I wish to know the VC-dimension of a range space $(X,\mathcal{R})$ constructed as follows:$X$ is the cylinder $\{(x,y,z)\in\mathbb{R}^3|x^2+y^2\leq 1\}$The ranges in $\mathcal{R}$ are formed by taking the union of circular disks such that:the plane containing the disk is orthogonal to the z axis (we stack the disks in the z direction)a disk is tangent to the cylinder boundary at the point $(1,0,z)$a disk has diameter $f(z)+1$, where $f(z)$ is bounded (strictly) by $-1<f(z)<1$, and strictly monotonically increasing, strictly monotonically decreasing, or constant.Any set constructed by rotating one of these ranges about the z axis by an arbitrary angle is also a range.Intuitively, imagine taking a set of coins (circular, of course) and sorting them by diameter, either decreasing or increasing. Then drop them carefully into a tube (the main cylinder) in that order, so each rests on the last. Now tip the tube slightly so that they all rest against the side of the cylinder. If our coins had zero thickness and we had one for every real number, this would be our range.I'm mostly interested in the case that $f(z)$ is sigmoid, like the error function or $\tanh$. Specifically, I'm interested in the cylindrical ranges formed by the family of functions $\tanh(\alpha(z-\beta))$, where $\alpha,\beta\in\mathbb{R}$. I know that this range space has at least VC-dim 4 (I can construct a set of four points that it shatters), but I'm interested in putting an upper bound on it and understanding why. I know that:Circular disks in $\mathbb{R}^2$ have VC-dim 3Subsets of the strip $\{-1\leq y\leq 1\}\subset\mathbb{R}^2$ that are bounded above or below by $\tanh(\alpha(z-\beta))$ have at least VC-dim 3, probably equal to 3, because the slope part of the $\tanh$ function acts much like a lineIs there any way to combine these facts to obtain an upper bound on the VC-dimension? Is there anything to say about general $f(z)$ that meet the criteria in (2)?
VC-dimension of Cylinders within a Cylinder
cg.comp geom;vc dimension