1. 1 Iteration Chapter 6 Fall 2006 CS 101 Aaron Bloomfield

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 New 2005 demotivatiors!

15. 15 End of lecture on 2 October 2006

16. 16 Converting text to lower case

17. 17 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); }

18. 18 Sample run A Ctrl+z was entered. It is the Windows escape sequence for indicating end-of-file An empty line was entered

19. 19 Program Demo LowerCaseDisplay.java LowerCaseDisplay.java

20. 20 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); }

21. 21 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

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 The For statement

29. 29 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

30. ForExpr Action truefalse ForInit PostExpr 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

31. 31 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

32. 32 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

33. 33 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

34. 34 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

35. 35 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

36. 36 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++; }

37. 37 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

38. 38 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++ ) {... }

39. 39 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

40. 40 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

41. 41 Another optical illusion

42. 42 do-while loops (we aren’t going over these this semester)

43. 43 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

44. 44 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

45. 45 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 );

46. 46 Program Demo GuessMyNumber.java GuessMyNumber.java

47. 47 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

48. 48 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; }

49. 49 Loop controls

50. 50 The continue keyword  The continue keyword will immediately start the next iteration of the loop The rest of the current loop is not executed 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

51. 51 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

52. 52 Today’s demotivators

53. 53 Four Hobos

54. 54 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  This problem is in section 6.10 of the text

55. 55 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 they work for 3 hours on 3 days (a total of 9 hours)  The smart hobo manages to draw the shortest straw  How many ways are there to split up such work?  Which one did the smart hobo choose?

56. 56 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

57. 57 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 + ")"); }

58. 58 Program Demo FourHobos.java FourHobos.java

59. 59 Results  The output: (2, 4, 6, 12) (6, 6, 8, 8)  Not surprisingly, the smart hobo picks the short straw of the first combination

60. 60 End of lecture on 4 October 2006

61. 61 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) )

62. 62 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 + ")");

63. 63 3 card poker

64. 64 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

65. 65 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.

66. 66 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.

67. 67 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

68. 68 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

69. 69 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()

70. 70 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

71. 71 Program Demo HandEvaluation.java HandEvaluation.java

72. 72 Becoming an IEEE author

73. 73 Triangle counting

74. 74 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

75. 75 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

76. 76 Program Demo TriangleDemo.java TriangleDemo.java

77. 77 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()

78. 78 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.

79. 79 Look at that them there code… TriangleDemo.java TriangleDemo.java

80. 80 Today’s demotivators

81. 81 End of this slide set?

82. 82 Fibonacci numbers

83. 83 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:

84. 84 Fibonacci sequence in nature 13 8 5 3 2 1

85. 85 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?

86. 86 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)

87. 87 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)

88. 88 Reproducing rabbits  Note the sequence: { 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … }  The Fibonacci sequence again

89. 89 Fibonacci sequence  Another application:  Fibonacci references from http://en.wikipedia.org/wiki/Fibonacci_sequence

90. 90 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

91. 91 The Golden Ratio

92. 92

93. 93 Number counting

94. 94 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

95. 95 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

96. 96 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

97. 97 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;

98. 98 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

99. 99 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”

100. 100 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);

101. 101 Look at that them there code… NumberGames.java NumberGames.java