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Jul 19, 2010 at 21:06 comment added Robin Saunders Well, that depends on what intuitive notion of "complexity" you're using. Under the surface-area definition, for example, a highly eccentric ellipsoid will look the same as a sphere with very shallow but convoluted wrinkles, since you're not taking variation in "diameter" into account.
Jul 19, 2010 at 20:46 comment added Joseph O'Rourke @Robin: Yes, this is one of the "ad hoc" measures I thought of. My concern is that two surfaces that differ in intuitive "complexity" might have the same normalized area. Which raises another question: What do all those surfaces with a given normalized area look like?
Jul 19, 2010 at 20:23 history undeleted Robin Saunders
Jul 19, 2010 at 20:23 history deleted Robin Saunders
Jul 19, 2010 at 20:22 history answered Robin Saunders CC BY-SA 2.5