A new exponential separation between quantum and classical one-way communication complexity

We present a new example of a partial boolean function whose one-way quantum communication complexity is exponentially lower than its one-way classical communication complexity. The problem is a natural generalisation of the previously studied Subgroup Membership problem: Alice receives a bit string x, Bob receives a permutation matrix M, and their task is to determine whether Mx=x or Mx is far from x. The proof uses Fourier analysis and an inequality of Kahn, Kalai and Linial.

💡 Research Summary

The paper introduces a new partial Boolean function that exhibits an exponential gap between one‑way quantum and one‑way classical communication complexity. The problem is a natural extension of the Subgroup Membership problem. Alice receives an n‑bit string x, while Bob receives an n × n permutation matrix M. Their task is to decide whether Mx = x (i.e., x is a fixed point of M) or whether Mx differs from x by at least εn bits in Hamming distance, where ε > 0 is a fixed constant. Inputs that do not fall into either case are undefined, making the function partial.

Quantum one‑way protocol.
Alice prepares the quantum state |ψ_x⟩ = (1/√n) ∑{i=1}^n x_i |i⟩, which can be encoded using O(log n) qubits via standard binary indexing. She sends this state to Bob. Bob applies his permutation M as a unitary that maps |i⟩ to |π(i)⟩, obtaining the state |ψ{Mx}⟩. He then performs a swap‑test (or equivalently a Hadamard‑based inner‑product measurement) between |ψ_x⟩ and |ψ_{Mx}⟩. In the “fixed‑point” case the two states are identical, giving inner product 1; in the “far” case the inner product is at most 1 − Ω(ε). Repeating the test a constant number of times reduces the error probability exponentially. Consequently the quantum one‑way communication cost is Θ(log n) qubits.

Classical one‑way lower bound.
For a classical protocol, Alice can only send a c‑bit message m to Bob. The authors model the decision function as a Boolean function f_M(x) ∈ {−1,1}, where f_M(x)=1 iff Mx = x and f_M(x)=−1 iff d_H(Mx,x) ≥ εn. They expand f_M in the Fourier basis: \