# How can I define a measure of similarity between two line segments in $\mathbb{R}^2$?

I have two segments in $\mathbb{R}^2$ and I would like to define a measure of similarity between the two segments.

My idea is that I can apply:

1. a scale transformation $s$ in order to equate the lenghts of the two segments;
2. a translation $\mathbf{t}$ in order to equate the center points of the two segments;
3. a 2D rotation of an angle $\alpha$ (described by a rotation matrix $\mathbf{R}$) in order to perfectly overlap the two segments.

and then the similarity measure would be based on

1. the distance of $s$ from 1;
2. the magnitude of $t$ (i.e. $\mathbf{t}^T\mathbf{t}$);
3. the distance between $\alpha$ and 0 (or between $\mathbf{R}$ and the identity matrix $\mathbf{I}$).

but how to combine the above contributes? In a linear way?

Is there any "classic" measure of similarity already defined?

Edit: The two segments are oriented so $AB$ and $BA$ are different.

• I would be much easier to compute the Hausdorff distance between the segments. If the vertices are $AB$ and $CD$, the distance is $\min(\max(|AC|,|BD|), \max(|AD|,|BC|))$. – user147263 Jul 8 '14 at 20:05
• what is the distance between $AB$ and $BA$? (it is $0$ for the Hausdorff distance, but, apparently, $\pi$ in your description). – sds Jul 8 '14 at 20:21
• @This: That's a great suggestion, but your formula can't be right. The Hausdorff distance between the two diagonals of a unit square is $1/\sqrt2$ but your formula gives $1$. I'd believe that it's only off by a bounded factor though. – Rahul Jul 8 '14 at 20:30
• @Rahul Thanks for noticing. I guess you found the worst case, when the formula is off by the factor of $\sqrt{2}$. uvts_cvs: I second the question by sds: are your line segments oriented? – user147263 Jul 8 '14 at 20:43
• Perhaps try the Fréchet distance, then (which I'm pretty sure is just $\max(|AC|,|BD|)$). – Rahul Jul 9 '14 at 6:38