The document discusses converting infix notation to postfix notation. It explains that postfix does not require parentheses and presents an algorithm that uses an operator precedence stack to convert an infix expression to postfix by pushing operators and operands to a stack according to precedence. The algorithm pops operators to add them to the postfix string when an operator with higher precedence is seen.

1. Class No.07 Data Structures http://ecomputernotes.com

2. Infix to Postfix Infix Postfix A + B A B + 12 + 60 – 23 12 60 + 23 – (A + B)*(C – D ) A B + C D – * A  B * C – D + E/F A B  C*D – E F/+ http://ecomputernotes.com

3. Infix to Postfix Note that the postfix form an expression does not require parenthesis. Consider ‘4+3*5’ and ‘(4+3)*5’. The parenthesis are not needed in the first but they are necessary in the second. The postfix forms are: 4+3*5 435*+ (4+3)*5 43+5* http://ecomputernotes.com

4. Evaluating Postfix Each operator in a postfix expression refers to the previous two operands. Each time we read an operand, we push it on a stack. When we reach an operator, we pop the two operands from the top of the stack, apply the operator and push the result back on the stack. http://ecomputernotes.com

5. Evaluating Postfix Stack s; while( not end of input ) { e = get next element of input if( e is an operand ) s.push( e ); else { op2 = s.pop(); op1 = s.pop(); value = result of applying operator ‘e’ to op1 and op2; s.push( value ); } } finalresult = s.pop(); http://ecomputernotes.com

6. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 http://ecomputernotes.com

7. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 http://ecomputernotes.com

8. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 http://ecomputernotes.com

9. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 http://ecomputernotes.com

10. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 http://ecomputernotes.com

11. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 http://ecomputernotes.com

12. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 8 6 5 1 1,3,8 http://ecomputernotes.com

13. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 8 6 5 1 1,3,8 2 6 5 1 1,3,8,2 http://ecomputernotes.com

14. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 8 6 5 1 1,3,8 2 6 5 1 1,3,8,2 / 8 2 4 1,3,4 http://ecomputernotes.com

15. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 8 6 5 1 1,3,8 2 6 5 1 1,3,8,2 / 8 2 4 1,3,4 + 3 4 7 1,7 http://ecomputernotes.com

16. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 8 6 5 1 1,3,8 2 6 5 1 1,3,8,2 / 8 2 4 1,3,4 + 3 4 7 1,7 * 1 7 7 7 http://ecomputernotes.com

17. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 8 6 5 1 1,3,8 2 6 5 1 1,3,8,2 / 8 2 4 1,3,4 + 3 4 7 1,7 * 1 7 7 7 2 1 7 7 7,2 http://ecomputernotes.com

18. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 8 6 5 1 1,3,8 2 6 5 1 1,3,8,2 / 8 2 4 1,3,4 + 3 4 7 1,7 * 1 7 7 7 2 1 7 7 7,2  7 2 49 49 http://ecomputernotes.com

19. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 8 6 5 1 1,3,8 2 6 5 1 1,3,8,2 / 8 2 4 1,3,4 + 3 4 7 1,7 * 1 7 7 7 2 1 7 7 7,2  7 2 49 49 3 7 2 49 49,3 http://ecomputernotes.com

20. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 8 6 5 1 1,3,8 2 6 5 1 1,3,8,2 / 8 2 4 1,3,4 + 3 4 7 1,7 * 1 7 7 7 2 1 7 7 7,2  7 2 49 49 3 7 2 49 49,3 + 49 3 52 52 http://ecomputernotes.com

21. Evaluating Postfix Evaluate 6 2 3 + - 3 8 2 / + * 2  3 + Input op1 op2 value stack 6 6 2 6,2 3 6,2,3 + 2 3 5 6,5 - 6 5 1 1 3 6 5 1 1,3 8 6 5 1 1,3,8 2 6 5 1 1,3,8,2 / 8 2 4 1,3,4 + 3 4 7 1,7 * 1 7 7 7 2 1 7 7 7,2  7 2 49 49 3 7 2 49 49,3 + 49 3 52 52 http://ecomputernotes.com

22. Converting Infix to Postfix Consider the infix expressions ‘A+B*C’ and ‘ (A+B)*C’. The postfix versions are ‘ABC*+’ and ‘AB+C*’. The order of operands in postfix is the same as the infix. In scanning from left to right, the operand ‘A’ can be inserted into postfix expression. http://ecomputernotes.com

23. Converting Infix to Postfix The ‘+’ cannot be inserted until its second operand has been scanned and inserted. The ‘+’ has to be stored away until its proper position is found. When ‘B’ is seen, it is immediately inserted into the postfix expression. Can the ‘+’ be inserted now? In the case of ‘A+B*C’ cannot because * has precedence. http://ecomputernotes.com

24. Converting Infix to Postfix In case of ‘(A+B)*C’, the closing parenthesis indicates that ‘+’ must be performed first. Assume the existence of a function ‘prcd(op1,op2)’ where op1 and op2 are two operators. Prcd(op1,op2) returns TRUE if op1 has precedence over op2, FASLE otherwise.

25. Converting Infix to Postfix prcd(‘*’,’+’) is TRUE prcd(‘+’,’+’) is TRUE prcd(‘+’,’*’) is FALSE Here is the algorithm that converts infix expression to its postfix form. The infix expression is without parenthesis.

26. Converting Infix to Postfix Stack s; While( not end of input ) { c = next input character; if( c is an operand ) add c to postfix string; else { while( !s.empty() && prcd(s.top(),c) ){ op = s.pop(); add op to the postfix string; } s.push( c ); } while( !s.empty() ) { op = s.pop(); add op to postfix string; }

27. Converting Infix to Postfix Example: A + B * C symb postfix stack A A

28. Converting Infix to Postfix Example: A + B * C symb postfix stack A A + A +

29. Converting Infix to Postfix Example: A + B * C symb postfix stack A A + A + B AB +

30. Converting Infix to Postfix Example: A + B * C symb postfix stack A A + A + B AB + * AB + *

31. Converting Infix to Postfix Example: A + B * C symb postfix stack A A + A + B AB + * AB + * C ABC + *

32. Converting Infix to Postfix Example: A + B * C symb postfix stack A A + A + B AB + * AB + * C ABC + * ABC * +

33. Converting Infix to Postfix Example: A + B * C symb postfix stack A A + A + B AB + * AB + * C ABC + * ABC * + ABC * +

34. Converting Infix to Postfix Handling parenthesis When an open parenthesis ‘(‘ is read, it must be pushed on the stack. This can be done by setting prcd(op,‘(‘ ) to be FALSE. Also, prcd( ‘(‘,op ) == FALSE which ensures that an operator after ‘(‘ is pushed on the stack.

35. Converting Infix to Postfix When a ‘)’ is read, all operators up to the first ‘(‘ must be popped and placed in the postfix string. To do this, prcd( op,’)’ ) == TRUE. Both the ‘(‘ and the ‘)’ must be discarded: prcd( ‘(‘,’)’ ) == FALSE. Need to change line 11 of the algorithm.

36. Converting Infix to Postfix if( s.empty() || symb != ‘)’ ) s.push( c ); else s.pop(); // discard the ‘(‘ prcd( ‘(‘, op ) = FALSE for any operator prcd( op, ‘)’ ) = FALSE for any operator other than ‘)’ prcd( op, ‘)’ ) = TRUE for any operator other than ‘(‘ prcd( ‘)’, op ) = error for any operator.