Update: A slightly different version of this question has been answered here.

As far as I can see, a major issue with Google's recent quantum supremacy claim is that it is hard to verify the results. If the quantum computer would be powerful enough to solve problems in NP (like factoring, using Shor's algorithm), it would be easy to verify the results, however, this computational power is not available yet. As an intermediate step,

what could be a problem within reach of today's quantum computers whose output can be verified efficiently with a witness?

Here efficiently need not stand for the problem itself being in NP, but rather it would be sufficient that the correctness of the solution is in NP. For example, consider a satisfiability task whose literals are $$x_i$$'s and $$y_j$$'s. If the quantum computer can produce all the $$x_i$$ that are part of a solution, then a witness for its correctness would be finding the $$y_j$$'s that make $$(x,y)$$ into a solution. The corresponding $$y$$ can be found with a classical algorithm, or set as a challenge, like in case of Bitcoin, to mint some quantum coin, with the computational power of all the miners looking for the witness. In fact, miners can be also rewarded for finding another $$(x',y')$$ solution of the original problem, which would determine whether $$x$$ is useful or not.

So what could be such a problem? Can Shor's algorithm or some other similar problem be broken into finding some $$x$$ with a quantum computer that can be extended into a witness with some (smaller) $$y$$?

• If a quantum computer output is verifiable in asymptotic polynomial time with classical computers, it is in NP (unless the verification algorithm is randomized, in which case it is in MA). But that's not what you should be asking. All we care about is the finite case, not the asymptotic one. Dec 29, 2019 at 15:13
• @Peter My question meant exactly to be about the state-of-the-art quantum computers (around 53 qubits). Is there a verifiable problem whose solution they can enhance? For example, if I want to find a hidden 100-digit factor, could a 53-qubit quantum algorithm tell me fast that the 42nd digit is 3 with >20% probability? I would consider this to be a verifiable (through a further exponential amount of computation) quantum advantage. Dec 30, 2019 at 19:58
• As I was trying to explain in my previous comments, "exponential" has no real meaning when you're talking about a 53-qubit machine. Everything is a constant-time computation. The question is whether it's a constant that is practical on our current computers or not. Jan 2, 2020 at 1:48
• @domotorp I think I was inspired by thinking about this to propose a very similar protocol on Aaronson's blog. For a random two-to-one oracle $f(x)$, a certificate of quantumness is split between a quantum pair $(d,y)$ and a classical pair $(x_1,x_2)$, with $y=f(x_1)=f(x_2)$ and $d\cdot(x_1\oplus x_2)=0$ - with the classical pair being found by mining rigs and a smart contract...(cont.) Sep 26, 2022 at 13:37
• @domotorp With Sam Jacques we also worked out some other details on QCSE - in particular, Sam pointed out that a 2-to-1 function may be amenable to a birthday attack, but a generic random oracle behaves according to the Poisson distribution and this can aid in verification. I initially proposed that $f(x)$ be the last $m$ qubits of SHA256, but that's much too difficult to implement now. Any other random 2-1 function may suffice, though. Sep 26, 2022 at 13:40

Brakerski, Christiano, Mahadev, Vazirani, and Vidick propose a scheme for verifiable quantum computational supremacy based on a strengthening of trap-door claw-free functions (TCFs).

In the above scheme:

• Vicky the classical verifier provides a description of a pair of functions $$f_0$$ and $$f_1$$ to Peggy the quantum prover, while saving the trapdoor to $$f_0$$ and $$f_1$$;
• Peggy prepares, measures, and reports the results of a register in a quantum state to provide a $$y$$ such that $$y=f_0(x_0)=f_1(x_1)$$, keeping a superposition of $$\vert b\rangle\vert x_b\rangle$$;
• Vicky asks Peggy to measure the superposition in either the computational basis to provide a bit $$b$$ and an $$x_b$$ such that $$f_b(x_b)=y$$, or in the Hadamard basis to provide a string $$d$$ orthogonal to $$x_0\oplus x_1$$; and
• Based on Vicky's possession of the trapdoor, Vicky can validate results (she uses the trapdoor to deduce both $$x_0$$ and $$x_1$$ from $$y$$ and Peggy's response above).

Thus Peggy proves that she was able to execute a function in quantum superposition (or she has broken the security of the TCF).

The authors instantiate their TCFs with learning-with-errors (LWE), which is a leading candidate for post-quantum security. That is, the functions they require Peggy to prepare and evaluate in superposition are associated with LWE.

My limited understanding is that the author's LWE scheme is out of reach of current and near-term noisy, intermediate-scale quantum computers. The authors appear somewhat optimistic that about 50 bits of security may correspond to a machine with about 2000 or so qubits, or perhaps even 200-500 noisy qubits if they could optimize their LWE scheme.

Nonetheless the scheme is interesting and I think close to the spirit of the question. For example, in this scheme Peggy does not actually find the claw by herself but only provides a proof that she possessed $$x_0$$ and $$x_1$$ in superposition.

Upon more consideration and to expand more of the comments above, I believe that the full scope of the OP's question can be realized, if not now then at least in the near future, for example with a quantum computer having less than a couple hundred qubits and capable of reliably running a relatively shallow depth circuit.

That is, I think we can separate a certificate of quantumness between a quantum part $$(y,d)$$ sampled with a quantum computer, and a classical part $$(x_1,x_2)$$ found with (cryptocurrency) mining. Start with the simple observation that for a random oracle $$f(x)$$ from $$n$$ bits to $$n$$ bits a quantum computer can find pairs $$(y,d)$$ such that for a large fraction there are preimages $$(x_1,x_2)$$ with:

$$f(x_1)=f(x_2)=y,\tag 1$$

and further that:

$$d\cdot (x_1\oplus x_2)=0.\tag 2$$

A classical computer cannot find the pair $$(d,y)$$ except by hunting for collisions.

For example a quantum computer (Peggy) can prepare $$|x\rangle|f(x)\rangle$$, and can measure the second register $$y=f(x)$$ in the computational basis, while measuring first register $$d$$ in the Hadamard basis. After Peggy broadcasts $$(d,y)$$, miners (Vicky) can hunt for all the preimages to $$y$$. If there are two preimages $$(x_1,x_2)$$ that the miners find and broadcast, then all the clients on the network can easily verify both $$(1)$$ and $$(2)$$ above. And, akin to the OP's suggestion, Peggy the quantum computer could offer a smart-contract as a way to motivate Vicky the miners to dedicate resources to finding such $$(x_1,x_2)$$ pairs. She puts her money where her mouth in claiming quantum computational advantage.

The function $$f(x)$$ need not satisfy Simon's promise nor even be two-to-one; I had initially envisioned $$f(x)$$ to be the last $$n$$ bits of SHA256. But SHA256 is cost-prohibitive in the NISQ era, probably requiring a circuit depth of several hundred thousand thousand thereabouts.

I think, however, that any shallow Boolean function $$f$$ may have the desirable properties of being easy to implement on a quantum computer but being hard to classically invert. I would guess that with maybe 150 qubits with all-to-all connectivity able to run a circuit of depth maybe 40-60 Boolean operations such as $$\mathsf{CCNOT}$$ or $$\mathsf{CSWAP}$$ would work.