Have three planes with curvatures c1, c2 and c3. Scraping 1 against 2 results in curvatures: c1' = (c1-c2)/2 and c2'=(c2-c1)/2 (proof is left as an exercise to the reader). The calculation is similar for scraping 1 against 3 and 2 against three. The three scraping operations define three linear transformations on the curvature vector c=(c1, c2, c3): A, B and C.
The article describes a cyclic permutations of these operations, so the final curvature is (A.B.C)^n*c. The absolute value of the largest eigenvalue of A.B.C determines the efficiency of this procedure, it's 1/sqrt(8). It means that by each cycle the slowest mode decays to its 1/sqrt(8) multiple. By each step it decays by (1/sqrt(8))^(1/3) = 1/sqrt(2) on average.
The article describes a cyclic permutations of these operations, so the final curvature is (A.B.C)^n*c. The absolute value of the largest eigenvalue of A.B.C determines the efficiency of this procedure, it's 1/sqrt(8). It means that by each cycle the slowest mode decays to its 1/sqrt(8) multiple. By each step it decays by (1/sqrt(8))^(1/3) = 1/sqrt(2) on average.
Question: Is there a more efficient procedure?