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Iteration. Chapter 4 Spring 2007 CS 101 Aaron Bloomfield. Java looping. Options while do-while for Allow programs to control how many times a statement list is executed. Averaging values. Averaging. Problem

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Iteration l.jpg

Iteration

Chapter 4

Spring 2007

CS 101

Aaron Bloomfield


Java looping l.jpg
Java looping

  • Options

    • while

    • do-while

    • for

  • Allow programs to control how many times a statement list is executed



Averaging l.jpg
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


Averaging5 l.jpg
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


Averaging6 l.jpg
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


Slide7 l.jpg

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

}

}


Slide8 l.jpg

int valuesProcessed = 0;

double valueSum = 0;

// set up the input

Scanner stdin = new Scanner (System.in);

// prompt user for values

System.out.println("Enterpositivenumbers1perline.\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 l.jpg
Program Demo

  • NumberAverage.java


While syntax and semantics l.jpg

Logical expression that

Action is either a single

determines whether Action

statement or a statement

is to be executed

list within braces

While syntax and semantics

Expression

Action

while

(

)


While semantics for averaging problem l.jpg

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

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();

}


While semantics l.jpg

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

While Semantics

Expression

false

true

Action


Execution trace l.jpg
Execution Trace

Suppose input contains: 4.50.51.3-1

Suppose input contains: 4.50.51.3 -1

Suppose input contains: 4.5 0.5 1.3 -1

Suppose input contains: 4.5 0.5 1.3 -1

Suppose input contains: 4.50.5 1.3 -1

valuesProcessed

1

0

3

2

4.5

valueSum

0

6.3

5.0

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);

value

1.3

4.5

-1

0.5

average

2.1


What do these pictures mean l.jpg
What do these pictures mean?

  • Light beer

  • Dandy lions

  • Assaulted peanut

  • Eggplant

  • Dr. Pepper

  • Pool table

  • Tap dancers

  • Card shark

  • King of pop

  • I Pod

  • Gator aide

  • Knight mare

  • Hole milk



Converting text to strictly lowercase l.jpg
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);

}


Sample run l.jpg

An empty line

was entered

A Ctrl+z was

entered. I

t is the

Windows escape

sequence for

indicating

end-of-file

Sample run


Program demo18 l.jpg
Program Demo

  • LowerCaseDisplay.java


Program trace l.jpg
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);

}


Program trace20 l.jpg

The append assignment operator updates the representation

of converted to include the current input line

Representation of lower case

Newline character is needed

conversion of current input line

because method nextLine()

"strips" them from the input

Program trace

converted += (currentConversion + "\n");




Loop design l.jpg
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?


Reading a file l.jpg

Same Scanner class!

filename is a String

The File class allows access to files

It’s in the java.io package

Reading a file

  • Background

Scanner fileIn = new Scanner (new File (filename) );


Reading a file25 l.jpg
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


Reading a file26 l.jpg
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 stream

Determine file name

Set up file stream

Process lines one by one

Get first line

Make sure got a line to process

Display current line

Get next line

Make sure got a line to process

If not, loop is done

Close the file stream




The for statement29 l.jpg

The body of the loop iterates

while the test expression is

Initialization step

true

is performed only

After each iteration of the

once -- just prior

body of the loop, the update

to the first

expression is reevaluated

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

The For Statement

int

currentTerm = 1;

for ( int i = 0; i < 5; ++i ) {

System.out.println(currentTerm);

currentTerm *= 2;

}


Slide30 l.jpg

Evaluated once

at the beginning

of the for

statements's

The ForExpr is

execution

evaluated at the

start of each

iteration of the

loop

If ForExpr is true,

Action is

executed

After the Action

If ForExpr is

has completed,

false, program

the

execution

PostExpression

continues with

is evaluated

next statement

After evaluating the

PostExpression, the next

iteration of the loop starts

ForInit

ForExpr

true

false

Action

ForUpdate


For statement syntax l.jpg

Logical test expression that determines whether the action and update step are

executed

Initialization step prepares for the

first evaluation of the test

Update step is performed after

expression

the execution of the loop body

The body of the loop iterates whenever

the test expression evaluates to true

for statement syntax

ForInit

ForExpression

ForUpdate

Action

for

(

;

;

)


For vs while l.jpg
for vs. while and update step are

  • 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


Variable declaration l.jpg
Variable declaration and update step are

  • 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


Variable declaration34 l.jpg
Variable declaration and update step are

  • 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


Execution trace35 l.jpg
Execution Trace and update step are

i

0

2

1

3

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

for (

int i = 0;

int i = 0;

i < 3;

i < 3;

