Local optimization (peephole) in Compiler Design - Time & Space Complexity

Time Complexity: Local optimization (peephole)

O(n)

Understanding Time Complexity

Local optimization, or peephole optimization, improves small parts of code to run faster.

We want to know how the time to optimize grows as the code size grows.

Scenario Under Consideration

Analyze the time complexity of the following peephole optimization code snippet.

for i in range(len(instructions) - window_size + 1):
    window = instructions[i:i+window_size]
    if can_optimize(window):
        optimized = optimize(window)
        instructions[i:i+window_size] = optimized

This code looks at small groups of instructions and tries to replace them with faster ones.

Identify Repeating Operations

• Primary operation: Sliding over the instruction list with a fixed-size window.
• How many times: Once for each instruction minus the window size plus one.

How Execution Grows With Input

As the number of instructions grows, the optimizer checks more windows, but each window is small and fixed.

Input Size (n)	Approx. Operations
10	About 10 checks
100	About 100 checks
1000	About 1000 checks

Pattern observation: The number of checks grows directly with the number of instructions.

Final Time Complexity

Time Complexity: O(n)

This means the time to optimize grows in a straight line as the code gets longer.

Common Mistake

[X] Wrong: "Peephole optimization takes constant time no matter how big the code is."

[OK] Correct: The optimizer must check each part of the code, so more code means more checks.

Interview Connect

Understanding how local optimization scales helps you explain compiler efficiency clearly and confidently.

Self-Check

"What if the window size grew with the input size? How would the time complexity change?"

Practice

1. What is the main goal of local optimization (peephole) in compiler design?

easy

A. To improve small sequences of instructions by removing redundancies

B. To rewrite the entire program in a different language

C. To increase the number of instructions for clarity

D. To change the program's output for optimization

Solution

1. Step 1: Understand local optimization scope

   Local optimization focuses on small instruction sequences, not the whole program.

2. Step 2: Identify the goal of peephole optimization

   Peephole optimization removes unnecessary or redundant instructions to make code faster and smaller without changing its meaning.

3. Final Answer:

   To improve small sequences of instructions by removing redundancies -> Option A

4. Quick Check:

   Local optimization = remove redundancies [OK]

Hint: Focus on small instruction sequences and redundancy removal [OK]

Common Mistakes:

• Thinking it rewrites the whole program
• Believing it changes program output
• Assuming it adds more instructions

2. Which of the following is a correct example of a peephole optimization?

easy

A. Replacing LOAD A; LOAD A with LOAD A

B. Changing variable names for readability

C. Adding comments to the code

D. Rearranging functions in the source file

Solution

1. Step 1: Identify peephole optimization examples

   Peephole optimization looks at small instruction sequences to remove redundancies.

2. Step 2: Evaluate each option

   Replacing LOAD A; LOAD A with LOAD A removes duplicate instructions, which is a classic peephole optimization. Options A, B, and D do not optimize instructions.

3. Final Answer:

   Replacing LOAD A; LOAD A with LOAD A -> Option A

4. Quick Check:

   Remove duplicate instructions = peephole optimization [OK]

Hint: Look for removing repeated instructions in short sequences [OK]

Common Mistakes:

• Confusing code readability changes with optimization
• Thinking comments affect optimization
• Assuming rearranging functions is peephole optimization

3. Consider this instruction sequence before peephole optimization:

LOAD R1, #5
ADD R1, R1, #0
STORE R1, 100(R2)

What will be the optimized sequence after applying peephole optimization?

medium

A. LOAD R1, #5
ADD R1, #0
STORE R1, 100(R2)

B. LOAD R1, #5
ADD R1, R1, #0
STORE R1, 100(R2)

C. LOAD R1, #5
STORE R1, 100(R2)

D. ADD R1, R1, #0
STORE R1, 100(R2)

Solution

1. Step 1: Identify redundant instructions

   The instruction ADD R1, R1, #0 adds zero, which does not change the value in R1, so it is redundant.

2. Step 2: Remove redundant instruction

   Removing the redundant ADD instruction leaves LOAD R1, #5 and STORE R1, 100(R2).

3. Final Answer:

   LOAD R1, #5 STORE R1, 100(R2) -> Option C

4. Quick Check:

   Remove add zero = optimized sequence [OK]

Hint: Remove instructions that do not change values (like add zero) [OK]

Common Mistakes:

• Keeping redundant add zero instruction
• Removing load instruction incorrectly
• Misunderstanding instruction effects

4. Identify the error in this peephole optimization attempt:

Original: LOAD R2, #10
SUB R2, R2, #0
STORE R2, 200(R3)

Optimized: LOAD R2, #10
SUB R2, R2, #10
STORE R2, 200(R3)

medium

A. The load instruction should be removed

B. No error, optimization is correct

C. The store instruction is missing

D. The subtraction of zero was incorrectly replaced with subtraction of ten

Solution

1. Step 1: Compare original and optimized subtraction

   Original subtracts zero, which does nothing. Optimized subtracts ten, which changes the value incorrectly.

2. Step 2: Identify the error in optimization

   Replacing subtraction by zero with subtraction by ten changes program meaning, which is incorrect for peephole optimization.

3. Final Answer:

   The subtraction of zero was incorrectly replaced with subtraction of ten -> Option D

4. Quick Check:

   Incorrect instruction change = error in peephole optimization [OK]

Hint: Ensure optimized instructions do not change program meaning [OK]

Common Mistakes:

• Changing instruction values incorrectly
• Removing necessary instructions
• Assuming all changes are optimizations

5. Given this code snippet:

LOAD R3, #0
ADD R3, R3, #5
ADD R3, R3, #0
STORE R3, 300(R4)

Which peephole optimization sequence correctly optimizes this code without changing its meaning?

hard

A. LOAD R3, #0
ADD R3, R3, #0
STORE R3, 300(R4)

B. LOAD R3, #5
STORE R3, 300(R4)

C. ADD R3, R3, #5
STORE R3, 300(R4)

D. LOAD R3, #5
ADD R3, R3, #0
STORE R3, 300(R4)

Solution

1. Step 1: Analyze the original instructions

   Loading zero into R3, then adding 5, then adding 0, is equivalent to loading 5 directly into R3.

2. Step 2: Apply peephole optimization

   Replace the sequence with a single instruction LOAD R3, #5 and then store. This removes redundant instructions without changing meaning.

3. Final Answer:

   LOAD R3, #5 STORE R3, 300(R4) -> Option B

4. Quick Check:

   Combine load and add zero to direct load [OK]

Hint: Combine load and add instructions when possible [OK]

Common Mistakes:

• Keeping redundant add zero
• Not combining load and add
• Removing load instruction incorrectly