1. COSC2007 Data Structures II
March 25, 2017
COSC Data Structures II
Chapter 5 Recursion as a
Problem-Solving Technique
Recursion as a Problem-Solving Technique

2. Topics 8-Queen problem Languages Algebraic expressions
March 25, 2017
Topics
8-Queen problem
Languages
Algebraic expressions
Prefix & Postfix notation & conversion
Recursion as a Problem-Solving Technique

3. Introduction Backtracking:
March 25, 2017
Introduction
Backtracking:
A problem-solving technique that involves guesses at a solution, backing up, in reverse order, when a dead-end is reached, and trying a new sequence of steps
Applications
8-Queen problems
Tic-Tac-Toe
Recursion as a Problem-Solving Technique

4. Eight-Queens Problem Given: Required:
March 25, 2017
Eight-Queens Problem
Given:
64 square that form 8 x 8 - rows x columns
A queen can attack any other piece within its row, column, or diagonal
Required:
Place eight queens on the chessboard so that no queen can attack any other queen
Recursion as a Problem-Solving Technique

5. Eight-Queens Problem Observations:
March 25, 2017
Eight-Queens Problem
Observations:
Each row or column contain exactly one queen
There are 4,426,165,368 ways to arrange 8 queens on the chessboard
Not all arrangements lead to correct solution
Wrong arrangements could be eliminated to reduce the number of solutions to the problem
8! = 40,320 arrangements
Recursion as a Problem-Solving Technique

6. Eight-Queens Problem Solution Idea:
March 25, 2017
Eight-Queens Problem
Solution Idea:
Start by placing a queen in the first square of column 1 and eliminate all the squares that this queen can attack
At column 5, all possibilities will be exhausted and you have to backtrack to search for another layout
When you consider the next column, you solve the same problem with one column less (Recursive step)
The solution combines recursion with backtracking
Recursion as a Problem-Solving Technique

7. March 25, 2017
Eight-Queens Problem
How can we use recursion for solving this problem?
For placing a queen in a column, assuming that previous queens were placed correctly
If there are no columns to consider  Finish – Base case
Place queen successfully in current column  Proceed to next column, solving the same problem on fewer columns – recursive step
No queen can be placed in current column  Backtrack
Base case:
Either we reach a solution to the problem with no columns left
 Success & finish
Or all previous queens will be under attack
 Failure & Backtrack
Recursion as a Problem-Solving Technique

8. Eight-Queens Problem Pseudocode PlaceQueens (in currColumn: integer)
March 25, 2017
Eight-Queens Problem
Pseudocode
PlaceQueens (in currColumn: integer)
// Places queens in columns numbered Column through 8
if (currColumn > 8)
Succes. Reached a solution // base case
else
{ while (unconsidered squares exist in the currColumn & problem is unsolved)
{determine the next square in column “currColumn” that is not under attack in earlier column
if (such a square exists)
{ Place a queen in the square
PlaceQuenns(currColumn + 1) // Try next column
if ( no queen possible in currColumn+1)
Remove queen from currColumn &
consider next square in that column
} // end if
} // end while
Recursion as a Problem-Solving Technique

9. Eight-Queens Problem Implementation: Classes: Data members QueenClass
March 25, 2017
Eight-Queens Problem
Implementation:
Classes:
QueenClass
Data members
board:
square
Indicates is the square has a queen or is empty
Recursion as a Problem-Solving Technique

10. Eight-Queens Problem Implementation: Methods needed: Public functions
March 25, 2017
Eight-Queens Problem
Implementation:
Methods needed:
Public functions
ClearBoard: Sets all squares to empty
DisplayBoard: Displays the board
PlaceQueens: Places the queens in board’s columns & returns either success or failure
Private functions
SetQueen: Sets a given square to QUEEN
RemoveQueen: Sets a given square to EMPTY
IsUnderAttack: Checks if a given square is under attack
Index: Returns the array index corresponding to a row or column
Recursion as a Problem-Solving Technique

11. Eight-Queens Problem:
March 25, 2017
Eight-Queens Problem:
public class Queens {
public static final int BOARD_SIZE = 8;
// squares per row or column
Public static final int EMPTY =0; //use to indicate an empty square
Public static final int QUEEN=1; //indicate square contains a queen
Private in board [][]; //chess board
public Queens() { // Creates an empty square board.
board = new int [BOARD_SIZE]{BOARD-SIZE]; }
public void clearBoard() {
} // Sets all squares to EMPTY.
public void displayBoard() { } // Displays the board.
public boolean placeQueens(int currColumn)
Recursion as a Problem-Solving Technique

