Hybrid Decision Trees: Longer Quantum Time is Strictly More Powerful

Abstract

In this paper, we introduce the hybrid query complexity, denoted as $\mathrm{Q}(f;q)$, which is the minimal query number needed to compute $f$, when a classical decision tree is allowed to call $q'$-query quantum subroutines for any $q'\leq q$. We present the following results: $\bullet$ There exists a total Boolean function $f$ such that $\mathrm{Q}(f;1) = \widetilde{\mathcal{O}}(\mathrm{R}(f)^{4/5})$. $\bullet$ $\mathrm{Q}(f;q) = \Omega(\mathrm{bs}(f)/q + \sqrt{\mathrm{bs}(f)})$ for any Boolean function $f$; the lower bound is tight when $f$ is the ${\rm O{\small R}}$ function. $\bullet$ $\mathrm{Q}(g \circ {\rm X{\small OR}}_{C \log n};1) = \widetilde{\Omega}(\sqrt{n})$ for some sufficiently large constant $C$, where $g := {\rm B{\small OOL}S{\small IMON}}_n$ is a variant of Simon's problem. Note that $\mathrm{Q}(g\circ {\rm X{\small OR}}_{C \log n}) = \mathcal{O}(\mathrm{polylog}\; n)$. Therefore an exponential separation is established. Furthermore, this open the road to prove the conjecture $\forall k,\,\mathrm{Q}(g \circ {\rm X{\small OR}}_{C \log^{k+1} n};\log^{k} n) = \widetilde{\Omega}(\sqrt{n})$, which would imply the oracle separation $\mathsf{HP}(\mathsf{QSIZE}(n^\alpha))^\mathfrak{O} \subsetneq \mathsf{BQP}^\mathfrak{O}$ for any $\alpha$, where $\mathsf{HP}(\mathsf{QSIZE}(n^\alpha))$ is a complexity class that contains $\mathsf{BQTIME}(n^\alpha)^{\mathsf{BPP}}$ and $\mathsf{BPP}^{\mathsf{BQTIME}(n^\alpha)}$ in any relativized world.