Depth-scaling fine-grained quantum supremacy based on SETH and qubit-scaling fine-grained quantum supremacy based on Orthogonal Vectors and 3-SUM

We first show that under SETH and its variant, strong and weak classical simulations of quantum computing are impossible in certain double-exponential time of the circuit depth. We next show that under Orthogonal Vectors, 3-SUM, and their variants, strong and weak classical simulations of quantum computing are impossible in certain exponential time of the number of qubits.

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I Introduction

It is known that acceptance probabilities of several sub-universal quantum computing models cannot be classically sampled in polynomial time within a multiplicative error

unless the polynomial-time 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 that

 |pacc−qacc|≤ϵpacc.

Classically sampling output probability distributions of quantum computing is also called a weak simulation. Several sub-universal models that exhibit such “quantum supremacy” have been found such as the depth-four model

TD , the Boson Sampling model BS , the IQP model IQP , the one-clean-qubit model KL ; MFF ; KobayashiPRL ; KobayashiICALP , and the HC1Q model HC1Q . All these quantum supremacy results, however, prohibit only polynomial-time classical simulations: these models could be classically simulated in exponential time. To exclude possibilities of classical super-polynomial-time simulations, the study of more “fine-grained” quantum supremacy has been started. In Ref. Huang ; Huang2 , impossibilities of some exponential-time strong simulations (i.e., classical calculations of acceptance probabilities of quantum computing) were shown based on the exponential-time hypothesis (ETH) and the strong exponential-time 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 SETH-like conjectures. Ref. MT showed similar results for the one-clean-qubit model and the HC1Q model. Refs. MT ; Huang2 also studied fine-grained quantum supremacy of Clifford- quantum computing, and Ref. MT studied Hadamard-classical quantum computing.

In this paper, we show the following results. First, all previous results Huang ; Huang2 ; Dalzell ; MT on fine-grained 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 double-exponential 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 fine-grained quantum supremacy results (in terms of the qubit-scaling) based on Orthogonal Vectors WilliamsSETHOV and its variant (Theorem 3 and Theorem 4), and 3-SUM 3-SUM and its variant (Theorem 5 and Theorem 6). Orthogonal Vectors and 3-SUM are other well-studied conjectures in fine-grained complexity.

Ii Depth-scaling based on SETH

In this section, we show impossibilities of classical simulations in the double-exponential 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

 #f≡∑x∈{0,1}nf(x).

Then, for any constant , there exists a constant such that holds for infinitely many .

Conjecture 2

Let be any non-deterministic -time algorithm such that the following holds: given (a description of) a CNF, , with at most clauses, accepts if and rejects if , where

 gap(f)≡∑x∈{0,1}n(−1)f(x).

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

 COPY(|x⟩⊗|0n⟩⊗m−1)=|x⟩⊗m

for any . As is shown in Fig. 1, it can be realized in the -depth circuit.

Second, let us define the -qubit state

 |Θs⟩≡1√2s[∑z∈{0,1}s∖{1s}|z⟩⊗|0⟩+|1s⟩⊗|1⟩].

As is shown in Fig. 2, can be generated in the postselection circuit with -depth and qubits.

Define

 |AND⟩ ≡ |Θm⟩, |OR⟩ ≡ X⊗n+1|Θn⟩.

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 depth-1 circuit that acts on as is shown in Fig. 3. It is easy to verify that

 (I⊗n⊗|0⟩⟨0|⊗n⊗I)Uj(|x⟩⊗|OR⟩)=1√2n(|x⟩⊗|0n⟩⊗|Cj(x)⟩)

for any .

Let us define the depth-1 circuit (which is nothing but the bit-wise CNOT) as is shown in Fig. 4. Then,

for any .

Let us consider the -depth -qubit circuit of Fig. 5. We can show that

 pacc≡|⟨0...0|W|0...0⟩|2=gap(f)22g(n,m), (1)

where is a certain function of and .

It is understood as follows. First,

 ∑x∈{0,1}n|x⟩⊗m⊗|OR⟩⊗m⊗|AND⟩

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

 ∑x∈{0,1}n|x⟩⊗m⊗(m⨂j=1|Cj(x)⟩)⊗|AND⟩,

which needs a single depth. Third, by applying and doing postselections (that can be postponed to the last), we obtain

 ∑x∈{0,1}n|x⟩⊗m⊗(m⨂j=1|Cj(x)⟩)⊗|f(x)⟩,

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

 d=⌈log2m⌉+2≤log2m+3≤log2(nc)+3=log2n+log2c+3,

we have

 n≥2d−log2c−3.

Therefore,

 2(1−a)n≥2(1−a)2d−log2c−3≡T.

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

 |pacc−qacc|≤ϵpacc.

If , then

 qacc≥(1−ϵ)pacc>0.

If , then

 qacc≤(1+ϵ)pacc=0.

It means that deciding or can be done in non-deterministic time , which contradicts to Conjecture 2. Hence Theorem 2 has been shown.

Iii Orthogonal Vectors

In this section, we show fine-grained 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 non-deterministic time . Here,

 gap≡|{(i,j) | ui⋅vj=0}|−|{(i,j) | ui⋅vj≠0}|.

It is known that SETH is reduced to OV WilliamsSETHOV :

Lemma 1

If Conjecture 1 is true, then Conjecture 3 is true.

Lemma 2

If Conjecture 2 is true, then Conjecture 4 is true.

