arXiv:2610.40886
Clifford synthesis is NP-complete
Søren Fuglede Jørgensen
Abstract
We show that the problem of synthesizing a quantum circuit for a given Clifford operator using at most a specified amount of resources is NP-complete for several resource models of practical interest. Specifically, we study synthesis of a given Clifford operator using single-qubit Clifford gates and two-qubit Clifford gates, with full qubit connectivity and no additional ancilla qubits. With single-qubit gates free and CNOT as the only two-qubit gate, we show that deciding whether a given target admits an implementation using at most a specified number of CNOT gates is NP-complete, and we show that the problem remains NP-complete when any two-qubit Clifford gate is allowed and the budget counts all such gates. Moreover, we show that both results continue to hold when SWAP gates may be inserted for free, and when single-qubit gates are also considered part of the budget. In each case, we build on recent work by Li and Guan and show that there are polynomial-time reductions from Hamiltonian path in grid graphs to these decision problems, using a number of qubits quadratic in the number of grid vertices.
AI Summary
The paper asks whether a specified Clifford operator admits an exact implementation within a gate-count budget when qubits have unrestricted connectivity and no additional ancillas are allowed. It extends recent NP-completeness results for CNOT-only synthesis to circuits that may use arbitrary single-qubit Clifford gates. Deciding whether a target can be implemented with at most $k$ CNOTs is NP-complete even when single-qubit Cliffords are free. Hardness already holds for linear reversible targets $\rho(A)=\operatorname{diag}(A,(A^{-1})^\top)$ with $A$ lower unitriangular and $(A-I)^3=0$. The result also holds when arbitrary two-qubit Clifford gates each count as one resource, when SWAPs are free, and when single- and two-qubit gates carry fixed nonnegative and positive rational costs, respectively, provided the two-qubit gate set contains CNOT. The reduction starts from Hamiltonian path in grid graphs and extends a recorder-qubit construction for CNOT synthesis. For a graph with $m$ vertices, it uses $r=6(m-1)+1$ recorders per vertex, $R=mr$ recorders total, $O(m^2)$ qubits, and threshold $k=R+m-1$. Tracking intersections of qubit symplectic planes with a Lagrangian subspace yields appearance-count bounds that prevent single-qubit Clifford shortcuts: any circuit meeting the threshold induces a CNOT-only implementation encoding a Hamiltonian path. Further counting handles output permutations and hence free SWAPs and arbitrary two-qubit Cliffords. Polynomial-size certificates establish membership in NP. The results concern exact gate-count optimization without extra ancillas, not depth optimization. They establish worst-case decision complexity, rather than approximation hardness or typical-instance difficulty.
Large language models were used during the preparation of the paper, in particular to prepare drafts of some of the proofs.
Written by gpt-6.1-sol from the paper's full text. Automated: it can be wrong.