The following code segment is executed on a processor which allows only register operands in its instructions. Each instruction can have atmost two source operands and one destination operand. Assume that all variables are dead after this code segment.c = a + b; d = c * a; e = c + a; x = c * c; if (x > a) { y = a * a; } else { d = d * d; e = e * e; }Suppose the instruction set architecture of the processor has only two registers. The only allowed compiler optimization is code motion, which moves statements from one place to another while preserving correctness. What is the minimum number of spills to memory in the compiled code?

To determine the minimum number of spills with 2 available registers ($R_1$ and $R_2$):

1. By code motion, the computations for $d$ and $e$ (d = c * a, e = c + a) can be moved entirely into the else block because neither $d$ nor $e$ is live or used inside the if block or branch condition.

2. Before the branch, we must compute c = a + b and x = c * c, and then evaluate the condition x > a.

   • Load $a$ into $R_1$ and $b$ into $R_2$.

   • Compute $c = a + b$ into $R_2$ ($b$ is now dead). Currently, $R_1 = a$ and $R_2 = c$.

   • To compute $x = c * c$, if we overwrite $R_2$ with $x$, the value of $c$ is lost. However, $c$ will be needed in the else block to compute $d$ and $e$.

   • Therefore, $c$ must be spilled to memory (1 spill).

   • Now compute $x = c * c$ into $R_2$, evaluate the condition x > a comparing $R_2$ and $R_1$.

3. In the if branch, compute $y = a * a$ into $R_2$ (0 additional spills).

4. In the else branch, reload $c$ from memory and compute $d$ and $e$ sequentially using the 2 registers (0 additional spills).

Thus, the minimum number of spills to memory is 1.