Q.Figure 3.4 shows the step-by-step process of evaluation of the postfix expression 7 8 2 * 4 / + using Algorithm 3.2.

Shape of Question

This is a plain theory question asking for a complete exam-ready explanation of the infix-to-postfix conversion concept, illustrated by the given figure showing postfix evaluation (not conversion). The figure demonstrates stack-based evaluation of 7 8 2 * 4 / +, which corresponds to the infix expression 7 + (8 * 2) / 4.

Infix to Postfix Conversion

What is Infix and Postfix?

Infix notation places operators between operands: A + B, (A + B) * C. It requires parentheses and precedence rules to resolve ambiguity.

Postfix notation (Reverse Polish Notation) places operators after their operands: A B +, A B + C *. No parentheses are needed; evaluation proceeds strictly left to right using a stack.

Why Convert Infix to Postfix?

Computers evaluate postfix expressions more efficiently. The stack-based algorithm is simple: push operands, pop two operands when an operator appears, apply the operator, push the result back. The figure demonstrates this for 7 8 2 * 4 / +:

Result: 11

Algorithm for Infix to Postfix Conversion

The conversion uses a stack to hold operators and follows operator precedence and associativity.

Steps:

1. Initialize an empty stack and an empty output string.
2. Scan the infix expression left to right, symbol by symbol.
3. For each symbol:
   • Operand (number/variable): Append directly to output.
   • Left parenthesis (: Push onto stack.
   • Right parenthesis ): Pop from stack to output until a ( is encountered; discard the (.
   • Operator (+, -, *, /, ^):
     • While the stack is not empty and the top is an operator with higher or equal precedence (for left-associative operators; strictly higher for right-associative like ^), pop to output.
     • Push the current operator onto the stack.
4. After scanning, pop all remaining operators from the stack to output.

Example: Convert A + B * C to Postfix

Postfix: A B C * + …

Why Postfix (and a Stack)?

Postfix notation—also called Reverse Polish Notation—places every operator after its operands. The beauty of postfix is that it needs no parentheses and no precedence rules: you simply scan left to right, and whenever you see an operator you know exactly which two values it acts on (the two most recent operands). A stack is the natural data structure because it gives you LIFO access: the last two numbers pushed are the first two popped when an operator arrives.

Infix (7 + ((8 * 2) / 4)) becomes postfix 7 8 2 * 4 / +. The postfix form encodes the same tree of operations but in an order a machine can execute in one linear pass.

Algorithm 3.2 (Postfix Evaluation)

1. Initialize an empty stack.
2. Scan the postfix expression from left to right, symbol by symbol.
3. For each symbol:
   • If it is an operand (a number), push it onto the stack.
   • If it is an operator (+, -, *, /, etc.):
     • Pop the top element → call it B (the second operand).
     • Pop the next top element → call it A (the first operand).
     • Compute A operator B.
     • Push the result back onto the stack.
4. After scanning all symbols, the stack contains exactly one element: the final answer. Pop and return it.

Step-by-Step Trace of 7 8 2 * 4 / +

We process each symbol in turn. The table below shows the action taken and the stack contents after that action.

The single remaining element is 11.

Python Implementation

Below is a complete function that implements Algorithm 3.2. It assumes the postfix expression is given as a string of space-separated tokens.

def evaluate_postfix(expression):
    """
    Evaluate a postfix expression given as a space-separated string.
    Returns the numeric result.
    """
    stack = []
    tokens = expression.split()
    
    for token in tokens:
        if token in {'+', '-', '*', '/'}:
            # Pop two operands (order matters!)
            b = stack.pop()  # right operand
            a = stack.pop()  # left operand
            
            # Apply the operator
            if token == '+':
                result = a + b
            elif token == '-':
                result = a - b
            elif token == '*':
                result = a * b
            elif token == '/':
                result = a / b  # use // for integer division if needed
            
            stack.append(result)
        else:
            # It's an operand; convert to number and push
            stack.append(float(token)) …