Beitr Algebra Geom
https://doi.org/10.1007/s13366-020-00523-6
ORIGINAL PAPER
Forbidden patterns in tropical plane curves
Michael Joswig 1,2 · Ayush Kumar Tewari 1
Received: 6 February 2020 / Accepted: 3 August 2020
© The Author(s) 2020
Abstract
T ropical curves in R 2 correspond to metric planar graphs b ut not all planar graphs
arise in this way . W e describe se veral ne w classes of graphs which cannot occur .
For instance, this yields a full combinatorial characterization of the tropically planar
graphs of genus at most five.
Keywords T riangulations · Splits · Moduli of tropical plane curves
Mathematics Subject Classification 52B20 · 05C10 · 14T15
1 Introduction
A tropical plane curve is a metric graph, G , embedded in R 2 , which is dual to a
regular subdi vision Δ of some lattice polygon P . In the smooth case Δ is a unimodular
triangulation of P , and the planar graph G is tri valent of genus g , where g agrees with
the number of interior lattice points of P . Such graphs ha ve been called tr opically
planar by Coles et al. ( 2019 ). Here we are concerned with the question: Which
graphs G occur in this way?
For a ny fixe d g ≥ 1, Castryck and V oight ga ve a procedure for finding finitely man y
lattice polygons with exactly g interior lattice points whose triangulations suf fice to
produce all such graphs (Castryck and V oight 2009 ). This was emplo yed in Brodsky
et al. ( 2015 ) and Coles et al. ( 2019 ) to computationally determine the tropically planar
graphs of genus g ≤ 7. Despite its success that approach is se verely limited by the
combinatorial explosion of the number of planar graphs and the number of triangu-
lations of the rele v ant polygons. While it is doubtful that this can be pushed much
B Michael Joswig
[email protected]
A yush Kumar T ew ari
te [email protected]
1 Chair of Discrete Mathematics/Geometry, T echnische Uni versität Berlin, Berlin, Germany
2 Max-Planck Institute for Mathematics in the Sciences, Leipzig, Germany
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Fig. 1 Kno wn forbidden patterns: graph with a sprawling node (left), a cro wded graph (center), and a
TIE-fighter graph (right). Each box represents some subgraph of positi ve genus
further , here we seek to find obstructions for a giv en triv alent graph to be realizable
as a tropical plane curve. It w as known pre viously that graphs with a sprawling node
(Cartwright et al. 2016 , Proposition 4.1), cr owded graphs (Morrison 2020 , Lemma
3.5) and TIE-fighter graphs (Coles et al. 2019 , Theorem 3.4) cannot occur; cf. Fig. 1 .
Here we add further forbidden patterns to this list. As an immediate consequence, for
the first time, we provide a complete combinatorial characterization of the tropically
planar graphs of genus g ≤ 5, in contrast to the proof in Brodsky et al. ( 2015 ) which
rests on substantial computer support. More importantly , we identify a particularly
interesting “boundary case”: if it is realizable, a graph with a heavy cycle forces a
number of geometric restrictions for the polygon P and the triangulation Δ .A sa n
example, this leads us to studying a natural f amily of triangulations, which we call
anti-hone ycombs . In this way we get a glimpse of tropically planar graphs of genus
g = 8 and beyond.
While this work is moti vated by the desire to understand tropical curv es and their
moduli, here we concentrate on the combinatorial aspects, which means we do not
take edge lengths of metric graphs into account. This comes with the adv antage that
our results also hold for triangulations which are not regular .
2 Lattice polygons and tropical plane curves
Let V ⊂ R d be a finite set of points. A polytopal subdivision of V is a polytopal
complex which co vers the con ve x hull con v V and uses (a subset of) the gi ven point
set V as its vertices. If that subdi vision is induced by a height function on V , it is called
r e gular . A subdivision which consists of simplices is a triangulation . A comprehensi ve
reference on the subject is the monograph (De Loera et al. 2010 ) by De Loera, Rambau
and Santos. Here we will be concerned with the follo wing case. Let P be a (con vex)
lattice polygon , i.e., P is the con ve x hull of finitely many points in Z 2 , and V = P ∩ Z 2
is the set of lattice points in P . A triangulation Δ of V is unimodular if each triangle
in Δ has normalized area one, i.e., Euclidean area 1
2 . This is the case if and only if Δ
uses all points in V . W e write ∂ P for the boundary of P and int P for its interior . The
number g ( P ) = # ( int P ) of interior lattice points is the genus of P ; cf. Figure 2 for
an example.
