Why won't Qiskit ccx gate accept registers?
$begingroup$
In Qiskit, they have the concept of a Toffoli gate (ccx). However, the following code throws an error:
from qiskit import QuantumCircuit, ClassicalRegister, QuantumRegister
from qiskit import execute, Aer
q1 = QuantumRegister(2)
qctrl = QuantumRegister(1)
c = ClassicalRegister(2)
qc = QuantumCircuit(q1, qctrl, c)
qc.x(q1)
qc.ccx(q1[0], q1[1], qctrl)
qc.measure(q1, c)
backend_sim = Aer.get_backend('qasm_simulator')
job_sim = execute(qc, backend_sim)
result_sim = job_sim.result()
print(result_sim.get_counts(qc))
According to my knowledge and research, this should produce the output 111 but instead I get an error message:
qiskit.qiskiterror.QiskitError: "QuantumRegister(1, 'q0') is not a tuple. A qubit should be formated as a tuple."
This seems like a bug with Qiskit to me. Am I wrong? And if so, how can I fix my code?
programming qiskit
edited 12 hours ago
Blue♦
5,68221354
asked 20 hours ago
Woody1193Woody1193
1184
New contributor
Woody1193 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
In Qiskit, they have the concept of a Toffoli gate (ccx). However, the following code throws an error:
from qiskit import QuantumCircuit, ClassicalRegister, QuantumRegister
from qiskit import execute, Aer
q1 = QuantumRegister(2)
qctrl = QuantumRegister(1)
c = ClassicalRegister(2)
qc = QuantumCircuit(q1, qctrl, c)
qc.x(q1)
qc.ccx(q1[0], q1[1], qctrl)
qc.measure(q1, c)
backend_sim = Aer.get_backend('qasm_simulator')
job_sim = execute(qc, backend_sim)
result_sim = job_sim.result()
print(result_sim.get_counts(qc))
According to my knowledge and research, this should produce the output 111 but instead I get an error message:
qiskit.qiskiterror.QiskitError: "QuantumRegister(1, 'q0') is not a tuple. A qubit should be formated as a tuple."
This seems like a bug with Qiskit to me. Am I wrong? And if so, how can I fix my code?
programming qiskit
edited 12 hours ago
Blue♦
5,68221354
asked 20 hours ago
Woody1193Woody1193
1184
New contributor
Woody1193 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In Qiskit, they have the concept of a Toffoli gate (ccx). However, the following code throws an error:
from qiskit import QuantumCircuit, ClassicalRegister, QuantumRegister
from qiskit import execute, Aer
q1 = QuantumRegister(2)
qctrl = QuantumRegister(1)
c = ClassicalRegister(2)
qc = QuantumCircuit(q1, qctrl, c)
qc.x(q1)
qc.ccx(q1[0], q1[1], qctrl)
qc.measure(q1, c)
backend_sim = Aer.get_backend('qasm_simulator')
job_sim = execute(qc, backend_sim)
result_sim = job_sim.result()
print(result_sim.get_counts(qc))
According to my knowledge and research, this should produce the output 111 but instead I get an error message:
qiskit.qiskiterror.QiskitError: "QuantumRegister(1, 'q0') is not a tuple. A qubit should be formated as a tuple."
This seems like a bug with Qiskit to me. Am I wrong? And if so, how can I fix my code?
programming qiskit
edited 12 hours ago
Blue♦
5,68221354
asked 20 hours ago
Woody1193Woody1193
1184
New contributor
Woody1193 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In Qiskit, they have the concept of a Toffoli gate (ccx). However, the following code throws an error:
from qiskit import QuantumCircuit, ClassicalRegister, QuantumRegister
from qiskit import execute, Aer
q1 = QuantumRegister(2)
qctrl = QuantumRegister(1)
c = ClassicalRegister(2)
qc = QuantumCircuit(q1, qctrl, c)
qc.x(q1)
qc.ccx(q1[0], q1[1], qctrl)
qc.measure(q1, c)
backend_sim = Aer.get_backend('qasm_simulator')
job_sim = execute(qc, backend_sim)
result_sim = job_sim.result()
print(result_sim.get_counts(qc))
According to my knowledge and research, this should produce the output 111 but instead I get an error message:
qiskit.qiskiterror.QiskitError: "QuantumRegister(1, 'q0') is not a tuple. A qubit should be formated as a tuple."
