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EECS 598-
Eigenanalysis for Doo-Sabin subdivision.
Igor Guskov
March 21, 2002
1 Doo-Sabin scheme
Doo-Sabin subdivision comes from regular tensor product quadratic B-splines. It is a dual scheme: the refinement happens via vertex splits. See Figure 1. The weights for the extraordinary case are given as α 0 = 1/4 + 5/ 4 N and αk = (3 + 2cos(2πk/N ))/ 4 N for k = 1... N − 1.
α N-1 α N-
α N-
Figure 1: Doo-Sabin subdivision scheme.
2 Eigenanalysis for Doo-Sabin
Consider a dual scheme at an extraordinary face with N sides. The one-ring subdivision matrix S is curculant with elements Slk = αk−l. That is, the
subdivision matrix is
α 0 α 1 α 2 · · · αN − 1 αN − 1 α 0 α 1 · · · αN − 2 · · · · · · · · · · · · · · · α 1 α 2 α 3 · · · α 0
Discrete Fourier transform is defined via multiplication with the matrix F where Fpq := (z)pq^ where z = e^2 πi/N^. Note that z depends on N and that zN^ = 1 (it is not the general symbolic z we use in z-transforms and masks). Also note that all the indices will be treated mod N below. The inverse Fourier transform is performed with the matrix F −^1 = (1/N )F ∗. Here F ∗^ is the conjugate matrix with elements F (^) pq∗ = z−pq. One can check that
N∑ − 1
q=
FpqF (^) qs∗ =
N∑ − 1
q=
zpqz−qs^ =
N∑ − 1
q=
z(p−s)q^ = N δp−s.
The last equality is easy to see graphically. The Fourier transform of a subdivision matrix produces a diagonal subdi- vision matrix Sˆ = F SF −^1. This operation does not affect eigenstructure of the matrix. Sˆ is diagonal with values on the diagonal being the eigenvalues and given via ˆαt =
q αqz qt. Indeed, we have
(F SF ∗)pt =
q,r
zpqαq−rz−rt^ =
q
z(p−t)q^
q′
αq′ zq ′t = N δp−t αˆt.
So the transformed matrix is
α ˆ 0 0 0 · · · 0 0 αˆ 1 0 · · · 0 0 0 αˆ 2 · · · 0 · · · · · · · · · · · · · · · 0 0 0 · · · αˆN − 1
The inverse transform gives us the α’s via:
αs =
N
αˆtz−ts.
Thus we can enforce any eigenstructure we want. The regular case has spectrum ˆα 0 = 1, αˆ 1 = ˆα 3 = 1/ 2 , αˆ 2 = 1/4.
