MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'd like to capture the intuitive notion that a Jordan curve $\gamma_2$ “follows” or “approximates” another Jordan curve $\gamma_1$, i.e. goes somehow “parallel” to it or “oscillates” around it.

Consider a differentiable Jordan curve $\gamma_1: [0,1] \rightarrow \mathbb{R}^2$ and its normals, seen as straight lines crossing the curve perpendicularly. Consider another Jordan curve $\gamma_2$ with the following properties:

  1. Each normal of $\gamma_1$ crosses $\gamma_2$ at least once. I.e., for each normal $n(s_1)$ of $\gamma_1$ there is an $s_2$ such that the point $\gamma_2(s_2)$ lies on $n(s_1)$.

  2. Furthermore when $s_1 < s_1'$ then there are unique $s_2 \le s_2'$ such that $\gamma_2(s_2)$ lies on $n(s_1)$ and $\gamma_2(s_2')$ lies on $n(s_1')$.

Do these conditions suffice to capture the notion described above? For which “pathological ” cases do they eventually fail? How then would they have to be adjusted to capture the notion?

Further questions:

  1. Under which extra conditions does “$\gamma_2$ follows $\gamma_1$” imply that $\gamma_1$ follows $\gamma_2$?

  2. When $\gamma_2$ follows $\gamma_1$, (how) can the area enclosed by $\gamma_1$ and $\gamma_2$ be calculated via the distance function $d(s_1) = |\gamma_1(s_1) - \gamma_2(s_2)|$
    ($s_2$ the unique parameter according to condition 2 above)?

(Let the area enclosed by $\gamma_1$ and $\gamma_2$ be the symmetric difference $X_1 \triangle X_2 = (X_1 \cup X_2) \setminus (X_1 \cap X_2)$ of the areas $X_1, X_2$ enclosed by $\gamma_1$ and $\gamma_2$.)

share|cite|improve this question
This looks like it would be easier to be made sense of if you tried to express it in terms of the function $f$ such that $\gamma_2(t)=\gamma_1(t)+f(t)\textbf{n}(t)$ with $\textbf n$ the normal vector to $\gamma_1$ (for some reparametrization of $\gamma_2$; this can be done, I think, because of your second condition) – Mariano Suárez-Alvarez Jan 6 '13 at 0:34
But you would have to impose some restrictions on $f(t)$. It must not get arbitrarily large to comply with the conditions, does it? But basically you are right, I have thought about this, too. – Hans Stricker Jan 6 '13 at 0:34
Well, what I meant is: if $\gamma_2$ can be expressed in that way (and it seems it can) then express your intution in terms of conditions on $f$. So, yes, you would have to impose conditions on $f$! – Mariano Suárez-Alvarez Jan 6 '13 at 0:51
The Fréchet distance might be an appropriate measure? Informally it is the shortest leash that allow a master walking on one curve to walk a dog who follows the other curve. – Joseph O'Rourke Jan 7 '13 at 1:18

I would use the multiscale flat norm. (I am biased of course -- see: this paper on the multiscale flat norm)

You still need the minimization over rigid motions, but the flat norm is close to what you have come up with above. It works in any ambient dimension on surfaces of any co-dimension. It also does not require the surfaces you are comparing to be boundaries. The paper I wrote with Simon Morgan (linked to above) was primarily the observation that in the codimension one boundary case, the flat norm is computed by the $L^1$TV image operator. That has lots of nice consequences, like fast algorithms to do the calculations and a very useful (though simple) generalization of the classical flat norm.

The basic idea is explained carefully (and intuitively) in the paper, but I will also explain it very briefly here.

To find the distance between two k-dimensional currents $T_1$, $T_2$ (think of currents as surfaces with orientations), you can

  1. decompose $T_1 - T_2$ into two pieces, $T_1 - T_2 = E + \partial S$, where $E$ is a k-current and $S$ is a k+1-current, so $\partial S$ is again a k-current,
  2. charge yourself $M(E) + M(S)$ for the decomposition, where $M$ is the mass, measuring the k-volume of $E$ and the k+1-volume of $S$, and
  3. minimize this cost over all k+1-currents $S$.
  4. The result is the flat norm of $T_1 - T_2$, $\Bbb{F}(T_1-T_2)$.

Collecting all of this into one equation, the flat norm of a current T (we were choosing $T = T_1 - T_2$ above) is given by:

$\hspace{1in}\Bbb{F}(T) = \min_{\text{ k-currents }S} ( M(T-\partial S) + M(S)$)

Here is an image illustrating this for the case $k = 1$:

alt text

For optimal $S$, $\Bbb{F}(T)$ is therefore the length of $T-\partial S$ plus the 2-dimensional area of $S$ (for the optimal $S$).

Finally, adding a parameter $\lambda$ permits us to controlling the tradeoff point between length and area. This gives us the $\color{blue}{\text{multiscale flat norm}}$:

$\hspace{1in}\Bbb{F}(T,\lambda) = \min_{\text{ k-currents }S} ( M(T-\partial S) + \lambda M(S)$)

If $\lambda$ is really big, we like length and try to avoid paying for area in $S$, if $\lambda$ is small then we prefer to replace cost of length with the cost of area whose boundary is used to cancel length. In the following image, for small $\lambda$'s we cancel all of the length cost in all but the largest circle:

alt text

$\color{blue}{\text{A cool result: }}$ for minimizing decompositions, the mean curvature of $T-\partial S$ is bounded above by $\lambda$.

share|cite|improve this answer
@Kevin: Thank you very much. This is a promising point of view I have not heard of before. – Hans Stricker Jan 6 '13 at 12:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.