This seems like a bug with Qiskit to me. Am I wrong? And if so, how can I fix my code?
programming qiskit
programming qiskit
edited 12 hours ago
Blue♦
5,68221354
asked 20 hours ago
Woody1193Woody1193
1184
New contributor
Woody1193 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 12 hours ago
Blue♦
5,68221354
asked 20 hours ago
Woody1193Woody1193
1184
New contributor
Woody1193 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 12 hours ago
Blue♦
5,68221354
edited 12 hours ago
Blue♦
5,68221354
edited 12 hours ago
Blue♦
5,68221354
5,68221354
asked 20 hours ago
Woody1193Woody1193
1184
New contributor
Woody1193 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 20 hours ago
Woody1193Woody1193
1184
asked 20 hours ago
Woody1193Woody1193
1184
1184
New contributor
Woody1193 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Woody1193 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Woody1193 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is not a bug. Qiskit makes the difference between quantum registers and quantum bits. In your case, qctrl is a QuantumRegister (as shown in your error) whereas the ccx gate expect a qubit.
For Qiskit, a qubit is defined as a Tuple of a QuantumRegister and an index. You can get a qubit from a QuantumRegister by using the indexing notation: qtrl[0] represents the first qubit (and the only one) of the QuantumRegister qctrl.
In summary, replace
qc.ccx(q1[0], q1[1], qctrl)
by
qc.ccx(q1[0], q1[1], qctrl[0])
and your issue should be fixed.
answered 19 hours ago
NelimeeNelimee
1,452226
$endgroup$
add a comment |
$begingroup$
Nelimee has provided the answer, but there's another point to note that is too big for a comment.
In Qiskit, it is possible to pass registers to gates instead of qubits. You do this in your example with
qc.x(q1)
This applied x to both qubits in the register q1, and so is equivalent to
for q in range(2):
qc.x(q1[n])
This should be regarded as a kind of shortcut notation. Providing single qubits is the 'proper way' to do it, and using registers is something that can be done to make your code a bit more elegant.
Like any shortcut, it should be used with care. Specifically, you have to remain consistent. So the problem in your case was supplying a single qubit for two arguments, and a single qubit register for the last. If you'd used single qubits for all, as in Nelimee's answer, then it would work. Using single qubit registers for all would also work. But mixing and matching does not.
answered 18 hours ago
James WoottonJames Wootton
6,2821943
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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votes
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oldest
votes
$begingroup$
It is not a bug. Qiskit makes the difference between quantum registers and quantum bits. In your case, qctrl is a QuantumRegister (as shown in your error) whereas the ccx gate expect a qubit.
For Qiskit, a qubit is defined as a Tuple of a QuantumRegister and an index. You can get a qubit from a QuantumRegister by using the indexing notation: qtrl[0] represents the first qubit (and the only one) of the QuantumRegister qctrl.
In summary, replace
qc.ccx(q1[0], q1[1], qctrl)
by
qc.ccx(q1[0], q1[1], qctrl[0])
and your issue should be fixed.
answered 19 hours ago
NelimeeNelimee
1,452226
$endgroup$
add a comment |
$begingroup$
It is not a bug. Qiskit makes the difference between quantum registers and quantum bits. In your case, qctrl is a QuantumRegister (as shown in your error) whereas the ccx gate expect a qubit.
For Qiskit, a qubit is defined as a Tuple of a QuantumRegister and an index. You can get a qubit from a QuantumRegister by using the indexing notation: qtrl[0] represents the first qubit (and the only one) of the QuantumRegister qctrl.
In summary, replace
qc.ccx(q1[0], q1[1], qctrl)
by
qc.ccx(q1[0], q1[1], qctrl[0])
and your issue should be fixed.
answered 19 hours ago
NelimeeNelimee
1,452226
$endgroup$
add a comment |
$begingroup$
It is not a bug. Qiskit makes the difference between quantum registers and quantum bits. In your case, qctrl is a QuantumRegister (as shown in your error) whereas the ccx gate expect a qubit.
For Qiskit, a qubit is defined as a Tuple of a QuantumRegister and an index. You can get a qubit from a QuantumRegister by using the indexing notation: qtrl[0] represents the first qubit (and the only one) of the QuantumRegister qctrl.
In summary, replace
qc.ccx(q1[0], q1[1], qctrl)
by
qc.ccx(q1[0], q1[1], qctrl[0])
and your issue should be fixed.
answered 19 hours ago
NelimeeNelimee
1,452226
$endgroup$
It is not a bug. Qiskit makes the difference between quantum registers and quantum bits. In your case, qctrl is a QuantumRegister (as shown in your error) whereas the ccx gate expect a qubit.
