Existence of Solution: Embedding from 2D Euclidean space to a circle

Given a real matrix $X$ with $n$ rows and 2 columns, can the matrix be transformed to a real matrix $Y$ such that all the points formed by the rows of $Y$ lie on a circle (2d) and their inter-point curved/geodesic distances on the circle approximate or are exactly equal to the the inter-point Euclidean distances (straight line distances) between the rows of $X$.

Question: Can all the distances be preserved with no error for all real matrices $X$ given that the points in the rows of $X$ are 2 dimensional (X has two columns) and the circle is a 2d curve (not a n-sphere in 3 or greater dimensional).

I believe the spherical distance between $d_s(Y_{i.},Y_{j.})=arccos(x_{i.},x_{j.}^T)$ for rows $i,j$?

• When you speak of the "inter-point distances," do you mean just between consecutive points or do you mean all distances. The distinction is that with $n$ points, you have $n-1$ distances ($n$ if you consider the list cyclically); whereas, you have $\binom{n}{2} = \frac{1}{2}n(n - 1)$ distances in all. – Sammy Black Apr 12 '13 at 17:04
• It is symmetric and hence you have $\frac{n}{2}(n-1)$ distances-say if you place the distances into a matrix, then the matrix would be hollow, i.e the diagonal would have zero's in it, as we are focussed on distance metrics, like that of the Euclidean for example.. – halms Apr 12 '13 at 17:06
• So you want to find $n$ points $\{y_1, \dots, y_n \}$ on a circle (of some radius) so that the arclengths between the various (not necessarily consecutive) pairs of points match the Euclidean distances from the original $n$ points $\{x_1, \dots, x_n\}$? – Sammy Black Apr 12 '13 at 17:09
• I'd say between 'all unique pairs' of points instead of 'various' and that would make it concrete-as to what I am looking for. I am also interested in the existence of 'perfectly matching', solutions when $x_i's$ are in 2d and the $y_i's$ on the circle (obviously) are 2d too. The existence of the solution equally interests me because I presently have one analytic solution to solve for it. There may be more elegant one's apart from my understanding of whether the problem is completely solvable with zero error or not being equally important. – halms Apr 12 '13 at 17:13
• A case where this wouldn't work is if one of the points in the plane were the center of a circle with three or more distinct points on the circle. No matter where you place the center in $Y$, there will only be two unique places to put an equidistant point so at least two of the rows of $X$ would get mapped to the same place. This violates the conservation of distance so it wouldn't work. – John Douma Apr 13 '13 at 6:51