COSC2007 Data Structures II

Chapter 6Recursion as a

Problem-Solving Technique

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Topics

8-Queen problem

Languages Algebraic expressions

Prefix & Postfix notation & conversion

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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

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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

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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

Any idea?

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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

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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 correctlya) If there are no columns to consider Finish – Base caseb) Place queen successfully in current column Proceed to next

column, solving the same problem on fewer columns – recursive step

c) 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

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Eight-Queens Problem Pseudocode

PlaceQueens (in currColumn: integer)// Places queens in columns numbered Column through 8if (currColumn > 8)

Success. Reached a solution // base caseelse{ 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 columnif ( no queen possible in currColumn+1)

Remove queen from currColumn & consider next square in that column

} // end if} // end while

} // end if

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Eight-Queens Problem

Implementation: Classes:

QueenClass Data members

board: How?

square Indicates that the square has a queen or is empty

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Eight-Queens Problem

Implementation: Methods needed:

Public functions (methods) 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 (methods)

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

Textbook provides partial implementation

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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 }

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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

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Languages

Example: Grammar for Java identifiersJava 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

Letter

Digit

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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 caseif (w is a letter)

return trueelse

return falseelse if

(the last character of w is a letter or a digit)return isId (w minus its last character)

// Point X else

return false

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Palindromes String that read the same from left-to-right as it does from

right-to-left

ExamplesRADAR

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:

? We need a recognition algorithm

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Palindromes

Recognition algorithm

isPal(in w: string) : boolean

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)

elsereturn false // Base-case failure

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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:

?

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An Bn Strings

Recognition algorithm

IsAnBn(in w: string): boolean

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

elsereturn false // Base–case failure

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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

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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 Postfixa + b + a b a b +a + ( b * c ) + a * b c a b c * +(a + b) * c * + a b c a b + c *

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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

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Prefix Expressions

How to Determine if a Given Expression is a Prefix Expression?

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Prefix Expressions

How to Determine if a Given Expression is a Prefix Expression:1. Construct a recursive valued function End-Pre (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

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Prefix Expressions End-Pre Function: (Tracing: Fig 6.5, p.309)endPre(first, last)

// Finds the end of a prefix expression, if one exists.// If no such prefix expression exists, endPre should return –1.

if ((First < 0) or (First > Last)) // string is emptyreturn –1

ch = character at position first of strExp if (ch is an identifier)

// index of last character in simple prefix expressionreturn First

else if (ch is an operator) { // find the end of the first prefix expression

firstEnd = endPre(First+1, last) // Point X// if the end of the first expression was found// find the end of the second prefix expression

if (firstEnd > -1)return endPre(firstEnd+1, last) // Point Y

elsereturn -1

} // end if else return -1

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Prefix Expressions

IsPre Function Uses endPre() to determine whether a string is a prefix or not

IsPre(): booleansize= length of expression strExplastChar = endpre(0, size-1)if (lastChar >= 0 and lastChar == strExp.length()-1)

return true else

return false

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Prefix Expressions EvaluatePrefix Function

E is the expression beginning at Index

evaluatePrefix( strExp: string)// Evaluates a prefix expression strExp{ Ch = first character of strExp

Delete the first character from strExpif (Ch is an identifier)

return value of the identifier // base caseelse if (Ch is an operator named op){ Operand1 = EvaluatePrefix(strExp) // Point A

Operand2 = EvaluatePrefix(strRxp) // Point Breturn Operand1 op Operand2

} // end if}

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Review ______ is a problem-solving technique

that involves guesses at a solution. Recursion Backtracking Iteration Induction

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Review In the recursive solution to the Eight

Queens problem, the problem size decreases by ______ at each recursive step. one square two squares one column two columns

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Review A language is a set of strings of ______.

numbers letters Alphabets symbols

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Review The statement:

JavaPrograms = {strings w : w is a syntactically

correct Java program} signifies that ______. all strings are syntactically correct Java programs all syntactically correct Java programs are strings the JavaPrograms language consists of all the

possible strings a string is a language

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Review The Java ______ determines whether a

given string is a syntactically correct Java program. applet editor compiler programmer

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Review If the string w is a palindrome, which of the

following is true? w minus its first character is a palindrome w minus its last character is a palindrome w minus its first and last characters is a

palindrome the first half of w is a palindrome the second half of w is a palindrome

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Review Which of the following strings is a

palindrome? “bottle” “beep” “coco” “r

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Review The symbol AnBn is standard notation for the

string that consists of ______. an A, followed by an n, followed by a B, followed

by an n an equal number of A’s and B’s, arranged in a

random order n consecutive A’s, followed by n consecutive B’s a pair of an A and a B, followed another pair of an

A and a B

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Review Which of the following is an infix

expression? / a + b c a b c + / a b / + c a / (b + c)

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Review What is the value of the prefix expression:

– * 3 8 + 6 5? 11 23 13 21