For Qiskit, a qubit is defined as a Tuple of a QuantumRegister and an index. You can get a qubit from a QuantumRegister by using the indexing notation: qtrl[0] represents the first qubit (and the only one) of the QuantumRegister qctrl.
In summary, replace
qc.ccx(q1[0], q1[1], qctrl)
by
qc.ccx(q1[0], q1[1], qctrl[0])
and your issue should be fixed.
answered 19 hours ago
NelimeeNelimee
1,452226
answered 19 hours ago
NelimeeNelimee
1,452226
answered 19 hours ago
NelimeeNelimee
1,452226
answered 19 hours ago
NelimeeNelimee
1,452226
1,452226
add a comment |
add a comment |
$begingroup$
Nelimee has provided the answer, but there's another point to note that is too big for a comment.
In Qiskit, it is possible to pass registers to gates instead of qubits. You do this in your example with
qc.x(q1)
This applied x to both qubits in the register q1, and so is equivalent to
for q in range(2):
qc.x(q1[n])
This should be regarded as a kind of shortcut notation. Providing single qubits is the 'proper way' to do it, and using registers is something that can be done to make your code a bit more elegant.
Like any shortcut, it should be used with care. Specifically, you have to remain consistent. So the problem in your case was supplying a single qubit for two arguments, and a single qubit register for the last. If you'd used single qubits for all, as in Nelimee's answer, then it would work. Using single qubit registers for all would also work. But mixing and matching does not.
answered 18 hours ago
James WoottonJames Wootton
6,2821943
$endgroup$
add a comment |
$begingroup$
Nelimee has provided the answer, but there's another point to note that is too big for a comment.
In Qiskit, it is possible to pass registers to gates instead of qubits. You do this in your example with
qc.x(q1)
This applied x to both qubits in the register q1, and so is equivalent to
for q in range(2):
qc.x(q1[n])
This should be regarded as a kind of shortcut notation. Providing single qubits is the 'proper way' to do it, and using registers is something that can be done to make your code a bit more elegant.
Like any shortcut, it should be used with care. Specifically, you have to remain consistent. So the problem in your case was supplying a single qubit for two arguments, and a single qubit register for the last. If you'd used single qubits for all, as in Nelimee's answer, then it would work. Using single qubit registers for all would also work. But mixing and matching does not.
answered 18 hours ago
James WoottonJames Wootton
6,2821943
$endgroup$
add a comment |
$begingroup$
Nelimee has provided the answer, but there's another point to note that is too big for a comment.
In Qiskit, it is possible to pass registers to gates instead of qubits. You do this in your example with
qc.x(q1)
This applied x to both qubits in the register q1, and so is equivalent to
for q in range(2):
qc.x(q1[n])
This should be regarded as a kind of shortcut notation. Providing single qubits is the 'proper way' to do it, and using registers is something that can be done to make your code a bit more elegant.
Like any shortcut, it should be used with care. Specifically, you have to remain consistent. So the problem in your case was supplying a single qubit for two arguments, and a single qubit register for the last. If you'd used single qubits for all, as in Nelimee's answer, then it would work. Using single qubit registers for all would also work. But mixing and matching does not.
answered 18 hours ago
James WoottonJames Wootton
6,2821943
$endgroup$
Nelimee has provided the answer, but there's another point to note that is too big for a comment.
In Qiskit, it is possible to pass registers to gates instead of qubits. You do this in your example with
qc.x(q1)
This applied x to both qubits in the register q1, and so is equivalent to
for q in range(2):
qc.x(q1[n])
This should be regarded as a kind of shortcut notation. Providing single qubits is the 'proper way' to do it, and using registers is something that can be done to make your code a bit more elegant.
Like any shortcut, it should be used with care. Specifically, you have to remain consistent. So the problem in your case was supplying a single qubit for two arguments, and a single qubit register for the last. If you'd used single qubits for all, as in Nelimee's answer, then it would work. Using single qubit registers for all would also work. But mixing and matching does not.
answered 18 hours ago
James WoottonJames Wootton
6,2821943
answered 18 hours ago
James WoottonJames Wootton
6,2821943
answered 18 hours ago
James WoottonJames Wootton
6,2821943
answered 18 hours ago
James WoottonJames Wootton
6,2821943
6,2821943
add a comment |
add a comment |
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Woody1193 is a new contributor. Be nice, and check out our Code of Conduct.
Woody1193 is a new contributor. Be nice, and check out our Code of Conduct.
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