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Fig. 2 Anti-honeycomb triangulation Δ ( − 2 , 4 ;− 2 , 4 ;− 2 , 4 ) of genus 4 (left), its dual graph (center), and the
corresponding skeleton (right)
Let Δ be a (not necessarily unimodular) triangulation of V .T h e dual gr aph Γ =
Γ( Δ ) is the abstract graph whose nodes are the triangles of Δ ; they form an edge in
Γ if two triangles share an edge in Δ . The dual graph is necessarily connected and
planar , and each node has degree at most three.
Next we describe a procedure to obtain a certain graph minor of Γ .F i r s t ,i ft h e r ei s
a node of degree one, we delete it together with the unique incident edge. W e repeat
this step until no nodes of degree one are left. The remaining edges are nonr edundant .
Second, if there is a node of degree tw o, we delete the node and join its neighbors by
an edge. Again we repeat until there are no more nodes of de gree two. The resulting
graph is the skeleton , which we denote G = G (Δ) . By construction the skeleton is
a tri valent planar graph, and it does not depend on the ordering in which the edge
deletions and contractions are performed (Gordon and McNulty 2012 , Section 3.1.2).
In this way each edge of G arises as an edge path in Γ . This yields a surjectiv e map,
which we denote η , from the nonredundant edges of Γ onto the edges of G . Note that
this contraction map η is undefined for an y edge which is redundant. This elementary
description of the skeleton is algorithmic in nature and serv es our purposes.
A split is a subdi vision of V with exactly two maximal cells; it is necessarily re gular
(Herrmann and Joswig 2008 , Lemma 3.5). If U and W are the tw o maximal cells of
a split, then the intersection U ∩ W is a common edge of the t w o con ve x polygons U
and W . It spans the corresponding split line . A set of splits of V is weakly compatible
if there is a triangulation which simultaneously refines them all. Moreov er , a set of
splits is (str ongly) compatible if any two split lines do not meet in the interior of P .
Compatibility implies weak compatibility . The split lines of two weakly compatible
splits which are not strongly compatible must meet in a point in V , i.e., an interior
lattice point of P . An edge e of the connected graph G is a cut edge if deleting e
creates two connected components. Otherwise e lies in some c ycle of G .
The follo wing three technical results are extracted from the proof of (Coles et al.
2019 , Theorem 3.4), where cut edges are called “bridges”.
Lemma 1 Let e be a cut edge of G . Then η − 1 ( e ) comprises a single cut edg e of Γ ,
which is dual to an edg e, s , of Δ . Mor eover , the vertices of s lie on the boundary ∂ P,
and s spans a split line of V .
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Here the “vertices” are the tw o endpoints of the edge s . I n the unimodular case these
are also the only lattice points contained in s . Our arguments belo w do not rely on the
uniqueness of the cut edge in η − 1 ( e ) . It suf fices to know that such edge (and its dual
edge s )e x i s t .
Lemma 2 Let e be an edg e between v 1 and v 2 in a cycle, C of G . Furthermor e, let T 1
and T 2 be triangles in Δ which corr espond to v 1 and v 2 on the path η − 1 ( e ) . Then T 1
and T 2 shar e an interior lattice point of P , and this is dual to C .
Lemma 3 Let z ∈ V be some lattice point in P with two incident triangles T 1 =
con v { z , a 1 , b 1 } and T 2 = con v { z , a 2 , b 2 } both of whic h ar e in Δ . Further , let L i be
the line spanned by a i and b i ,f o ri = 1 , 2 . Suppose that L 1 and L 2 meet in some
point, say w , such that a i is closer to w than b i ,f o ri = 1 , 2 . Then the interior of the
quadrangle con v { z , a 1 ,w , a 2 } does not contain a point in V , unless a 1 = a 2 = w .
W ith this we can take a small first step to our main results. Recall that Lemma 1
associates a split of V to each cut edge of G .
Lemma 4 Splits corr esponding to distinct cut edges ar e compatible.
Proof It suf fices to consider pairs of splits. From Lemma 1 the cut edges e 1 and e 2
yields two split lines, S 1 and S 2 , which may not be unique. Unless S 1 and S 2 are
compatible, they must meet in a point of the point configuration V , which does not lie
on ∂ P , i.e., an interior lattice point of P . Y et, by Lemma 1 the line S 1 (and similarly
S 2 ) contains precisely tw o points of V , and neither lies in the interior .  
