Cross-posted on Mathematics SE.

Since I have to implement an algorithm in the language of linear algebra, I'm trying to understand K-Truss Graphs which are defined as such

The k-truss is a subset of the graph with the same number of vertices, where each edge appears in at least $𝑘 − 2$ triangles in the original graph.

Given this example:


The $4$-Truss should be $2-1-4$ since we want edges present in at least $2$ triangles of the original graph:

  • edge $1-2$ is present in triangle $0-1-2$ and triangle $1-2-4.$
  • edge $1-4$ is present in triangle $1-3-4$ and triangle $1-2-4.$

Is the solution given on my example correct?

  • 1
    $\begingroup$ I don't think your example is correct. From this description, a 'truss' is a subgraph of some original graph. A k-truss requires that every edge $e$ in your truss, is supported by at least k-2 edges that are also part of the truss and that form triangles with edge $e$. In the example that you gave for the subgraph induced by vertices {1,2,4}, edge (0,1) cannot be a supporting edge as it is not part of the subgraph. $\endgroup$ Aug 19, 2021 at 0:25
  • 1
    $\begingroup$ The subgraph induced by vertices {1,2,4} is however a 3-truss because every edge {(1,4),(2,4),(1,2)} is part of at least 1 triangle made up of nodes that are part of the truss. $\endgroup$ Aug 19, 2021 at 0:29
  • $\begingroup$ Thanks for the answer Joris, so, {0,1,2} and {1,3,4} are also 3-truss, is that correct? $\endgroup$
    – karalis1
    Aug 19, 2021 at 10:01
  • 2
    $\begingroup$ Welcome to OR.SE. In future, please add a cross-posting link to your question when you cross-post. $\endgroup$
    – TheSimpliFire
    Aug 19, 2021 at 14:47
  • $\begingroup$ @TheSimpliFire sorry didn't know this rule, thanks for the edit $\endgroup$
    – karalis1
    Aug 19, 2021 at 14:48

1 Answer 1


The definition of a k-truss you are working with seems to deviate from the 'standard' definition. See below for a few different definitions that boil down to the same thing.

A k-truss of a graph $G$ is the largest subgraph of $G$ such that each edge is contained in at least $k−2$ triangles in this subgraph.

source: Li, Z., Lu, Y., Zhang, WP. et al. Discovering Hierarchical Subgraphs of K-Core-Truss. Data Sci. Eng. 3, 136–149 (2018)

A k-truss is an inclusion-maximal subgraph $H$ in which each edge belongs to at least $k - 2$ triangles inside $H$

source: A. Conte, D. De Sensi, R. Grossi, A. Marino and L. Versari, "Truly Scalable K-Truss and Max-Truss Algorithms for Community Detection in Graphs," in IEEE Access, vol. 8, pp. 139096-139109, 2020

As I stated in the comments earlier, the subgraph induced by vertex set $\{2,1,4\}$ is not a 4-truss because that would require that every edge in your induced subgraph, is contained in 2 different triangles which must also be contained in the subgraph. In your case, edge $(1,2)$ is only contained in 1 triangle.

Following the above definitions, the subgraph induced by vertex set $\{2,1,4\}$ is not a 3-truss either, because it is not maximal. A 3-truss of your graph, would be the entire graph, since that is the largest subgraph in which every edge is part of at least 1 triangle in your subgraph.

Your graph does not contain a k-truss with $k\geq 4$.

Edit: as often in mathematics, there are inconsistencies in definitions. The definitions I cited above, explicitly require a k-truss to be the largest subgraph or an inclusion-maximal subgraph $H$ s.t. every edge in $H$ is covered by at least $k-2$ triangles in $H$. Note that there's a difference between largest (read: maximum) and maximal subgraph. Moreover, some works seem to drop this 'maximal' requirement and refer to a k-truss as any (not necessarily maximal) subgraph $H$ s.t. every edge in $H$ is covered by at least $k-2$ triangles in $H$, see e.g.: https://louridas.github.io/rwa/assignments/finding-trusses/

  • $\begingroup$ perfectly explained, thank you again! $\endgroup$
    – karalis1
    Aug 19, 2021 at 21:54

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