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Circuit construction for execution of conditional statements using least significant bit

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4

$begingroup$

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

Now an $n$ bit integer, say $x$, is encoded as a state $|x_0x_1....x_n-1rangle$ 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. $x_n-2x_n-1$. 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 $|3rangle=|11rangle$ and $|0rangle=|00rangle $, should this be the circuit for the control?

algorithm circuit-construction

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edited May 18 at 23:29

Sanchayan Dutta

7,44841660

asked May 18 at 18:22

UpstartUpstart

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$endgroup$

add a comment | 

4

$begingroup$

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

Now an $n$ bit integer, say $x$, is encoded as a state $|x_0x_1....x_n-1rangle$ 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. $x_n-2x_n-1$. 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 $|3rangle=|11rangle$ and $|0rangle=|00rangle $, should this be the circuit for the control?

algorithm circuit-construction

share|improve this question

edited May 18 at 23:29

Sanchayan Dutta

7,44841660

asked May 18 at 18:22

UpstartUpstart

34019

$endgroup$

add a comment | 

$begingroup$

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

Now an $n$ bit integer, say $x$, is encoded as a state $|x_0x_1....x_n-1rangle$ 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. $x_n-2x_n-1$. 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 $|3rangle=|11rangle$ and $|0rangle=|00rangle $, should this be the circuit for the control?

algorithm circuit-construction

share|improve this question

edited May 18 at 23:29

Sanchayan Dutta

7,44841660

asked May 18 at 18:22

UpstartUpstart

34019

$endgroup$

share|improve this question

edited May 18 at 23:29

Sanchayan Dutta

7,44841660

asked May 18 at 18:22

UpstartUpstart

34019

share|improve this question

edited May 18 at 23:29

Sanchayan Dutta

7,44841660

asked May 18 at 18:22

UpstartUpstart

34019

edited May 18 at 23:29

Sanchayan Dutta

7,44841660

edited May 18 at 23:29

Sanchayan Dutta

7,44841660

asked May 18 at 18:22

UpstartUpstart

34019

asked May 18 at 18:22

UpstartUpstart

34019

1 Answer
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5

$begingroup$

I would use a circuit that looks like:

Where $|0rangle$ 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 $|x_n-1rangle=|1rangle$ and $|x_n-2rangle=|1rangle$ 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, $|x_n-1rangle=|0rangle$ and $|x_n-2rangle=|0rangle$. 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 $|1rangle$, are those for which the two least significant bits were either $|00rangle$ or $|11rangle$. Satisfying the requirements of your question.

Please let me know if I can clarify further.

share|improve this answer

edited May 18 at 23:21

answered May 18 at 23:11

Arthur-1Arthur-1

1266

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add a comment | 

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5

$begingroup$

I would use a circuit that looks like:

Where $|0rangle$ 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 $|x_n-1rangle=|1rangle$ and $|x_n-2rangle=|1rangle$ 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, $|x_n-1rangle=|0rangle$ and $|x_n-2rangle=|0rangle$. 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 $|1rangle$, are those for which the two least significant bits were either $|00rangle$ or $|11rangle$. Satisfying the requirements of your question.

Please let me know if I can clarify further.

share|improve this answer

edited May 18 at 23:21

answered May 18 at 23:11

Arthur-1Arthur-1

1266

$endgroup$

add a comment | 

5

$begingroup$

I would use a circuit that looks like:

Where $|0rangle$ 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 $|x_n-1rangle=|1rangle$ and $|x_n-2rangle=|1rangle$ 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, $|x_n-1rangle=|0rangle$ and $|x_n-2rangle=|0rangle$. 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 $|1rangle$, are those for which the two least significant bits were either $|00rangle$ or $|11rangle$. Satisfying the requirements of your question.

Please let me know if I can clarify further.

share|improve this answer

edited May 18 at 23:21

answered May 18 at 23:11

Arthur-1Arthur-1

1266

$endgroup$

add a comment | 

$begingroup$

I would use a circuit that looks like:

Where $|0rangle$ 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 $|x_n-1rangle=|1rangle$ and $|x_n-2rangle=|1rangle$ 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, $|x_n-1rangle=|0rangle$ and $|x_n-2rangle=|0rangle$. 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 $|1rangle$, are those for which the two least significant bits were either $|00rangle$ or $|11rangle$. Satisfying the requirements of your question.

Please let me know if I can clarify further.

share|improve this answer

edited May 18 at 23:21

answered May 18 at 23:11

Arthur-1Arthur-1

1266

$endgroup$

share|improve this answer

edited May 18 at 23:21

answered May 18 at 23:11

Arthur-1Arthur-1

1266

answered May 18 at 23:11

Arthur-1Arthur-1

1266

answered May 18 at 23:11

Arthur-1Arthur-1

1266