A novel quantum algorithm for converting between one-hot and binary encodings

Abstract

In the domain of quantum computing, two widely employed techniques for encoding a normalized vector of length N, denoted as $\{ \alpha _i \}$, are one-hot encoding and binary encoding. The one-hot encoding state is represented as $\vert \psi _{OH}^{(N)} \rangle$ and can be expressed as: $\vert \psi _{OH}^{(N)} \rangle =\sum _{i=0}^{N-1} \alpha _i \vert 0 \rangle ^{\otimes N-i-1} \vert 1 \rangle \vert 0 \rangle ^{\otimes i}$. On the other hand, the binary encoding state is symbolized as $\vert \psi _{BI}^{(N)} \rangle$ and is defined as: $\vert \psi _{BI}^{(N)} \rangle =\sum _{i=0}^{N-1} \alpha _i \vert b_i \rangle$, where $b_i$ corresponds to the binary representation of i. In this paper, we introduce a method for converting between the one-hot encoding state and the binary encoding state, utilizing the Domain Wall state as an intermediary. The Domain Wall state, denoted as $\vert \psi _{DW}^{(N)} \rangle$, is defined as: $\vert \psi _{DW}^{(N)} \rangle =\sum _{i=0}^{N-1} \alpha _i \vert 0 \rangle ^{\otimes N-i-1} \vert 1 \rangle ^{\otimes i}$. Our proposed circuit achieves a depth of $O(\log ^2 N)$ and a size of O(N).Kindly check and confirm that the corresponding author mail id is correctly identified.

Appendices

The procedure in $U_O^{(4)}$ and $U_O^{(5)}$

Here, we give the transformation of the quantum state in $U_O^{(4)}$ and $U_O^{(5)}$.

$U_O^{(4)}$:

$$\begin{aligned} \vert 000\rangle \xrightarrow {\otimes \vert 1\rangle } \vert 0001\rangle \xrightarrow {CNOT \text { 1 to 2, 3 to 4}} \vert 0001\rangle \xrightarrow {U_O^{(2)} \text { 1 to 3}} \vert 0001\rangle \xrightarrow {CNOT \text { 2 to 3}} \vert 0001\rangle \\ \vert 001\rangle \xrightarrow {\otimes \vert 1\rangle } \vert 0011\rangle \xrightarrow {CNOT \text { 1 to 2, 3 to 4}} \vert 0010\rangle \xrightarrow {U_O^{(2)} \text { 1 to 3}}\vert 0010\rangle \xrightarrow {CNOT \text { 2 to 3}} \vert 0010\rangle \\ \vert 011\rangle \xrightarrow {\otimes \vert 1\rangle } \vert 0111\rangle \xrightarrow {CNOT \text { 1 to 2, 3 to 4}} \vert 0110\rangle \xrightarrow {U_O^{(2)} \text { 1 to 3}}\vert 0110\rangle \xrightarrow {CNOT \text { 2 to 3}} \vert 0100\rangle \\ \vert 111\rangle \xrightarrow {\otimes \vert 1\rangle } \vert 1111\rangle \xrightarrow {CNOT \text { 1 to 2, 3 to 4}} \vert 1010\rangle \xrightarrow {U_O^{(2)} \text { 1 to 3}}\vert 1000\rangle \xrightarrow {CNOT \text { 2 to 3}} \vert 1000 \rangle .\\ \end{aligned}$$

As for $U_O^{(5)}$, first tensor $\vert 1\rangle$ with the quantum state.