The connection to tropical geometry comes about as follows. See the book by
Maclagan and Sturmfels ( 2015 ) for the general context and Brodsk y et al. ( 2015 )
for a detailed analysis of the case which is of interest here. Let Φ be a bi variate
tropical polynomial. Its v anishing locus, the tr opical hypersurface T = T (Φ ) ,i sa
tropical plane curve in R 2 . The Ne wton polygon of Φ is a lattice polygon, and the
terms of Φ correspond to lattice points in P . Moreover , the coef ficients of Φ induce a
regular subdi vision, Δ , on those lattice points which are dual to T , i.e., T essentially
agrees with Γ( Δ ) . The tropical plane curve T is smooth if its dual subdi vision Δ
is unimodular . Con versely , each unimodular regular triangulation of P arises in this
way . In that case the edges of G (with their lengths, which we do not need here)
describe T as a point in the moduli space M planar
g of tropical plane curves of genus g .
Skeleta of unimodular re gular triangulations of lattice polygons are tr opically planar
or “troplanar” graphs in Coles et al. ( 2019 ).
Example 1 W e consider the anti-hone ycomb triangle
A ( − 2 , 4 ;− 2 , 4 ;− 2 , 4 ) = con v { ( 2 , 2 ), ( − 2 , 0 ), ( 0 , − 2 ) } , (1)
which occurs as Q ( 4 )
3 in Brodsky et al. ( 2015 ). The genus is g ( A ( − 2 , 4 ;− 2 , 4 ;− 2 , 4 ) ) = 4.
W e call the unimodular triangulation Δ ( − 2 , 4 ;− 2 , 4 ;− 2 , 4 ) , sho wn in Fig. 2 (left), the anti-
hone ycomb triangulation of A ( − 2 , 4 ;− 2 , 4 ;− 2 , 4 ) . Its skeleton, sho wn in Fig. 2 (right) and
called ( 303 ) in Brodsky et al. ( 2015 ), features three cut edges which correspond
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to three compatible splits of A ( − 2 , 4 ;− 2 , 4 ;− 2 , 4 ) . The triangulation Δ ( − 2 , 4 ;− 2 , 4 ;− 2 , 4 )
is re gular , whence it defines a moduli cone of tropical plane curves. That cone is 7-
dimensional, while the entire moduli space M planar
4 has dimension nine; see Brodsk y et
al. ( 2015 , T able 4). See Sect. 5 for a more comprehensi ve discussion of anti-hone ycomb
polygons, their triangulations and the notation ( 1 ).
W e no w sketch the forbidden patterns which are kno wn already . A node in G is
sprawling if its deletion lea ves three connected components; cf. Fig. 1 (left). This
obstruction to tropical planarity was identified in Cartwright et al. ( 2016 , Proposi-
tion 4.1) and Brodsky et al. ( 2015 , Proposition 8.3). Note that graphs with a spra wling
node were called “spra wling” in Brodsky et al. ( 2015 ) and Coles et al. ( 2019 ). A
planar embedding of a graph G is called cr owded if either: there exist tw o bounded
regions sharing at least tw o edges; or , there exists a bounded re gion sharing an edge
with itself. If all planar embeddings of G are cro wded, then G itself is said to be
cr owded . In Morrison ( 2020 , Lemma 3.5), it is shown that cro wded graphs can nev er
be tropically planar . Additionally , Morrison ( 2020 , Corollary 3.7) describes a family
of cro wded graphs, and this is depicted in Fig. 1 (center). A graph is a TIE-fighter if it
looks like the one in Fig. 1 (right). TIE-fighter graphs can ne ver be tropically planar;
this was sho wn in (Coles et al. 2019 , Theorem 3.4).
3 Heavy cycles and sprawling triangles
Let P be a lattice polygon with precisely g interior lattice points. Moreo ver , let Δ be a
unimodular triangulation of P with dual graph Γ = Γ( Δ ) and skeleton G (Δ) = G .
The contraction map η sends nonredundant edges of Γ to edges of G .T h e g interior
lattice points of P bijecti vely correspond to the bounded re gions of the planar graph
G . By Euler’ s formula we ha ve g = m − n + 1, where m and n are the numbers of
edges and nodes of G , respectiv ely . Our ar guments in this section do not require Δ
to be re gular . That is, our results e xtend to a class of planar graphs, which is slightly
more general than tropical plane curves.
