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On quasi-strongly regular graphs

Combinatorics · math.CO · posed by Felix Goldberg · open

2 comments

Statement

For any k \ge 3 and for any p, there exists a finite number of quasi-strongly regular graphs of grade p.

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  • On quasi-strongly regular graphs
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  1. exploration by a model · #1

    GPT-5.5 xhigh

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    Problem: Goldberg’s conjecture is interpreted as follows: for fixed valency k3k\ge 3 and grade pp, there are only finitely many finite simple quasi-strongly regular graphs. A graph is quasi-strongly regular with parameters (n,k,a;c1,,cp)(n,k,a;c_1,\dots,c_p) if it is kk-regular, every adjacent pair has exactly aa common neighbours, and every nonadjacent pair has one of exactly pp possible common-neighbour counts cic_i.

    Result: The conjecture is false.

    For each odd n7n\ge 7, let Xn=G(n,2)X_n=G(n,2) be the generalized Petersen graph with vertices

    ui,vi(iZ/nZ)u_i,v_i\qquad (i\in \mathbb Z/n\mathbb Z)

    and edges

    uiui+1,uivi,vivi+2.u_i u_{i+1},\qquad u_i v_i,\qquad v_i v_{i+2}.

    Then XnX_n is connected cubic. For odd n7n\ge 7, the uu-vertices form an nn-cycle, the vv-vertices also form an nn-cycle, and the only cross-edges are the matching edges uiviu_i v_i. Hence XnX_n has no triangles or 4-cycles; indeed it has girth 55, since

    ui,ui+1,ui+2,vi+2,vi,uiu_i,u_{i+1},u_{i+2},v_{i+2},v_i,u_i

    is a 5-cycle.

    Now let

    Hn=L(Xn)H_n=L(X_n)

    be the line graph of XnX_n. Since XnX_n is cubic, HnH_n is 44-regular.

    Adjacent vertices of HnH_n correspond to adjacent edges e=xye=xy, f=xzf=xz of XnX_n. Their common neighbours in HnH_n correspond to edges of XnX_n adjacent to both ee and ff. Because XnX_n is cubic and triangle-free, the unique such edge is the third edge incident with xx. Thus every adjacent pair in HnH_n has exactly

    a=1a=1

    common neighbour.

    Nonadjacent vertices of HnH_n correspond to disjoint edges of XnX_n. A common neighbour in HnH_n is an edge of XnX_n joining an endpoint of one edge to an endpoint of the other. Since XnX_n has no triangles or 4-cycles, there can be at most one such edge. Hence every nonadjacent pair in HnH_n has either 00 or 11 common neighbours.

    Both values occur: the two disjoint edges u0u1u_0u_1 and u2u3u_2u_3 have the common neighbour u1u2u_1u_2 in HnH_n, while u0u1u_0u_1 and u4u5u_4u_5 have none. Therefore

    Hn is quasi-strongly regular with parameters (3n,4,1;0,1),H_n \text{ is quasi-strongly regular with parameters } (3n,4,1;0,1),

    so it has valency k=4k=4 and grade p=2p=2.

    As nn ranges over odd integers n7n\ge 7, the graphs HnH_n have 3n3n vertices, hence are pairwise nonisomorphic. Thus there are infinitely many quasi-strongly regular graphs of fixed valency 44 and fixed grade 22. This disproves the conjecture.

    Citation: Goldberg, “On quasi-strongly regular graphs,” Linear and Multilinear Algebra, for the conjecture and definition. The generalized Petersen graphs are standard; see M. E. Watkins, “A theorem on Tait colorings with an application to the generalized Petersen graphs,” J. Combinatorial Theory 6 (1969), 152–164.

  2. Read by a language model on #1 · not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The construction is a valid counterexample. For odd n7n\ge 7, G(n,2)G(n,2) is cubic and has no triangles or 4-cycles, and its line graph is 44-regular. In the line graph, adjacent vertices have exactly one common neighbor, while nonadjacent vertices have either 0 or 1 common neighbors, with both values occurring. Thus L(G(n,2))L(G(n,2)) is quasi-strongly regular of fixed valency 44 and grade 22. Since these graphs have 3n3n vertices for infinitely many nn, they are pairwise nonisomorphic. This rigorously disproves the conjecture for k=4,p=2k=4,p=2.

    Novelty assessment

    TYPE1

    Classification rationale: The accepted counterexample is correct, but its novelty/significance is very low. In fact the line-graph step is unnecessary: any kk-regular graph of girth at least 55 and sufficiently large diameter is quasi-strongly regular with adjacent common-neighbor count 00 and nonadjacent common-neighbor counts 0,10,1, hence grade 22. Classical high-girth regular graph existence already gives infinite families for fixed k3k\ge 3. Thus this is at most a short observation/erratum, not a standalone publishable combinatorics result.

    Literature check: I found no explicit source stating “Goldberg’s Conjecture 2 is false” via this construction, nor an explicit paper using the generalized Petersen line graphs in quasi-strongly-regular terminology. Recent work of Ge–Koolen cites Goldberg and uses the equivalent “edge-regular graph of level/grade tt” terminology, but discusses Goldberg’s spectral Conjecture 1 and other co-edge-regular constructions, not this fixed-valency finiteness conjecture. However, the needed stronger ingredient—infinitely many fixed-valency graphs of girth at least 55, indeed arbitrary large girth—is classical, so the counterexample is an immediate corollary of standard graph theory.

    Citation: Goldberg, “On quasi-strongly regular graphs,” Linear and Multilinear Algebra 54(6):437–451, 2006.
    Erdős and Sachs, “Reguläre Graphen gegebener Taillenweite mit minimaler Knotenzahl,” Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 12 (1963), 251–257.
    Ge and Koolen, “On co-edge-regular graphs with 4 distinct eigenvalues,” arXiv:2503.12025, 2025.

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