Common subexpression elimination in Compiler Design - Time & Space Complexity

Time Complexity: Common subexpression elimination

O(n²)

Understanding Time Complexity

When a compiler removes repeated calculations, it saves time. We want to know how the time to optimize grows as the program size grows.

How does the work of finding repeated expressions change when the code gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

for each basic block in program:
  for each expression in block:
    if expression seen before in block:
      replace with previous result
    else:
      record expression

This code finds repeated expressions inside each block and replaces duplicates with stored results.

Identify Repeating Operations

Look for loops or repeated checks that take most time.

• Primary operation: Checking each expression against previously seen ones in the same block.
• How many times: For each expression in each block, this check happens once per previously seen expression.

How Execution Grows With Input

As the number of expressions grows, the number of checks grows too.

Input Size (n expressions)	Approx. Operations
10	About 55 checks
100	About 5,050 checks
1000	About 500,500 checks

Pattern observation: The number of checks grows roughly like the square of the number of expressions.

Final Time Complexity

Time Complexity: O(n²)

This means if you double the number of expressions, the work to find repeats roughly quadruples.

Common Mistake

[X] Wrong: "Finding repeated expressions only takes time proportional to the number of expressions."

[OK] Correct: Because each new expression is compared to all previous ones, the total checks add up much faster than just counting expressions.

Interview Connect

Understanding how optimization steps grow with program size helps you explain compiler efficiency clearly. It shows you can think about how tools work behind the scenes.

Self-Check

"What if we used a fast lookup structure like a hash table to track expressions? How would the time complexity change?"

Practice

1. What is the main purpose of Common Subexpression Elimination in compiler design?

easy

A. To find repeated calculations and compute them only once

B. To remove unused variables from the code

C. To reorder instructions for parallel execution

D. To translate code into machine language

Solution

1. Step 1: Understand the concept of repeated calculations

   Common subexpression elimination looks for expressions that appear multiple times in code.

2. Step 2: Identify the optimization goal

   It computes these expressions once and reuses the result to save time and work.

3. Final Answer:

   To find repeated calculations and compute them only once -> Option A

4. Quick Check:

   Repeated calculations = compute once [OK]

Hint: Look for repeated expressions to compute once [OK]

Common Mistakes:

• Confusing with dead code elimination
• Thinking it changes instruction order
• Assuming it translates code

2. Which of the following is the correct way to represent a common subexpression elimination in code optimization?

easy

A. Recompute the same expression every time it appears

B. Remove the expression completely from the code

C. Store the result of the expression in a temporary variable and reuse it

D. Replace the expression with a random value

Solution

1. Step 1: Identify the optimization technique

   Common subexpression elimination stores the result of repeated expressions in a temporary variable.

2. Step 2: Understand reuse of stored value

   Later uses of the expression reuse this stored value instead of recomputing.

3. Final Answer:

   Store the result of the expression in a temporary variable and reuse it -> Option C

4. Quick Check:

   Store once, reuse many [OK]

Hint: Store repeated results once to reuse [OK]

Common Mistakes:

• Thinking expressions are removed
• Believing expressions are recomputed every time
• Replacing expressions with unrelated values

3. Consider the code snippet:

a = b + c
d = b + c + e

After applying common subexpression elimination, what is the optimized code?

medium

A. temp = b + c
a = temp
d = temp + e

B. a = b + c
d = b + c + e

C. a = b + c + e
d = b + c + e

D. temp = b + c + e
a = temp
d = temp

Solution

1. Step 1: Identify common subexpression

   The expression b + c appears twice in the code.

2. Step 2: Store and reuse the common expression

   Compute b + c once, store in temp, then use temp for both a and part of d.

3. Final Answer:

   temp = b + c a = temp d = temp + e -> Option A

4. Quick Check:

   Common part stored once =

   temp = b + c
   a = temp
   d = temp + e

   [OK]

Hint: Extract repeated parts into temp variable [OK]

Common Mistakes:

• Not storing the common part
• Storing the entire expression incorrectly
• Duplicating computations

4. Identify the error in the following code after applying common subexpression elimination:

temp = a * b
c = temp + d
temp = a * b + e

medium

A. No error, code is optimized correctly

B. Reassigning temp overwrites previous value causing incorrect reuse

C. The expression a * b is not a common subexpression

D. temp should be declared as a constant

Solution

1. Step 1: Analyze temp variable usage

   The variable temp is assigned a * b first, then reassigned a * b + e.

2. Step 2: Understand impact on reuse

   Reassigning temp overwrites the original value, so reuse of temp later may be incorrect.

3. Final Answer:

   Reassigning temp overwrites previous value causing incorrect reuse -> Option B

4. Quick Check:

   Overwriting temp breaks reuse [OK]

Hint: Avoid overwriting temp holding common subexpression [OK]

Common Mistakes:

• Ignoring reassignment effects
• Assuming temp is constant automatically
• Thinking all expressions are common

5. Given the code:

x = (a + b) * c
y = (a + b) * d
z = (a + b) * c + (a + b) * d

How can common subexpression elimination be applied to optimize this code?

hard

A. temp = (a + b) * (c + d)
x = temp
y = temp
z = temp + temp

B. x = (a + b) * c
y = (a + b) * d
z = x + y

C. temp1 = a + b
z = temp1 * (c + d)
x = z - temp1 * d
y = z - temp1 * c

D. temp = a + b
x = temp * c
y = temp * d
z = x + y

Solution

1. Step 1: Identify common subexpression

   The expression a + b repeats in all calculations.

2. Step 2: Store common subexpression and reuse

   Compute temp = a + b once, then use temp * c and temp * d for x and y. Finally, z = x + y.

3. Final Answer:

   temp = a + b x = temp * c y = temp * d z = x + y -> Option D

4. Quick Check:

   Extract common sum, reuse in products [OK]

Hint: Extract repeated sums first, then multiply [OK]

Common Mistakes:

• Not extracting (a + b) once
• Incorrectly combining terms
• Duplicating calculations