T wo lattice polygons, P and Q ,a r e unimodularly equivalent if there is a lattice
vector α ∈ Z 2 and an inte ger linear transformation τ ∈ SL ( 2 , Z ) such that P =
α + τ( Q ) ; cf. (De Loera et al. 2010 , Section 9.3). In that case, we ha ve P ∩ Z 2 =
α + τ( Q ∩ Z 2 ) , i.e., the lattice points are transformed alike. The map x  → α + τ( x )
is a unimodular transformation .
Lemma 5 Assume that P contains a unimodular triangle with vertices a , b , zs u c h
that neither a nor b is a verte x of P , and z is an interior lattice point. If a and b lie
on ∂ P then either a and b lie on a common edge of P or the lattice point a + b − zi s
contained in P .
Proof Up to unimodular equi valence we may assume a = ( 1 , 0 ) , b = ( 0 , 1 ) and
z = ( 0 , 0 ) .L e t L and M be the lines spanned by the edges of P containing a and b ,
respecti vely . Suppose that ( 1 , 1 ) is not contained in P . Since P is a lattice polygon,
then the intersection of P with the positi ve orthant agrees with the unit triangle abz .
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Fig. 3 Graph with the hea vy
cycle C
v 2 v 1
C
e 2 e 1
G 3
G 2 G 1
This entails that L and M agree, which means that L = M defines an edge of P , and
this contains a as well as b .  
Definition 1 W e say that a cycle C in a planar graph G is heavy ,i f
(i) it has two nodes, v 1 and v 2 , such that v i is incident with a cut edge e i connecting
v i with a subgraph G i of positiv e genus;
(ii) and there is a third subgraph, G 3 , also of positi ve genus, which shares at least one
node with the cycle C ; cf. Fig. 3 .
In particular , a graph with a heavy c ycle has genus at least four . From the classification
of hyperelliptic graphs in Morrison ( 2020 ), we infer that a graph with a hea vy cycle is
not hyperelliptic. Moreo ver , it follows from Lemma 1 that there are split lines, S 1 and
S 2 , dual to the edges e 1 and e 2 of G . While the split lines may not be unique we just
pick some. By Lemma 4 the splits S 1 and S 2 are compatible, and thus P decomposes
into a union of three lattice polygons P 1 , P 2 and P  such that Δ induces triangulations
of all three. In this way , we get triangulations Δ 1 , Δ 2 and Δ  such that the component
G i is the skeleton of Δ i for i = 1 , 2, and G 3 ∪ C is the skeleton of Δ  . The triangles
in Δ  which are dual to v 1 and v 2 are denoted T 1 and T 2 , respectiv ely . W e refer to the
polygon P  as the heavy component of P , and lik e wise Δ  is the heavy component of
Δ . Expressed in the language of Coles et al. ( 2019 ), the heavy component Δ  arises
from Δ via “bridge-reduction”.
The follo wing lemma is the technical core of this paper . Its proof is a bit cumber-
some, with se veral cases to distinguish. Ho we ver , it is rather powerful as it delineates a
fine border between tri valent planar graphs which are realizable as sk eleta of tropically
planar curves and those which are not.
Lemma 6 Suppose that G has a heavy cycle with cut edges e 1 and e 2 as in Fig. 3 .
Then the triangles T 1 and T 2 in Δ shar e an edge [ z ,w ] , wher e z is the interior lattice
point dual to C , and the split lines S 1 and S 2 intersect in w , which is a verte x of P  ,
and which lies in the boundary of P .
Before we enter the proof, we sketch an outline. The o verall strate gy is indirect,
i.e., we will assume that T 1 and T 2 share a verte x b ut not an edge. Now the split lines
from Lemma 4 are either parallel or not; cf. Fig. 4 . In both cases we can exploit the
con ve xity of the four lattice polygons P 1 , P 2 , P  and P to arriv e at contradictions.
The situation with intersecting split lines is the more dif ficult one, as it ramifies into
sub-cases.
