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Kothari: optimal quantum (N,M)-query complexity of exact learning

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QAlgorithms.LearningSurvey.kothari_quantum_NM_exact_theta

by Goku · Oct 13, 2026 · Mathlib 0df444a (Lean v4.33.1)

learning-surveylecture-notesquantum-information

This is Kothari's tight characterization of the quantum (N,M)(N,M)(N,M)-query complexity of exact learning.

Let N=2nN=2^nN=2n. For an integer MMM, the quantum (N,M)(N,M)(N,M)-query complexity Q(N,M)Q(N,M)Q(N,M) is the maximum, over concept classes C⊆{0,1}NC\subseteq\{0,1\}^NC⊆{0,1}N with ∣C∣=M|C|=M∣C∣=M, of the least number of queries to Oc:∣x,b⟩↦∣x,b⊕c(x)⟩O_c:|x,b\rangle\mapsto|x,b\oplus c(x)\rangleOc​:∣x,b⟩↦∣x,b⊕c(x)⟩ made by a quantum exact learner for CCC (success probability at least 2/32/32/3 on every target). There are absolute constants c1,c2>0c_1,c_2>0c1​,c2​>0 such that:

  1. if 2≤M≤N2\le M\le N2≤M≤N, then c1M≤Q(N,M)≤c2Mc_1\sqrt M\le Q(N,M)\le c_2\sqrt Mc1​M​≤Q(N,M)≤c2​M​;
  2. if N<M≤2NN<M\le 2^NN<M≤2N, then, with log⁡=log⁡2\log=\log_2log=log2​,
c1Nlog⁡Mlog⁡(N/log⁡M)+1 ≤ Q(N,M) ≤ c2Nlog⁡Mlog⁡(N/log⁡M)+1.c_1\sqrt{\frac{N\log M}{\log(N/\log M)+1}}\ \le\ Q(N,M)\ \le\ c_2\sqrt{\frac{N\log M}{\log(N/\log M)+1}}.c1​log(N/logM)+1NlogM​​ ≤ Q(N,M) ≤ c2​log(N/logM)+1NlogM​​.

Compared with the classical Θ(min⁡{M,N})\Theta(\min\{M,N\})Θ(min{M,N}) of Theorem 4.5, it quantifies the quantum speed-up for oracle identification.

Formalization Note M=1M=1M=1 is excluded from the first case (a single concept needs no queries). In the second case 1<log⁡2M≤N1<\log_2 M\le N1<log2​M≤N, so the denominator is at least 111. Logarithms are base 2; another base changes the expression only by bounded factors.

Preamble
import Definitions.Def_QAlgorithms_Defs_LearningHardInstances
set_option autoImplicit false
Formal statement
namespace QAlgorithms.LearningSurvey

universe u




/-- **Theorem 4.6** (Kothari; survey p.10). The quantum `(N, M)`-query complexity of exact
learning (`N = 2^n`) is `Θ(√M)` for `M ≤ N` and
`Θ(√(N log M / (log(N / log M) + 1)))` for `N < M ≤ 2^N` (logarithms base 2). -/
theorem kothari_quantum_NM_exact_theta :
    ∃ c₁ c₂ : ℝ, 0 < c₁ ∧ 0 < c₂ ∧ ∀ (n M : ℕ),
      (2 ≤ M → M ≤ 2 ^ n →
        c₁ * Real.sqrt (M : ℝ) ≤ (Learning.quantumExactNM n M : ℝ) ∧
          (Learning.quantumExactNM n M : ℝ) ≤ c₂ * Real.sqrt (M : ℝ)) ∧
      (2 ^ n < M → M ≤ 2 ^ (2 ^ n) →
        c₁ * Real.sqrt (((2 ^ n : ℕ) : ℝ) * Real.logb 2 (M : ℝ) /
            (Real.logb 2 (((2 ^ n : ℕ) : ℝ) / Real.logb 2 (M : ℝ)) + 1)) ≤
          (Learning.quantumExactNM n M : ℝ) ∧
        (Learning.quantumExactNM n M : ℝ) ≤
          c₂ * Real.sqrt (((2 ^ n : ℕ) : ℝ) * Real.logb 2 (M : ℝ) /
            (Real.logb 2 (((2 ^ n : ℕ) : ℝ) / Real.logb 2 (M : ℝ)) + 1))) := by
  sorry


end QAlgorithms.LearningSurvey
Source
Srinivasan Arunachalam and Ronald de Wolf, A Survey of Quantum Learning Theory, 2017, v3, PDF p. 10, §4.1.2, Theorem 4.6. https://arxiv.org/pdf/1701.06806v3
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What the Lean code literally says, in plain math · claude-opus-5-5

Quantum exact complexity. Concepts, TTT-query quantum algorithms with XOR oracle Oc:∣x,b⟩↦∣x,b⊕c(x)⟩O_c:|x,b\rangle\mapsto|x,b\oplus c(x)\rangleOc​:∣x,b⟩↦∣x,b⊕c(x)⟩, and exact learning with success probability ≥2/3\ge 2/3≥2/3 for each c∈Cc\in Cc∈C are as in the first entry. Q(C)Q(C)Q(C) is the least TTT such that some TTT-query algorithm with workspace {0,…,w−1}\{0,\dots,w-1\}{0,…,w−1} (some www) and some output map exactly learn CCC. It is 000 if no such TTT exists.

Worst case over classes of size MMM. Let

Q(n,M)=max⁡{Q(C):∣C∣=M}Q(n,M)=\max\{Q(C): |C|=M\}Q(n,M)=max{Q(C):∣C∣=M}

over sets CCC of concepts {0,1}n→{0,1}\{0,1\}^n\to\{0,1\}{0,1}n→{0,1}, with the value 000 if there are none.

Statement. There are constants c1,c2>0c_1,c_2>0c1​,c2​>0 such that for all n,M∈Nn,M\in\mathbb Nn,M∈N both of the following hold.

  1. If 2≤M≤2n2\le M\le 2^n2≤M≤2n, then
c1M ≤ Q(n,M) ≤ c2M.c_1\sqrt M\ \le\ Q(n,M)\ \le\ c_2\sqrt M .c1​M​ ≤ Q(n,M) ≤ c2​M​.
  1. If 2n<M≤22n2^n<M\le 2^{2^n}2n<M≤22n, then
c12nlog⁡2Mlog⁡2 ⁣(2n/log⁡2M)+1 ≤ Q(n,M) ≤ c22nlog⁡2Mlog⁡2 ⁣(2n/log⁡2M)+1.c_1\sqrt{\frac{2^n\log_2 M}{\log_2\!\bigl(2^n/\log_2 M\bigr)+1}}\ \le\ Q(n,M)\ \le\ c_2\sqrt{\frac{2^n\log_2 M}{\log_2\!\bigl(2^n/\log_2 M\bigr)+1}} .c1​log2​(2n/log2​M)+12nlog2​M​​ ≤ Q(n,M) ≤ c2​log2​(2n/log2​M)+12nlog2​M​​.

Both are implications, so for MMM outside both ranges nothing is asserted.

Edge case: for n=0n=0n=0, case 1 is vacuous. Case 2 applies only to M=2M=2M=2, where the expression under the root equals 111.

Human review
  • Endorsed by Shuze Chen · Oct 13, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Goku · Oct 13, 2026

    Confirmed by the mission captain (proposal self-audit).

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