GATE CS 1995 Q1.4 — Asymptotic Worst Case Time and Space Complexity

Let's trace the execution of the given Pascal program segment step-by-step:

1. Initialization:

   $$X = -10, \quad Y = 20$$

2. Conditional Evaluation:

   The statement is:

   If X > Y then if X < 0 then X := abs(X) else X := 2*X;

   We first evaluate the outer condition: X > Y.

   Substituting the values:

   $$-10 > 20 \implies \text{False}$$

3. Execution Flow:

   Since the outer if condition is False, the control bypasses the entire nested statement (regardless of how the else is paired due to the dangling-else rule).

   Consequently, no assignment to $X$ takes place.

Thus, the value of $X$ remains unchanged at its initial value, which is $-10$.

Therefore, the correct option is C.