < 10; i++ ) {... }  Will perform the action with i being value 0 through 9, but not 10  To do a for loop from 1 to 10, it would look like this: for ( int i = 1; i <= 10; i++ ) {... }

40. 40 Nested loops int m = 2; int n = 3; for (int i = 0; i < n; ++i) { System.out.println("i is " + i); for (int j = 0; j < m; ++j) { System.out.println(" j is " + j); } i is 0 j is 0 j is 1 i is 1 j is 0 j is 1 i is 2 j is 0 j is 1

41. 41 Nested loops int m = 2; int n = 4; for (int i = 0; i < n; ++i) { System.out.println("i is " + i); for (int j = 0; j < i; ++j) { System.out.println(" j is " + j); } i is 0 i is 1 j is 0 i is 2 j is 0 j is 1 i is 3 j is 0 j is 1 j is 2

42. 42 How well do you understand for loops? 1. Very well! This stuff is easy! 2. Fairly well – with a little review, I’ll be good 3. Okay. It’s not great, but it’s not horrible, either 4. Not well. I’m kinda confused 5. Not at all. I’m soooooo lost

43. 43 From Dubai

44. 44 do-while loops

45. 45 The do-while statement  Syntax do Action while (Expression)  Semantics Execute Action If Expression is true then execute Action again Repeat this process until Expression evaluates to false  Action is either a single statement or a group of statements within braces Action true false Expression

46. 46 Picking off digits  Consider System.out.print("Enter a positive number: "); int number = stdin.nextInt(); do { int digit = number % 10; System.out.println(digit); number = number / 10; } while (number != 0);  Sample behavior Enter a positive number: 1129 9 2 1 1

47. 47 Guessing a number  This program will allow the user to guess the number the computer has “thought” of  Main code block: do { System.out.print ("Enter your guess: "); guessedNumber = stdin.nextInt(); count++; } while ( guessedNumber != theNumber );

48. 48 Program Demo GuessMyNumber.java GuessMyNumber.java

49. 49 while vs. do-while  If the condition is false: while will not execute the action do-while will execute it once while ( false ) { System.out.println (“foo”); } do { System.out.println (“foo”); } while ( false ); never executed executed once

50. 50 while vs. do-while  A do-while statement can be translated into a while statement as follows: do { Action; } while ( WhileExpression );  can be translated into: boolean flag = true; while ( WhileExpression || flag ) { flag = false; Action; }

51. 51 How well do you understand do-while loops? 1. Very well! This stuff is easy! 2. Fairly well – with a little review, I’ll be good 3. Okay. It’s not great, but it’s not horrible, either 4. Not well. I’m kinda confused 5. Not at all. I’m soooooo lost

52. 52 Today’s demotivators

53. 53 End of lecture on 26 February 2007

54. 54 Loop controls

55. 55 The continue keyword  The continue keyword will immediately start the next iteration of the loop The rest of the current loop is not executed  But the ForUpdate part is, if continue is in a for loop for ( int a = 0; a <= 10; a++ ) { if ( a % 2 == 0 ) { continue; } System.out.println (a + " is odd"); }  Output:1 is odd 3 is odd 5 is odd 7 is odd 9 is odd

56. 56 The break keyword  The break keyword will immediately stop the execution of the loop Execution resumes after the end of the loop for ( int a = 0; a <= 10; a++ ) { if ( a == 5 ) { break; } System.out.println (a + " is less than five"); }  Output:0 is less than five 1 is less than five 2 is less than five 3 is less than five 4 is less than five

57. 57 Four Hobos

58. 58 Four Hobos  An example of a program that uses nested for loops  Credited to Will Shortz, crossword puzzle editor of the New York Times And NPR’s Sunday Morning Edition puzzle person

59. 59 Problem  Four hobos want to split up 200 hours of work  The smart hobo suggests that they draw straws with numbers on it  If a straw has the number 3, then that hobo works for 3 hours on 3 days each (a total of 9 hours)  How many combinations are there to split up such work?  The smart hobo draws the shortest straw  Which combination did the smart hobo choose?