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

 Ua≡(d⨂j=1Xaj⊕1⊗I)GTd+1(d⨂j=1Xaj⊕1⊗I),

where

 GTd+1≡|1d⟩⟨1d|⊗X+(I⊗d−|1d⟩⟨1d|)⊗I

is the generalized TOFFOLI gate on qubits. Then it is clear that

 Ua(|x⟩⊗|b⟩)=|x⟩⊗|b⊕δx,a⟩

for any and , where if for all , and otherwise.

For a given vectors, , let us define the -qubit state by

 |Ψu⟩ ≡ (n∏j=1Uuj)(H⊗d⊗I)|0d+1⟩ = 1√2d∑z∈{0,1}d|z⟩⊗|fu(z)⟩,

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

 |Ψu⟩⊗|Ψv⟩⊗|0d⟩⊗|0⟩=12d∑z∈{0,1}d∑y∈{0,1}d|z⟩⊗|fu(z)⟩⊗|y⟩⊗|fv(y)⟩⊗|0d⟩⊗|0⟩.
• Apply bit-wise TOFFOLI on the 1st, 3rd, and 5th registers to generate

 12d∑z∈{0,1}d∑y∈{0,1}d|z⟩⊗|fu(z)⟩⊗|y⟩⊗|fv(y)⟩⊗|z1y1,z2y2,...,zdyd⟩⊗|0⟩.
• Flip the 6th register if and only if the 5th register is :

 12d∑z∈{0,1}d∑y∈{0,1}d|z⟩⊗|fu(z)⟩⊗|y⟩⊗|fv(y)⟩⊗|z1y1,z2y2,...,zdyd⟩⊗|δz⋅y,0⟩≡|Φ⟩,

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

 pacc≡|⟨+d1+d1+d−|Φ⟩|2=gap225d+1.

Let . Assume that can be classically exactly calculated in time

 T≡2(2−δ)(N−4)3c=n2−δ.

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 non-deterministic time , which contradicts to Conjecture 4. Hence Theorem 4 has been shown.

Iv 3-Sum

Finally, in this section, we show quantum supremacy results in terms of the qubit scaling based on 3-SUM and its variant. Let us consider the following two conjectures:

Conjecture 5 (3-Sum)

Given a set of size , deciding or cannot be done in time for any . Here,

 s≡|{(a,b,c)∈S×S×S | a+b+c=0}|.
Conjecture 6

Given a set of size , deciding or cannot be done in non-deterministic time for any . Here,

 gap≡|{(a,b,c)∈S×S×S | a+b+c=0}|−|{(a,b,c)∈S×S×S | a+b+c≠0}|.

There is no known reduction between SETH and 3-SUM. 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 non-negative integer , let us denote its binary representation by

 B[a]≡(a0,...,ar−1)∈{0,1}r,

where

 a=r−1∑j=02jaj.

We also define the integer representation of an -bit string by

 I[x]≡r−1∑j=02jxj.

For any -bit non-negative integer , let us define the -qubit unitary operator by

 Ua≡(r−1⨂j=0Xaj⊕1⊗I)GTr+1(r−1⨂j=0Xaj⊕1⊗I),

where

 GTr+1≡|1r⟩⟨1r|⊗X+(I⊗r−|1r⟩⟨1r|)⊗I

is the generalized TOFFOLI on qubits. Then it is clear that

 Ua(|x0,...,xr−1⟩⊗|b⟩)=|x0,...,xr−1⟩⊗|b⊕δx,a⟩

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

 A(|0⟩⊗|a0,...,ar−1⟩⊗|b0,...,br−1⟩⊗|0⟩)=|0⟩⊗|a0,...,ar−1⟩⊗|s0,...,sr−1⟩⊗|sr⟩

for any non-negative 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

 S′≡{e′1,e′2,...,e′n},

where for all . Then, all elements of are non-negative integers, and if and only if . Let be the smallest integer such that . Let us define the -qubit state by

 |Ψ⟩ ≡ (n∏j=1Ue′j)(H⊗r⊗I)|0r+1⟩ = 1√2r∑x∈{0,1}r|x⟩⊗|f(x)⟩,

where if and if .

Let us consider the following quantum computing (Fig. 7):

• Generate

 |Ψ⟩⊗3⊗|0⟩=1√23r∑x,y,z|x⟩⊗|f(x)⟩⊗|y⟩⊗|f(y)⟩⊗|z⟩⊗|f(z)⟩⊗|0⟩.
• Apply the addition circuit between the first and third registers:

 1√23r∑x,y,z|x⟩⊗|f(x)⟩⊗|B[I[x]+I[y]]⟩⊗|f(y)⟩⊗|z⟩⊗|f(z)⟩⊗|0⟩.
• Apply the addition circuit between the third and fifth registers:

 1√23r∑x,y,z|x⟩⊗|f(x)⟩⊗|B[I[x]+I[y]]⟩⊗|f(y)⟩⊗|B[I[x]+I[y]+I[z]]⟩⊗|f(z)⟩⊗|0⟩.
• 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:

 1√23r∑x,y,z|x⟩⊗|f(x)⟩⊗|B[I[x]+I[y]]⟩⊗|f(y)⟩⊗|B[I[x]+I[y]+I[z]]⟩ ⊗|f(z)⟩⊗|δI[x]+I[y]+I[z],3n3+η⟩≡|Φ⟩.

Then

 pacc≡|⟨+r1+r+11+r+21−|Φ⟩|2=gap226r+4.

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,

 T≡2(2−δ)(N−15)3(3+η)≤n2−δ.

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 non-deterministic 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 non-negative 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 .