# How to determine 2D coordinates of points given only pairwise distances

I have a set of 2D points in which each pair has a known Euclidean distance between them. How can I go about determining an arrangement of them?

I understand there is not a unique solution in general, but for the sake of my question, assume one point is fixed at the origin.

Mathematically, we have $P = \{p \vert p \in R^2\}$ and $D = \{d \in P \times P \vert \| d\| \text{is known} \}$. How can I find a valid arrangement? (Forgive my rustiness with proper set notation)

Note: I feel like least squares may be the best solution. This is related to bundle adjustment in photogrammetry.

• I may not be understanding, but even if you fix one of the points at the origin, there is still no unique solution since only knowing the distance will give you a circle about the origin. – Carser Sep 23 '16 at 13:50
• @Carser: That's fine. I guess I should have specified that. Or left out the fixed point. Or assume that a second point is fixed along the X axis as well (i.e. (X, 0)) – marcman Sep 23 '16 at 13:52
• How many points do you have? – Paul Sep 23 '16 at 14:03
• @Paul: >=3 points – marcman Sep 23 '16 at 14:19
• Notice that you can fix two points, not just one. If you have the first point at the origin, you can also a set a second point at $(d_{12},0)$ where $d_{12}$ is the distance from the first and second points. If there are only 3 points, there are 2 locations that 3rd point can be (if there are any). For more than 3 points, you can iterate this process, although there is no guarantee of any solution. – Paul Sep 23 '16 at 15:45

Notice that you can fix two points, not just one. If you have the first point at the origin, you can also a set a second point at $(d_{12},0)$ where $d_{12}$ is the distance from the first and second points. If there are only 3 points, there are 2 locations that 3rd point can be (if there are any). For more than 3 points, you can iterate this process, picking a pair of points to locate the third. Since for each step there are 2 possible locations, you may run into a situation that is impossible for a previous choice, so you have to go back and make the opposite choice.
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