60. 60 Analysis  We are looking for integer solutions to the formula: a 2 +b 2 +c 2 +d 2 = 200 Where a is the number of hours & days the first hobo worked, b for the second hobo, etc.  We know the following: Each number must be at least 1 No number can be greater than 200 = 14 That order doesn’t matter  The combination (1,2,1,2) is the same as (2,1,2,1) Both combinations have two short and two long straws  We will implement this with nested for loops

61. 61 Implementation public class FourHobos { public static void main (String[] args) { for ( int a = 1; a <= 14; a++ ) { for ( int b = 1; b <= 14; b++ ) { for ( int c = 1; c <= 14; c++ ) { for ( int d = 1; d <= 14; d++ ) { if ( (a <= b) && (b <= c) && (c <= d) ) { if ( a*a+b*b+c*c+d*d == 200 ) { System.out.println ("(" + a + ", " + b + ", " + c + ", " + d + ")"); }

62. 62 Program Demo FourHobos.java FourHobos.java

63. 63 Results  The output: (2, 4, 6, 12) (6, 6, 8, 8)  Not surprisingly, the smart hobo picks the short straw of the first combination

64. 64 Today’s demotivators

65. 65 Alternate implementation  We are going to rewrite the old code in the inner most for loop: if ( (a <= b) && (b <= c) && (c <= d) ) { if ( a*a+b*b+c*c+d*d == 200 ) { System.out.println ("(" + a + ", " + b + ", " + c + ", " + d + ")"); } }  First, consider the negation of ( (a <= b) && (b <= c) && (c <= d) ) It’s ( !(a <= b) || !(b <= c) || !(c <= d) ) Or ( (a > b) || (b > c) || (c > d) )

66. 66 Alternate implementation  This is the new code for the inner-most for loop: if ( (a > b) || (b > c) || (c > d) ) { continue; } if ( a*a+b*b+c*c+d*d != 200 ) { continue; } System.out.println ("(" + a + ", " + b + ", " + c + ", " + d + ")");

67. 67 3 card poker

68. 68 3 Card Poker  This is the looping HW from a previous fall  The problem: count how many of each type of hand in a 3 card poker game  Standard deck of 52 cards (no jokers) Four suits: spades, clubs, diamonds, hearts 13 Faces: Ace, 2 through 10, Jack, Queen, King  Possible 3-card poker hands Pair: two of the cards have the same face value Flush: all the cards have the same suit Straight: the face values of the cards are in succession Three of a kind: all three cards have the same face value Straight flush: both a flush and a straight

69. 69 The Card class  A Card class was provided Represents a single card in the deck  Constructor: Card(int i) If i is in the inclusive interval 1... 52 then a card is configured in the following manner  If 1 <= i <= 13 then the card is a club  If 14 <= i <= 26 then the card is a diamond  If 27 <= i <= 39 then the card is a heart  If 40 <= i <= 52 then the card is a spade  If i % 13 is 1 then the card is an Ace;  If i % 13 is 2, then the card is a 2, and so on.

70. 70 Card class methods  String getFace() Returns the face of the card as a String  String getSuit() Returns the suit of the card as a String  int getValue() Returns the value of the card  boolean equals(Object c) Returns whether c is a card that has the same face and suit as the invoking card  String toString() Returns a text representation of the card. You may find this method useful during debugging.

71. 71 The Hand class  A Hand class was (partially) provided Represents the three cards the player is holding  Constuctor: Hand(Card c1, Card c2, Card c3) Takes those cards and puts them in sorted order

72. 72 Provided Hand methods  public Card getLow() Gets the low card in the hand  public Card getMiddle() Gets the middle card in the hand  public Card getHigh() Gets the high card in the hand  public String toString() We’ll see the use of the toString() method later  public boolean isValid() Returns if the hand is a valid hand (no two cards that are the same)  public boolean isNothing() Returns if the hand is not one of the “winning” hands described before

73. 73 Hand Methods to Implement  The assignment required the students to implement the other methods of the Hand class We haven’t seen this yet  The methods returned true if the Hand contained a “winning” combination of cards public boolean isPair() public boolean isThree() public boolean isStraight() public boolean isFlush() public boolean isStraightFlush()

74. 74 Class HandEvaluation  Required nested for loops to count the total number of each hand  Note that the code for this part may not appear on the website

75. 75 Program Demo HandEvaluation.java HandEvaluation.java

76. 76 How well do you understand 3-card poker? 1. Very well! This stuff is easy! 2. Fairly well – with a little review, I’ll be good 3. Okay. It’s not great, but it’s not horrible, either 4. Not well. I’m kinda confused 5. Not at all. I’m soooooo lost

