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This paper establishes a precise equivalence between computing arbitrary symmetric Boolean functions and implementing specific fanout operations within the $\mathsf{QAC}^0$ quantum circuit complexity class. By linking these tasks to the transition radius of the function, the authors provide new insights into the limits of constant-depth quantum circuits with ancilla qubits, particularly regarding the unresolved status of parity computation. The results suggest that if parity cannot be computed in this model, then functions with high approximate degrees must necessarily lack symmetry.
Sources coveringarXiv, Quantum Zeitgeist
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Fanout Complexity of Symmetric Boolean Functions in $\mathsf{QAC}^0$

arXiv2 min read

arXiv:2609.05153v1 Announce Type: new Abstract: Whether $\mathsf{QAC}^0$ can compute $\mathtt{PARITY}_n$ remains open. Computing $\mathtt{PARITY}_n$ is equivalent to implementing $\mathtt{FANOUT}_n$ under $\mathsf{QAC}^0$ reductions. This raises a more general question: for an arbitrary symmetric Boolean function $f:\{0,1\}^n\to\{0,1\}$, what fanout size is necessary and sufficient for computing $f$ in $\mathsf{QAC}^0$? We show that the answer is exactly the transition radius $\rho(f)$: computing $f$ and implementing $\mathtt{FANOUT}_{\rho(f)}$ are equivalent under $\mathsf{QAC}^0$ reductions. In particular, if $\rho(f)\ge n^\delta$ for some constant $\delta>0$, then computing $f$ is $\mathsf{QAC}^0_{\mathrm{f}}$-complete. Combined with Paturi's theorem, our characterization implies that if $\mathtt{PARITY}_n \notin \mathsf{QAC}^0$, then any Boolean function in $\mathsf{QAC}^0$ of approximate degree $n^{1/2+\Omega(1)}$ must be nonsymmetric.

Abstract:Whether $\mathsf{QAC}^0$ can compute $\mathtt{PARITY}_n$ remains open. Computing $\mathtt{PARITY}_n$ is equivalent to implementing $\mathtt{FANOUT}_n$ under $\mathsf{QAC}^0$ reductions. This raises a more general question: for an arbitrary symmetric Boolean function $f:\{0,1\}^n\to\{0,1\}$, what fanout size is necessary and sufficient for computing $f$ in $\mathsf{QAC}^0$? We show that the answer is exactly the transition radius $\rho(f)$: computing $f$ and implementing $\mathtt{FANOUT}_{\rho(f)}$ are equivalent under $\mathsf{QAC}^0$ reductions. In particular, if $\rho(f)\ge n^\delta$ for some constant $\delta>0$, then computing $f$ is $\mathsf{QAC}^0_{\mathrm{f}}$-complete. Combined with Paturi's theorem, our characterization implies that if $\mathtt{PARITY}_n \notin \mathsf{QAC}^0$, then any Boolean function in $\mathsf{QAC}^0$ of approximate degree $n^{1/2+\Omega(1)}$ must be nonsymmetric.

arXiv:2609.05153 [quant-ph]

(or arXiv:2609.05153v1 [quant-ph] for this version)

https://doi.org/10.48550/arXiv.2609.05153 Focus to learn more arXiv-issued DOI via DataCite (pending registration)

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Full text under the publisher's open licence · 193 words

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