I Introduction
It is known that acceptance probabilities of several subuniversal quantum computing models cannot be classically sampled in polynomial time within a multiplicative error
unless the polynomialtime hierarchy collapses to the second level. Here, we say that an acceptance probability is classically sampled in time within a multiplicative error if there exists a time classical probabilistic algorithm that accepts with probability such thatClassically sampling output probability distributions of quantum computing is also called a weak simulation. Several subuniversal models that exhibit such “quantum supremacy” have been found such as the depthfour model
TD , the Boson Sampling model BS , the IQP model IQP , the onecleanqubit model KL ; MFF ; KobayashiPRL ; KobayashiICALP , and the HC1Q model HC1Q . All these quantum supremacy results, however, prohibit only polynomialtime classical simulations: these models could be classically simulated in exponential time. To exclude possibilities of classical superpolynomialtime simulations, the study of more “finegrained” quantum supremacy has been started. In Ref. Huang ; Huang2 , impossibilities of some exponentialtime strong simulations (i.e., classical calculations of acceptance probabilities of quantum computing) were shown based on the exponentialtime hypothesis (ETH) and the strong exponentialtime hypothesis (SETH) SETH1 ; SETH2 ; SETH3 . Ref. Dalzell showed that acceptance probabilities of the IQP model, the QAOA model QAOA , and the Boson Sampling model cannot be classically sampled in some exponential time within a multiplicative error under some SETHlike conjectures. Ref. MT showed similar results for the onecleanqubit model and the HC1Q model. Refs. MT ; Huang2 also studied finegrained quantum supremacy of Clifford quantum computing, and Ref. MT studied Hadamardclassical quantum computing.In this paper, we show the following results. First, all previous results Huang ; Huang2 ; Dalzell ; MT on finegrained quantum supremacy are impossibilities of classical simulations in exponential times of the number of qubits or the number of certain gates (such as the Hadamard or .) In Sec. II, we show impossibilities of strong and weak classical simulations in the doubleexponential time of the circuit depth based on SETH and its variant (Theorem 1 and Theorem 2). Second, all previous results Huang ; Huang2 ; Dalzell ; MT are based on ETH, SETH, or their variants that are conjectures for SAT. In Sec. III and Sec. IV, we show finegrained quantum supremacy results (in terms of the qubitscaling) based on Orthogonal Vectors WilliamsSETHOV and its variant (Theorem 3 and Theorem 4), and 3SUM 3SUM and its variant (Theorem 5 and Theorem 6). Orthogonal Vectors and 3SUM are other wellstudied conjectures in finegrained complexity.
Ii Depthscaling based on SETH
In this section, we show impossibilities of classical simulations in the doubleexponential time of the circuit depth based on SETH and its variant. We consider the following two conjectures:
Conjecture 1 (Seth)
Let be any deterministic time algorithm such that the following holds: given (a description of) a CNF, , with at most clauses, accepts if and rejects if , where
Then, for any constant , there exists a constant such that holds for infinitely many .
Conjecture 2
Let be any nondeterministic time algorithm such that the following holds: given (a description of) a CNF, , with at most clauses, accepts if and rejects if , where
Then, for any constant , there exists a constant such that holds for infinitely many .
Based on these two conjectures, we can show the following two results:
Theorem 1 (Strong simulation)
Assume that Conjecture 1 is true. Then, for any constant and for infinitely many , there exists a constant and a depth quantum circuit whose acceptance probability cannot be classically exactly calculated in time .
Theorem 2 (Weak simulation)
Assume that Conjecture 2 is true. Then, for any constant and for infinitely many , there exists a constant and a depth quantum circuit whose acceptance probability cannot be classically sampled in time within a multiplicative error .
Proof. First, let us define the COPY operation by
for any . As is shown in Fig. 1, it can be realized in the depth circuit.
Second, let us define the qubit state
As is shown in Fig. 2, can be generated in the postselection circuit with depth and qubits.
Define
The state is generated in the depth qubit postselection circuit. The state is generated in the depth qubit postselection circuit.
Let be a CNF with clauses. Let be the th clause of . Each contains at most literals, because if both and are in , such a clause is trivially satisfied. For each , let us define the depth1 circuit that acts on as is shown in Fig. 3. It is easy to verify that
for any .
Let us define the depth1 circuit (which is nothing but the bitwise CNOT) as is shown in Fig. 4. Then,
for any .
Let us consider the depth qubit circuit of Fig. 5. We can show that
(1) 
where is a certain function of and .
It is understood as follows. First,
can be generated in the depth and qubits with postselection (that can be postponed to the last). Second, by applying and doing postselections (that can be postponed to the last), we obtain
which needs a single depth. Third, by applying and doing postselections (that can be postponed to the last), we obtain
which needs a single depth. Finally, apply on the last qubit to obtain
and apply on all qubits that are not postselected. Then, it is clear that we obtain Eq. (1).
Since
we have
Therefore,
Assume that of Eq. (1) can be classically exactly calculated in time . Then, can be obtained in time , which contradicts to Conjecture 1. Therefore Theorem 1 has been shown. Assume that can be classically sampled within a multiplicative error in time . It means that there exists a classical probabilistic time algorithm that accepts with probability such that
If , then
If , then
It means that deciding or can be done in nondeterministic time , which contradicts to Conjecture 2. Hence Theorem 2 has been shown.
Iii Orthogonal Vectors
In this section, we show finegrained quantum supremacy in terms of the qubit scaling based on the Orthogonal Vectors and its variant. Let us introduce the following two conjectures:
Conjecture 3 (Orthogonal Vectors)
For any , there is a such that deciding whether or for given vectors, , with cannot be done in time . Here .
Conjecture 4
For any , there is a such that deciding whether or for given vectors, , with cannot be done in nondeterministic time . Here,
It is known that SETH is reduced to OV WilliamsSETHOV :
Proofs of Lemma 1 and Lemma 2 are given in Ref. WilliamsSETHOV . For the convenience of readers, we provide a proof in Appendix A. On the other hand, no reduction is known from OV to SETH. Based on the above two conjectures, we can show the following two results:
Theorem 3 (Strong simulation)
Assume that Conjecture 3 is true. Then, for any , there is a such that there exists an qubit quantum circuit whose acceptance probability cannot be classically exactly calculated in time .
Theorem 4 (Weak simulation)
Assume that Conjecture 4 is true. Then, for any , there is a such that there exists an qubit quantum circuit whose acceptance probability cannot be classically sampled within a multiplicative error in time .
Proof. For any , let us define the qubit unitary operator by
where
is the generalized TOFFOLI gate on qubits. Then it is clear that
for any and , where if for all , and otherwise.
For a given vectors, , let us define the qubit state by
where if and if . For a given vectors, , let us also define the qubit state in a similar way.
Let us consider the following quantum computing (Fig. 6):

