A Quantum Network Flow Puzzle 23 points by Strilanc on May 1, 2015 | hide | past | web | favorite | 7 comments

 Yup, that solution also works. I completely missed it. It's even a bit more flexible, in that the A-to-Receiver edge could be flipped or turned into a broadcast-from-third-party without breaking the solution.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'm using plain ol' Firefox on Linux. I just refreshed and it rendered correctly. Go figure.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.
 > 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 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.pngGlad you liked the puzzle.
 I think if I split the cleaner node into two nodes, the standard solution has to change (not sure if there is one or not). 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. ┌───────┐ | | | A ==> S | || | | || | | vv v | B ──> C | |^╲ | | | ╲╲ | | | ╲╲ | | v ╲vv └>D────>R  Spot any solutions?
 I'm not even going to pretend I understood any of that
 This is a pretty common comment. Unfortunately, explaining the background details turns any quantum computing post into an "intro to quantum computing" post.My current compromise is to just slip in links to the video series Quantum Computing for the Determined [1] and/or a good textbook with a free version available [2], so people at least have the opportunity to get a toe-hold (if they want).

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