COSC2007 Data Structures II

1. COSC2007 Data Structures II Chapter 6Recursion as a Problem-Solving Technique

2. Topics • 8-Queen problem • Languages • Algebraic expressions • Prefix & Postfix notation & conversion

3. 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

4. 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

5. 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?

6. 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

7. 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

8. Eight-Queens Problem • Pseudocode PlaceQueens (in currColumn: integer) // Places queens in columns numbered Column through 8 if (currColumn > 8) Success. 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 } // end if

9. Eight-Queens Problem • Implementation: • Classes: • QueenClass • Data members • board: • How? • square • Indicates that the square has a queen or is empty

10. 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

11. 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 }

12. 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

13. Letter Letter Digit 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

14. 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 else return false

15. Palindromes • String that read the same from left-to-right as it does from right-to-left • Examples RADAR 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

16. 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) else return false // Base-case failure

17. 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: • ?

18. 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 else return false // Base–case failure

19. 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

20. 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 *

21. 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

22. Prefix Expressions • How to Determine if a Given Expression is a Prefix Expression?

23. 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

24. 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 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, 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 else return -1 } // end if else return -1

25. Prefix Expressions • IsPre Function • Uses endPre() to determine whether a string is a prefix or not IsPre(): boolean size= length of expression strExp lastChar = endpre(0, size-1) if (lastChar >= 0 and lastChar == strExp.length()-1) return true else return false

26. 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 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 }

27. Review • ______ is a problem-solving technique that involves guesses at a solution. • Recursion • Backtracking • Iteration • Induction

28. 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

29. Review • A language is a set of strings of ______. • numbers • letters • Alphabets • symbols

30. 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

31. Review • The Java ______ determines whether a given string is a syntactically correct Java program. • applet • editor • compiler • programmer

32. 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

33. Review • Which of the following strings is a palindrome? • “bottle” • “beep” • “coco” • “r

34. 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

35. Review • Which of the following is an infix expression? • / a + b c • a b c + / • a b / + c • a / (b + c)

36. Review • What is the value of the prefix expression: – * 3 8 + 6 5? • 11 • 23 • 13 • 21