Loop optimization (invariant code motion) in Compiler Design - Step-by-Step Execution

Concept Flow - Loop optimization (invariant code motion)

Start Loop

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Identify Invariant Code

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Move Invariant Code Outside Loop

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Execute Loop Body Without Invariant Code

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Loop End

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Repeat or Exit

The process finds code inside a loop that does not change and moves it outside to run once, improving efficiency.

Execution Sample

Compiler Design

for i in range(3):
    x = 5 * 2
    print(i + x)

This loop calculates 5*2 inside each iteration, but since 5*2 is always 10, it can be moved outside the loop.

Analysis Table

Iteration	i	Expression '5 * 2' Computed?	Value of x	Action	Output
1	0	Yes	10	Calculate x inside loop	0 + 10 = 10
2	1	Yes	10	Calculate x inside loop	1 + 10 = 11
3	2	Yes	10	Calculate x inside loop	2 + 10 = 12
				Optimization: Move 'x = 5 * 2' outside loop
1	0	No (moved outside)	10	Use precomputed x	0 + 10 = 10
2	1	No (moved outside)	10	Use precomputed x	1 + 10 = 11
3	2	No (moved outside)	10	Use precomputed x	2 + 10 = 12
				Loop ends after 3 iterations

💡 Loop ends after 3 iterations as i reaches 3, condition i < 3 is False

State Tracker

Variable	Start	After 1	After 2	After 3	Final
i	undefined	0	1	2	3 (loop exit)
x	undefined	10	10	10	10 (constant after optimization)

Key Insights - 3 Insights

Why can 'x = 5 * 2' be moved outside the loop?

Because '5 * 2' always equals 10 and does not depend on the loop variable 'i', so computing it once before the loop saves repeated work (see execution_table rows 1-3 vs 5-7).

Does moving invariant code outside the loop change the output?

No, the output remains the same because the value of 'x' is constant and used in the same way inside the loop (compare outputs in execution_table rows 1-3 and 5-7).

What happens if the code depends on the loop variable?

If the code uses the loop variable or changes each iteration, it cannot be moved outside because its value changes every time (not invariant).

Visual Quiz - 3 Questions

Look at the execution_table, at iteration 2 before optimization, is '5 * 2' computed inside the loop?

A No, it is computed only once

B Yes, it is computed every iteration

C It depends on the value of i

D It is never computed

💡 Hint

Check execution_table row 2 under 'Expression 5 * 2 Computed?' column

At which iteration does the loop stop executing according to the execution_table?

A After iteration 2

B After iteration 4

C After iteration 3

D It never stops

💡 Hint

Look at the exit_note and the 'i' variable values in variable_tracker

If 'x = 5 * i' instead of 'x = 5 * 2', can 'x' be moved outside the loop?

A No, because x depends on i which changes each iteration

B Yes, because multiplication is fast

C Yes, but only for the first iteration

D No, because 5 * 2 is different

💡 Hint

Refer to key_moments about dependency on loop variable

Concept Snapshot

Loop optimization (invariant code motion):
- Find code inside loops that does not change each iteration.
- Move that code outside the loop to run once.
- This reduces repeated work and speeds up execution.
- Only safe if code does not depend on loop variables.
- Output remains the same after optimization.

Full Transcript

Loop optimization by invariant code motion means moving calculations or code inside a loop that do not change during the loop outside it. For example, in a loop running three times, if a calculation like 5 times 2 is done inside each iteration, it can be moved before the loop to run once. This saves time because the calculation is constant. The execution table shows that before optimization, the expression is computed every iteration, but after optimization, it is computed once and reused. Variables like 'i' change each iteration, but 'x' remains constant after moving outside. This optimization does not change the output but makes the loop faster. If the code depends on the loop variable, it cannot be moved outside safely.