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Craig Gidney
@CraigGidney
Programmer turned research scientist on Google's quantum computing team. Maker of Quirk, a fun drag-and-drop quantum circuit simulator ( algassert.com/quirk ).
Santa Barbara, CAalgassert.comJoined July 2013

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Suppose as a hypothetical that quantum mechanics is wrong, in a way that breaks quantum computers. In that world, how do you expect quantum computers to break *first*? The simplest task that couldn't be made to work sufficiently well, even with error correction, would be...
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On a plane, the blossom algorithm dual variables form visual proofs the matching is optimal. A better solution would correspond to applying an augmenting cycle or path. But the augment crosses each non-overlapping circle or blossom-layer *twice*. So it can only move or add costs.
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In other words, prove that entanglement distillation stop working at 50% depolarization, instead of only at 75% depolarization, by showing the state you get after applying 50% depolarization to |00>+|11> can be created while using only non-entangling operations.
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Quantum circuit puzzle: Alice and Bob share a Bell pair corrupted by 50% depolarization on Alice's qubit (max mixing is 75%). Show this error destroyed all entanglement by finding a circuit that makes the same mixed state using local operations and classical communication.
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The color code decoding problem is similar to a game of LightsOut on a hex grid, with the lights on faces and buttons on nodes. The goal is to turn the lights out in as few moves as possible. AFAIK the QEC community doesn't know an efficient way to solve it exactly. Does anyone?
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Although, now that I think about it more, maybe the visual diff can be run the other way. If turning the square prism into a triangle prism worked turned [[7,1,3]] into [[5,1,3]], maybe turning the five sided pyramid into a seven sided pyramid will turn [[5,1,3]] into [[7,1,3]]?
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Thanks to this flawless poll with no issues whatsoever, we finally know the nature of reality: amplitudes are continuous. Unless you account for vote splitting, in which case they're discretized, most likely as double precision floating point numbers. Phew. Glad that's resolved.
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It also included another view of the sphere. Did you know that, from the correct perspective, every sphere has a triforce? Neither did I! (On the other hand, it's amazing it can make any text diagram at all in the first place.)
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It's funny how ChatGPT fails when it fails. I asked it for a text diagram of a sphere and it made an upside down ice cream cone shape. But don't worry; it helpfully labelled it "sphere" in case that was confusing.
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arxiv.org/abs/2212.12372 sure does spend a lot of space not mentioning the expected number of circuit shots it requires. It's critical to the entire premise of the paper to have a small bound on this number, and as far as I can tell they just don't talk about it. Very bad sign.
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Last year, a group of geeks convened on Discord in response to a Twitter thread about California’s woes with the vaccine rollout. They very quickly found themselves running a critical piece of national public health infrastructure. Here is our story:
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Advent of code 2022 starts tonight! A bunch of us playing along has become a minor yearly tradition on google's quantum team. See y'all on the puzzlefield.
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Rule of thumb: avoid big stabilizers. By default big stabilizers hurt detection fractions cubically: more qubits, more ops per measurement, and spreadier hook errors. It's REAL BAD to use big stabilizers. It's a common mistake to pump code distance by using bigger stabilizers.
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Quantum circuit challenge: take as input one S state |0> + i|1> and output two such states, using the gate set {ResetX,ResetZ,MeasureX,MeasureZ,CNOT}. Or prove that it's not possible.
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Most likely this is because of the stabilizer generator corresponding to the observable being perturbed by the intermediate measurements. In a way that requires a lot of iterations before the specific choice of basis being used from iteration to iteration starts repeating.
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Something that surprised me: the tableau representation of the honeycomb code's state varies from round to round. I tried optimizing stim's measurement sampling by watching for states recurring during loops. It works, technically... but not on the case I wanted it to improve!
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Fun fact: this is also basically the qec equivalent of three interlocked gears unable to turn with the words "working together" written next to it😂. See if you can fix it making as few changes as possible.
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Office hours complete! Ended up seeing two people. I think they each went well, in that they conveyed random important details that aren't well explained in papers but are absolutely crucial to making something that works.
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The underlying reason is that everything only matters up to Paulis, so there's not really a distinction between period 2 and period 4. Also, I sort of lied when I said it's 4:2 and 2:4; really there are also examples of 4:4 and 2:2 it's just the cheapest ones are 4:2 and 2:4.
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You'd expect the period of single qubit Cliffords to correspond to the periods of surface code corner permutations implementing them, like S has period 4 so it would do a period 4 permutation. But actually the period correspondence is 4:2, 3:3, 2:4.
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I'm doing another Stim office hours, where I help y'all with QEC/coding problems. 30 minute slots wednesday November 9th from 1pm to 5pm PST. If you want a slot, find my email from a paper and email me with subject STIM OFFICE HOURS.
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Probably the main way stim gains convenience functions is someone asks a question on stack exchange, I write demo code for it, and I'm not happy at how long the demo code is.
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Here are quantum solutions in quantum flytrap and in quirk. Here's the saved JSON of the flytrap setup gist.github.com/Strilanc/3a6d5 and here's the link to the Quirk circuit algassert.com/quirk#circuit=
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