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Iteration Chapter 6 Fall 2006 CS 101 Aaron Bloomfield 1 Java looping Options while do-while for Allow programs to control how many times a statement list is executed 2 Averaging values 3 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 4 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 5 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 -1 Average 2.1 6 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 } } 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"); } Program Demo NumberAverage.java 9 While syntax and semantics while ( Expression ) Action Logical expression that determines whether Action is to be executed Action is either a single statement or a statement list within braces 10 While semantics for averaging problem Test expression is evaluated at the start of each iteration of the loop. // 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(); } If test expression is true, these statements are executed. Afterward, the test expression is reevaluated and the process repeats 11 While Semantics Expression is evaluated at the start of each iteration of the loop Expression If Expression is true, Action is executed true Action false If Expression is false, program execution continues with next statement 12 Suppose input contains: 4.5 0.5 1.3 -1 Execution Trace valuesProcessed int valuesProcessed = 0; double valueSum = 0; 0 1 2 3 valueSum 4.5 6.3 5.0 0 value 0.5 1.3 4.5 -1 average 2.1 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"); } 13 New 2005 demotivatiors! 14 Converting text to lower case 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 Sample run An empty line was entered A Ctrl+z was entered. tI is the Windows escape sequence for indicating end-of-file 18 Program Demo LowerCaseDisplay.java 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); 20 Program trace The append assignment operator updates the representation of converted to include the current input line converted += (currentConversion + "\n"); Representation of lower case conversion of current input line Newline character is needed because method nextLine() "strips" them from the input 21 Loop Design & Reading From a File 22 Loop design Questions to consider in loop design and analysis What initialization expression? is necessary for the loop’s test 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? 23 Reading a file Background Same Scanner class! filename is a String Scanner fileIn = new Scanner (new File (filename) ); The File class allows access to files It’s in the java.io package 24 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 25 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(); } Close Determine Set Process Display Get Make up first next sure the standard file current lines line line file stream got fileone stream name aline input line by one tostream process If not, loop is done 26 Today’s demotivators 27 The For statement 28 The For Statement The body of the loop iterates while the test expression is true Initialization step is performed only After each iteration of the once -- just prior int currentTerm = 1; body of the loop, the update to the first expression is reevaluated evaluation of the for ( int i = 0; i < 5; ++i ) { test expression System.out.println(currentTerm); currentTerm *= 2; 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 29 Evaluated once at the beginning of the for statements's execution If ForExpr is true, Action is executed After the Action has completed, the PostExpression is evaluated ForInit ForExpr true Action PostExpr After evaluating the PostExpression, the next iteration of the loop starts The ForExpr is evaluated at the start of each iteration of the loop false If ForExpr is false, program execution continues with next statement for statement syntax Logical test expression that determines whether the action and update step are executed Initialization step prepares for the first evaluation of the test expression for ( ForInit ; ForExpression Update step is performed after the execution of the loop body ; ForUpdate ) Action The body of the loop iterates whenever the test expression evaluates to true 31 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 32 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 33 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 34 Execution Trace i for ( int i = 0; i < 3; ++i ) { 0 3 2 1 System.out.println("i is " + i); } System.out.println("all done"); i is 0 i is 1 i is 2 all done Variable i has gone out of scope – it is local to the loop 35 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++; } 36 for vs. while An example when a for loop CANNOT be directly translated into a while loop: only difference for ( int count = 0; count < 10; count++ ) { System.out.println (count); } count is NOT defined here Would (mostly) translate as: int count = 0; while (count < 10) { System.out.println (count); count++; } count IS defined here 37 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++ ) { ... } 38 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 39 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 i is 2 j is j is i is 3 j is j is j is 40 0 0 1 0 1 2 Another optical illusion 41 Loop controls 49 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 3 5 7 9 is is is is is odd odd odd odd odd 50 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 1 2 3 4 is is is is is less less less less less than than than than than five five five five five 51 Today’s demotivators 52 Four Hobos 53 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 54 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? 55 Analysis We are looking for integer solutions to the formula: a2+b2+c2+d2 = 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 56 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 + ")"); } } } } } } } } 57 Program Demo FourHobos.java 58 Results The output: (2, 4, 6, 12) (6, 6, 8, 8) Not surprisingly, the smart hobo picks the short straw of the first combination 59 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) ) 61 Alternate implementation This is the new code for the inner-most for loop: if ( (a > b) || (b > c) continue; } if ( a*a+b*b+c*c+d*d != continue; } System.out.println ("(" + c || (c > d) ) { 200 ) { + a + ", " + b + ", " + ", " + d + ")"); 62 3 card poker 63 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 64 Straight flush: both a flush and a straight 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. 65 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. 66 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 67 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 68 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() 69 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 70 Program Demo HandEvaluation.java 71 Becoming an IEEE author 72 Triangle counting 73 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 74 Sample execution Enter n: 5 (1,1,1) (1,2,2) (1,3,3) (1,4,4) (1,5,5) (2,2,2) (2,2,3) (2,3,3) (2,3,4) (2,4,4) (2,4,5) isosceles equilateral isosceles isosceles isosceles isosceles isosceles equilateral isosceles isosceles scalene isosceles scalene (2,5,5) (3,3,3) (3,3,4) (3,3,5) (3,4,4) (3,4,5) (3,5,5) (4,4,4) (4,4,5) (4,5,5) (5,5,5) isosceles isosceles equilateral isosceles isosceles isosceles right scalene isosceles isosceles equilateral isosceles isosceles isosceles equilateral 75 Program Demo TriangleDemo.java 76 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() 77 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 78 others are not. Look at that them there code… TriangleDemo.java 79 Today’s demotivators 80 End of this slide set? 81 Fibonacci numbers 82 Fibonacci sequence Sequences can be neither geometric or arithmetic Fn = Fn-1 + Fn-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: 1 5 1 5 F ( n) n 5 2n n 83 Fibonacci sequence in nature 13 8 5 3 2 1 84 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? 85 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) 86 Reproducing rabbits Sixth month: 8 pairs All 3 new rabbit pairs are pregnant, as well pregnant in the last month (2) Seventh month: 13 pairs All 5 new rabbit pairs are pregnant, as well pregnant in the last month (3) Eighth month: 21 pairs All 8 new rabbit pairs are pregnant, as well pregnant in the last month (5) Ninth month: 34 pairs All 13 new rabbit pairs are pregnant, as well pregnant in the last month (8) Tenth month: 55 pairs All 21 new rabbit pairs are pregnant, as well pregnant in the last month (13) as those not as those not as those not as those not as those not 87 Reproducing rabbits Note the sequence: { 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … } The Fibonacci sequence again 88 Fibonacci sequence Another application: Fibonacci references from http://en.wikipedia.org/wiki/Fibonacci_sequence 89 Fibonacci sequence As the terms increase, approaches 1.618 the ratio between successive terms F (n 1) 5 1 1.618933989 n F ( n) 2 lim 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 90 The Golden Ratio 91 92 Number counting 93 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 94 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 95 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 96 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; 97 Program Demo NumberGames.java Note what happens when you enter 100 With the Fibonacci numbers With the powers of 2 98 BigIntegers An int can only go up to 2^31 or about 2*109 A long can only go up to 2^63, or about 9*1018 What if we want to go higher? 2100 = 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 100th term It “flowed over” the limit for ints – called “overflow” 99 BigInteger usage BigIntegers are in the java.math library import java.math.*; To get nn: 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); 100 Look at that them there code… NumberGames.java 101