Question

Using ADT Stack: Evaluating infix expressions by converting them to postfix expressions Postfix notation: In a postfix expression, a binary operation follows its two opperands. The order of the operands in a infix expression is the same as in the corresponding postfix expression but the order of the operators might change based on the precedence of the operators and the existing of paranthses. Infix Postfix a + b a b + (a + b) * c a b + c * a + b * c a b c * +

1. Infix-to-postfix convertion To convert an infix to postfix expression we can use a stack to store temporary operators and open and close parantheses. Take the following actions which depend on the currently processed symbol from the infix expression (scanned from left to right). Construct the output (resulting) expression, by appending to its end the correct items (operands or operators). If Operand: Append the operand to the end of the output expression. If Operator ^: Push it onto the stack. If Operator +, -, *, /: - If the stack is empty or its top entry is an operator that has a lower precedence than the new operator, push the new operator onto the stack (Note: ‘*’ and ‘/’ have higher precedence than ‘+’ and ‘-‘.) - Otherwise pop operators from the stack and append them to the output expression until the stack is empty, or its top entry has a lower precedence than the new operator. Then push the new operator onto the stack. If Open paranthesis: Push ( onto the stack. If Close paranthesis: Pop operators from the stack and append them to the output expression until an open paranthesis is poped. Then discard both parantheses.

Problem 1 ) Convert the following infix expressions to postfix expressions using the rules given above: (a – b * c ) / (d * e + (d – c))

2.Evaluating postfix expressions To evaluate a postfix expression: scan the expression and take the following actions depending on the currently processed symbol: If Operand: push it onto the stack If Operator: - pop an operand from the stack (this will be the second operand, which has been pushed last); - pop another operand from the stack (this will be the first operand) - perform the operation and push the result back onto the stack.

Problem 2) Using the rules given above, evaluate the following postfix expression, if a = 2, b = 3, c = 4, d = 5: a b c - / d *

The language is Java

Answer 1

================ code for InfixToPostfix ================================java used============

import java.util.Stack;

class InfixToPostFix
{
////////////////////////////////////////////////
   // calculating precedence order
   static int PrecedenceOrder(char ch)
   {
       switch (ch)
       {
       case '+':
       case '-':
           return 1;
  
       case '*':
       case '/':
           return 2;
  
       case '^':
           return 3;
       }
       return -1;
   }
////////////////////////////////////////////////////  
   // Method that converts infix to postfix expression
   static String myInToPostfix(String exp)
   {
      
       String result = new String("");
      
       // a empty stack
       Stack<Character> stack = new Stack<>();
      
       for (int i = 0; i<exp.length(); ++i)
       {
           char c = exp.charAt(i);
          
           // scanning character is lettre or digit
           if (Character.isLetterOrDigit(c))
               result += c;
          
           // character is an '(' do it
           else if (c == '(')
               stack.push(c);
          
           // character is ) loop until ( found
           else if (c == ')')
           {
               while (!stack.isEmpty() && stack.peek() != '(')
                   result += stack.pop();
              
               if (!stack.isEmpty() && stack.peek() != '(')
                   return "Expression is not valid---";                  
               else
                   stack.pop();
           }
           else // operator found
           {
               while (!stack.isEmpty() && PrecedenceOrder(c) <= PrecedenceOrder(stack.peek())){
                   if(stack.peek() == '(')
                       return "Expression is not valid--";
                   result += stack.pop();
           }
               stack.push(c);
           }
  
       }
  
       // poping the operators
       while (!stack.isEmpty()){
           if(stack.peek() == '(')
               return "Expression is invalid----";
           result += stack.pop();
       }
       return result;
   }
   ////////////////////////////////////////////////////
     
   public static void main(String[] args)
   {
       String exp = "a+b*(c^d-e)^(f+g*h)-i";
       System.out.println(myInToPostfix(exp));
exp = "a + b a b + (a + b) * c a b + c * a + b * c a b c * +";
System.out.println(myInToPostfix(exp));
   }
}

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Screenshots of the code and the output are:

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