77. 77 All your base are belong to us Flash animation Flash animation Flash animation Flash animation Reference: http://en.wikipedia.org/wiki/All_your_base_are_belong_to_us Reference: http://en.wikipedia.org/wiki/All_your_base_are_belong_to_us

78. 78 End of lecture on 28 Feb 2007  Class was optional on 2 March 2007  Then spring break…

79. 79 The Halting Problem

80. 80 What’s wrong with this program? public class LoopsForever { public static void main (String args[]) { while ( true ) { System.out.println (); } } }  Given a more complicated program, how do we tell if it gets stuck in an infinite loop? Such as when an application “hangs”?

81. 81

82. 82 The Halting problem  Given a Java program P, and input I Let P be a filename for a program file on a disk somewhere Let I be a filename for a file that contains all the input the program takes in  Will the program P with input I ever terminate? Meaning will program P with input I loop forever or halt?  Can a computer program determine this? Can a human?  First shown by Alan Turing in 1936 Before digital computers existed! (I’m ignoring which way he showed it for now)

83. 83 A few notes  To “solve” the halting problem means we have a method Oracle.CheckHalt (String P, String I) Let Oracle be a class that can give lots of (truthful) answers  Oracle.PredictFuture(), Oracle.GetNextLotteryNumbers(), etc. P is the (filename of the) program we are checking for halting I is the (filename of the) input to that program  And it will return “loops forever” or “halts” As a boolean: true means “loops forever”, false means “halts”  Note it must work for any (Java) program, not just some programs Or simple programs

84. 84 Take your best guess – do you think it’s possible to solve the halting problem? 1. Yes 2. No 3. I don’t understand what the halting problem is

85. 85 Can a human determine if a program halts?  Given a program of 10 lines or less, can a human determine if it halts? Assuming no tricks – the program is completely understandable And assuming the computer works properly, of course  And we ignore the fact that an int will max out at 4 billion As there are ways we can get around this…  For the sample programs on the next page: Assume that the code is in a proper main() method in a proper class Assume “…print” stands for “System.out.print”  Likewise for “…println”

86. 86 Halting problem examples: will they halt?  First sample program:...println (“Alan Turing”);...println (“was a genius”); System.exit();  Second sample program: for (int n = 0; n < 10; n++)...println (n); System.exit();  Third sample program while (true)...println (“hello world”); System.exit();  Fourth sample program: int x = 10; while ( x > 0 ) {...println (“hello world”); x = x + 1; } System.exit();

87. 87 Take your best guess – do you think it’s possible to solve the halting problem? 1. Yes 2. No 3. I don’t understand what the halting problem is

88. 88 Perfect numbers  Numbers whose divisors (not including the number) add up to the number 6 = 1 + 2 + 3 28 = 1 + 2 + 4 + 7 + 14  The list of the first 10 perfect numbers: 6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128, 2658455991569831744654692615953842176, 1915619426082361072947933780843036381309973215 48169216 The last one was 54 digits!  All known perfect numbers are even; it’s an open (i.e. unsolved) problem if odd perfect numbers exist  Sequence A000396 in OEIS Sequence A000396 in OEIS

89. 89 Odd perfect number search  Will this program ever halt? int n = 1; // arbitrary-precision integer while (true) { int sumOfFactors = 0; for ( int factor = 1; factor < n; factor++ ) if ( n % factor == 0 ) // factor is a factor of n sumOfFactors = sumOfFactors + factor; if (sumOfFactors == n) then break; n = n + 2; } System.out.exit();  Adapted from http://en.wikipedia.org/wiki/Halting_problem http://en.wikipedia.org/wiki/Halting_problem

90. 90 Take your best guess – do you think it’s possible to solve the halting problem? 1. Yes 2. No 3. I don’t understand what the halting problem is

91. 91 Where does that leave us?  If a human can’t figure out how to do the halting problem, we can’t make a computer do it for us  It turns out that it is impossible to write such a CheckHalt() method But how to prove this?

