Distance of a non-standard topological code

Distance of a non-standard topological code Ask Question Asked yesterday Modified yesterday Viewed 78 times 1 While trying to construct a distance 5 magic state cultivation circuit according to Gidney et al.'s "Magic state cultivation: growing T states as cheap as CNOT gates paper." I inspected the detector slice diagram of the circuit after the escape stage and wanted to verify the code distance. Is verifying the code distance of a topological code like this possible using visual inspection or some other property of topological codes? Or is it necessary to brute-force enumerate errors of all lower weight until one doesn't commute with the stabilizers? detector slice diagram: error-correctionstabilizer-codesurface-codemagic-states Share Improve this question Follow asked yesterday PrimeSoup 474 4 bronze badges Add a comment 1 Answer Sorted by: Highest score (default) Date modified (newest first) Date created (oldest first) 2 You can do better than brute force search by knowing that for an undetectable logical error it must be the case that any stabilizer that gets flipped by an error must also get un-flipped by some other error. So after you pick your first error to insert, you can pick one of your excited stabilizers and pick between errors that would flip it again (rather than picking between all possible errors). Keep doing that until all the stabilizers are cleared but the logical operator is still flipped and you have a logical error. A second optimization you can apply that is always safe is to only start the search from errors that flip the logical operator. A logical error must include at least one physical error that flipped the logical operator. A third optimization is to remember states you have seen (i.e. the same set of stabilizers and logical operators being flipped) to avoid exploring them again if they are encountered via a different path. In some situations (which happen to include the case you showed) these three optimizations are sufficient to reduce the problem into a linear time shortest path search rather than an exponential time search. In particular, if you have a surface code where the logical operator is defined along a boundary and you are only considering bit flip errors (or phase flip errors) then finding an error is equivalent to a shortest path problem from one boundary to another. Unfortunately, in general, things are harder than linear time. For example, in the toric code the shortest logical errors correspond to loops rather than paths and there's no boundary errors to seed the search with a single excitation rather than two excitations (which can introduce quadratic overheads). Share Improve this answer Follow answered yesterday Craig Gidney 51.8k1 1 gold badge 54 54 silver badges 135 135 bronze badges Ok this is quite helpful but what about the case I showed lets you know it is a shortest path search? Or is this only known because you checked? –  PrimeSoup Commented 22 hours ago @PrimeSoup you can tell because it's matchable (each qubit touched by at most 2 blue and 2 red shapes) and has logical operators only running along boundaries. –  Craig Gidney Commented 10 hours ago Add a comment Your Answer Sign up or log in Sign up using Google Sign up using Email and Password Post as a guest Name Email Required, but never shown Post Your Answer By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions error-correctionstabilizer-codesurface-codemagic-states See similar questions with these tags. The Overflow Blog What it takes to be a player in the international AI... Making the OWASP top ten in the vibe code... 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