$$\begin{aligned} \vert 0000\rangle \xrightarrow {\otimes \vert 1\rangle } \vert 00001\rangle \xrightarrow {U_O^{(4)}} \vert 00001\rangle \xrightarrow {CNOT \text { 1 TO 2}} \vert 00001\rangle \\ \vert 0001\rangle \xrightarrow {\otimes \vert 1\rangle } \vert 00011\rangle \xrightarrow {U_O^{(4)}} \vert 00010\rangle \xrightarrow {CNOT \text { 1 TO 2}} \vert 00010\rangle \\ \vert 0011\rangle \xrightarrow {\otimes \vert 1\rangle } \vert 00111\rangle \xrightarrow {U_O^{(4)}} \vert 00100\rangle \xrightarrow {CNOT \text { 1 TO 2}} \vert 00100\rangle \\ \vert 0111\rangle \xrightarrow {\otimes \vert 1\rangle } \vert 01111\rangle \xrightarrow {U_O^{(4)}} \vert 01000\rangle \xrightarrow {CNOT \text { 1 TO 2}} \vert 01000\rangle \\ \vert 1111\rangle \xrightarrow {\otimes \vert 1\rangle } \vert 11111\rangle \xrightarrow {U_O^{(4)}} \vert 11000\rangle \xrightarrow {CNOT \text { 1 TO 2}} \vert 10000\rangle .\\ \end{aligned}$$

The procedure of $U_B^{(7)}$

Here, we give the transformation of the quantum state in $U_B^{(7)}$

$U_B^{(7)}$:

$$\begin{aligned} \vert 000000\rangle \xrightarrow {U_B^{(4)} \otimes U_B^{(4)}} \vert 000000\rangle \xrightarrow {adder-plus-1} \vert 000001\rangle \xrightarrow {adder} \vert 001001\rangle \\ \xrightarrow {Toffoli} \vert 001001\rangle \xrightarrow {CNOT} \vert 000001\rangle \xrightarrow {adder-minus-1} \vert 000000\rangle \\ \vert 000001\rangle \xrightarrow {U_B^{(4)} \otimes U_B^{(4)}} \vert 000001\rangle \xrightarrow {adder-plus-1} \vert 000010\rangle \xrightarrow {adder} \vert 010010\rangle \\ \xrightarrow {Toffoli} \vert 010010\rangle \xrightarrow {CNOT} \vert 000010\rangle \xrightarrow {adder-minus-1} \vert 000001\rangle \\ \vert 000011\rangle \xrightarrow {U_B^{(4)} \otimes U_B^{(4)}} \vert 000010\rangle \xrightarrow {adder-plus-1} \vert 000011\rangle \xrightarrow {adder} \vert 011011\rangle \\ \xrightarrow {Toffoli} \vert 011011\rangle \xrightarrow {CNOT} \vert 000011\rangle \xrightarrow {adder-minus-1} \vert 000010\rangle \\ \vert 000111\rangle \xrightarrow {U_B^{(4)} \otimes U_B^{(4)}} \vert 000011\rangle \xrightarrow {adder-plus-1} \vert 000100\rangle \xrightarrow {adder} \vert 100100\rangle \\ \xrightarrow {Toffoli} \vert 100100\rangle \xrightarrow {CNOT} \vert 000100\rangle \xrightarrow {adder-minus-1} \vert 000011\rangle \\ \vert 001111\rangle \xrightarrow {U_B^{(4)} \otimes U_B^{(4)}} \vert 001011\rangle \xrightarrow {adder-plus-1} \vert 001100\rangle \xrightarrow {adder} \vert 101100\rangle \\ \xrightarrow {Toffoli} \vert 101101\rangle \xrightarrow {CNOT} \vert 000101\rangle \xrightarrow {adder-minus-1} \vert 000100\rangle \\ \vert 011111\rangle \xrightarrow {U_B^{(4)} \otimes U_B^{(4)}} \vert 010011\rangle \xrightarrow {adder-plus-1} \vert 010100\rangle \xrightarrow {adder} \vert 110100\rangle \\ \xrightarrow {Toffoli} \vert 110110\rangle \xrightarrow {CNOT} \vert 000110\rangle \xrightarrow {adder-minus-1} \vert 000101\rangle \\ \vert 111111\rangle \xrightarrow {U_B^{(4)} \otimes U_B^{(4)}} \vert 011011\rangle \xrightarrow {adder-plus-1} \vert 011100\rangle \xrightarrow {adder} \vert 111100\rangle \\ \xrightarrow {Toffoli} \vert 111111\rangle \xrightarrow {CNOT} \vert 000111\rangle \xrightarrow {adder-minus-1} \vert 000110\rangle \\ \end{aligned}$$

The adder-plus-d gate

We can leverage the Fourier gate to construct the adder-plus-d gate, as depicted in Fig. 12.

