qubit

What also depends on the encoding is the number M of logical qubits per physical qubit.

C superdense codes the two bits it just decoded into the entangled qubit it got from Receiver and sends the coded qubit to Receiver.

You want to figure out which of the 2^n axes in the n-qubit register your black box negates.

However, being close to the thresholds means that you need to be on the pessimistic side of how many physical qubits you need per logical qubit.

For example, it would be useful to have a 1024 qubit machine to simultaneously represent all 1024 bit states.

Brady and Anderson's point is that it's not a qubit until you can calculate with it.

It prevents the qubit needed to encode the information for the receiver from being in the same place as the qubit needed to decode the information from the sender, so no re-encoding.

The same applies to quantum computing -- there are many ways to make a qubit.

N] This generates a [N, N/3, 2] code, with N/3 logical qubits per physical qubit.

Sender superdense codes two bits and sends the resulting qubit to the receiver via C.

Sender superdense codes the remaining two bits and sends the coded qubit to C.

For performing a quantum algorithm two sets of gates are needed, for instance single qubit rotations and a square root of swap gate.

Is the "!" showing in some of the gate definitions a post-selection operation on the corresponding qubit?

I think there are state of the art codes ("surface codes") that can get the error threshold down to 10^2 or so, which is close to what's achievable in systems with small numbers of superconducting qubits (which is what IBM and Google are looking at.) But, the number M of physical qubits per logical qubit for a surface code is very, very high, I think on the order of 10^5.