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Bypassing orthogonalization in the quantum DPP sampler

Published 7 Mar 2025 in quant-ph, cs.LG, and stat.CO | (2503.05906v2)

Abstract: Given an $n\times r$ matrix $X$ of rank $r$, consider the problem of sampling $r$ integers $\mathtt{C}\subset {1, \dots, n}$ with probability proportional to the squared determinant of the rows of $X$ indexed by $\mathtt{C}$. The distribution of $\mathtt{C}$ is called a projection determinantal point process (DPP). The vanilla classical algorithm to sample a DPP works in two steps, an orthogonalization in $\mathcal{O}(nr2)$ and a sampling step of the same cost. The bottleneck of recent quantum approaches to DPP sampling remains that preliminary orthogonalization step. For instance, (Kerenidis and Prakash, 2022) proposed an algorithm with the same $\mathcal{O}(nr2)$ orthogonalization, followed by a $\mathcal{O}(nr)$ classical step to find the gates in a quantum circuit. The classical $\mathcal{O}(nr2)$ orthogonalization thus still dominates the cost. Our first contribution is to reduce preprocessing to normalizing the columns of $X$, obtaining $\mathsf{X}$ in $\mathcal{O}(nr)$ classical operations. We show that a simple circuit inspired by the formalism of Kerenidis et al., 2022 samples a DPP of a type we had never encountered in applications, which is different from our target DPP. Plugging this circuit into a rejection sampling routine, we recover our target DPP after an expected $1/\det \mathsf{X}\top\mathsf{X} = 1/a$ preparations of the quantum circuit. Using amplitude amplification, our second contribution is to boost the acceptance probability from $a$ to $1-a$ at the price of a circuit depth of $\mathcal{O}(r\log n/\sqrt{a})$ and $\mathcal{O}(\log n)$ extra qubits. Prepending a fast, sketching-based classical approximation of $a$, we obtain a pipeline to sample a projection DPP on a quantum computer, where the former $\mathcal{O}(nr2)$ preprocessing bottleneck has been replaced by the $\mathcal{O}(nr)$ cost of normalizing the columns and the cost of our approximation of $a$.

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