REVIEW 2 major objections 4 minor 41 references
A quantum divide-and-conquer algorithm solves general TSP in O*(1.866^n) time, the first to beat the classical 2^n barrier.
Reviewed by Pith at T0; open to challenge. T0 means a machine referee read the full paper against a public rubric. the ladder, T0–T4 →
T0 review · grok-4.5
2026-07-12 14:52 UTC pith:2HPCT2T7
load-bearing objection Solid query-complexity improvement and a clean Ambainis correction; the O*(1.866^n) time claim is model-dependent on QRAM depth-only accounting. the 2 major comments →
Quantum Divide-and-Conquer for the Traveling Salesman Problem: Surpassing the 2^n Barrier
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
Core claim
The first quantum algorithm that solves general TSP with both query and time complexity O*(1.866^n), obtained by partitioning the vertex set into four subsets of optimal relative sizes, classically pre-computing all short paths inside subsets of that size, and searching the resulting labelled partitions by quantum minimum finding after preparing a structured set-partition superposition.
What carries the argument
The set-partition state: a uniform superposition over all ways to colour the n vertices with k labels and mark an origin and end inside each colour class, prepared by a sequence of Dicke-state circuits of O(n^{2}) gates and O(n) depth that lets the oracle evaluate tour lengths in quantum parallel without a classical bijection.
Load-bearing premise
The claim that a memory device needing exponentially many qubits can still be counted as having only polynomial cost because its circuit depth is linear in n.
What would settle it
Either a corrected counting argument that pushes the optimal four-subset exponent back to at least 1, or an explicit classical construction showing that the set-partition state preparation itself requires super-polynomial resources once realistic memory access is accounted for.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript corrects a structural counting error in the Ambainis et al. (SODA 2019) hybrid quantum algorithm for TSP, proving that the claimed O*(1.727^n) query complexity is in fact Omega*(2^n) once the oracle is required to concatenate complete solutions over the full recursive tree (Theorem 1). It then introduces a parameterized quantum divide-and-conquer framework: classically precompute shortest paths inside subsets of size at most alpha_1 n via Held-Karp DP, prepare a uniform superposition over all labeled k-partitions (the set-partition state), and run quantum minimum finding. Optimizing the classical-quantum trade-off yields O*(1.890^n) for k=3 and the global optimum O*(1.866^n) for k=4 with alpha_i approximately 0.3157 (Theorem 2). An explicit O(n^2)-gate, O(n)-depth circuit based on Dicke-state preparation constructs the set-partition state (Theorem 3); the oracle is realized via bucket-brigade QRAM, converting the query bound into a claimed total time complexity of O*(1.866^n). Qiskit simulations for n=6,7 (k=3) report high success probabilities.
Significance. A genuine asymptotic improvement over the classical Held-Karp O*(2^n) barrier for general TSP would be a landmark result in quantum algorithms for NP-hard problems. The correction of the SODA 2019 claim is carefully derived and valuable on its own. The optimization of the classical-quantum balance (balancing binary-entropy precomputation against the square-root of the multinomial partition count) is thorough and free of circularity, and the explicit Dicke-based circuit for the set-partition state is a concrete technical contribution that addresses a common oracle-construction bottleneck. The framework is cleanly parameterized and potentially portable to other partition-based combinatorial problems. These strengths stand even if the strongest time-complexity claim requires model clarification.
major comments (2)
- Abstract and Section III.C: the conversion of the O*(1.865666...^n) query bound into a total time complexity of O*(1.866^n) rests on counting only the O(n) circuit depth of each bucket-brigade QRAM query while explicitly acknowledging that the QRAM itself requires exponentially many qubits and gates. Under ordinary quantum circuit accounting (total gate count, or space-time product), that exponential resource is not absorbed into the O* notation; the claimed time advantage therefore holds only in a non-standard model that charges solely for depth. The manuscript should either adopt an explicit depth-only model with a clear statement of the exponential space cost, or weaken the headline claim to the (already solid) query complexity plus polynomial-depth oracles under exponential space.
- Section III.C (select(f_i) construction): the Boolean-function selectors that load the classical DP table are stated to use O*(2^{4n}) ancillary qubits. Even if each individual QRAM query has depth O(n), the preparation and uncomputation of these exponential ancillas must be accounted for in any total-resource claim; otherwise the same model inconsistency that affects the QRAM tree reappears. A precise resource table (qubits, gates, depth) for one full oracle call under the chosen model is needed.
minor comments (4)
- Table I and the abstract both list 1.865666...n / 1.866^n; a single consistent numerical presentation (or an explicit statement that the two differ only by rounding) would avoid reader confusion.
- Figure 1 caption and the surrounding text in Section II.A: the recursive tree is clear, but labeling the eight leaves that must be concatenated would make the counting argument of Theorem 1 immediately visual.
- Section IV: the simulations replace QRAM by a QFT shortcut and fix the starting vertex; a short remark that these experiments validate only the amplitude-amplification and state-preparation components, not the full QRAM oracle, would set expectations correctly.
- Equation (15) and the subsequent gradient calculation: the induction establishing g_k is clean, but a one-line reminder that the multinomial coefficient identity underlies the closed form would help readers who skip the induction.
