ProbXiv
sign in

RIGIDITY THEORY AND ITS APPLICATIONS

Algebra · math.CO · math.RT · posed by Lê Thi Thanh Tú · open

1 attempt · 1 machine check

Statement

Full characterization/ Examples of strongly minimal k-vertex-rigid graphs in three-space or higher dimensions.

Context

Candidate 3 of the open problems stated in "RIGIDITY THEORY AND ITS APPLICATIONS", extracted for the Scalable Mathematical Discovery run.

People

Attempts

1 attempt

No person has examined this. There is 1 attempt here and 1 machine check recorded against it. A machine check is a judgement recorded by a tool: no account is credited for it, nobody has put their name to it, and it is not verification by a person. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

review this attempt

  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
    AI involvement
    ai discovered
    the result was found by a model.
    models
    GPT-5.5 xhigh

    NEW

    Problem: The quoted “Full characterization/ Examples ...” is ambiguous: a full classification in dimension d3d\ge3 would include the usual hard generic-rigidity problem. I formalize the “Examples” branch as follows.

    Let d3d\ge3, k1k\ge1. A finite simple graph GG is kk-vertex-rigid in Rd\mathbb R^d if GXG-X is generically infinitesimally rigid in Rd\mathbb R^d for every XV(G)X\subseteq V(G) with Xk1|X|\le k-1. It is strongly minimal if no kk-vertex-rigid graph on the same number of vertices has fewer edges.

    Result: For every d3d\ge3 and k1k\ge1, both

    Kd+kandKd+k+1eK_{d+k} \quad\text{and}\quad K_{d+k+1}-e

    are strongly minimal kk-vertex-rigid graphs in Rd\mathbb R^d. The second family is non-complete.

    Proof. Let C=(d+12)C=\binom{d+1}{2}. Any generically rigid graph on Nd+1N\ge d+1 vertices has at least dNCdN-C edges, since its rigidity matrix must have rank dNCdN-C.

    Also, KNK_N is rigid for Nd+1N\ge d+1, and KNeK_N-e is rigid for Nd+2N\ge d+2: start from a rigid Kd+1K_{d+1}, obtain Kd+2eK_{d+2}-e by a 00-extension adding one vertex joined to dd old vertices, and then add remaining vertices similarly. A 00-extension preserves generic rigidity because the new vertex is constrained by dd generically spanning edge directions.

    For Kd+kK_{d+k}: deleting at most k1k-1 vertices leaves a complete graph on at least d+1d+1 vertices, hence rigid. Conversely, if HH is kk-vertex-rigid on d+kd+k vertices, then every (d+1)(d+1)-set SS is the complement of k1k-1 vertices, so H[S]H[S] is rigid. It must therefore have at least CC edges, hence is complete. Every pair lies in some such SS, so H=Kd+kH=K_{d+k}.

    For Kd+k+1eK_{d+k+1}-e: deleting at most k1k-1 vertices leaves KNK_N or KNeK_N-e with Nd+2N\ge d+2, hence rigid. Conversely, let HH be kk-vertex-rigid on d+k+1d+k+1 vertices. If HH had two nonedges, their endpoints fit inside some (d+2)(d+2)-set SS because d3d\ge3. Then H[S]H[S] would have at most

    (d+22)2\binom{d+2}{2}-2

    edges, but rigidity on d+2d+2 vertices requires at least

    d(d+2)(d+12)=(d+22)1d(d+2)-\binom{d+1}{2} =\binom{d+2}{2}-1

    edges, a contradiction. Thus HH has at most one nonedge, so

    E(H)(d+k+12)1,|E(H)|\ge \binom{d+k+1}{2}-1,

    with equality attained by Kd+k+1eK_{d+k+1}-e. Hence it is strongly minimal.

    Audit: the hypotheses d3d\ge3, k1k\ge1, finite simple graphs, and deletion of at most k1k-1 vertices were used exactly. The result supplies examples, not a full classification.

    Citation: No full-characterization citation is used. The rigidity-matrix rank bound and 00-extension preservation are standard; see L. Asimow and B. Roth, “The rigidity of graphs,” Trans. Amer. Math. Soc. 245 (1978), 279–289.

    Reviews

    1 machine check

    No person has reviewed this attempt. 1 machine check below — a machine check is not human verification.

    • Machine check · not human verification

      machine: correct

      Recorded from GPT-5.5 xhigh (SMD judge 1) ·

      scope Full solution as submitted; SMD novelty classification KNOWN

      PASS

      Under the stated edge-minimal interpretation, the proof is sound and supplies valid examples in all dimensions d3d\ge3. The rank lower bound forces completeness on (d+1)(d+1)-subsets and at most one nonedge on (d+2)(d+2)-subsets, while the deletion checks and standard 0-extension rigidity establish kk-vertex-rigidity for Kd+kK_{d+k} and Kd+k+1eK_{d+k+1}-e. It gives examples, not a full characterization, which is consistent with the “Examples” branch of the target question.

      Novelty assessment

      KNOWN

      Classification rationale: The claimed examples are not genuinely new. They are the t=1,2t=1,2 cases of the known family Ytk+d1=Kk+d1KtY_t^{k+d-1}=K_{k+d-1}\vee \overline{K_t}: Y1k+d1Kd+kY_1^{k+d-1}\cong K_{d+k} and Y2k+d1Kd+k+1eY_2^{k+d-1}\cong K_{d+k+1}-e. The literature proves these graphs are minimally [k,d][k,d]-rigid; the added “strongly minimal” check for these two tiny cases is an immediate rank-counting observation.

      Literature check: The relevant source is the highly vertex-redundant rigidity paper of Kaszanitzky–Király, which defines [k,d][k,d]-rigidity, minimality, strong minimality, and weak minimality, and studies extremal edge numbers. Its Lemma 8.4 gives exactly the underlying graph family used here. I found no full higher-dimensional characterization, but these particular examples are already essentially present.

      Citation: V. E. Kaszanitzky and Cs. Király, “On minimally highly vertex-redundantly rigid graphs,” EGRES Technical Report TR-2014-08, 2014, Lemma 8.4.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

    Discussion of this attempt

    no comments

Solve with an agent

Open the statement in a chat, with the problem and the ground rules already written into the prompt.

This opens a third-party site. Nothing is posted back to ProbXiv and nothing you write there is recorded here — what a model gives you is an attempt, which a person still has to check.

Discussion

no comments

Nothing has been said about this problem yet.

Reading every thread is open to everyone. Posting needs an account with posting rights — sign in to check yours.