Shift-reduce parsing in Compiler Design - Time & Space Complexity

Time Complexity: Shift-reduce parsing

O(n)

Understanding Time Complexity

When we analyze shift-reduce parsing, we want to understand how the time it takes grows as the input gets longer.

We ask: How does the parser's work increase when the input size increases?

Scenario Under Consideration

Analyze the time complexity of the following simplified shift-reduce parsing steps.

while (input not empty) {
  if (can reduce top of stack) {
    reduce();
  } else {
    shift();
  }
}
while (stack can be reduced) {
  reduce();
}
    

This code shows the parser shifting input symbols onto a stack or reducing them by grammar rules until the input is fully processed.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Shifting input symbols and reducing stack elements repeatedly.
• How many times: Each input symbol is shifted once, and reductions happen as needed, but total steps relate to input length.

How Execution Grows With Input

As input size grows, the parser processes each symbol mostly once, with some extra steps for reductions.

Input Size (n)	Approx. Operations
10	About 20-30 steps
100	About 200-300 steps
1000	About 2000-3000 steps

Pattern observation: The total steps grow roughly in proportion to the input size, maybe a few times more due to reductions.

Final Time Complexity

Time Complexity: O(n)

This means the parsing time grows linearly with the length of the input; doubling input roughly doubles the work.

Common Mistake

[X] Wrong: "Shift-reduce parsing always takes quadratic time because of repeated reductions."

[OK] Correct: Each input symbol is shifted once, and reductions are limited by grammar rules, so total steps grow linearly, not quadratically.

Interview Connect

Understanding how shift-reduce parsing scales helps you explain parser efficiency clearly and shows you grasp practical compiler design concepts.

Self-Check

"What if the grammar allowed ambiguous reductions causing backtracking? How would the time complexity change?"

Practice

1. What is the main purpose of shift in shift-reduce parsing?

easy

A. To move the next input symbol onto the stack

B. To remove symbols from the stack

C. To check the grammar rules

D. To accept the input string

Solution

1. Step 1: Understand the role of shift

   In shift-reduce parsing, shifting means taking the next input symbol and placing it on the stack.

2. Step 2: Differentiate shift from reduce

   Reduce means replacing symbols on the stack using grammar rules, while shift just moves input symbols onto the stack.

3. Final Answer:

   To move the next input symbol onto the stack -> Option A

4. Quick Check:

   Shift = move input symbol to stack [OK]

Hint: Shift means push next input symbol onto stack [OK]

Common Mistakes:

• Confusing shift with reduce
• Thinking shift removes symbols
• Assuming shift checks grammar

2. Which of the following is the correct sequence of actions in shift-reduce parsing?

easy

A. Accept, Shift, Reduce

B. Reduce, Shift, Accept

C. Shift, Reduce, Accept

D. Shift, Accept, Reduce

Solution

1. Step 1: Recall the parsing process order

   Shift-reduce parsing first shifts symbols onto the stack, then reduces them using grammar rules, and finally accepts if input is valid.

2. Step 2: Match the correct sequence

   The correct order is Shift, then Reduce, then Accept.

3. Final Answer:

   Shift, Reduce, Accept -> Option C

4. Quick Check:

   Shift -> Reduce -> Accept [OK]

Hint: Remember parsing order: shift first, then reduce [OK]

Common Mistakes:

• Mixing up reduce and shift order
• Thinking accept happens before reduce
• Assuming reduce happens before any shift

3. Given the grammar rules:
S -> aA
A -> b
And input string ab, what is the correct sequence of stack contents during shift-reduce parsing?

medium

A. [ ] -> [a] -> [b] -> [S]

B. [ ] -> [a] -> [S] -> [S, b]

C. [ ] -> [b] -> [a, b] -> [S]

D. [ ] -> [a] -> [a, b] -> [S]

Solution

1. Step 1: Shift input symbols onto stack

   First, shift 'a' onto the stack: stack = [a]. Then shift 'b': stack = [a, b].

2. Step 2: Apply reductions using grammar rules

   Reduce 'b' to 'A' using A -> b, then reduce 'aA' to 'S' using S -> aA. Final stack is [S].

3. Final Answer:

   [ ] -> [a] -> [a, b] -> [S] -> Option D

4. Quick Check:

   Shift a, shift b, reduce to S [OK]

Hint: Shift all input, then reduce using grammar rules [OK]

Common Mistakes:

• Reducing too early before all symbols shifted
• Mixing order of symbols on stack
• Skipping reduction steps

4. In a shift-reduce parser, if the parser tries to reduce a symbol that does not match any grammar rule, what is the likely problem?

medium

A. The parser is shifting symbols incorrectly

B. The input string is invalid or contains errors

C. The parser is correctly accepting the input

D. The grammar rules are too simple

Solution

1. Step 1: Understand reduction failure

   If no grammar rule matches the symbols on the stack, the parser cannot reduce, indicating a problem with the input.

2. Step 2: Interpret the failure meaning

   This usually means the input string does not follow the grammar rules and is invalid or contains errors.

3. Final Answer:

   The input string is invalid or contains errors -> Option B

4. Quick Check:

   Failed reduce = invalid input [OK]

Hint: No matching reduce means input error [OK]

Common Mistakes:

• Assuming parser is always correct
• Thinking reduce failure means grammar is simple
• Confusing shift errors with reduce errors

5. Consider the grammar:
E -> E + T | T
T -> int
For the input string int + int + int, which sequence of actions correctly shows the shift-reduce parsing steps to fully reduce the input?

hard

A. Shift int, reduce T, reduce E, shift +, shift int, reduce T, reduce E, shift +, shift int, reduce T, reduce E

B. Shift int, shift +, reduce T, reduce E, shift int, shift +, reduce T, reduce E, shift int, reduce E

C. Shift int, reduce T, shift +, reduce E, shift int, reduce T, shift +, reduce E, shift int, reduce T

D. Shift int, reduce E, shift +, shift int, reduce T, reduce E, shift +, shift int, reduce T, reduce E

Solution

1. Step 1: Shift and reduce first int

   Shift 'int', reduce to T, then reduce T to E (since E -> T).

2. Step 2: Process first plus and second int

   Shift '+', shift 'int', reduce 'int' to T, then reduce 'E + T' to E.

3. Step 3: Process second plus and third int

   Shift '+', shift 'int', reduce 'int' to T, then reduce 'E + T' to E again, fully reducing input.

4. Final Answer:

   Shift int, reduce T, reduce E, shift +, shift int, reduce T, reduce E, shift +, shift int, reduce T, reduce E -> Option A

5. Quick Check:

   Shift and reduce each int, then reduce E + T repeatedly [OK]

Hint: Shift int, reduce T, reduce E, repeat for each + int [OK]

Common Mistakes:

• Reducing E too early or too late
• Skipping shifts of '+' symbols
• Confusing order of reduce steps