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Suppose that we have a plane $P$ whose equatoin is unknown.

However, we know some pairwise Euclidean distances $|p_1q_1|, \dots, |p_nq_n|$ between some points $p_1, \dots, p_n$ on P to some other points $q_1, \dots, q_n$ on $z=0$ plane.
Moreover, we know all the pairwise distances in the set $p_1, \dots, p_n$ (we can assign relative coordinates to them in $2D$.

For which value of $n$ can we find the equation of $P$?

I approach this problem as follows.

  1. Draw $n$ spheres $S_1, \dots, S_n$ where each $S_i$ is centered at $q_i$, and its radius is $|p_iq_i|$.
  2. Take two arbitrary spheres $S_i$ and $S_j$ with radii $r_i, r_j$ respectively.
  3. Find all the point pairs $(a,b)$ such that $a \in S_i,\ b \in S_j$ and $|ab| = |p_ip_j|$ (remember that we know all pairwise distances between $p_i,p_j|\ i,j \in \{1,\dots,n\}$.
  4. We obtain a (possibly non linear) equation $f(n)$.
  5. Now, take another sphere $S_k$ and repeat steps 2 and 3 for $S_i, S_k$ and $S_j, S_k$

After step 5, I am stuck. I cannot tell if the solution space expands or not, and if there is an lower bound for $n$ that allows us to solve the above problem.

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  • $\begingroup$ @amd This comment really helps me. But out of curiosity, how did you guess that I will work with noise in the future? :) $\endgroup$ – padawan Sep 6 '17 at 23:35
  • $\begingroup$ There’s always some sort of error due to noise and other causes in real-world data. Theoretically-correct solution sometimes fail miserably as a result. $\endgroup$ – amd Sep 7 '17 at 5:41
  • $\begingroup$ @amd I think the plane is not necessarily a tangent to the spheres. It can also intersect them. $\endgroup$ – padawan Sep 7 '17 at 15:27
  • $\begingroup$ You’re right. I was thinking of a different situation. $\endgroup$ – amd Sep 7 '17 at 17:53
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I don’t believe that this process will converge in general. Suppose, for instance, that $|p_iq_i|=1$ for all $i$. Any plane parallel to the $z=0$ plane and within a distance of $1$ from it satisfies these point-pair distance constraints, and many others do as well. I suspect that the best that you’re going to be able to do with this pair distance data is to limit the possibilities somewhat.

For each of the points $q_i=(x_i,y_i,0)$, the plane must intersect the sphere with radius $r_i=|p_iq_i|$ centered at $q_i$. Working in homogeneous coordinates, this sphere can be represented by the matrix $$S=\begin{bmatrix}1&0&0&-x_i\\0&1&0&-y_i\\0&0&1&0\\-x_i&-y_i&0&x_i^2+y_i^2-r_i^2\end{bmatrix}.$$ If we represent the plane by the vector $(\lambda,\mu,\nu,\tau)^T$, then this intersection condition can be stated as $$P^TS^{-1}P\le0.$$ Using the adjugate $S^{\tiny\triangle}$ instead of $S^{-1}$ might be more convenient since it eliminates division by $r^2$. The matrix $S^{\tiny\triangle}$ represents a quadric surface (the dual to the constraining sphere, in fact), so the constraint can be interpreted as requiring the vector that represents the plane to lie within a region bounded by this quadric surface, and thus the solution space consists of the intersection of these regions. The solution space certainly gets no larger with each additional point-distance pair, but there’s no particular reason that I can see to expect that the intersection will consist of a single point, either.

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  • $\begingroup$ Thank you! And thank you again. This is a very nice explaination that pushes research further. $\endgroup$ – padawan Sep 7 '17 at 20:13

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