Circuit construction for execution of conditional statements using least significant bit

Question

Suppose I have integers and encode them as binary strings for example: $$|0\rangle=|00\rangle,|1\rangle= |01\rangle,|2\rangle=|10\rangle,|3\rangle=|11\rangle.$$

Now an $n$ bit integer, say $x$, is encoded as a state $|x_0x_1....x_{n-1}\rangle$ with $x_0$ being the most significant bit. To check the remainder by $4$ we only care about the two least significant bit i.e. $x_{n-2}x_{n-1}$. If the remainder is $0$ or $2$ then the least significant bit is $0$, and hence for a controlled operation that executes whether the remainder is $2$ or $0$ can be triggered only by the least significant bit being $0$. But my question is if I want to execute a controlled operation conditioned on the remainder being $3$ or $1$ what should be done? Since $|3\rangle=|11\rangle$ and $|0\rangle=|00\rangle $, should this be the circuit for the control?

Answer 1

I would use a circuit that looks like:

Where $|0\rangle$ is an ancillary output bit. Assume the initial state of the system is in the standard basis.

Here, the first $\land_2(X)$ (CCNOT) applies the $X$ gate to the target qubit for all states that have their two least significant bits set to 1. Note that if none of the states have $|x_{n-1}\rangle=|1\rangle$ and $|x_{n-2}\rangle=|1\rangle$ then the $X$ gate is never applied to the ancilla.

We then negate both of the control bits, so that if they were previously 1 they are then set to 0, and vice versa. This results in the next $\land_2(X)$ gate only being applied if originally, $|x_{n-1}\rangle=|0\rangle$ and $|x_{n-2}\rangle=|0\rangle$. The final two $X$ gates uncompute the original negation, so that the original states of the two least significant bits are preserved.

Therefore, the only states that will have the ancillary bit set to $|1\rangle$, are those for which the two least significant bits were either $|00\rangle$ or $|11\rangle$. Satisfying the requirements of your question.

Please let me know if I can clarify further.