Fig. 12

The Fourier adder-plus-d Gate

An adder-plus-d gate, comprised of n qubits (where $n=\lceil \log _2 \frac{N}{2} \rceil +1$ in our approach), is assembled using a sequence involving a quantum Fourier transformation (QFT) gate, n phase gates, and an inverse quantum Fourier transformation (IQFT) gate. When initiated with the state $\vert b_j \rangle$, it evolves as follows:

$$\begin{aligned} \frac{1}{2^{n/2}} \mathop {\otimes } \limits _{k=0}^{n-1}(\vert 0 \rangle +e^{2\pi ij \frac{2^k}{2^n}}\vert 1 \rangle ) \end{aligned}$$

(C1)

after undergoing the QFT gate. Subsequently, the collective effect of phase gates $\mathop \otimes _{k=0}^{n-1} U(2d\pi \frac{2^k}{2^n})$, where $U(\lambda )$ is defined as

$$\begin{aligned} U(\lambda )= \begin{bmatrix} 1 &{}0 \\ 0 &{}e^{i\lambda } \end{bmatrix}, \end{aligned}$$

transforms the state into:

$$\begin{aligned} \frac{1}{2^{n/2}} \mathop {\otimes } \limits _{k=0}^{n-1}(\vert 0 \rangle +e^{2\pi i(j+d) \frac{2^k}{2^n}}\vert 1 \rangle ), \end{aligned}$$

(C2)

Ultimately, after the application of the IQFT gate, the state transitions to $\vert b_{j+d} \rangle$.

The Fourier adder-plus-d gate exhibits a computational depth of O(n) and a size of $O(n \log n)$. For those seeking reduced complexity, the quantum carry lookahead adder (QCLA) method [28] offers a viable alternative. By introducing approximately O(n) ancillary qubits, QCLA can implement the adder with a computational depth of $O(\log n)$ and a size of O(n).

The recursion converter

In this section, we introduce the recursion converter as presented by Plesch et al. [23]. While Plesch et al. demonstrated that the circuit initially possesses a depth of $O(N\log ^2 N)$ and a size of $O(N\log ^2 N)$, it is noteworthy that the circuit depth can be optimized to $O(N\log N)$ following a scheme proposed by Saeedi et al. This optimization stems from the decomposition of n-qubit Toffoli gates in linear depth.

In Plesch’s framework, the converter is initially designed to facilitate conversion between the one-hot encoding state and the binary encoding state. With a slight modification, the same circuit can be adapted to convert between the Domain Wall state and the binary state. A visual representation of the $N=6$ qubit converter is presented in Fig. 13.

Fig. 13

The converter with binary calculation

Assuming we have already prepared an $N-2$ qubit converter, denoted as $U_B^{(N-1)}$, and initialized the state $\vert \psi _{DW}^{(N)} \rangle$, applying $U_B^{(N-1)}$ to the last $N-2$ qubits yields the quantum state:

$$\begin{aligned} \alpha _{N-1} \vert 1\rangle \vert 0\rangle ^{\otimes N-1-\lceil \log _2 N \rceil } \vert b_{N-2}\rangle + \vert 0\rangle \sum _{i=0}^{N-2} \alpha _{i} \vert 0\rangle ^{\otimes N-1-\lceil \log _2 N \rceil } \vert b_i\rangle . \end{aligned}$$

(D3)

Next, a bitwise XOR operation, denoted as $c=c_1 \ldots c_{\lceil \log _2 N \rceil }=b_{N-1} \oplus b_{N-2}$, is computed. For all i satisfying $c_i=1$, CNOT gates are applied between the first qubit and the $N-\lceil \log _2 N \rceil +i$ th qubit. Subsequently, a multiple-qubit Toffoli gate is employed to flip the first qubit if the last $\lceil \log _2 N \rceil$ qubits are in the state $\vert b_{N-1}\rangle$. As a result, the final state is transformed into the binary state. Notably, in the original scheme by Plesch et al. [23], c can be replaced with $b_{N-1}$ to prepare $\vert \psi _{BI}^{(N-1)}\rangle$ from $\vert \psi _{OH}^{(N-1)}\rangle$.