92. 92 CheckHalt()’s non-existence  Consider a program P with input I  Suppose that a method Oracle.CheckHalt(P,I) exists Tests if P(I) will either “loop forever” or “halt”  A program is a series of bits And thus can be considered data as well  Thus, we can call CheckHalt(P,P) It’s using the bytes of program P as the input to program P

93. 93 CheckHalt()’s non-existence  Consider a new program: public class Test { public static void main (String args[]) { if ( Oracle.CheckHalt(“Test.java”, “Test.java”) ) // if Test.java loops forever System.exit();// then halt else// else if Test.java halts while (true) { }// then loop forever } }  Do we agree that class Test is a valid program?

94. 94 CheckHalt()’s non-existence  A (somewhat condensed) version of class Test: public class Test { … main … (String args[]) { if ( Oracle.CheckHalt (“Test.java”, “Test.java”) ) System.exit(); else while (true) { } } }  Two possibilities:  Either class Test halts… Then CheckHalt(Test,Test) returns true (“loops forever”)… Which means that class Test loops forever Contradiction!  Or class Test loops forever… Then CheckHalt(Test,Test) returns false (“halts”)… Which means that class Test halts Contradiction!

95. 95 How well do you understand the halting problem? 1. Very well! This stuff is easy! 2. Fairly well – with a little review, I’ll be good 3. Okay. It’s not great, but it’s not horrible, either 4. Not well. I’m kinda confused 5. Not at all. I’m soooooo lost

96. 96 Why do we care about the halting problem?  It was the first algorithm that was shown to not be able to exist by a computer You can prove something exists by showing an example (a correct program) But it’s much harder to prove that a program can never exist  First shown by Alan Turing in 1936 Before digital computers existed!

97. 97 New 2005 demotivatiors!

98. 98 Not going over any more slides in this slide set

99. 99 Triangle counting

100. 100 The programming assignment  This was the looping HW from two springs ago  List all the possible triangles from (1,1,1) to (n,n,n) Where n is an inputted number In particular, list their triangle type  Types are: equilateral, isosceles, right, and scalene

101. 101 Sample execution Enter n: 5 (1,1,1) isosceles equilateral (1,2,2) isosceles (1,3,3) isosceles (1,4,4) isosceles (1,5,5) isosceles (2,2,2) isosceles equilateral (2,2,3) isosceles (2,3,3) isosceles (2,3,4) scalene (2,4,4) isosceles (2,4,5) scalene (2,5,5) isosceles (3,3,3) isosceles equilateral (3,3,4) isosceles (3,3,5) isosceles (3,4,4) isosceles (3,4,5) right scalene (3,5,5) isosceles (4,4,4) isosceles equilateral (4,4,5) isosceles (4,5,5) isosceles (5,5,5) isosceles equilateral

102. 102 Program Demo TriangleDemo.java TriangleDemo.java

103. 103 The Triangle class  That semester we went over classes by this homework So they had to finish the class We will be seeing class creation after spring break  Methods in the class: public Triangle() public Triangle (int x, int y, int z) public boolean isTriangle() public boolean isRight() public boolean isIsosceles() public boolean isScalene() public boolean isEquilateral() public String toString()

104. 104 The TriangleDemo class  Contained a main() method that tested all the triangles  Steps required: Check if the sides are in sorted order (i.e. x < y < z)  If not, then no output should be provided for that collection of side lengths Create a new Triangle object using the current side lengths Check if it is a valid triangle  If it is not, then no output should be provided for that collection of side lengths Otherwise, indicate which properties the triangle possesses  Some side length values will correspond to more than 1 triangle  e.g., (3, 3, 3) is both isosceles and equilateral  Thus, we can’t assume that once a property is present, the others are not.

105. 105 Look at that them there code… TriangleDemo.java TriangleDemo.java

106. 106 How well do you understand triangle counting? 1. Very well! This stuff is easy! 2. Fairly well – with a little review, I’ll be good 3. Okay. It’s not great, but it’s not horrible, either 4. Not well. I’m kinda confused 5. Not at all. I’m soooooo lost

107. 107 Fibonacci numbers

108. 108 Fibonacci sequence  Sequences can be neither geometric or arithmetic F n = F n-1 + F n-2, where the first two terms are 1  Alternative, F(n) = F(n-1) + F(n-2) Each term is the sum of the previous two terms Sequence: { 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … } This is the Fibonacci sequence Full formula:

109. 109 Fibonacci sequence in nature 13 8 5 3 2 1

110. 110 Reproducing rabbits  You have one pair of rabbits on an island The rabbits repeat the following:  Get pregnant one month  Give birth (to another pair) the next month This process repeats indefinitely (no deaths) Rabbits get pregnant the month they are born  How many rabbits are there after 10 months?