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T 1
T 2
z p 1
p 2
q
r
s 1
s 2
P 
T 1
T 2
z p 1
p 2 w
s 1
s 2
P 
Fig. 4 T wo possibilities for S 1 and S 2 , which are ruled out a posteriori in the proof of Lemma 6 . Left: S 1
and S 2 are parallel. Right: S 1 and S 2 intersect in a point
Proof By Lemma 2 the triangles T 1 and T 2 s ha re a ve rt ex z , which is the interior
lattice point in P dual to the heavy c ycle C . Up to a unimodular transformation we
may assume that z = ( 0 , 0 ) , and T 1 = con v { z ,( 1 , 0 ), ( 0 , 1 ) } .
W e consider the point ( 1 , 1 ) . W e notice that if ( 1 , 1 )/ ∈ P , then by con ve xity of
P , the positi ve genus component attached to the split edge s 1 =[ ( 1 , 0 ), ( 0 , 1 ) ] is
squeezed between the lines y = 0 and y = 1, which has no interior lattice points and
we get a contradiction. Therefore the point ( 1 , 1 ) lies in P . W e want to sho w that ( 1 , 1 )
is an interior lattice point of P 1 (and P ). First, since ( 1 , 0 ) is in the boundary and z is
in the interior of P , there are no interior lattice points on the ray ( 1 , 0 ) + pos { ( 1 , 0 ) } .
Similarly , there are no interior lattice points on ( 0 , 1 ) + pos { ( 0 , 1 ) } . Howe ver , s ince
G 1 has positi ve genus at least one interior lattice point of P 1 e xists, and it must lie
in the translated orthant ( 1 , 1 ) + pos { ( 1 , 0 ), ( 0 , 1 ) } .A s ( 1 , 0 ) and ( 0 , 1 ) lie in P 1 it
follo ws from the con ve xity of P 1 that ( 1 , 1 ) must be an interior lattice point. Now
consider the triangle T 2 = con v { z ,( α ,β ) ,( γ ,δ ) } where α, β , γ , δ ∈ Z with
αδ − βγ = 1 . (2)
Suppose T 1 and T 2 do not share an edge. W e are aiming for a contradiction, which
then establishes a proof of this lemma. Since ( 1 , 1 ) ∈ int P 1 the horizontal line ( 0 , 1 ) +
R ( 1 , 0 ) intersects P  only in ( 0 , 1 ) . Similarly , the vertical line ( 1 , 0 ) + R ( 0 , 1 ) intersects
P  only in ( 1 , 0 ) .A s (α, β ) , (γ , δ ) , ( 1 , 0 ) , and ( 0 , 1 ) are pairwise distinct this yields
α, β , γ , δ ≤ 0.
First, suppose the split line S 2 is parallel to S 1 .A s T 2 is a unimodular triangle S 2 is
the line through ( − 1 , 0 ) and ( 0 , − 1 ) . Then, S 2 and α, β , γ , δ ≤ 0 forces α = δ =− 1
and β = γ = 0.
W e consider the lattice points q = ( − 1 , 1 ) and r = ( 1 , − 1 ) .A s P  has positiv e
genus, it follo ws from Lemma 5 that either q or r is an interior lattice point of P  .B y
symmetry we may assume r ∈ int P  . Then, the point q can either lie on the boundary
of P or is not in P . No w the interv al [ ( 1 , 1 ), ( 1 , − 1 ) ] is the con ve x hull of tw o interior
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lattice points of P , b ut it contains the boundary point p 1 in its relati ve interior . Hence
we obtain a contradiction.
W e conclude that the split lines S 1 and S 2 intersect in some point w = (ψ , ω) ,
as depicted in the Fig. 4 . Since T 1 is fixed, T 1 and T 2 do not share an edge and
α, β , γ , δ ≤ 0 , we obtain ω ≤ 0. W e use the labels from Fig. 4 and intend to sho w
that w lies in ∂ P . Suppose the contrary . Then, due to Lemma 4 , the point w must
lie outside P . The three points p 1 := ( 1 , 0 ) , w and ( 0 , 1 ) are collinear with p 1 in
the middle; hence w is closer to p 1 than to ( 0 , 1 ) with respect to Euclidean distance.
From ( 2 ) it follo ws that w is closer to p 2 := (γ , δ ) than to (α, β ) . Applying Lemma 3
we infer that the interior of the quadrangle con v { z , p 1 ,w , p 2 } does not contain an
interior lattice point of P . W e realize that since the triangles T 1 and T 2 do not share
an edge, and since p 1 , p 2 ∈ ∂ P therefore in this case [ p 1 , p 2 ] is a boundary edge and
the triangle con v { z , p 1 , p 2 } belongs to Δ .
