# java lect 20 by SanjuDudeja

VIEWS: 4 PAGES: 19

• pg 1
```									Introduction to Recursion

Manish Sinha
Topics

   Two Parts to Recursion:
–   Solves easy problem in one step
–   Divide hard problem in smaller ones, and solve small problems
   Examples in Recursion:
–   Walk a Distance
–   Smashing a Rock
   Recursion in JAVA
   Quiz
How to cross a Parking Lot

   It is Sunday evening and the only parking spot you can
find at R-Mall is far from the entrance (horror!). How do
you get from your car to the mall
   A journey of 1000 yards begins with a single step ---
Lao Tse
   Apply Lao’s instruction to our problem
   What do you do next?
   How do you know when the “crossing the parking lot”
problem has been solved?
Two Parts to Recursion

   If the problem is easy, solve it immediately
   If the problem can not be solved immediately,
divide it into smaller problems:
–   Solve the smaller problems by applying this
procedure to each of them
Recursion on Parking lot problem
   If you are one step from the mall, take that step and
you are done
   If you are further than one step from the mall, divide
the distance into two parts:
–   a single step, and
–   the remaining distance
–   Now take a step and then cross the remaining distance
   Say that you have a small rock and you have to break
it into smaller pieces with a hammer. How can you do
this?
Pounding a rock to Dust
Recursion on the Rocks

   When a piece is small, don’t pound it any
further
   To destroy a large rock, hit it with a hammer.
The rock shatters, leaving smaller and large
pieces
–   Apply this procedure to each of the pieces
String Equality

   Lets forget that there is an equal() method that is part
of class String.
   Here are some equal strings:
–   “abc” equals “abc”
–   “abc de” equals “abc de”
   Here are not equal strings:
–   “ab” !equals “abc”
–   “abC” !equals “abc”
–   “abc ” !equals “aBc”
Rules for string equality
   The symbols x stands for a single character, as does y.
The symbol X stands for a string of characters, as does
Y. The symbol + stands for concatenation.
   Rule 1: equals(“” ,“”) = true
   Rule 2: equals(“”, X) = false if X is not empty string
   Rule 3: equals(X, “”) = false if X is not empty string
   Rule 4: equals(x+X,y+Y) = false if x != y
   Rule 5: equals(x+X,y+Y) = true if x == y and
equals(X,Y)
String Equality Examples

   equals(“bat”, “radio”) = false // rule 4
   equals(“rat”, “rat”) = equals( "at", "at") // rule 5
–   equals( "at", "at" ) = equals( "t", "t") // rule 5
–   equals( "t", "t" ) = equals( "", "") // rule 5
–   equals( "", "" ) = true // rule 1
   equals( "rat", "ra" ) = equals( "at", "a") // rule 5
–   equals( "at", "a" ) = equals( "t", "") // rule 5
–   equals( "t", "" ) = false // rule 3
String Equality Examples …

   equals( "rAt", "rat" ) = equals( "At", "at") // rule5
–   equals( "At", "at" ) = false // rule 4
   Definition of Base case: A base case is a
problem that can solved immediately.
   The base cases are:
–   equals( "", "" ) = true
–   equals( "", X ) = false if X is not the empty string
–   equals( X, "" ) = false if X is not the empty string
–   equals( x+X, y+Y ) = false if x != y
Translation into JAVA
boolean equals( String strA, String strB ) {
// 1. equals( "", "" ) = true
if ( strA.length() ________ 0 && strB.length() ________ 0 ) return true;
// 2. equals( "", X ) = false if X is not the empty string
else if ( strA.length() ________ 0 && strB.length() ________ 0 )
return false;
// 3. equals( X, "" ) = false if X is not the empty string
else if ( strA.length() ________ 0 && strB.length() ________ 0 )
return false;
// 4. equals( x+X, y+Y ) = false if x != y
else if ( strA.charAt(0) ________ strB.charAt(0) ) return false;
// 5. equals( x+X, y+Y ) = true if x == y and equals( X, Y )
else return ________( strA.substring(1), strB.substring(1) ); }
boolean equals( String strA, String strB ) {
if ( strA.length() == 0 && strB.length() == 0 )
return true;
else if ( strA.length() == 0 && strB.length() != 0 )
return false;
else if ( strA.length() != 0 && strB.length() == 0 )
return false;
else if ( strA.charAt(0) != strB.charAt(0) )
return false;
else return equals( strA.substring(1), strB.substring(1) );
}
Quiz

   What are the two parts in recursion?
A. (1) If the problem is easy, solve it immediately,
and (2) If the problem can't be solved
immediately, divide it into smaller problems.
B. 1) Divide the problem into smaller problems,
and (2) give immediate solutions for the hard
problems.
C. (1) Discard the hard cases , and (2) solve the
easy easy cases.
D. (1) Solve the problem by asking it to solve
itself, (2) Solve the easy cases in one step.
Quiz…

   How can you drink an entire keg of root
beer?
A. (1) take one swallow, then (2) take another
swallow.
B. (1) If the keg is empty do nothing, otherwise
(2) take one swallow, then drink the rest of the
keg.
C. (1) take one enormous gulp, and (2) wish you
D. (1) drink one keg, and (2) drink another keg.
Quiz…

   How do you study a text book?
A. (1) Read the book on day 1, and (2) read it
again each day of the semester.
B. (1) If you have reached the end of the book
you are done, else (2) study one page, then
study the rest of the book.
C. (1) Divide the book in two, and (2) study each
half.
D. (1) Cram all the pages in one horrible session,
and (2) forget everything the next night.
Quiz…

   How does detective solve a mystery?
A. (1) Examine one clue, and (2) examine the
remaining clues.
B. (1) Question one witness, and (2) question the
victim.
C. (1) Eliminate one witness, and (2) eliminate the
remaining witnesses.
D. (1) When one suspect remains, that is who did
it. (2) Examine the evidence to eliminate one
suspect, then eliminate the remaining suspects.
Quiz…

   How does a Web crawler visit every Web
page at a Web site?