111. 111 Reproducing rabbits  First month: 1 pair The original pair  Second month: 1 pair The original (and now pregnant) pair  Third month: 2 pairs The child pair (which is pregnant) and the parent pair (recovering)  Fourth month: 3 pairs “Grandchildren”: Children from the baby pair (now pregnant) Child pair (recovering) Parent pair (pregnant)  Fifth month: 5 pairs Both the grandchildren and the parents reproduced 3 pairs are pregnant (child and the two new born rabbits)

112. 112 Reproducing rabbits  Sixth month: 8 pairs All 3 new rabbit pairs are pregnant, as well as those not pregnant in the last month (2)  Seventh month: 13 pairs All 5 new rabbit pairs are pregnant, as well as those not pregnant in the last month (3)  Eighth month: 21 pairs All 8 new rabbit pairs are pregnant, as well as those not pregnant in the last month (5)  Ninth month: 34 pairs All 13 new rabbit pairs are pregnant, as well as those not pregnant in the last month (8)  Tenth month: 55 pairs All 21 new rabbit pairs are pregnant, as well as those not pregnant in the last month (13)

113. 113 Reproducing rabbits  Note the sequence: { 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … }  The Fibonacci sequence again

114. 114 Fibonacci sequence  Another application:  Fibonacci references from http://en.wikipedia.org/wiki/Fibonacci_sequence

115. 115 Fibonacci sequence  As the terms increase, the ratio between successive terms approaches 1.618  This is called the “golden ratio” Ratio of human leg length to arm length Ratio of successive layers in a conch shell  Reference: http://en.wikipedia.org/wiki/Golden_ratio

116. 116 The Golden Ratio

117. 117

118. 118 Number counting

119. 119 The programming assignment  This was the looping HW from last fall  Get an integer i from the user  The homework had four parts Print all the Fibonacci numbers up to i Print all the powers of 2 up to i Print all the prime numbers up to i Time the previous three parts of the code

120. 120 Sample execution Input an integer i: 10 The 10th Fibonacci number is 55 Computation took 1 ms 2 3 5 7 11 13 17 19 23 29 The 10th prime is 29 Computation took 0 ms The 10th power of 2 is 1024 Computation took 6 ms 2 4 8 16 32 64 128 256 512 1024 BigInteger: The 10th power of 2 is 1024 Computation took 2 ms

121. 121 Background: Prime numbers  Remember that a prime number is a number that is ONLY divisible by itself and 1  Note that 1 is not a prime number! Thus, 2 is the first prime number  The first 10 prime numbers: 2 3 5 7 11 13 17 19 23 29  The easiest way to determine prime numbers is with nested loops

122. 122 How to time your code  Is actually pretty easy: long start = System.currentTimeMillis(); // do the computation long stop = System.currentTimeMillis(); long timeTakenMS = stop-start;  This is in milliseconds, so to do the number of actual seconds: double timeTakenSec = timeTakenMS / 1000.0;

123. 123 Program Demo NumberGames.java NumberGames.java Note what happens when you enter 100 Note what happens when you enter 100 With the Fibonacci numbers With the Fibonacci numbers With the powers of 2 With the powers of 2

124. 124 BigIntegers  An int can only go up to 2^31 or about 2*10 9  A long can only go up to 2^63, or about 9*10 18  What if we want to go higher?  2 100 = 1267650600228229401496703205376  To do this, we can use the BigInteger class It can represent integers of any size  This is called “arbitrary precision” Not surprisingly, it’s much slower than using ints and longs  The Fibonacci number part didn’t use BigIntegers That’s why we got -980107325 for the 100 th term It “flowed over” the limit for ints – called “overflow”

125. 125 BigInteger usage  BigIntegers are in the java.math library import java.math.*;  To get n n : BigInteger bigN = new BigInteger (String.valueOf(n)); BigInteger biggie = new BigInteger (String.valueOf(1)); for ( int i = 0; i < n; i++ ) biggie = biggie.multiply (bigN); System.out.println (biggie);

126. 126 Look at that them there code… NumberGames.java NumberGames.java