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Home/ Questions/Q 7168659
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T14:49:27+00:00 2026-05-28T14:49:27+00:00

This is a question about an algorithm to solve a graph theory problem. Let

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This is a question about an algorithm to solve a graph theory problem.

Let us say we are given two generically globally rigid graphs F1 and F2. We also have a list E of edges that connect vertex pairs (i,j), where i is always a vertex in F1 and j is always a vertex in F2. Eren et al. (see citation below) proved that, in two dimensions, the graph F created by merging F1, F2 and E is also generically globally rigid if the following two conditions are satisfied:

  1. E contains at least three distinct vertices in F1, and at least three distinct vertices in F2.
  2. There are at least four edges in E.

It is relatively straightforward to check that the above two conditions hold, for any given F1, F2 and E. What I am interested in is providing a fix for the case where the above conditions are not satisfied. Specifically, I’m looking for an algorithm to find the smallest set of new edges E’ that, when merged with E, satisfy the conditions above. Does anybody have an idea on how I should go about doing this?

I have given the problem a stab, and I know how to check whether F1 and F2 meet Condition (1). I am then able to randomly select new vertices from each graph to use in E+E’. What I’m stuck on is how to decide on the connectivity between the new and old vertices.

Citation (Eren et al.) http://www.cs.columbia.edu/techreports/cucs-022-05.pdf

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  1. Editorial Team
    Editorial Team
    2026-05-28T14:49:28+00:00Added an answer on May 28, 2026 at 2:49 pm

    I think you should just be able to

    • Pick some arbitrary sets of vertices V1 and V2 (not incident to any edge in E, of course) until we have enough to satisfy (1)
    • Pair off elements of V1 and V2 to form edges in E’ until one set is exhausted
    • Connect the remainder to arbitrary vertices in the opposite set
    • Throw in one more arbitrary edge between F1 and F2 if E+E’ still doesn’t satisfy (2)
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