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From Wikipedia

The definition of the Hausdorff distance can be derived by a series of natural extensions of the distance function $d(x, y)$ in the underlying metric space $M$, as follows:[4]

  • Define a distance function between any point $x$ of $M$ and any non-empty set $Y$ of $M$ by: $$ d(x,Y)=\inf \{ d(x,y) | y \in Y \}\ . $$

  • Define a distance function between any two non-empty sets $X$ and $Y$ of $M$ by: $$ d(X,Y)=\sup \{ d(x,Y) | x \in X \}\ . $$

  • If $X$ and $Y$ are compact then $d(X,Y)$ will be finite; $d(X,X)=0$; and $d$ inherits the triangle inequality property from the distance function in $M$. As it stands, $d(X,Y)$ is not a metric because $d(X,Y)$ is not always symmetric, and $d(X,Y) = 0$ does not imply that $X = Y$ (It does imply that $X \subseteq Y$). However, we can create a metric by defining the Hausdorff distance to be: $$ d_{\mathrm H}(X,Y) = \max\{d(X,Y),d(Y,X) \} \, . $$

I was wondering why in the three steps, $\inf$ and $\sup$ are as they are? What if some or all of they are flipped between $\inf$ and $\sup$?

For example, the first step defines the distance between a point and a set by $\inf$, but the second step defines the distance between two sets by $\sup$ and the third step uses $\max$ again. Why are they not all using $\inf$ or $\min$ (or $\sup$ or $\max$, respectively), which I think would make the three steps more agreeable, since in the first step, a point can be seen as a singleton set.

Thanks and regards!

Tim
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  • Have you tried experimenting with your own versions to see if they satisfy the triangle inequalities? It's a nice exercise. (Do it with $M=\mathbb R^2$ to keep things concrete and easily visualized.) –  Feb 24 '13 at 23:21
  • @5pm: Thanks! Is making it a metric the reason that the definition uses $\inf$ and $\sup$ in such a obscure way? – Tim Feb 24 '13 at 23:23
  • Wait, maybe I should go to bed, but isn't it $d(X,Y)=\inf_x d(x,Y)$? – Julien Feb 24 '13 at 23:31
  • @julien: Do you mean Wikipedia is wrong? WHat is your source? – Tim Feb 24 '13 at 23:33
  • @julien: Then any two sets with nonempty intersection would have distance zero. That's a reasonable definition of distance in terms of minimum separation between shapes, but it's not Hausdorff distance. –  Feb 24 '13 at 23:33
  • I don't know, I haven't touched that in a long time. So it's a kind of reflex. But just drawing a picture seems to indicate that $\inf $ would make more sense. Otherwise, with unbounded $X$, you would often get $+\infty$. – Julien Feb 24 '13 at 23:35
  • @ℝⁿ. Oh, I see. Thanks. I was not really paying attention to the last step. I just focused on $d(X,Y)$ and an old habit came back. – Julien Feb 24 '13 at 23:37
  • @Tim Of course, Wikipedia is not wrong. I am! It is just that usually $d(X,Y)$ without further context, means $\inf_X\inf_Y d(x,y)$. I forgot the Hausdorff distance context. Sorry... – Julien Feb 24 '13 at 23:39
  • @julien: I also have the impression as the one "without further context". That is exactly why I don't understand Hausdorff distance is defined that way. – Tim Feb 24 '13 at 23:41
  • @Tim Yes, that's very different. The usual $d(X,Y)$ tells you the distance between the two sets. While the Hausdorff distance measures how far they are from being equal as sets. – Julien Feb 24 '13 at 23:43
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    $d$ measures how far the sets are from touching each other; $d_{H}$ measures how far they are from being the same. Such a sensual subject, these distances. –  Feb 24 '13 at 23:48
  • @5pm: Thanks, that makes some sense. Your first $d$ is $d(X,Y):=\inf_x \inf_y d(x,y)$? – Tim Feb 24 '13 at 23:50
  • Yes, that's what I meant. I should write $\operatorname{dist}(X,Y)$ for this, since this notation seems to be uniform in the literature : it always means $\inf\inf$. –  Feb 24 '13 at 23:56
  • @5pm: In Wikipedia, I also saw Hausdorff distance measure difference between the shapes of two sets. BUt I don't quite understand why it can measure shape difference, given that the two sets can be translated arbitrarily? – Tim Feb 25 '13 at 00:01
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    @Tim The (plain) Hausdorff distance is not a good way to measure shape difference, since it is indeed affected by translation. It only works when the two sets to be compared are accurately placed "on top of each other". To account for this "placement", the notion of Gromov-Hausdorff distance is introduced. It has one more $\inf$ in it, which is taken over all possible ways to place $X$ and $Y$ together. –  Feb 25 '13 at 00:05

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