W e no w look at possible coordinates of the point p 2 = (γ , δ ) . Since the triangle
con v { z , p 1 , p 2 } is unimodular and as δ ≤ 0w eh a v e δ =− 1. Hence p 2 lies on the
ray { (γ , δ ) : γ ≤ 0 ,δ =− 1 } . W e consider some v alues of γ starting with the case of
γ = 0. Then ( 2 ) implies that (α, β ) is on the line x =− 1, i.e., α =− 1. W e explore
the possible v alues for β .
For β =− 2 the split edge on the line S 2 would be s 2 =[ ( 0 , − 1 ), ( − 1 , − 2 ) ] .T h i s
cannot be as then S 2 would contain p 1 as a third boundary point, a contradiction to
Lemma 1 .F o r β =− 1 and s 2 =[ ( 0 , − 1 ), ( − 1 , − 1 ) ] , we realize that the polygon P 2
is bounded in the rectangular strip between the parallel lines y = x and y = x − 1. Y et
there is no interior lattice point between them, and this contradicts G 2 to ha ve positi ve
genus. Any other v alue of β< − 2 contradicts the con vexity of P at z . This rules out
the possibility γ = 0.
The ar guments for excluding the other v alues of γ are very similar . W e summarize
them briefly . For γ =− 1 we obtain β = α + 1. Then either s 2 =[ ( − 1 , − 1 ), ( − 1 , 0 ) ] ,
and we obtain ω = 2 ≥ 0, which is absurd. Or s 2 =[ ( − 1 , − 1 ), ( − 2 , − 1 ) ] , whence
P 2 is squeezed between the lines 2 y = x and 2 y = x − 1, which does not leav e space
for an interior lattice point. Or s 2 =[ ( − 1 , − 1 ), ( − 3 , − 2 ) ] , which then lies on the
line spanned by the boundary edge [ p 1 , p 2 ] . Again, any other v alue of β on the line
y = x + 1, contradicts the con ve xity of P at z .
Similarly , γ =− 2 implies 2 β = α + 1. Then either s 2 =[ ( − 2 , − 1 ), ( − 1 , 0 ) ] , and
we obtain ω = 1 ≥ 0, which is absurd. Or s 2 =[ ( − 2 , − 1 ), ( − 3 , − 1 ) ] , whence P 2
is squeezed between the lines 3 y = x and 3 y = x − 1, which does not lea ve space
for an interior lattice point. Or s 2 =[ ( − 2 , − 1 ), ( − 5 , − 2 ) ] , which then lies on the
line spanned by the boundary edge [ p 1 , p 2 ] . Again, any other v alue of β on the line
2 y = x + 1, contradicts the con ve xity of P at z .
The case γ ≤− 3 is left. Then the point (α, β ) lies on the line − γ y = x + 1. The
follo wing three possibilities occur for the split edge s 2 . Either s 2 =[ ( − 1 , 0 ), (γ , − 1 ) ] ,
and we obtain ω ≥ 0, forcing ω = 0. In this case we see that there is no edge to join
between the points ( − 1 , 0 ) and ( 0 , 1 ) such that the polygon P remains con ve x, which
gi ves us a contradiction. Or s 2 =[ (γ − 1 , − 1 )(γ , − 1 ) ] , whence P 2 is squeezed
between the lines ( 1 − γ) y = x and ( 1 − γ) y = x − 1, which does not leav e space for
an interior lattice point. Or s 2 =[ ( 2 γ − 1 , − 2 ), (γ , − 1 ) ] , which then lies on the line
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Fig. 5 A graph with a sprawling
triangle; e 1 , e 2 , e 3 are cut edges;
G 1 , G 2 , G 3 are subgraphs of
positi ve genus
e 2
e 3
e 1
G 3
G 2 G 1
spanned by the boundary edge [ p 1 , p 2 ] . Any other v alue of β on the line − γ y = x + 1,
contradicts the con ve xity of P at z .
Therefore, finally , we conclude that there is no lattice point p 2 = (γ , δ ) such that
the triangle con v { z , p 1 , p 2 } belongs to Δ . Hence, our initial assumption was wrong,
and the triangles T 1 and T 2 do share the edge [ z ,w ] , where w = p 1 = p 2 is the
intersection of S 1 and S 2 . Consequently , s 1 and s 2 are two distinct edges of P  , and so
w is a v ertex of P  .  