12. Eight-Queens Problem:
March 25, 2017
Eight-Queens Problem:
// Calls: isUnderAttack, setQueen, removeQueen.
{ if (currColumn > BOARD_SIZE)
return true; // base case
else
{ boolean queenPlaced = false;
int row = 1; // number of square in column
while ( !queenPlaced && (row <= BOARD_SIZE) )
{ // if square can be attacked
if (isUnderAttack(row, currColumn))
++row; // then consider next square in currColumn
else // else place queen and consider next
{ // column
setQueen(row, currColumn);
queenPlaced = placeQueens(currColumn+1);
// if no queen is possible in next column,
if (!queenPlaced)
{ // backtrack: remove queen placed earlier
// and try next square in column
removeQueen(row, currColumn);
++row;
} // end if
} // end while
return queenPlaced;
} // end placeQueens
Recursion as a Problem-Solving Technique

13. Eight-Queens Problem: Java Code
March 25, 2017
Eight-Queens Problem: Java Code
private void setQueen(int row, int column) { }
// Sets the square on the board in a given row and column to QUEEN.
private void removeQueen(int row, int column) { }
// Sets the square on the board in a given row and column to EMPTY.
private boolean isUnderAttack(int row, int column)
// Determines whether the square on the board at a given row and column is under
// attack by any queens in the columns 1 through column-1.
// Precondition: Each column between 1 and column-1 has a queen placed in a
// square at a specific row. None of these queens can be attacked by any other queen.
// Postcondition: If the designated square is under attack, returns true;
// otherwise, returns false.
private int index(int number)
// Returns the array index that corresponds to
// a row or column number.
// Precondition: 1 <= number <= BOARD_SIZE.
// Postcondition: Returns adjusted index value.
} // end class Queens
Recursion as a Problem-Solving Technique

14. Languages Language: A set of strings of symbols
March 25, 2017
Languages
Language:
A set of strings of symbols
Can be described using syntax rules
Examples:
Java program
program =
{ w : w is a syntactically correct java program}
Algebraic expressions
Algebraic Expression = { w : w is an algebraic expression }
Recursion as a Problem-Solving Technique

15. Languages Language recognizer: Recognition algorithm:
March 25, 2017
Languages
Language recognizer:
A program (or machine) that accepts a sentence and determines whether the sentence belongs to the language
Recognition algorithm:
An algorithm that determines whether a given string belongs to a specific language, given the language grammar
If the grammar is recursive, we can write straightforward recursive recognition algorithms
Recursion as a Problem-Solving Technique

16. Languages Example: Grammar for Java identifiers Grammar Syntax graph
March 25, 2017
Languages
Example: Grammar for Java identifiers
Java Ids = {w:w is a legal Java identifier}
Grammar
< identifier > = < letter > | < identifier > < letter >
| < identifier > < digit >
< letter > = a | b | | z | A | B | | Z | _
< digit > = 0 | 1 | | 9
Syntax graph
Observations
The definition is recursive
Base case:
A letter ( length = 1)
We need to construct a recognition algorithm
Letter
Digit
Recursion as a Problem-Solving Technique

17. Languages Java Identifiers’ Recognition algorithm
March 25, 2017
Languages
Java Identifiers’ Recognition algorithm
isId (in w: string): boolean
// Returns true if w is a legal Java identifier;
// otherwise returns false.
if (w is of length 1 ) // base case
if (w is a letter)
return true
else
return false
else if
(the last character of w is a letter or a digit)
return isId (w minus its last character)
// Point X
Recursion as a Problem-Solving Technique

18. March 25, 2017
Palindromes
String that read the same from left-to-right as it does from right-to-left
Examples
RADAR DEED
WOW ADA
Madam I’m Adam (without the blanks and quotes)
Language:
Palindromes = { w : w reads the same from left to right
as from right to left }
Grammar:
<pal > = empty string | < ch > | a <pal > a | | z <pal> z |
A < pal > A | | Z < pal > Z
< ch > = a | b | | z | A | B | | Z
The definition is recursive
Base cases:
Empty string or one character
We need a recognition algorithm
Recursion as a Problem-Solving Technique

19. Palindromes Recognition algorithm isPal(in w: string) : boolean
March 25, 2017
Palindromes
Recognition algorithm
isPal(in w: string) : boolean
// Checks if a string w is a palindrome;
// If palindrome, it returns true, false otherwise
If (w is of length 0 or 1)
return true // Base-case success
else
if (first and last characters are the same)
return IsPal( w minus first and last characters)
// Point A
return false // Base-case failure
Recursion as a Problem-Solving Technique

20. An Bn Strings Language: Grammar: The definition is recursive
March 25, 2017
An Bn Strings
Language:
L = { w : w is of the form An Bn for some n  0 }
Grammar:
< legal word > = empty string | A < legal word > B
The definition is recursive
Base case:
empty string
Recursion as a Problem-Solving Technique