++i

++i

) {

Variable i has gone out of scope – it

is local to the loop


For vs while36 l.jpg
for vs. while and update step are

  • 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++;

    }


For vs while37 l.jpg
for vs. while and update step are

  • 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++;

    }

only difference

count is NOT defined here

count IS defined here


For loop indexing l.jpg
for loop indexing and update step are

  • 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++ ) { ... }


Nested loops l.jpg

i is 0 and update step are

j is 0

j is 1

i is 1

j is 0

j is 1

i is 2

j is 0

j is 1

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);

}

}


Nested loops40 l.jpg

i is 0 and update step are

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

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);

}

}


How well do you understand for loops l.jpg
How well do you understand for loops? and update step are

  • Very well! This stuff is easy!

  • Fairly well – with a little review, I’ll be good

  • Okay. It’s not great, but it’s not horrible, either

  • Not well. I’m kinda confused

  • Not at all. I’m soooooo lost


From dubai l.jpg
From Dubai and update step are


Do while loops l.jpg

do-while loops and update step are


The do while statement l.jpg

Action and update step are

true

Expression

false

The do-while statement

  • Syntax

    doAction

    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


Picking off digits l.jpg
Picking off digits and update step are

  • 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


Guessing a number l.jpg
Guessing a number and update step are

  • 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 );


Program demo47 l.jpg
Program Demo and update step are

  • GuessMyNumber.java


While vs do while l.jpg
while vs. do-while and update step are

  • 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


While vs do while49 l.jpg
while vs. do-while and update step are

  • 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;

    }


How well do you understand do while loops l.jpg
How well do you understand do-while loops? and update step are

  • Very well! This stuff is easy!

  • Fairly well – with a little review, I’ll be good

  • Okay. It’s not great, but it’s not horrible, either

  • Not well. I’m kinda confused

  • Not at all. I’m soooooo lost


Today s demotivators51 l.jpg
Today’s demotivators and update step are


Loop controls l.jpg

Loop controls and update step are


The continue keyword l.jpg
The continue keyword and update step are

  • 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


The break keyword l.jpg
The break keyword and update step are

  • 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


Four hobos l.jpg

Four Hobos and update step are


Four hobos56 l.jpg
Four Hobos and update step are

  • 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


Problem l.jpg
Problem and update step are

  • 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?


Analysis l.jpg
Analysis and update step are

  • 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


Implementation l.jpg
Implementation and update step are

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 + ")");

}

}

}

}

}

}

}

}


Program demo60 l.jpg
Program Demo and update step are

  • FourHobos.java


Results l.jpg
Results and update step are

  • The output:

    (2, 4, 6, 12)

    (6, 6, 8, 8)

  • Not surprisingly, the smart hobo picks the short straw of the first combination


Today s demotivators62 l.jpg
Today’s demotivators and update step are


Alternate implementation l.jpg
Alternate implementation and update step are

  • 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) )


Alternate implementation64 l.jpg
Alternate implementation and update step are

  • 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 + ")");


How well do you understand four hobos l.jpg
How well do you understand four hobos? and update step are

  • Very well! This stuff is easy!

  • Fairly well – with a little review, I’ll be good

  • Okay. It’s not great, but it’s not horrible, either

  • Not well. I’m kinda confused

  • Not at all. I’m soooooo lost


The 2006 ig nobel prizes l.jpg
The 2006 Ig Nobel Prizes and update step are

  • Ornithology

  • Nutrition

  • Peace

  • Acoustics

  • Mathematics

  • Literature

  • Medicine

  • Physics

  • Chemistry

  • Biology

For explaining why woodpeckers don’t get headaches

For showing that Kuwaiti dung beetles are finicky eaters

For development of a high-pitched electronic teen-ager repellent (and, later, ring tones)

For experiments to determine why people don’t like the sound of fingernails scraping on a blackboard

For calculating the number of photos you must take to ensure that (almost) nobody in a group will have their eyes closed

For a report entitled, “Consequences of Erudite Vernacular Utilized Irrespective of Necessity: Problems with Using Long Words Needlessly.“

For a medical case report titled, “"Termination of Intractable Hiccups with Digital Rectal Massage“

For studying why dry spaghetti breaks into multiple pieces

For a study entitled, “Ultrasonic Velocity in Cheddar Cheese as Affected by Temperature,"

For showing that the female malaria mosquito is equally attracted to the smells of limburger cheese and human feet


3 card poker l.jpg

3 card poker and update step are


3 card poker68 l.jpg
3 Card Poker and update step are

  • 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


The card class l.jpg
The Card class and update step are

  • 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.