Generation of the binomial distribution state

As Fig. 6 shows, we first implement $R_Y(\theta )^{\otimes N}$ on $\vert 0\rangle ^{\otimes N}$, where $\theta =2\arcsin \sqrt{p}$ and $R_Y(\theta )=\begin{bmatrix} \cos \frac{\theta }{2} &{} -\sin \frac{\theta }{2}\\ \sin \frac{\theta }{2} &{} \cos \frac{\theta }{2}\end{bmatrix}$ and we will obtain

$$\begin{aligned} R_Y(\theta )^{\otimes N}(\vert 0\rangle ^{\otimes N}) = \sum _{k=0}^N \sqrt{f(k,p)}\vert D_k^N \rangle , \end{aligned}$$

(E4)

where $\vert D_k^{N} \rangle =\left( {\begin{array}{c}N\\ k\end{array}}\right) ^{-\frac{1}{2}} \sum _{x \in \{0,1\}^N, wt(x)=k} \vert x \rangle$ is the Dicke state. Second, we implement the inverse gate of $U_{N,N-1}$ presented in [24] and the state becomes

$$\begin{aligned} U_{N,N-1} \vert 0 \rangle ^{\otimes N-k} \vert 1 \rangle ^{\otimes k} = \vert D_k^{N} \rangle \end{aligned}$$

(E5)

for each $k \in \{0, 1, \ldots , N\}$, so $U_{N,N-1}^{-1}$ satisfies that

$$\begin{aligned} U_{N,N-1}^{-1} \sum _{k=0}^N \sqrt{f(k,p)}\vert D_k^N \rangle = \sum _{i=0}^{N} \sqrt{f(i, p)} \vert 0 \rangle ^{\otimes N-i} \vert 1 \rangle ^{\otimes i}. \end{aligned}$$

(E6)

$U_{N,N-1}$ can be decomposed to Split & Cyclic Shift gate ($SCS_{n,k}$) that is in the form of

$$\begin{aligned} U_{N, N-1}:=\prod _{\ell =2}^{N-1} (S C S_{\ell , \ell -1} \otimes I^{\otimes N-\ell } ) \cdot (I^{\otimes N-k-1} \otimes S C S_{N, N-1} ), \end{aligned}$$

(E7)

where $SCS_{n,k}$ satisfies that

$$\begin{aligned}&S C S_{n, k} \vert 0\rangle ^{\otimes k+1} =\vert 0\rangle ^{\otimes k+1}, \nonumber \\&S C S_{n, k}\vert 0\rangle ^{\otimes k+1-\ell }\vert 1\rangle ^{\otimes \ell } =\sqrt{\frac{\ell }{n}}\vert 0\rangle ^{\otimes k+1-\ell }\vert 1\rangle ^{\otimes \ell }+\sqrt{\frac{n-\ell }{n}}\vert 0\rangle ^{\otimes k-\ell }\vert 1\rangle ^{\otimes \ell }\vert 0\rangle , \nonumber \\&S C S_{n, k}\vert 1\rangle ^{\otimes k+1} =\vert 1\rangle ^{\otimes k+1} . \end{aligned}$$

(E8)

As proved in [24], $SCS_{n,k}$ can be decomposed to 1 two-qubit gates and $k-1$ three-qubit gates shown as in Figs. 14 and 15.

