Definition and Existence of Spanning Face-Trees
Maxime Gagnebin
March 7, 2016
Abstract
We introduce the notion of spanning face-tree (SFT ) and prove that they exist on the n × n
torus if and only if n is not a multiple of 3.
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Introduction
The notion of spanning tree in a graph has been extensively studied. A spanning tree is a subset of
edges with the property of being a tree and of covering all sites. It has seen use in many domains such
as graph theory, combinatorics or algorithm theory see [1]. Every connected graph has a spanning tree.
The proof of the latter fact is easy, and the main question is "how many spanning tree can you find
on a given graph?" We introduce a new object with similar behavior to the spanning tree but whose
properties are harder to prove.
Let G = (V, E, F) be an embedded planar graph with no loops, and let F be the set of faces of G.
Definition 1.1. We say that two faces f1 , f2 ∈ F are connected if f1 ∩ f2 6= ∅.
Definition 1.2. A subset T ⊂ F is a face-tree if
• for all pair of distinct faces f1 , f2 ∈ T either f1 ∩ f2 = ∅ or f1 ∩ f2 = v for some v ∈ V .
• the set T is simply connected.
Definition 1.3. A subset T ⊂ F is a spanning face-tree ( SFT) if T is a face-tree, and for all v ∈ V
there is some f ∈ T with v ∈ f .
Figure 1: An example of a face-tree on a portion of Z2 (the connection between the faces is highlighted
in red) and an SFT on a generic graph.
We will be interested in the question: On which graph is it possible to find an SFT? To the author’s
knowledge, this question hasn’t been invested and no general criteria is know for the existence of an
SFT. In the next section, we answer the question when the graph in the n × m square torus Tn×m .
The solution is non-trivial (we prove existence for some n, m and non-existence for others) however
the technics used are elementary.
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Main results
Theorem 2.1. On the n × n torus Tn×n , it is possible to construct a spanning face-tree if and only if
three does not divide n.
Proof. The proof relies on the following Lemmas. For n not a multiple of 3, we can write n = 2k m
where m is odd. Lemma 2 implies that finding a SFT on Tm×m is sufficient. Now lets look at the
modulo classes, m ≡ i mod 6, the case 0, 2, 3, 4 are done in Lemma 3
Lemma 1. If n is a multiple of three, then no spanning face-tree exists on Tn×n . If T is a spanning
face tree on Tn×n then |T | = (n2 − 1)/3.
Proof. Suppose there exists T an SFT on Tn×n There exist a face f ∈ T such that f is a leaf of the
tree, ie only one of the vertex of f is also used in another face f˜ ∈ T . We can remove f and its three
vertices and the remaining faces T \f will still be a face-tree. The operation can be repeated for every
face in the three, except the last face. We see that each face was removed with 3 vertices and that the
last face is left with its 4 vertices. Hence, 3 (|T | − 1) + 4 = |V |. This implies that |V | is congruent to
1 modulo 4 and that |T | = (|V | − 1)/3.
lem:n_to_2n
Lemma 2. If a spanning face-tree exists on the n × n torus, then one also exists on the 2n × 2n torus.
Proof. The proof of this Lemma is constructive. We will need to set coordinates on the torus and
we’ll choose the coordinates obtained from Z2 where one face is (0, 0) and all others have non-negative
coordinates. For a Let T0 be an STF on Tn×n . Lets construct T a SFT on T2n×2n . First let all faces
of the form (2k, 2k) ∈ T , those faces are already spanning (but far from being connected). Next use
the mapping τ : (k, j) 7→ (2k + 1, 2j + 1), to include all the faces of the form τ (k, j) where (k, j) ∈ T0 .
It remains to show that we have constructed a SFT. The fact that T is covering is clear, the
even faces already did this. The number of faces in T is n2 /4 + |T0 | = n2 /4 + (n2 /4 − 1)/3 =
(3n2 + n2 − 4)/12 = (n2 − 1)/3, which is the right number of faces. It suffices then to show either
connectivity or non-existence of loops to conclude. Lets show the latter, suppose T has a loop. On
that loop, we alternate between even and odd faces. Removing the even faces and applying τ −1 to the
rest implies that this loop already existed in T0 , which is impossible by assumption.
⇒
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Figure 2: An example of a SFT on T10×10 constructed from the one on T5×5 .
lem:mod_1_or_6
Lemma 3. If n is congruent to 1 or 5 modulo 6 then one can construct a spanning face-tree on Tn×n .
Proof. The proof is based on the construction of a pattern that can be repeatedly added to the SFT
and that keeps all the features intact.
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References
[1] B. Bollobás, Modern graph theory, vol. 184, Springer Science & Business Media, 2013.
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