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So I'm reading "Introduction to Machine Learning" 2nd edition, by Bishop, et. all. On page 27 they discuss the Vapnik-Chervonenkis Dimension which is,

"The maximum number of points that can be shattered by H [the hypothesis class] is called the Vapnik-Chervonenkis (VC) Dimension of H, is denoted VC(H) and measures the capacity of H."

Whereas "shatters" indicates a hypothesis $h \in H$ for a set of N data points such that it separates the positive examples from the negative. In such an example it is said that "H shatters N points".

So far I think I understand this. However, the authors lose me with the following:

"For example, four points on a line cannot be shattered by rectangles."

There must be some concept here I'm not fully understanding, because I cannot understand why this is the case. Can anyone explain this to me?

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    $\begingroup$ Call the four points $p,q,r,s$ in order along the line. There is no rectangle that contains $p$ and $r$ but excludes $q$ and $s$. $\endgroup$
    – JeffE
    May 18, 2012 at 20:48
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    $\begingroup$ Yes, but there are rectangles that can contain $p$ and $q$, excluding $r$ and $s$; or contain $q$ and $r$ and exclude $p$ and $s$. Are you saying that each combination must be possible for the points to be shattered, and if so WHY IS THIS NOT AN ANSWER :P ? $\endgroup$ May 18, 2012 at 21:03
  • $\begingroup$ I don't get the answer and I believe your point is right bro @BrotherJack can you explain why I am wrong ? $\endgroup$ Feb 11, 2022 at 9:33

1 Answer 1

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The definition of "a set $P$ can be shattered by rectangles" is that for every subset of $P$, there is a rectangle that contains precisely that subset and excludes the rest of $P$. Equivalently, every labeling of the points as positive and negative is consistent with at least one hypothesis in $H$.

Now consider four points $p,q,r,s$ along a line in the plane. Since there is no rectangle that contains $p$ and $r$ but excludes $q$ and $s$, these four points cannot be shattered by rectangles.

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  • $\begingroup$ There we go. Much better to have this as an answer, no? $\endgroup$ May 19, 2012 at 21:49

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