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Finite length-eight fingerprint overlap

Statement

Let four inputs be assigned codewords 00000000, 00001111, 11110000, and 11111111. Distinct codewords have distance at least four. For the first two, the fingerprint overlap is 1/2 and one swap test declares unequal with probability 3/8; equal inputs yield anti probability zero. Each fingerprint uses four qubits.

Facts & Assumptions

Given: Assign $00, 01, 10, 11$ respectively to $u_{0} = 00000000$ , $u_{1} = 00001111$ , $u_{2} = 11110000$ , and $u_{3} = 11111111$ .

[F1]

Fingerprint states are normalized and encode agreement as overlap: The fingerprint overlap is the fraction of agreeing code coordinates.

[F2]

Swap-test measurement and pure-state outcome law: The antisymmetric probability on pure unit states is $(1 - ∣ ⟨ ψ ∣ ϕ ⟩ ∣^{2}) / 2$ .

Proof

technique · direct finite calculation

1.1givenF1algebra

Direct coordinate comparison gives the six pairwise distances $d (u_{0}, u_{1}) = 4$ , $d (u_{0}, u_{2}) = 4$ , $d (u_{0}, u_{3}) = 8$ , $d (u_{1}, u_{2}) = 8$ , $d (u_{1}, u_{3}) = 4$ , and $d (u_{2}, u_{3}) = 4$ . In particular $u_{0}, u_{1}$ agree in their first four positions and differ in the last four; [F1] gives $⟨ h_{00} ∣ h_{01} ⟩ = 4 / 8 = 1 / 2$ .

2.1step 1.1F1F2algebra∎

By [F2], one swap test declares unequal with probability $(1 - (1 / 2)^{2}) / 2 = 3 / 8$ . For any equal codeword pair the overlap is one and the same formula gives probability zero. The index needs $⌈ \log_{2} 8 ⌉ = 3$ qubits and the data bit adds one, so each fingerprint uses four qubits.

Depends on

Used by

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Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Harry Buhrman, Richard Cleve, John Watrous, Ronald de Wolf, Quantum Fingerprinting (standard reference, not scraped)