Use σ(n) = Σ εⱼ bⱼ(n) mod 4, z₀ = 0, zₙ₊₁ = zₙ + i^σ(n), and Cambie’s offsets (c₀,c₁,c₂,c₃) = (0,−1,−1+i,−i). Then Qₙ = (Re(2zₙ+cσ(n)), Im(2zₙ+cσ(n)), 4n+σ(n)). Translate Q₀ to zero if desired (here it already is zero).

A transition r → s gives (Re(2iʳ+cₛ−cᵣ), Im(2iʳ+cₛ−cᵣ), 4+s−r). With k trailing ones, the state change is εₖ − Σⱼ<ₖ εⱼ mod 4. For a period-p sign word, four periods exhaust this change orbit; bits above the carry suffix realize every starting state. Equal vectors are merged in the list. This exact finite calculation counts step types; it is not itself a proof of infinite triple avoidance.