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Let $X = \{x_1, \dots, x_n\}$ be a finite set of points in $\mathbb{R}^d$. We can associate to $X$ its multiset of distances $$ D_X := \{ \lVert x_i - x_j \rVert : 1 \le i,j \le n \} \qquad \text{(read as a multiset)} $$ where $\lVert \cdot \rVert$ denotes the Euclidean norm, and each pair $(i,j)$ counts once in the multiset, so that $\#D_X = n^2$. A natural question is to ask whether $X$ can be uniquely reconstructed from $D_X$ up to Euclidean isometry (translation, rotation, and reflection). The answer turns out to be no, and there are even infinitely many counterexamples in dimension $d=1$.

However, suppose we restrict the points $x_i$ to lie on the unit sphere $S^{d-1} \subset \mathbb{R}^d$. Is this constraint sufficient to ensure that $X$ is uniquely determined by $D_X$ up to Euclidean isometry?

David Zhang
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1 Answers1

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Even for $d = 2$ there are such sets of points. Thus, for $ n = 4: \lbrace 0,1,2,5 \rbrace $ and $\lbrace 0,1,5,6 \rbrace \mod 8$. You can also specify homometric systems with equal sets of distances. For $n = 6$, you can find at least 5, and for $n = 12$ - at least 18 homometric sets of points on the circle with equal sets of distances.

Yog Urt
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  • To clarify, by ${0,1,2,5} \mod 8$ you mean to place $8$ equally spaced points $p_0, \dots, p_7$ around a circle, and take ${p_0, p_1, p_2, p_5}$? – David Zhang Jul 11 '17 at 22:29
  • @David Zhang, Yes, in this case we take the regular octagon, enumerate the vertices clockwise and put into the two sets of points vertices with the indicated numbers. – Yog Urt Jul 11 '17 at 22:45