A graph is chordal if every cycle on four or more vertices contains a chord i.e. an edge between non-adjacent vertices of the cycle. A triangulation (or chordalization) of a graph $G=(V,E)$ is the addition of ``fill-in'' edges $F$ to $G$ such that the graph $H = (V, E \cup F)$ is chordal. A triangulation is minimal if, for every $F' \subset F$, $(V, E \cup F')$ is not chordal.

Is the following a known property of minimal triangulations? It seems highly plausible, but I have struggled to locate a proof in the literature.

  • If $H=(V,E \cup F)$ is a minimal triangulation of a graph $G$, then every fill-in edge in $F$ is a chord of some cycle $C$ in the original graph $G$.

Informally, this says that the addition of fill-in edges cannot create completely new cycles, not present in the original graph, that in turn require extra fill-in edges.

I am aware that there is a deep literature on minimal triangulations linking them to (amongst other objects) minimal separators, but I have struggled to pin down a direct statement and proof of this intuitively reasonable claim. Can anybody help with a proof and/or reference? Thank you!

  • 1
    $\begingroup$ It should follow from the fact that every fill-edge $uv$ is a subset $\{u,v\} \subseteq S$ of a minimal separator $S$ of the original graph. $\endgroup$
    – Laakeri
    Commented Feb 15 at 13:09

2 Answers 2


As @Laakeri commented, the connection between triangulations and minimal separators can be used to show this property.

Based on the definitions:

  • A subset $S \subseteq V$ is an $a, b$-separator of $G$ if $a, b \in V$ are in different components of $G \setminus S$.
  • An $a, b$-separator $S$ is minimal if no proper subset of $S$ separates $a$ and $b$.
  • A subset $S$ is a minimal separator of $G$ if $S$ is minimal $a, b$-separator for some $a, b \in V$.

And from the propositions:

  • Graph $H = (V, E \cup F)$ is a minimal triangulation of $G$ if and only if $H$ can be obtained from $G$ by completing a maximal set of pairwise non-crossing minimal separators into cliques.
  • Furthermore, every minimal separator in $H$ is also a minimal separator in $G$.

It immediately follows that:

  • For every edge $uv \in F$ there exists a minimal separator $S$ of both $H$ and $G$ such that $u, v \in S$.

  • To sum up, for every edge $uv \in F$, there exist some $a, b \in V$, such that $a, u, b, v$ are part of a cycle in $G$. As a result, the edge $uv$ is a cord of a cycle in $G$.

Lastly, a reference that contains proofs of the propositions:


I often wondered about this question so I appreciate the comments and answers of Laakeri and Marko.

I don't know if it helps anybody but as someone who is not an expert on minimal separators I attempted to construct a proof, leveraging lemmas and theorems from the paper [1] "Minimal triangulations of graphs: A survey" by Heggenes (Discrete Mathematics, Volume 306, Issue 3, 2006, Pages 297-317) as a black box, but also supplementing some of the concepts with a few more details.

  • Claim 1: every fill-in edge $\{u,v\}$ is a subset of a minimal separator $S$ of the original graph $G$. Proof: first I show that $\{u,v\}$ is a subset of a minimal separator $S$ of the chordal graph $H$. The claim will then follow because every minimal separator of $H$ is a minimal separator of $G$ (see Lemma 5.1 of [1]). Continuing: $H$ is chordal so it can be represented as a tree decomposition in which the bags are exactly the maximal cliques of $H$. By Theorem 3.7 of [1], the minimal separators of a chordal graph $H$ are in bijection with the edges of the tree decomposition in the following sense: if $S$ is a minimal separator of $H$, then there exist two adjacent bags $X$ and $Y$ in the tree decomposition such that $X \cap Y = S$. So to complete the proof of the claim, we need to find two adjacent bags $X, Y$ such that $\{u,v\}$ is a subset of both of them. If this holds, we are done. If not, then the subtree of the tree decomposition induced by bags containing $u$, and the subtree induced by bags containing $v$, overlap in exactly one bag $X$. We replace $X$ with two adjacent bags $(X \setminus \{u\})$ and $(X \setminus \{v\})$. This is a tree decomposition for the chordal graph $H'$ obtained by deleting edge $\{u,v\}$ from $H$, and the new bags are in bijection with the maximal cliques of $H'$. Hence, $H'$ is a triangulation of the original graph $G$ with a subset of the original fill-in edges, yielding a contradiction on the assumption that $H$ was a minimal triangulation of $H$.

  • Claim 2: If fill-in edge $\{u,v\}$ is part of a minimal separator $S$ of the original graph $G$, then it must be a chord of some cycle of the original graph $G$. Proof: $G \setminus S$ is a separator so splits what remains of $G$ into connected components $C_1, C_2, ...$. By minimality, $G\setminus (S -u)$ is not a separator, so in $G$ there must be an edge from $C_1$ to $u$, from $C_2$ to $u$, and so on. Symmetrically, there must be an edge from $C_1$ to $v$, from $C_2$ to $v$, and so on. Hence, we know that the edges $\{p_1,u\}, \{q_1, v\}$ exist, where $p_1, q_1 \in C_1$ and we know that the edges $\{p_2, u\}, \{q_2, v\}$ exist where $p_2, q_2 \in C_2$. In $C_1$ there is a (shortest) path from $p_1$ to $q_1$ and in $C_2$ there is a path from $p_2$ to $q_2$. These paths, together with the four named edges, form a cycle on at least 4 vertices, and $\{u,v\}$ is a chord of this cycle.

  • $\begingroup$ Nice, just a small comment: In the explanation of Claim 2, you miss a subtlety of the definition of minimal separators: We define that $S$ is a minimal separator if there exists $a,b \in V(G) \setminus S$ so that $S$ is a minimal $a,b$-separator. This means that it does not hold necessarily for every component $C$ of $G \setminus S$ that $C$ is adjacent to every $u \in S$, but instead there exists two components $C_1,C_2$ so that $C_1$ and $C_2$ are adjacent to every $u \in S$. But two components is all you need for the argument to go through anyway. $\endgroup$
    – Laakeri
    Commented Feb 21 at 23:19
  • $\begingroup$ Ah, yes, thanks for pointing that out :-) $\endgroup$ Commented Feb 22 at 12:58

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