Fig. 14

Two-qubit gates $SCS_2$

Fig. 15

Three-qubit gates $SCS_3$

Finally, we apply $U_B^{(N+1)}$ or $U_O^{(N+1)}$ to transform the Domain Wall state to the one-hot encoding state or the binary encoding state,⁴ which is denoted as

$$\begin{aligned}{} & {} U_O^{(N+1)} \left( \sum _{i=0}^{N} \sqrt{f(i, p)} \vert 0 \rangle ^{\otimes N-i} \vert 1 \rangle ^{\otimes i} \otimes \vert 1 \rangle \right) =\sum _{i=0}^{N} \sqrt{f(i, p)} \vert 0 \rangle ^{\otimes N-i} \vert 1\rangle \vert 0\rangle ^{\otimes i}\end{aligned}$$

(E9)

$$\begin{aligned}{} & {} U_B^{(N+1)} \left( \sum _{i=0}^{N} \sqrt{f(i, p)} \vert 0 \rangle ^{\otimes N-i} \vert 1 \rangle ^{\otimes i}\right) =\sum _{i=0}^{N} \sqrt{f(i, p)} \vert 0\rangle ^{\otimes N-\lceil \log _2 N \rceil } \vert b_i \rangle . \end{aligned}$$

(E10)

Proof in Sect. 2.4

1.1 F.1 The solution of Eq. (17)

We utilize the binary representation $b_N=c_n c_{n-1}\cdots c_1$ to denote the number N. Referring to Eq. (17), we can express this as follows:

$$\begin{aligned} d_o(c_n c_{n-1}\cdots c_1)&=d_o(c_n c_{n-1}\cdots c_{2} 0)\\&=d_o(c_n c_{n-1}\cdots c_{2})+2\\&=d_o(c_n c_{n-1}\cdots c_{3})+2*2\\&=d_o(c_n c_{n-1})+2(n-2)\\&=2n-1=O(\log n) \end{aligned}$$

1.2 F.2 The solution of Eq. (18)

We utilize the binary representation $b_N=c_n c_{n-1}\cdots c_1$ to denote the number N. Referring to Eq. (18), we can express this as follows:

$$\begin{aligned} s_o(c_n c_{n-1}\cdots c_1)&=s_o(c_n c_{n-1}\cdots c_{2} 0)+c_1\\&=s_o(c_n c_{n-1}\cdots c_{2})+N-1\\&=s_o(c_n c_{n-1}\cdots c_{3})+N+\lfloor \frac{N}{2} \rfloor -2\\&=s_o(c_n c_{n-1})+\sum _{i=0}^{n-3} \lfloor \frac{N}{2^i} \rfloor -n+2\\&=c_{n-1}+\sum _{i=0}^{n-3} \lfloor \frac{N}{2^i} \rfloor -n+3 = O(N) \end{aligned}$$

1.3 F.3 The solution of Eq. (19)

When $2^{m-1}+1 <N \le 2^m+1$ and $O(\log N)=O(m)$, in expanding-to-$2^k$, we have

$$\begin{aligned} d_b(N)&=d_b(2^m+1)=d_b(2^{m-1}+1)+O(m)\\ {}&=d_b(2^{m-2}+1)+O(m)+O(m-1)\\&=\sum _{i=1}^m O(i)\\&=O(m^2) = O(\log ^2 N) \end{aligned}$$

$$\begin{aligned} s_b(N)&= s_b(2^m+1)=2s_b(2^{m-1}+1)+O(m)\\ {}&=4s_b(2^{m-2}+1)+O(m)+2O(m-1)\\&=\sum _{i=0}^{m-1} 2^i O(m-i)\\&=O(2^m)=O(N) \end{aligned}$$

1.4 F.4 The solution of Eq. (20)

Let’s note that $M=N-1$, and $d(x)=d_b(x+1)$. We represent M in binary as $b_M=c_n c_{n-1}\cdots c_1$. Therefore, we have the following relationships:

$$\begin{aligned} d(M)&=d\left( \frac{M}{2}\right) +O(n), \text {if }M\text { is even.}\\ d(M)&=d_b(M+1), \text {if }M\text { is odd.} \end{aligned}$$

