Algorithm for converting an infix expression to a postfix expression using a stack structure:
- Scan the infix expression from left to right, one token at a time (where a token can be an operand, operator, or parenthesis)
- For each token in the input expression, do the following:
1. If the token is an operand (i.e. a number), append it to the output string
2. If the token is an open parenthesis ' \( ( \) ', push it onto the stack
3. If the token is a closing parenthesis ')', pop operators from the stack and append them to the output string until an open parenthesis is encountered. Pop and discard the '(' from the stack.
4. If the token is an operator, pop operators from the stack and append them to the output string while they have higher or equal precedence compared to the current operator. Then push the current operator onto For instance, if the stack contains "s" and the current operator is "+", pop "s" and push "+", leaving only one operator in the stack (i.e. [+<-).
However, if the stack contains " + " and the current operator is " "*", simply push "s" onto the stack, resulting in two operators in the stack (i.e. [+, *<-).
- After processing all tokens, pop any remaining operators from the stack and append them to the output string.
For example, consider the infix expression " \( 1+2 \) ". The steps are:
- append 1
- push +
- append 2
- pop and append +
Now, convert the infix expression " \( 6-8 \) " 7 " to postfix and write the steps in the answer box below.
Answer: (penalty regime: \( 0,0,5,10,15,20,25,30,35,40,45,50 \% \) )
\begin{tabular}{l|l|}
\hline Step 1: & append 6 \\
Step 2: & push - \\
Step 3: & append 8 \\
Step 4: & push \( * \) \\
Step 5: & append 7 \\
step 6: & end of expression \\
Step 7: & pop * \\
\hline
\end{tabular}
Check
\begin{tabular}{|l|l|l|}
\hline Step & Feedback & \\
\hline 1 & correct & \( \checkmark \) \\
\hline 2 & Your solution: push - is incorrect. & \( \times \) \\
\hline 3 & Correct & \( \checkmark \) \\
\hline 4 & Correct & \( \checkmark \) \\
\hline 5 & Your solution: append 7 is incorrect. & \( \times \) \\
\hline 6 & Your solution: end of expression is incorrect. & \( \times \) \\
\hline 7 & Your solution: pop * is incorrect. & \( \times \) \\
\hline
\end{tabular}
Incorrect.
Incorrect
Marks for this submission: \( 0.00 / 1.00 \).
