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CS2006 - Data Structures I Chapter 7 Stacks III. 2 Topics Applications Infix to postfix expression Evaluate postfix expression.

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Presentation on theme: "CS2006 - Data Structures I Chapter 7 Stacks III. 2 Topics Applications Infix to postfix expression Evaluate postfix expression."— Presentation transcript:

1 CS Data Structures I Chapter 7 Stacks III

2 2 Topics Applications Infix to postfix expression Evaluate postfix expression

3 3 Infix Expressions Binary operators appear between operands: W - X / Y – Z (4+3)*2 (2+3)/(9-4) Order of evaluation is determined by: precedence rules parentheses association (L to R or R to L)

4 4 Application: Algebraic Expressions Infix Expression Evaluation To evaluate an infix expression Convert the infix expression into postfix Evaluate the postfix expression using stacks

5 5 Postfix Expressions Binary operators appear after both operands: X + Y in postfix form is: X Y + Order of operations is completely determined by the expression no parentheses or precedence required for example: A * B - C becomes A B * C -

6 6 Postfix Expression InfixPostfix 6-1 (4+3)*2 (2+3)/(9-4)

7 7 Postfix Expression Evaluation Assumptions: The string is syntactically correct postfix expression No unary operators are present No exponentiation operators are present

8 8 Postfix Expression Evaluation Evaluation of postfix expressions makes use of an operand stack. Algorithm: Parse expression from left to right When an operand is encountered, push it onto stack when an operator is encountered, pop the last two operands off the stack and apply the operation, and push the result onto the stack when the expression is completely scanned, the final result will be on the stack

9 9 Postfix Expression Evaluation Pseudocode: for ( each character ch in the string ) { if (ch is an operand ) Push value that operand ch represents onto stack else/ / ch is an operator named op {/ / evaluate and push the result operand2 = top of stack Pop the stack operand1 = top of stack Pop the stack result = operand1 op operand2 Push Result onto stack } / / end if }/ / end for

10 10 Example Evaluate * * + *

11 11 Converting Infix into Postfix Since postfix expressions are easily evaluated, the easiest way to evaluate infix is to convert from infix to postfix and then evaluate. This conversion uses an operator stack since low precedence operators must be saved and applied after high precedence ones.

12 12 Converting Infix into Postfix The operands always stay in the same order with respect to one another An operator will move only to “ the right” with respect to the operands If in infix expression, the operand x precedes the operator op, in the postfix, the operand x also precedes the operator x All parentheses are removed

13 13 Converting Infix into Postfix Scan infix string from left to right Each time an operand is encountered, copy it to output When a bracket is encountered, check its orientation. Push left brackets onto stack If it is a right bracket, pop all operators on the stack and copy them to output until a matching left bracket is encountered. Discard the left bracket

14 14 Converting Infix into Postfix When an operator is encountered, check the top item of the stack If the priority is >= the current operator, pop the top operator and copy it to output Continue until an operator of lower priority is encountered or until the stack is empty Finally, push current operator onto the stack When the end of the expression is reached, copy the remaining stack contents to output in the order popped

15 15 Converting Infix into Postfix Precedence Table OperatorNamePrecedence when on stack Precedence when on input (Opening parentheses always stack 0 7 )Closing parenthesesnever stacked0 ^Exponentiation65 * /Multiply & divide43 + -Add & Subtract21

16 16 Conversion Example W - X /Y + Z '-' is pushed onto operator stack '/' has higher precedence than '-' on stack - it gets pushed '+' has lower precedence than '/' on stack - pop '/' and output - then '-' is on top of stack - since subtraction associates left to right - pop '-' off stack and output - push '+' onto stack when end of expression is reached pop remaining operator and output resulting expression is: W X Y /- Z +

17 17 Converting Infix into Postfix stack.push(EOL marker) for (each character ch in the infix expression) { switch (ch) { case operand: //append ch to output postfix expression postfixExpr = postfixExpr + ch; case '(': stack.push(ch); // push ch onto operator stack case ')': while (top of stack is not '(' ) { postfixExpr = postfixExpr + stack.pop(); // pop and append expr. } stack.pop(); case operator: while (!stack.isEmpty() && inputPrec(ch) <= stackPrec(stack.top())) { postfixExpr = postfixExpr + stack.pop(); } stack.push(ch); } // end for while (!stack.isEmpty() { // pop all remaining operators off stack and append postfixExpr = postfixExpr + stack.pop(); }

18 18 Conversion Example * [(7-5)/2] copy into output ‘+' is pushed onto operator stack 3 copy into output ‘*' has higher precedence than ‘+' on stack - it gets pushed ‘[‘ is left bracket, push it onto stack ‘(‘is left bracket, push it onto stack 7 is copied into output ‘-' has higher precedence than ‘(‘, push onto stack

19 19 Conversion Example * [(7-5)/2]

20 20 Prefix Expressions InfixPrefix (4+3)*2* (2+3)/(9-4)/ – 9 4

21 21 Review Which of the following is NOT true about converting infix expressions to postfix expressions? the operands always stay in the same order with respect to one another the operators always stay in the same order with respect to one another an operator will move only “to the right” with respect to the operands all parentheses are removed

22 22 Review Which of the following is the postfix form of the infix expression: (a + b) * c / d a b + c * d / a b * c / d + a + b * c / d a b + c d * /

23 23 Review StackInterface provides the specifications for ______. only the array-based implementation of a stack only the reference-based implementation of a stack only an implementation of a stack that uses the ADT list all the implementations of a stack

24 24 Review In the StackInterface class, the push method accepts as its parameter an item that is an instance of ______. Integer Double String Object


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