Generate

Apply bitwise TOFFOLI on the 1st, 3rd, and 5th registers to generate

Flip the 6th register if and only if the 5th register is :
where .
Note that this quantum computing needs qubits. The reason is as follows: first, it is clear that qubits are needed. Second, the above quantum computing uses generalized TOFFOLI gates. A generalized TOFFOLI can be decomposed into a linear number of TOFFOLI gates with a single ancilla qubit that can be reused without any initialization Barenco . Therefore, an additional single ancilla qubit is needed. Hence in total, qubits are necessary.
Then
Let . Assume that can be classically exactly calculated in time
Then, or can be decided in time , which contradicts to Conjecture 3. Hence Theorem 3 has been shown. Next assume that can be classically sampled within a multiplicative error in time . Then, or can be decided in nondeterministic time , which contradicts to Conjecture 4. Hence Theorem 4 has been shown.
Iv 3Sum
Finally, in this section, we show quantum supremacy results in terms of the qubit scaling based on 3SUM and its variant. Let us consider the following two conjectures:
Conjecture 5 (3Sum)
Given a set of size , deciding or cannot be done in time for any . Here,
Conjecture 6
Given a set of size , deciding or cannot be done in nondeterministic time for any . Here,
There is no known reduction between SETH and 3SUM. Based on these two conjectures, we can show the following results:
Theorem 5 (Strong simulation)
Assume that Conjecture 5 is true. Then for any , there exists an qubit quantum circuit whose acceptance probability cannot be classically exactly calculated in time.
Theorem 6 (Weak simulation)
Assume that Conjecture 6 is true. Then for any , there exists an qubit quantum circuit whose acceptance probability cannot be classically sampled within a multiplicative error in time .
Proof. For any bit nonnegative integer , let us denote its binary representation by
where
We also define the integer representation of an bit string by
For any bit nonnegative integer , let us define the qubit unitary operator by
where
is the generalized TOFFOLI on qubits. Then it is clear that
for any bit string and any bit . Here, if for all , and otherwise.
There are quantum circuits that can do the addition. For example, in Ref. addition , the circuit such that
for any nonnegative integers and is introduced, where and , and with . (For details, see Appendix B.)
For a given set of size , let us define the set by
where for all . Then, all elements of are nonnegative integers, and if and only if . Let be the smallest integer such that . Let us define the qubit state by
where if and if .
Let us consider the following quantum computing (Fig. 7):

Generate

Apply the addition circuit between the first and third registers:

Apply the addition circuit between the third and fifth registers:

Flip the last register if the fifth register encodes . More precisely, let with . Then, first apply on the fifth register, and then apply the generalized TOFFOLI on the fifth register and the last register:
Then
This quantum computing needs qubits, because of the following reasons: first, it is clear that qubits are necessary. Second, the generalized TOFFOLI gates need a single ancilla qubit. Third, the first addition circuit needs two ancilla qubits. One of then can be reused for the second addition circuit. The second addition circuit needs three ancilla qubits. Hence in total qubits are necessary. Let . Then,
Assume that is classically exactly calculated in time . Then, or can be decided in time , which contradicts to Conjecture 5. Hence Theorem 5 has been shown. Next assume that is classically sampled within a multiplicative error in time . Then, or can be decided in nondeterministic time , which contradicts to Conjecture 6. Hence Theorem 6 has been shown.
Appendix A Reduction from SETH to OV
The reduction is shown in Ref. WilliamsSETHOV . For the convenience of readers, we provide a proof here.
Proof. Let be a CNF with clauses, where is even and is a constant. For each , define the vector whose th element is decided as follows for each :

From the th clause of , construct the clause by replacing all literals in with 0.

If is not satisfied by the assignment , then .

If is satisfied by the assignment , then .
For each , define the vector whose th element is decided as follows for each :

From the th clause of , construct the clause by replacing all literals in with 0.

If is not satisfied by the assignment , then .

If is satisfied by the assignment , then .
Then it is clear that if and only if the assignment satisfies . If we define , . Therefore . Assume that Conjecture 1 is true. Let be a CNF with . Then, for any , there exists a constant such that deciding or needs time. Therefore, for any , there exists such that deciding or for given vectors, needs time with , which is Conjecture 3. The reduction from Conjecture 2 to Conjecture 4 is the same.
Appendix B Addition circuit
Here we explain the addition circuit of Ref. addition . Let and be two nonnegative integers, where and . Let us define the MAJ gate and the UMA gate as is shown in Fig. 8. Here, and for , and for all and . The sum of and is , where .
In Fig. 9, we provide an example of the addition circuit for .
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