Circularity Check
No circularity: the O*(1.866^n) bound is obtained by independently balancing classical binary-entropy precomputation against the square-root of a multinomial partition count and solving a low-dimensional numerical optimization.
full rationale
The paper's central claim (Theorem 2) equates the classical cost O*(2^{n H(\alpha1)}) with the quantum cost O*(sqrt(n! / \prod mi!)), rewrites the latter via the closed-form entropy identity proved by induction in Eq. (15), and numerically solves f(\alpha1) = g*_4(\alpha1) for the unique balancing point \alpha1 = \alpha2 = \alpha3 \approx 0.315742. Neither expression is defined in terms of the other; the numerical value is simply the intersection of two independently derived monotone functions. The correction of Ambainis et al. (Theorem 1) is a pure counting argument that does not feed into the new bound. Structured state preparation (Theorem 3) and the QRAM oracle are constructive circuit arguments that convert the query bound into a claimed time bound; they introduce no fitted parameters or self-referential definitions. Self-citations are limited to ordinary background (Dicke states, quantum minimum finding) and are not load-bearing for the complexity claim. The derivation is therefore self-contained against external benchmarks and exhibits no circular step of any of the seven enumerated kinds.
Axiom & Free-Parameter Ledger
free parameters (1)
- optimal subset proportions alpha_1=alpha_2=alpha_3 =
0.315742...
axioms (5)
- standard math Quantum minimum finding (Durr-Hoyer / Boyer et al.) finds the minimum among N candidates with O(sqrt(N)) queries to a coherent oracle.
- standard math Classical Held-Karp DP computes all shortest paths through subsets of size at most m in O*(2^{n H(m/n)}) time.
- standard math Dicke states |D_n^k> can be prepared deterministically with O(kn) gates and O(n) depth without ancillas (Bartschi-Eidenbenz).
- domain assumption A bucket-brigade QRAM can load classical data into a quantum register with circuit depth linear in the address length, even though the number of qubits and gates is exponential.
- domain assumption The optimal Hamiltonian cycle always arises as the concatenation of shortest paths on some labeled k-partition (Lemma 1).
invented entities (1)
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set partition state
independent evidence
read the original abstract
The traveling salesman problem (TSP) is a classic NP-hard problem. Held--Karp dynamic programming~\cite{held1962dynamic, bellman1962dynamic} solves it exactly in $O(n^2 2^n)$ time, a barrier that has stood for over six decades. Whether quantum computing can surpass $O^*(2^n)$ is a central open question. The authors of~\cite{ambainis2019quantum}\ (SODA~2019) claimed a query complexity of $O^*(1.727^n)$, but we identify a structural counting error: when corrected, their scheme requires $\Omega^*(2^n)$ queries and offers no advantage over classical Held--Karp. We design a quantum divide-and-conquer framework: partition $n$-vertex set into $k$ subsets, classically precompute shortest paths within each, then search over all $k$-partitions via quantum minimum finding. We prove $k=3$ achieves $O^*(1.890^n)$, and $k=4$ attains the global optimum $O^*(1.866^n)$, the first quantum algorithm to surpass $O^*(2^n)$ for general TSP. To convert the query bound into a time-complexity advantage, we overcome the oracle-construction bottleneck by preparing a set partition state, requiring $O(n^2)$ gates and $O(n)$ depth. Leveraging structured state preparation, we achieve a total time complexity of $O^*(1.866^n)$. Qiskit simulations on $n=6,7$ achieve $98.9\%$ and $100\%$ accuracy.
Figures
Reference graph
Works this paper leans on
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[1]
Fork >2, letα 1 < 1 2, thenf(α 1) is monotonically increasing
= 1, meaning that the complexity cannot be better thanAlgorithm 1fork= 2. Fork >2, letα 1 < 1 2, thenf(α 1) is monotonically increasing. By (15), the gradient ofg k(⃗ α) is ∇gk(⃗ α) =[− 1 2 ln 2(lnα j + 1−ln 1− k−1X i=1 αi ! −1)] k−1 j=1 =(−1 2 log2 αj 1− Pk−1 i=1 αi )k−1 j=1 .(16) We have 1 2 > α 1 ≥...≥α k−1 ≥1− Pk−1 i=1 αi >0 forn > m 1 ≥m 2 ≥...≥m k >...
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[2]
Finally, we calculate the optimal query complexity
Moreover, in this case, sincefis fixed for a givenα 1, and the optimization ofg k eventually drives its value down to that ofg r, it suffices to optimizeg r wherer <5. Finally, we calculate the optimal query complexity. Fork= 3, we haveα 1 ≥ 1
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atα 1 =α 2 = 1 3, which is below the mini- mum value off, i.e.,f( 1
It is easy to see thatg 3 attains its global maximum of 0.792481. . .atα 1 =α 2 = 1 3, which is below the mini- mum value off, i.e.,f( 1
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[4]
The corresponding query complexity isO ∗(20.918295...n) = O∗(1.889881
= 0.918295. . .The corresponding query complexity isO ∗(20.918295...n) = O∗(1.889881. . . n). Fork= 4, we haveα 1 ≥ 1
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qubits” denotes the required number of qubits, “num
Ifα 1 < 1 3, we haveg 4(⃗ α)≥g∗ 4(α1). Sinceg ∗ 4 is mono- tonically decreasing whilefis monotonically increasing, the optimal complexity is attained when f(α 1) =g ∗ 4(α1), where numerical evaluation givesf(0.315742. . .) =g ∗ 4(0.315742. . .) = 0.899691. . . Ifα 1 ≥ 1 3, the complexity cannot be lower becausef(α 1)≥f( 1 3). 11 Therefore, the optimal que...
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