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; }What is the minimum number of registers needed in the instruction set architecture of the processor to compile this code segment without any spill to memory? Do not apply any optimization other than optimizing register allocation.

Without code motion or other optimizations, we determine the maximum number of simultaneously live variables at any point during execution:

1. Initially, variables $a$ and $b$ are live (2 variables).

2. After c = a + b: $b$ is dead, so live variables are $\{a, c\}$ (2 variables).

3. After d = c * a: live variables are $\{a, c, d\}$ (3 variables).

4. After e = c + a: live variables are $\{a, c, d, e\}$ (4 variables).

5. After x = c * c: $c$ is now dead, so live variables are $\{a, d, e, x\}$ (4 variables).

6. At if (x > a): $x$ is consumed, leaving $\{a, d, e\}$ live (3 variables).

7. In the if block, $y = a * a$ is computed.

8. In the else block, $d$ and $e$ are used and updated.

The maximum number of simultaneously live variables at any point is 4 (specifically, after computing $e$ and $x$). Hence, at least 4 registers are needed to compile the code without any memory spills.