Sure, I can help with that. Here is a Python program that converts infix expressions to postfix notation:

```python
class Conversion:
def __init__(self, capacity):
self.top = -1
self.capacity = capacity
self.array = []
self.precedence = {'+':1, '-':1, '*':2, '/':2, '^':3}
self.result = []

def isEmpty(self):
return True if self.top == -1 else False

def peek(self):
return self.array[-1]

def pop(self):
if not self.isEmpty():
self.top -= 1
return self.array.pop()
else:
return "$"

def push(self, op):
self.top += 1
self.array.append(op)

def isOperand(self, ch):
return ch.isalpha() or ch.isdigit()

def notGreater(self, i):
try:
a = self.precedence[i]
b = self.precedence[self.peek()]
return True if a

Question

Sure, I can help with that. Here is a Python program that converts infix expressions to postfix notation:

```python
class Conversion:
    def __init__(self, capacity):
        self.top = -1
        self.capacity = capacity
        self.array = []
        self.precedence = {'+':1, '-':1, '*':2, '/':2, '^':3}
        self.result = []
 
    def isEmpty(self):
        return True if self.top == -1 else False
 
    def peek(self):
        return self.array[-1]
 
    def pop(self):
        if not self.isEmpty():
            self.top -= 1
            return self.array.pop()
        else:
            return "$"
 
    def push(self, op):
        self.top += 1
        self.array.append(op)
 
    def isOperand(self, ch):
        return ch.isalpha() or ch.isdigit()
 
    def notGreater(self, i):
        try:
            a = self.precedence[i]
            b = self.precedence[self.peek()]
            return True if a <= b else False
        except KeyError:
            return False
 
    def infixToPostfix(self, exp):
        for i in exp:
            if self.isOperand(i):
                self.result.append(i)
            elif i  == '(':
                self.push(i)
            elif i == ')':
                while self.peek() != '(':
                    self.result.append(self.pop())
                self.pop()
            else:
                while (not self.isEmpty() and self.notGreater(i)):
                    self.result.append(self.pop())
                self.push(i)
        while not self.isEmpty():
            self.result.append(self.pop())
        print("".join(self.result))
 
exp = "3+4*5/(6-2)"
obj = Conversion(len(exp))
obj.infixToPostfix(exp)
```

This program uses a stack to hold operators and parentheses during the conversion process. It scans the infix expression from left to right, and does the following for every scanned character:

- If the scanned character is an operand, it is added to the output.
- If the scanned character is an operator, then it is pushed onto the stack.
- If the scanned character is an '(', it is pushed onto the stack.
- If the scanned character is a ')', then it pops and outputs from the stack until an '(' is encountered.
- If the stack is not empty and the precedence of the scanned operator is less than or equal to the precedence of the operator in the stack, it pops and adds to the output all the operators from the stack which are greater than or equal to in precedence the scanned operator. Then it pushes the scanned operator to the stack.

Knowee AI · Accepted Answer

Sure, I can help with that. Here is a Python program that converts infix expressions to postfix notation:

```python
class Conversion:
    def __init__(self, capacity):
        self.top = -1
        self.capacity = capacity
        self.array = []
        self.precedence = {'+':1, '-':1, '*':2, '/':2, '^':3}
        self.result = []
 
    def isEmpty(self):
        return True if self.top == -1 else False
 
    def peek(self):
        return self.array[-1]
 
    def pop(self):
        if not self.isEmpty():
            self.top -= 1
            return self.array.pop()
        else:
            return "$"
 
    def push(self, op):
        self.top += 1
        self.array.append(op)
 
    def isOperand(self, ch):
        return ch.isalpha() or ch.isdigit()
 
    def notGreater(self, i):
        try:
            a = self.precedence[i]
            b = self.precedence[self.peek()]
            return True if a <= b else False
        except KeyError:
            return False
 
    def infixToPostfix(self, exp):
        for i in exp:
            if self.isOperand(i):
                self.result.append(i)
            elif i  == '(':
                self.push(i)
            elif i == ')':
                while self.peek() != '(':
                    self.result.append(self.pop())
                self.pop()
            else:
                while (not self.isEmpty() and self.notGreater(i)):
                    self.result.append(self.pop())
                self.push(i)
        while not self.isEmpty():
            self.result.append(self.pop())
        print("".join(self.result))
 
exp = "3+4*5/(6-2)"
obj = Conversion(len(exp))
obj.infixToPostfix(exp)
```

This program uses a stack to hold operators and parentheses during the conversion process. It scans the infix expression from left to right, and does the following for every scanned character:

- If the scanned character is an operand, it is added to the output.
- If the scanned character is an operator, then it is pushed onto the stack.
- If the scanned character is an '(', it is pushed onto the stack.
- If the scanned character is a ')', then it pops and outputs from the stack until an '(' is encountered.
- If the stack is not empty and the precedence of the scanned operator is less than or equal to the precedence of the operator in the stack, it pops and adds to the output all the operators from the stack which are greater than or equal to in precedence the scanned operator. Then it pushes the scanned operator to the stack.

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