Convert the following expressions from infix to Reverse Polish Notation (postfix). Think of Squareroot and tan as unary operators: ((2 - 1) (4 * 3)) + 15 -b+ Squareroot b^- 4ac/2a tan alpha + tan beta/1 - tan alpha tan beta Solution To convert an expression from infix to postfix, the method is: 1. If it is an operand at it to output expression. 2. If it is an operator, if stack is empty, or if the operator on top of stack is of lower precedence than this, push it onto stack. 3. Else, pop operators from the stack, till the operators on stack are of higher precedence than the one in hand, and add them to the output. Then, add this operator to the stack. So, applying this logic to the expressions: 1. ((2-1) / (4 / 3)) + 15 (, Push it onto stack. (, Push it onto stack. 2 Add it to output, so output expression now is: 2 - Push it onto stack. 1 Add it to output, so output expression now is: 2 1 ) Now, pop elements and add it to output, till a open brace. So, output expression is: 2 1 -, and stack will now hold (. / Push it onto stack. ( Push it onto stack. 4 Add it to output, so output expression now is: 2 1 - 4 * Push it onto stack. 3 Add it to ouput, so output expression now is: 2 1 - 4 3 ) Now, pop elements and add it to output, till a open brace. So, output expression is: 2 1 - 4 3 *, and stack will now hold ( /. ) Now, pop elements and add it to output, till a closed brace. So, output expression is: 2 1 - 4 3 * /, and stack is now empty. + Push it onto stack. 15 Add it to output, so output expression now is: 2 1 - 4 3 * / 15. As it is the end of expression, pop all elements from stack, and add it to output: So, now the expression is: 2 1 - 4 3 * / 15 +. Finally, the postfix equivalent of the given expression is: 2 1 - 4 3 * / 15 +..

1. Convert the following expressions from infix to Reverse Polish Notation (postfix). Think of
Squareroot and tan as unary operators: ((2 - 1) (4 * 3)) + 15 -b+ Squareroot b^- 4ac/2a tan
alpha + tan beta/1 - tan alpha tan beta
Solution
To convert an expression from infix to postfix, the method is:
1. If it is an operand at it to output expression.
2. If it is an operator, if stack is empty, or if the operator on top of stack is of lower precedence
than this, push it onto stack.
3. Else, pop operators from the stack, till the operators on stack are of higher precedence than the
one in hand, and add them to the output. Then, add this operator to the stack.
So, applying this logic to the expressions:
1. ((2-1) / (4 / 3)) + 15
(, Push it onto stack.
(, Push it onto stack.
2 Add it to output, so output expression now is: 2
- Push it onto stack.
1 Add it to output, so output expression now is: 2 1
) Now, pop elements and add it to output, till a open brace. So, output expression is: 2 1 -, and
stack will now hold (.
/ Push it onto stack.
( Push it onto stack.
4 Add it to output, so output expression now is: 2 1 - 4
* Push it onto stack.
3 Add it to ouput, so output expression now is: 2 1 - 4 3
) Now, pop elements and add it to output, till a open brace. So, output expression is: 2 1 - 4 3 *,
and stack will now hold ( /.
) Now, pop elements and add it to output, till a closed brace. So, output expression is: 2 1 - 4 3 *
/, and stack is now empty.
+ Push it onto stack.
15 Add it to output, so output expression now is: 2 1 - 4 3 * / 15.
As it is the end of expression, pop all elements from stack, and add it to output: So, now the
expression is: 2 1 - 4 3 * / 15 +.
Finally, the postfix equivalent of the given expression is:
2 1 - 4 3 * / 15 +.