Card class methods l.jpg
Card class methods and update step are

  • 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.


The hand class l.jpg
The Hand class and update step are

  • 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


Provided hand methods l.jpg
Provided Hand methods and update step are

  • 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


Hand methods to implement l.jpg
Hand Methods to Implement and update step are

  • 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()


Class handevaluation l.jpg
Class HandEvaluation and update step are

  • 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


Program demo75 l.jpg
Program Demo and update step are

  • HandEvaluation.java


How well do you understand 3 card poker l.jpg
How well do you understand 3-card poker? and update step are

  • Very well! This stuff is easy!

  • Fairly well – with a little review, I’ll be good

  • Okay. It’s not great, but it’s not horrible, either

  • Not well. I’m kinda confused

  • Not at all. I’m soooooo lost


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All your base are belong to us and update step are

  • Flash animation

  • Reference: http://en.wikipedia.org/wiki/All_your_base_are_belong_to_us


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The Halting Problem and update step are


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What’s wrong with this program? and update step are

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”?


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The Halting problem and update step are

  • 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)


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A few notes and update step are

  • 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


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Take your best guess – do you think it’s possible to solve the halting problem?

  • Yes

  • No

  • I don’t understand what the halting problem is


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Can a human determine if a program halts? solve the halting problem?

  • 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”


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First sample program: solve the halting problem?

...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();

Halting problem examples: will they halt?


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Take your best guess – do you think it’s possible to solve the halting problem?

  • Yes

  • No

  • I don’t understand what the halting problem is


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Perfect numbers solve the halting problem?

  • 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, 191561942608236107294793378084303638130997321548169216

    • 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


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Odd perfect number search solve the halting problem?

  • 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


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Take your best guess – do you think it’s possible to solve the halting problem?

  • Yes

  • No

  • I don’t understand what the halting problem is


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Where does that leave us? solve the halting problem?

  • 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?


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CheckHalt()’s non-existence solve the halting problem?

  • 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


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CheckHalt()’s non-existence solve the halting problem?

  • 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?


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A (somewhat condensed) version of class Test: solve the halting problem?

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!

CheckHalt()’s non-existence


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How well do you understand solve the halting problem?the halting problem?

  • Very well! This stuff is easy!

  • Fairly well – with a little review, I’ll be good

  • Okay. It’s not great, but it’s not horrible, either

  • Not well. I’m kinda confused

  • Not at all. I’m soooooo lost


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Why do we care about the halting problem? solve 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!


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New 2005 demotivatiors! solve the halting problem?



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Triangle counting solve the halting problem?


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The programming assignment solve the halting problem?

  • 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


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Enter n: solve the halting problem?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

Sample execution


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Program Demo solve the halting problem?

  • TriangleDemo.java


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The Triangle class solve the halting problem?

  • 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()


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The TriangleDemo class solve the halting problem?

  • 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.


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Look at that them there code… solve the halting problem?

  • TriangleDemo.java


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How well do you understand triangle counting? solve the halting problem?

  • Very well! This stuff is easy!

  • Fairly well – with a little review, I’ll be good

  • Okay. It’s not great, but it’s not horrible, either

  • Not well. I’m kinda confused

  • Not at all. I’m soooooo lost


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Fibonacci numbers solve the halting problem?


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Fibonacci sequence solve the halting problem?

  • 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:


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Fibonacci sequence in nature solve the halting problem?

13

8

5

3

2

1


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Reproducing rabbits solve the halting problem?

  • 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?


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Reproducing rabbits solve the halting problem?

  • 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)


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Reproducing rabbits solve the halting problem?

  • 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)


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Reproducing rabbits solve the halting problem?

  • Note the sequence:

    { 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … }

  • The Fibonacci sequence again


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Fibonacci sequence solve the halting problem?

  • Another application:

  • Fibonacci references from http://en.wikipedia.org/wiki/Fibonacci_sequence


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Fibonacci sequence solve the halting problem?

  • 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


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The Golden Ratio solve the halting problem?


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Number counting solve the halting problem?


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The programming assignment solve the halting problem?

  • 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


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Sample execution solve the halting problem?

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


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Background: Prime numbers solve the halting problem?

  • 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


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How to time your code solve the halting problem?

  • 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;


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Program Demo solve the halting problem?

  • NumberGames.java

  • Note what happens when you enter 100

    • With the Fibonacci numbers

    • With the powers of 2


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BigIntegers solve the halting problem?

  • 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”


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BigInteger usage solve the halting problem?

  • 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);


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Look at that them there code… solve the halting problem?

  • NumberGames.java


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