Available expressions analysis in Compiler Design - Time & Space Complexity

Time Complexity: Available expressions analysis

O(n * b * i)

Understanding Time Complexity

Analyzing time complexity helps us understand how the cost of available expressions analysis grows as the program size increases.

We want to know how the number of operations changes when the program has more expressions or statements.

Scenario Under Consideration

Analyze the time complexity of the following pseudocode for available expressions analysis.

for each basic block B in the program:
  IN[B] = empty set
  OUT[B] = all expressions

repeat
  for each basic block B:
    IN[B] = intersection of OUT[P] for all predecessors P of B
    OUT[B] = GEN[B] union (IN[B] - KILL[B])
until IN and OUT sets do not change
    

This code computes which expressions are available at each point by repeatedly updating sets until they stabilize.

Identify Repeating Operations

Look for loops and repeated updates in the code.

• Primary operation: The repeated loop that updates IN and OUT sets for each basic block.
• How many times: This loop runs until no changes occur, which depends on the number of expressions and blocks.

How Execution Grows With Input

As the number of basic blocks and expressions grows, the number of updates increases.

Input Size (n = number of expressions)	Approx. Operations
10	Few hundred updates
100	Thousands of updates
1000	Hundreds of thousands of updates

Pattern observation: The work grows roughly with the product of expressions and blocks times the number of iterations until stabilization.

Final Time Complexity

Time Complexity: O(n * b * i)

This means the time grows with the number of expressions (n), the number of basic blocks (b), and the number of iterations (i) needed to reach stability.

Common Mistake

[X] Wrong: "The analysis finishes in just one pass over the blocks."

[OK] Correct: The sets depend on each other and usually require multiple passes until no changes happen.

Interview Connect

Understanding this time complexity shows you can reason about iterative data flow analyses, a key skill in compiler design and optimization.

Self-Check

"What if we used a worklist algorithm that only updates changed blocks? How would the time complexity change?"

Practice

1. What is the main purpose of available expressions analysis in compiler design?

easy

A. To find expressions that have already been computed and not changed

B. To detect syntax errors in the source code

C. To allocate memory for variables

D. To generate machine code from source code

Solution

1. Step 1: Understand the goal of available expressions analysis

   This analysis identifies expressions that have been computed before and whose values have not changed since.

2. Step 2: Compare with other compiler tasks

   Unlike syntax checking or code generation, available expressions focus on optimization by reusing computed values.

3. Final Answer:

   To find expressions that have already been computed and not changed -> Option A

4. Quick Check:

   Available expressions = computed and unchanged expressions [OK]

Hint: Focus on expressions reused without recomputation [OK]

Common Mistakes:

• Confusing with syntax error detection
• Thinking it allocates memory
• Mixing with code generation tasks

2. Which of the following correctly represents the GEN set in available expressions analysis?

easy

A. All variables used in the program

B. Variables assigned new values in a block

C. Expressions computed in a block and not killed within it

D. Expressions that are never computed in the program

Solution

1. Step 1: Define the GEN set

   GEN set contains expressions that are computed in a block and remain valid (not overwritten) within that block.

2. Step 2: Eliminate incorrect options

   Variables assigned new values relate to KILL sets, not GEN. Expressions never computed or all variables are unrelated.

3. Final Answer:

   Expressions computed in a block and not killed within it -> Option C

4. Quick Check:

   GEN = computed and valid expressions [OK]

Hint: GEN = expressions generated and still valid in block [OK]

Common Mistakes:

• Mixing GEN with KILL sets
• Confusing variables with expressions
• Selecting unrelated program elements

3. Consider a basic block with the following statements:

a = b + c
d = a + e
b = 5

Which expressions are available at the end of this block?

medium

A. {a + e}

B. {} (empty set)

C. {b + c, a + e}

D. {b + c}

Solution

1. Step 1: Identify expressions computed

   The expressions computed are b + c and a + e.

2. Step 2: Check if expressions remain valid

   b + c is killed because b is assigned later (b = 5). a + e remains available since neither a nor e is reassigned after its computation.

3. Final Answer:

   {a + e} -> Option A

4. Quick Check:

   Expressions available if no operand reassigned after computation [OK]

Hint: Check operand reassignments after each computation [OK]

Common Mistakes:

• Ignoring variable reassignment kills expressions
• Assuming all computed expressions remain available
• Confusing direct operands with transitive dependencies

4. Identify the error in this available expressions analysis step:

GEN = {x + y}
KILL = {x = 3}
Available expressions at block end = GEN ∪ (IN - KILL)

What is wrong with this calculation?

medium

A. Using union (∪) instead of intersection (∩) for available expressions

B. Confusing KILL set with GEN set

C. Not updating IN set before calculation

D. Including variables instead of expressions in KILL

Solution

1. Step 1: Recall available expressions formula

   The formula Available_out = GEN ∪ (IN - KILL) is correct when GEN and KILL contain expressions.

2. Step 2: Identify the error in the sets

   KILL = {x = 3} includes an assignment instead of killed expressions (e.g., those with x as operand).

3. Final Answer:

   Including variables instead of expressions in KILL -> Option D

4. Quick Check:

   KILL = expressions killed by assignments [OK]

Hint: KILL contains killed expressions, not assignments [OK]

Common Mistakes:

• Misrepresenting KILL set contents
• Mixing variables and expressions in sets
• Ignoring proper IN set update

5. Given two basic blocks with:

Block 1: GEN = {a + b}, KILL = {a = 4}
Block 2: GEN = {a + b, c + d}, KILL = {b = 2}

If the IN set for Block 2 is {a + b}, what is the OUT set for Block 2 after available expressions analysis?

hard

A. {a + b, c + d}

B. {c + d}

C. {a + b}

D. {} (empty set)

Solution

1. Step 1: Calculate OUT set formula

   OUT = GEN ∪ (IN ∩ ¬KILL). Here, GEN = {a + b, c + d}, IN = {a + b}, KILL kills expressions involving b.

2. Step 2: Remove killed expressions from IN

   a + b is killed (involves b), so IN ∩ ¬KILL = {}.

3. Step 3: Compute OUT set

   OUT = GEN ∪ {} = {a + b, c + d}, but since a + b is killed by KILL, it should not be in OUT. Therefore, OUT = {c + d}.

4. Final Answer:

   {c + d} -> Option B

5. Quick Check:

   OUT = GEN ∪ (filtered IN) with killed expressions removed [OK]

Hint: Filter IN with ¬KILL, then union GEN [OK]

Common Mistakes:

• Applying KILL to GEN expressions
• Not intersecting IN with complement of KILL
• Assuming all GEN expressions remain available