ITERATION

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ITERATION

• CSC 171 FALL 2004
• LECTURE 10

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Simple Branching
If (test) A
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Backward Branching
while(test) A
A
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Example what is the output?

• int x 3
• int sum 0
• while (x gt 0)
• sum sum x
• x x 1
• System.out.print(x x)
• System.out.println(sum sum)

x 3, sum 0
x 2, sum 3
x 1, sum 5
x 0, sum 6
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Example what is the output?

• int x 3
• int sum 0
• while (x lt 5)
• sum sum x
• x x 1
• System.out.print(x x)
• System.out.println(sum sum)

x 3, sum 0
x 2, sum 3
x 1, sum 5
x 0, sum 6
x -1, sum 6
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Infinite Loops

• A very common fault in program design
• System compiles ok
• Then appears to hang

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Shorthand Operators

• a a b
• a a b
• a a b
• a a / b
• a a b

• a b
• a - b
• a b
• a / b
• a b

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More Shorthand

• int x
• x x 1
• x 1
• x x 1
• x - 1

• x
• x--

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Shorthand Operators x vs. x

• int x 0
• System.out.println(x x)
• System.out.println(x x)
• System.out.println(x x--)
• System.out.println(x --x)

x0
x2
x2
x0
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Loop invariants

• Intro to mathematical program analysis
• In order to verify loops we often establish an
  assertion (boolean expression) that is true each
  time we reach a specific point in the loop.
• We call this assertion, a loop invariant

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Assertions

• When ever the program reaches the top of the
  while loop, the assertion is true

INIT
BODY
INVARIANT
TEST
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Example

• Write a method to compute an
• public static int power(int a , int n)
• Positive values only ok
• You have 5 minutes

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Possible solution

• public static int power(int a, int n)
• int r 1 int b a int i n
• while (igt0)
• r b
• return r

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Possible solution (Euclid)

• public static double power(int a, int n)
• int r 1 int b a int i n
• while (igt0)
• if (i2 0) b b b i i / 2
• else r r b i--
• return r

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Does it work?

• SURE! TRUST ME!
• Well, look at a100 if you dont believe me!
• Note, less loops!
• Can you prove that it works?

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What is the loop invariant?

• At the top of the while loop, it is true that
• rbi an
• It is?
• Well, at the top of the first loop
• r1
• ba
• in

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So, if its true at the start

• Even case
• rnew rold
• bnew (bold)2
• inew(iold)/2
• Therefore,
• rnew (bnew)inew rold ((bold)2)iold/2
• rold
  (bold)iold
• an

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So, if its true at the start II

• Odd case
• rnew roldbold
• bnew bold
• inewiold-1
• Therefore,
• rnew (bnew)inew rold bold (bold)iold-1
• rold
  (bold)iold
• an

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So,

• If its true at the start
• And every time in the loop, it remains true
• Then, it is true at the end
• rbi an
• And, i 0 ( the loop ended)
• What do we know?

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Correctness Proofs

• Proof are more valuable than testing
• Tests demonstrate limited correctness
• Proofs demonstrate correctness for all inputs
• For some time, people hoped that all formal logic
  would replace programming
• The naïve idea that programming is a form of
  math proved to be an oversimplification

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Correctness Proofs

• Unfortunately, in practice, these methods never
  worked very well.
• Instead of buggy programs,
• people wrote buggy logic
• Nonetheless, the approach is useful for program
  analysis

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The take away message?

• In the end, engineering and (process) management
  are at least as important as mathematics and
  logic for the successful completion of large
  software projects

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Do while

• A useful variant of the while loop

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Backward Branching
while(test) A
A
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Backward Branching
do A while (test)
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YOU WRITE THE CODE
start
X 1 y 1
X 2 x y
y lt 3
end
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Use of a sentinel

• A variable that keeps track of the ending
  condition

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Example

• static int myGetInt()
• boolean done false
• String locIn ""
• int returnVal 0
• InputStreamReader reader
• new InputStreamReader(System.in)
• BufferedReader console
• new BufferedReader(reader)

//sentinal
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• do
• try
• locIn console.readLine()
• returnValue Integer.ParseInt(locIn)
• done true
• catch (Exception e)
• System.out.println(e trouble")
• done false
• while (!done)
• return returnVal

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For loops

• Sometimes, we want the loop to repeat for a set
  (definite) number of times.
• A type of while loop
• That became so common
• It was given its own syntax

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TYPES OF ITERATION

• Indefinite Iteration
• We dont know exactly how many times we want the
  loop to repeat
• while loop
• Definite Iteration
• We know how many time we want the loop to repeat
• for loop

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While loop
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While loop Code

• int year 0
• while (balance lt 2 initialBalance)
• balance balance interest
• year

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Example of a for loop as a while
start
int count count 0 while(count lt 10)
A count
count0
count
A
end
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Example of a for loop as a for
start
for(int count 0 count lt 10 count)
A
count0
count
A
end
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Generic for loop
start
for(init test update) body
init
update
body
end
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For loop
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For loop Code

• for (int year 1yearlt20year)
• balance balance interest

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For Loop

• Initialization
• Test
• Body
• Increment
• for(initialzationtestincrement)
• //Body

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Off by one errors

• If you want n iterations
• Start at 0 and go to x lt n
• Or
• Start at 1 and go to x lt n

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Nested Loops

• Sometimes, we want to perform 2D operations
• Tables
• Addition
• Multiplication
• Interest rates

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Multiplication Table

• What is the output?
• int size 5
• for(int i 0 iltsizei)
• for(int j 0 jltsizej)
• System.out.println(String.toString(ij))

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Multiplication Table Fixed

• int size 5
• for(int i 0 iltsizei)
• for(int j 0 jltsizej)
• System.out.print(String.toString(ij) )
• System.out.println()

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Enumerations

• Integers are nice, because we always have a clear
  idea of what the next one is.
• But sometimes, we have an orderd set or list of
  things that arent numbers
• Hearts, Spades, Diamonds, Clubs
• Bob, Carol, Ted, Alice
• We would like to go through them one at a time

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public interface Enumeration

• An object that implements the Enumeration
  interface generates a series of elements, one at
  a time. Successive calls to the nextElement
  method return successive elements of the series.
• For example, to print all elements of a vector v

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General Case

• Enumeration e
• while(e.hasMoreElements()) System.out.println(e
  .nextElement())

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Our Fave StringTokenizer

• A String Tokenizer breaks strings up into tokens
  (surprize!)
• String Hello CSC 171, How are you
• Tokens
• Hello,CSC,171,,How,are,you
• StringTokenizer tokenizer
• new StringTokenizer(inputLine)

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Using String Tokenizer

• import java.util.StringTokenizer
• public class Split
• public static void main(String args)
• boolean done false
• while (!done)
• String inputLine myGetString()
• if (inputLine null) done true
• else
• // break input line into words

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String Tokeizer II

• StringTokenizer tokenizer
• new StringTokenizer(inputLine)
• while (tokenizer.hasMoreTokens())
• String word tokenizer.nextToken()
• System.out.println(word)