Skip to main content

Infix, prefix and postfix notations

Mathematical expressions can be written in 3 different methods - infix, postfix and prefix.

Infix notation

All these days we have been using infix notation. In this operators are written between the operands. Some operators are having higher priorities than others. If some expressions are enclosed in brackets, then they are evaluated first.

e.g.
    (A+B)*C
Here + operator will operate on A and B as it is between them. Though * has higher priority, as A+B is enclosed in brackets, that is evaluated before multiplication.

Postfix notation

In a postfix notation - also called Reverse Polish Notation, operator is written after the operands. The operator operates on two operands to its immediate left. And expressions are always evaluated from left to right. Brackets do not have any effect on this.

e.g.
(A+B)*C in postfix is  AB+C*
A+B*C+D in postfix is ABC*+D+

Prefix notation

In a prefix notation, operator is written before the operands. And just like postfix, expressions are evaluated from left to right and brackets do not have any effect.

e.g.
(A+B)*C is prefix is   *+ABC
A+B*C+D in prefix is   ++A*BCD

Postfix notation is quite useful in processing of expression as it can be evaluated in single parsing.

Next let us consider how to convert from one notation to another.

Conversion of an infix expression to postfix expression

In order to convert an infix expression to postfix, we need to use an operator stack.

While traversing all the operands are transferred to output expression. When  an operator is encountered it is pushed to stack. But if the stack already has higher or equal priority operators, these are popped out and added to output expression. If there is a bracket - opening bracket, that too is pushed to stack. But when there is a closing bracket, then all the operators in the stack up to opening brackets are to be popped out of stack and added to output expression. In the end, if the stack has any more operators, they are popped and added to output expression.

Here is the algorithm

  •  Create an empty stack called opstack for keeping operators. Create an empty array for output.
  • Scan the token list from left to right.
    • If the token is an operand, append it to the end of the output string.
    • If the token is a left parenthesis, push it on the opstack.
    • If the token is a right parenthesis, pop the opstack until the corresponding left parenthesis is removed. Append each operator to the end of the output string.
    • If the token is an operator, *, /, +, or -, push it on the opstack. However, first remove any operators already on the opstack that have higher or equal precedence and append them to the output string.
  •  When the input expression has been completely processed, check the operator stack. Any operators still on the stack can be removed and appended to the end of the output string.
Here is the function for conversion. 

int is_operator(char ch)
{
   switch(ch)
   {
      case '+':
      case '-':
      case '/':
      case '*':
      case '%':
      case '^':
                return 1;
      default : return 0;
   }
}
int priority(char ch)
{
 switch(ch)
   {
      case '+':
      case '-':return 1;
      case '/':
      case '*':
      case '%':return 2;
      case '^':return 3;
                 
      default : return 0;
   }
}

void convert(char *input, char *output)
{
   struct stack s1;
   s1.top = -1;
   while(*input)
   {
 char ch = *input++;
        char ch2;
 if(ch=='(')
    push(&s1,ch);
        else if(ch==')')
    {
              while(!is_empty(&s1)  && ( (ch=pop(&s1))!='('))
                  *output++ = ch;
            }
        else if(is_operator(ch))
           {
              while(!is_empty(&s1)  &&  priority(peek(&s1))>=priority(ch))
                    *output++ = pop(&s1);
              push(&s1,ch);
           }
        else
           *output++ = ch;
    }
    while(!is_empty(&s1))
       *output++ = pop(&s1);
    *output = 0;
}

Here is the complete program

#include<stdio.h>
#include"stack.h"
int is_operator(char ch)
{
   switch(ch)
   {
      case '+':
      case '-':
      case '/':
      case '*':
      case '%':
      case '^':
                return 1;
      default : return 0;
   }
}
int priority(char ch)
{
 switch(ch)
   {
      case '+':
      case '-':return 1;
      case '/':
      case '*':
      case '%':return 2;
      case '^':return 3;
                 
      default : return 0;
   }
}

void convert(char *input, char *output)
{
   struct stack s1;
   s1.top = -1;
   while(*input)
   {
 char ch = *input++;
        char ch2;
 if(ch=='(')
    push(&s1,ch);
        else if(ch==')')
    {
              while(!is_empty(&s1)  && ( (ch=pop(&s1))!='('))
                  *output++ = ch;
            }
        else if(is_operator(ch))
           {
              while(!is_empty(&s1)  &&  priority(peek(&s1))>=priority(ch))
                    *output++ = pop(&s1);
              push(&s1,ch);
           }
        else
           *output++ = ch;
    }
    while(!is_empty(&s1))
       *output++ = pop(&s1);
    *output = 0;
}
                   
int main()
{
   char infix[40],postfix[40];
   printf("Enter infix expression:");
   scanf("%s",infix);
   convert(infix,postfix);
   printf("the postfix expression is %s\n",postfix);
   return 0;
}

The program uses a header file stack.h which has declarations of push, pop, is_empty and peek. It also has struct stack declaration and MAX.