W e notice that the anti-hone ycomb triangulation of genus 4 in Example 1 and Fig. 2
is a positi ve e xample for the result in Lemma 6 . For the three triangular cells containing
the split edges and sharing a common v ertex, each pair of them has a common edge. In
fact, the specific shape of the hea vy component in that e xample gi ves rise to another
concept, which cov ers a special case of Definition 1 .
Definition 2 W e say that the planar graph G has a sprawling triangle if it has a cycle
with three nodes, each of which is also incident with a unique cut edge, such that at
the other end there is a subgraph of positi ve genus; cf. Fig. 5 .
It turns out that Example 1 is the only case where this occurs:
Theorem 1 If G has a sprawling triangle then g = 4 , and, up to unimodular equiva-
lence, we have
P = A ( − 2 , 4 ;− 2 , 4 ;− 2 , 4 ) and Δ = Δ ( − 2 , 4 ;− 2 , 4 ;− 2 , 4 ) .
Proof W e use the notation from Fig. 5 .L e t s 1 , s 2 and s 3 be the split edges in Δ
corresponding to the cut edges e 1 , e 2 and e 3 . Using Lemma 6 on these three splits,
we obtain that each pair of them intersects in a point on ∂ P . This gi ves three lattice
points a , b , c in the boundary of P such that each pair is joint by one of the three
edges s 1 , s 2 , s 3 .L e t Δ | ab c be the subcomplex of Δ restricted to the lattice triangle
a bc . Its skeleton is the spra wling triangle, whose genus is one. Hence there is a unique
interior lattice point, z ,o f P contained in a bc . It follo ws that the lattice triangle ab c
has normalized area three, and its unimodular triangulation into abz , ac z , and bc z
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forms the induced subcomplex Δ | ab c . Those maximal cells of Δ are dual to the three
nodes of the sprawling triangle of G .
As in the proof of Lemma 5 we may assume that a = ( 1 , 0 ) , b = ( 0 , 1 ) and
z = ( 0 , 0 ) . It then follo ws that c = ( − 1 , − 1 ) . Directly from Definition 2 it follows
that the line ab is dual to a cut edge in G and thus induces a split of Δ . All the interior
lattice points corresponding to a region of the subgraph G 1 must lie in the halfspace
defined by ab which does not contain z . Since z is the only interior lattice point of P
which does not correspond to a region of G 1 ∪ G 2 ∪ G 3 , the lattice points a , b and c
must lie in the boundary of P .
No w Definition 2 requires the subgraphs G 1 , G 2 and G 3 to hav e positiv e genus.
This excludes the possibility that one of the three points a , b , c is a v ertex of P . Thus
they must lie on pairwise distinct edges of P . Now we can apply Lemma 5 three times
to learn that the three lattice points
( 1 , 0 ) + ( 0 , 1 ) − ( 0 , 0 ) = ( 1 , 1 )
( 1 , 0 ) + ( − 1 , − 1 ) − ( 0 , 0 ) = ( 0 , − 1 )
( 0 , 1 ) + ( − 1 , − 1 ) − ( 0 , 0 ) = ( − 1 , 0 )
are contained in P .
Let L , M , N be the three lines spanned by the edges of P through a , b , c , respec-
ti vely . Suppose that ( 1 , 1 ) is a boundary point of P . Then, since P is lattice polygon,
( 1 , 1 ) is the intersection point of L and M and thus a vertex of P . This contradicts the
fact that G 1 has positiv e genus, and it follo ws that ( 1 , 1 ) is an interior lattice point of
P . By symmetry , also ( 0 , − 1 ) and ( − 1 , 0 ) are interior lattice points of P .
No w the lattice point ( 1 , 2 ) is not contained in P because it would then need to lie
on the line M , together with ( − 1 , 0 ) . But this cannot happen as ( − 1 , 0 ) was already
identified as an interior point of P . Similarly , ( 2 , 1 ) is not contained in P either . Y et
this implies that the intersection of the lines L and M lies strictly between the two
parallel lines ( 1 , 0 ) + λ( 1 , 1 ) and ( 0 , 1 ) + λ( 1 , 1 ) . Thus there is at least one v ertex of
P , which must be a lattice point, on the line ( 0 , 0 ) + λ( 1 , 1 ) . This forces ( 2 , 2 ) to lie
in P .
Again the situation is symmetric, which is why also ( − 2 , 0 ) and ( 0 , − 2 ) lie in P .