21. An Bn Strings Recognition algorithm IsAnBn(in w: string): boolean
March 25, 2017
An Bn Strings
Recognition algorithm
IsAnBn(in w: string): boolean
// Checks if a string w is of the form An Bn;
If (w is of length 0)
return true // Base–case success
else if (w begins with ‘A’ & ends with ‘B’)
return IsAnBn( w minus first & last characters) // Point A
else
return false // Base–case failure
Recursion as a Problem-Solving Technique

22. Algebraic Expressions
March 25, 2017
Algebraic Expressions
Algebraic expression
A fully or partially parenthesized infix arithmetic expression
Examples
A + ( B – C ) * D { partially parenthesized}
( A + ( ( B – C ) * D )) {Fully parenthesized}
Solution
Find a way to get rid of the parentheses completely
Recursion as a Problem-Solving Technique

23. Algebraic Expressions
March 25, 2017
Algebraic Expressions
Infix expression:
Binary operators appear between their operands
Prefix (Polish) expression:
Binary operators appear before their operands
Postfix (Reverse Polish) expression:
Binary operators appear after their operands
Examples:
Infix Prefix Postfix
a + b + a b a b +
a + ( b * c ) a * b c a b c * +
(a + b) * c * + a b c a b + c *
Recursion as a Problem-Solving Technique

24. Algebraic Expressions
March 25, 2017
Algebraic Expressions
Converting or evaluating partially parenthesized infix expression are complicated (discussed in chapter 6-Stacks)
Postfix & prefix expressions can be evaluated easily by the computer
Recursion as a Problem-Solving Technique

25. Prefix Expressions Language: Grammar: Definition is recursive
March 25, 2017
Prefix Expressions
Language:
L = { w : w is a prefix expression }
Grammar:
< prefix > = <identifier>
| < operator > < prefix1> < prefix2 >
< operator > = + | - | * | /
< identifier > = A | B | | Z
Definition is recursive
Size of prefix1 & prefix2 is smaller than prefix, since they are subexpressions
Base case:
When an identifier is reached
Recursion as a Problem-Solving Technique

26. March 25, 2017
Prefix Expressions
How to Determine if a Given Expression is a Prefix Expression:
1. Construct a recursive valued function End-Pre (S , First, Last ) that returns either the index of the end of the prefix expression beginning at S(First), or -1 , if no such expression exists in the given string
2. Use The previous function to determine whether a character string S is a prefix expression
Recursion as a Problem-Solving Technique

27. Prefix Expressions End-Pre Function: (Tracing: Fig 5.5, p.229)
March 25, 2017
Prefix Expressions
End-Pre Function: (Tracing: Fig 5.5, p.229)
+EndPre(in First)
// Finds the end of a prefix expression, if one exists.
// Precondition: The substring of strExp from index first through
// the end of the string contains no blank characters.
// Postcondition: Returns the index of the last character in the
// prefix expression that begins at index first of strExp.
// If no such prefix expression exists, endPre should return –1.
last = strExp.length –1
if ((First < 0) or (First > Last)) // string is empty
return –1
ch = character at position first of strExp
if (ch is an identifier)
// index of last character in simple prefix expression
return First
else if (ch is an operator)
{ // find the end of the first prefix expression
FirstEnd = endPre(First+1) // Point X
// if the end of the first expression was found
// find the end of the second prefix expresion
if (FirstEnd > -1)
return EndPre(firstEnd+1) // Point Y
else
return -1
} // end if
Recursion as a Problem-Solving Technique

28. Prefix Expressions IsPre Function IsPre(): boolean
March 25, 2017
Prefix Expressions
IsPre Function
Uses End-Pre() to determine whether a string is a prefix or not
IsPre(): boolean
// Determines whether an expression i a prefix expression.
// Precondition: The class has a data member strExp that
// contains a string with no blank characters.
// Postcondition: Returns true if the expression is in prefix form;
// otherwise return false.
lastChar = endpre(0)
return (LastChar >= 0) and (LastChar == strExp.length()-1)
Recursion as a Problem-Solving Technique

29. Prefix Expressions EvaluatePrefix Function
March 25, 2017
Prefix Expressions
EvaluatePrefix Function
E is the expression beginning at Index
evaluatePrefix( in strExp: string)
// Evaluates a prefix expression strExp
// Precondition: strExp is a string containing a valid prefix expression
// with no blanks
// PostCondition: Returns the value of the prefix expression
{ Ch = first character of strExp
Delete the first character from strExp
if (Ch is an identifier)
return value of the identifier // base case
else if (Ch is an operator named op)
{ Operand1 = EvaluatePrefix(strExp) // Point A
Operand2 = EvaluatePrefix(strRxp) // Point B
return Operand1 op Operand2
} // end if
Recursion as a Problem-Solving Technique