Classes of circle rims

Question

The input space is a unit circle, $\mathcal{X} = \mathbb{S}^1 \subset \mathbb{R}^2$. There is class $\mathcal{F}$ of arcs on $\mathbb{S}^1$, where a point is labeled 1 if it is on the arc, and 0 otherwise. We want to find the VC dimension of $\mathcal{F}$

I think the answer is 2. Any two points can be shattered $(++, -+, +-, --)$. But if we have three points $\{(x_1,y_1),\dots,(x_3,y_3)\}$ that all have the label 1, with radius $r_1 = r_2 = 1$, and $r_3 = 0$, it is impossible to shatter them. Is my intuition correct?

Answer 1

In fact, the VC dimension is 3.

To show this, we need to show that there is a set of 3 points which is shattered, and that no set of 4 points is shattered.

All sets of 3 points behave the same, but for definiteness let us choose three points $x,y,z$ which form the corners of an equilateral triangle. By symmetry, there are only 4 cases to consider:

• $\emptyset$: choose some arc which avoids all points.
• $\{x,y,z\}$: choose the complement arc.
• $\{x\}$: choose a small arc around $x$.
• $\{y,z\}$: choose the complement arc.

This shows that the VC dimension is at least 3.

To show that the VC dimension is at most 3, let us consider some arbitrary set of points $x,y,z,w$, presented in clockwise order starting at an arbitrary point. An arc that contains both $x$ and $z$ must contain either $y$ or $w$, and in particular there is no arc that contains exactly $\{x,z\}$.