Here's how I'd approach solving it. First, I'd split it into two subproblems, each sending two bits.
The first two bits are straightforward. B makes a Bell pair and sends half to A and half to Sender. A forwards its entangled qubit directly to the receiver. Sender superdense codes two bits and sends the resulting qubit to the receiver via C. The receiver decodes the two bits the usual way. This costs one qubit each on the B->A, A->Receiver, B->Sender, Sender->C, and C->Receiver links.
Now for the other two bits. B generates a Bell pair and sends half to C via A and half to Sender. Sender superdense codes the remaining two bits and sends the coded qubit to C. C decodes the two bits. Receiver generates a Bell pair, keeps half, and sends half to C via A. C superdense codes the two bits it just decoded into the entangled qubit it got from Receiver and sends the coded qubit to Receiver. Receiver decodes it. This sends one qubit each B->A, B->Sender, Sender->C, C->Receiver, and Receiver->A as well as two qubits A->C.
Adding it up, this uses exactly the allowed capacity on all links. Problem solved using standard techniques.
Edit: fixed a typo
I did generate the puzzle by trying to find a network that forced the superdense bell pair coding + cleanup, so it's a bit embarassing that there's another solution. Maybe it's just always possible to flip the cleanup into an intermediate decoding and re-encoding.
As for the rendering: do you use noscript? The latex is rendered with mathjax. I put a note at the top of the page when scripts are disabled... but it might fail to show up in some cases. Knowing those cases would be useful (maybe if you unblocked the blog but not the mathjax cdn?).
I'll try to find some time this afternoon to re-read the flat coding thing now that the math is there. But I'm a bit confused -- if I have a one-qubit state that's restricted to the set of states that are linear combinations of ±|0> ± |1>, then are only two such states up to global phase, and those states are orthogonal, so I can just measure without loss and treat it as a classical bit. Am I missing something?
It's a neat puzzle, though, and I imagine that there are variants with very interesting solutions.
It's not a one-qubit state limited to ±|0> ± |1>, it's a two-qubit state limited to a|00> + b|01> + c|10> d|11> where a^2+b^2+c^2+d^2 = 1 and a,b,c,d are real. Also the state can be entangled with other qubits, as long the non-180-degree phase information is not between the various a's, b's, c's, and d's.
For example, if you have the state (1/2 |000> - 1/sqrt(2) |010> + i/2 |101>)(|00> + |11>), then you can copy the first two qubits into the bell pair on one side then pull them out on the receiving side to end up in the state 1/2 |00000> - 1/sqrt(2) |01001> + i/sqrt(2) |10110>. E.g. see this circuit doing just that: http://i.imgur.com/wlcVAZG.png
Glad you liked the puzzle.
| A ==> S
| || |
| || |
| vv v
| B ──> C
| |^╲ |
| | ╲╲ |
| | ╲╲ |
| v ╲vv
My current compromise is to just slip in links to the video series Quantum Computing for the Determined  and/or a good textbook with a free version available , so people at least have the opportunity to get a toe-hold (if they want).