Download presentation

Presentation is loading. Please wait.

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

Similar presentations

OK

© 2006 Pearson Addison-Wesley. All rights reserved 6-1 Chapter 6 Recursion as a Problem- Solving Technique.

© 2006 Pearson Addison-Wesley. All rights reserved 6-1 Chapter 6 Recursion as a Problem- Solving Technique.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google