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Next: 5. Computing the topological zone segmentation Up: 4. Criticality tree and topological zones Previous: 4.2 Comparison to the contour tree

4.3 Properties of topological zones

Theorem 4   Let p be a criticality with parent q with a=f(p) and b=f(q).

1) For each satisfying $a \geq \tau > b$, $B (O_p ( \tau ))$ is contained in and each component M of is contained in a single component of .

2) For any pair $a \geq \tau_1 > \tau_2 > b$, $B (O_p ( \tau_1 ))$ and $B (O_p ( \tau_2 ))$ are homeomorphic.

3) For all , if O is any component of not containing p, then B (O) and are disjoint.

4) For any such that $\tau > a$ or $\tau \leq b$, and are disjoint.

Proof: A point $r \in \zeta (p)$ if and only if $f(q) \leq f(r) < f(p)$ and $r \in O_p ( f(q) )$. For $f(p) > \tau \geq f(q)$, each point $r \in B (O_p ( \tau ))$ satisfies $f(r) = \tau$ (by continuity of f) and by monotonicity of (as is decreased) each such , so that that is contained in . The connectivity of M insures that it is contained in a single component of , establishing claim 1. By the definitions of the criticality tree and the zones, contains no critical point with value between f(p) and f(q); this establishes claim 2. Claim 3 follows from the connectivity of and the fact that boundary manifolds are pairwise disjoint by property 1 of admissable functions. Claim 4 follows immediately from the proof of claim 1. Q.E.D.


next up previous
Next: 5. Computing the topological zone segmentation Up: 4. Criticality tree and topological zones Previous: 4.2 Comparison to the contour tree
Dr. Jim Cox
1999-12-14