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Consider the following ZX-diagram:

enter image description here

As you can notice, there are some nodes, belonging the same qubit, which are not connected by any edge (neither blue or black).

What is the meaning of that (during circuit extraction)? Can I assume a black edge?

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    $\begingroup$ Imagine a diagram for a SWAP. Then you would have the first qubit connected to the second, and the second to the first. A lack of edge doesn't really 'mean' anything, it just means there just happens to not be an edge needed there. $\endgroup$
    – John
    Commented Jun 8, 2023 at 12:17
  • $\begingroup$ I see. Please consider the example at bit.ly/zx_cross Is this a possible scenario in your normal form? How could we relate it to standard gates (e.g. SWAP)? $\endgroup$ Commented Jun 8, 2023 at 18:54
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    $\begingroup$ That particular setting is not possible in the unitary setting for the simple reason that it does not implement a unitary map: if you consider the biadjacency matrix of the left-hand side to the right-hand side, you will see that it is a 2x2 matrix of all 1's. This matrix is not full rank, and hence not invertible, so that the linear map it implements is also not invertible. $\endgroup$
    – John
    Commented Jun 10, 2023 at 15:50

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If there's no edge there's no edge. It doesn't mean there is an edge, it means there's no edge.

The ZX calculus doesn't have a special case for horizontal edges vs other edges. All edges are the same, regardless of orientation. (I guess there's blue ones and black ones in your specific diagram, and that is a relevant distinction, but you get my point.)

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  • $\begingroup$ Yes, I got your point $\endgroup$ Commented Jul 16, 2022 at 11:26

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