The program used array based stack implementation. Here is charstack.c which has to be compiled along with the above program


#include<stdio.h>
#include"stack.h"
void push(struct stack *s, char val)
{
   if( is_full(s))
      printf("Stack overflow");
   else
   {
 (s->top)++;
 s->arr[s->top]=val;
   }    
}
int is_empty(struct stack *s)
{
    return s->top==-1;
}
int is_full(struct stack *s)
{
    return s->top>=MAX;
}
char pop(struct stack *s)
{
    int val = 0;
    if(is_empty(s))
      printf("Stack empty");
    else
     {
 char t = s->arr[s->top];
        s->top--;
        val = t;
      }
    return val;
}
char peek(struct stack *s)
{
    if(is_empty(s))
         printf("Stack empty");
    else
       return s->arr[s->top];
}

e.g.
Let me take simplest example
(A+B)*C

Character      Stack       Output
(                      (
A                     (               A
+                     (+            A
B                      (+           AB
)                       (              AB+
                                        AB+
*                       *              AB+
C                      *              AB+C
--------------------------------------------------------End of expression
                                        AB+C*


So now we have our postfix expression. Try this out for lengthier expressions yourself


                                    
Labels:

Comments

Post a Comment

Popular posts from this blog

Linked list in C++

A linked list is a versatile data structure. In this structure, values are linked to one another with the help of addresses. I have written in an earlier post about how to create a linked list in C.  C++ has a library - standard template library which has list, stack, queue etc. data structures. But if you were to implement these data structures yourself in C++, how will you implement? If you just use new, delete, cout and cin, and then claim it is your c++ program, you are not conforming to OOPS concept. Remember you have to "keep it together". Keep all the functions and variables together - in a class. You have to have class called linked list in which there are methods - append, delete, display, insert, find, find_last. And there will also be a data - head. Defining node We need a structure for all these nodes. A struct can be used for this purpose, just like C. struct node { int val; struct node * next; }; Next we need to define our class. W
Post a Comment
Read more

Swap nodes of a linked list

Qn: Write a function to swap the adjacent nodes of a singly linked list.i.e. If the list has nodes as 1,2,3,4,5,6,7,8, after swapping, the list should be 2,1,4,3,6,5,8,7 Image from: https://tekmarathon.com Though the question looks simple enough, it is tricky because you don't just swap the pointers. You need to take care of links as well. So let us try to understand how to go about it. Take two adjacent nodes p1 and p2 Let prevnode be previous node of p1 Now link prevnode to p2 Link p2 to p1 Link p1 to next node of p2 So the code will be prevnode -> next = p2; p1 -> next = p2 -> next; p2 -> next = p1; But what about the start node or head? head node does not have previous node If we swap head with second node, modified head should be sent back to caller  To take care of swapping first and second nodes, we can write p1 = head; p2 = head -> next; p1 -> next = p2 -> next; p2 -> next = p1; head = p2;  Now we are read
Post a Comment
Read more

Binary tree deletion - non-recursive

In the previous post we have seen how to delete a node of a binary search tree using recursion. Today we will see how to delete a node of BST using a non-recursive function. Let us revisit the 3 scenarios here Deleting a node with no children just link the parent to NULL Deleting a node with one child link the parent to  non-null child of node to be deleted Deleting a node with both children select the successor of node to be deleted copy successor's value into this node delete the successor In order to start, we need a function to search for a node in binary search tree. Did you know that searching in  a BST is very fast, and is of the order O(logn). To search Start with root Repeat until value is found or node is NULL If the search value is greater than node branch to right If the search value is lesser than node branch to left.  Here is the function NODEPTR find_node (NODEPTR root,NODEPTR * parent, int delval) { NODEPTR nd = root; NODEPTR pa = root; if (root -> v
Post a Comment
Read more