Hence,

$$\begin{aligned} d_b(N)&=d(c_n c_{n-1}\cdots c_1)\\&=d(c_n c_{n-1}\cdots c_{2} + c_1)+O(n)\\&=d(c_n c_{n-1}\cdots c_{3} + c_{2} \vee c_1)+O(n)+O(n-1)\\&=d(c_n c_{n-1}+c_{n-2}\vee \cdots \vee c_1)+\sum _{i=0}^{n-3} O(n-i) = O(\log ^2 N) \end{aligned}$$

1.5 F.5 The solution of Eq. (21)

Let us note that $M=N-1$, and $s(x)=s_b(x+1)$. We represent M in binary as $b_M=c_n c_{n-1}\cdots c_1$. Therefore, we have the following relationships:

$$\begin{aligned} s(M)&=2s\left( \frac{M}{2}\right) +O(n^2), \text {if }M\text { is even.}\\ s(M)&=s_b(M+1), \text {if }M\text { is odd.} \end{aligned}$$

Hence,

$$\begin{aligned} s_b(N)&=s(c_n c_{n-1}\cdots c_1)\\&=2^1s(c_n c_{n-1}\cdots c_{2} + c_1)+O(n^2)\\&=2^2s(c_n c_{n-1}\cdots c_{3} + c_2 \vee c_1)+O(n^2)+2O((n-1)^2)\\&=2^{n-2}s(c_n c_{n-1}+c_{n-2}\vee \cdots \vee c_1)+\sum _{i=0}^{n-3} 2^i O((n-i)^2)\\&=O(2^n)+2^n\sum _{k=3}^{n}2^{-k}O(k^2)=O(2^n)-O(n^2)=O(N) \end{aligned}$$

1.6 F.6 The solution of Eq. (22)

Let us note that $M=N-1$, and $d(x)=d_b(x+1)$. We represent M in binary as $b_M=c_n c_{n-1}\cdots c_1$. Therefore, we have the following relationships:

$$\begin{aligned} d(M)&=d\left( \frac{M}{2}\right) +O(n), \text {if }M\text { is even.}\\ d(M)&=d_b(M-1)+O(n), \text {if }M\text { is odd.} \end{aligned}$$

Hence,

$$\begin{aligned} d_b(N)&=d(c_n c_{n-1}\cdots c_1)\\&=d(c_n c_{n-1}\cdots c_{2} 0)+c_1 O(n)\\&=d(c_n c_{n-1}\cdots c_{2})+(c_1+1) O(n)\\&=d(c_n c_{n-1}\cdots c_{3})+(c_1+1) O(n)+(c_2+1) O(n-1)\\&=d(c_n c_{n-1})+\sum _{i=1}^{n-2}(c_i+1) O(n-i+1) \\&<2 O(n^2)=O(\log ^2 N)\\ \end{aligned}$$

1.7 F.7 The solution of Eq. (23)

Let us note that $M=N-1$, and we denote $s(x)=s_b(x+1)$. We represent M in binary as $b_M=c_n c_{n-1}\cdots c_1$. Therefore, we have the following relationships:

$$\begin{aligned} s(M)&=2s\left( \frac{M}{2}\right) +O(n^2), \text {if }M\text { is even.}\\ s(M)&=s_b(M-1)+O(n^2), \text {if }M\text { is odd.} \end{aligned}$$

Hence,

$$\begin{aligned} s_b(N)&=s(c_n c_{n-1}\cdots c_1)\\&=s(c_n c_{n-1}\cdots c_{2} 0)+c_1 O(n^2)\\&=2s(c_n c_{n-1}\cdots c_{2})+(c_1+1) O(n^2)\\&=2^2s(c_n c_{n-1}\cdots c_{3})+(c_1+1) O(n^2)+2(c_2+1)O((n-1)^2)\\&=2^{n-2}s(c_n c_{n-1})+\frac{1}{2}\sum _{i=1}^{n-2}2^i(c_i+1) O((n-i+1)^2) \\&<O(2^n)+\sum _{i=1}^{n-2}2^iO((n-i)^2)=O(2^n) +2^n\sum _{k=2}^{n-1}2^{-k}O(k^2)\\&=O(2^n)-O(n^2)=O(N) \end{aligned}$$