The single choice left is that ( 2 , 2 ) as well as ( − 2 , 0 ) and ( 0 , − 2 ) are vertices of P ,
and L , M , N are the only facets. This establishes P = con v { ( − 2 , 0 ), ( 0 , − 2 ), ( 2 , 2 ) } .
The claim about Δ follo ws as there is only one way to e xtend the triangulation from
the triangle ab c to all of P .  
Remark 1 Contracting a sprawling triangle in a planar graph yields a graph with a
sprawling node, and this cannot be tropically planar . Therefore, Example 1 sho ws that
the class of tropically planar graphs is not minor closed; this was observ ed before
(Coles et al. 2019 , Figure 6).
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Fig. 6 Heavy c ycle with two
loops
v 2 v 1
e 2 e 1
C
G 3
4 Graph with a heavy cycle and two loops
Graphs with a sprawling triangle form special cases of graphs with a hea vy cycle.
Thus our main technical result, Lemma 6 , allo wed us to deriv e decisi ve structural
constraints for triangulations whose skeleton has a spra wling triangle in Theorem 1 .
No w we are looking into another special class of groups with a hea vy cycle, aiming
for a second structural result on unimodular triangulations of lattice polygons.
Definition 3 W e say that a connected triv alent planar graph G has a heavy cycle with
two loops if it has the form as described in Fig. 6 , where the shaded region represents
a subgraph of positi ve genus.
The latter is the special case of Definition 1 , where g ( G 1 ) = 1 = g ( G 2 ) . This type
of skeleton does actually occur .
Example 2 The quadrangle Q ( 5 )
4 of genus 5, cf. (Brodsky et al. 2015 , Figure 22), admits
a (re gular) unimodular triangulation whose skeleton features a hea vy cycle with tw o
loops, cf. (Brodsky et al. 2015 , Figure 23).
Our aim in this section is to establish the follo wing,
Theorem 2 Suppose G is a gr aph with a heavy cycle C and two loops with cut edges,
e 1 and e 2 ,a si n Fig. 6 . Then the heavy component P  can have at most thr ee interior
lattice points, and these lie on the line spanned by the edge [ z ,w ]∈ Δ , wher e z is the
interior lattice point dual to C , and w is the intersection point of the split edg es s 1
and s 2 . In particular , P  is hyper elliptic and g ≤ 5 .
Proof It follo ws from Lemma 6 that the two triangles share the edge [ z ,w ] , where
w ∈ ∂ P is the point where the two split edges meet. W e will first show that the interior
lattice points of P  lie on the line spanned by [ z ,w ] .
As pre viously , we fix T 1 = con v { ( 0 , 0 ), ( 0 , 1 ), ( 1 , 0 ) } and use the labels from Figure
7 ; in particular , we may assume that w = ( 0 , 1 ) . Using Lemma 6 and the unimodularity
of T 2 , we realize that p 2 must lie on the line x =− 1; so let p 2 = ( − 1 , − k ) for some
integer k . W e use Lemma 5 on the points z = ( 0 , 0 ) , w = ( 0 , 1 ) and p 1 = ( 1 , 0 ) and
infer that the point r := ( 1 , 1 ) is an interior lattice point of P ; moreov er it is the unique
interior lattice point of P 1 . Similarly , we infer that the point r  := ( − 1 , − k + 1 ) is
the unique interior lattice point of P 2 . By considering the lines connecting the interior
point r with the boundary points p 1 and w , respecti vely , we see that k ≥ 0. The same
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Beitr Algebra Geom
T 1
z
wr
s 1
s 2
r 
P 
p 2
p 1
T 1
z
wr
q
s 1
p 1
Fig. 7 This illustrates Theorem 2 : general sketch (left) and the case when g ( P  ) ≥ 4 (right), which is
impossible
ar gument sho ws that the entire polygon P  is squeezed between the lines x = 1 and
x =− 1. In particular , the interior lattice points of P  lie on x = 0, which is the line
spanned by z and w .
Next we will sho w that g ( P  ) ≤ 3. W e assume the contrary , i.e., g ( P  ) ≥ 4 Then,
since all interior lattice points of P  lie on the line x = 0, the point ( 0 , − 3 ) must be
an interior lattice point of P .
The vertical line x = 1 contains the point p 1 , which is a boundary point, and r ,
which is an interior lattice point. It follo ws that there is a lattice point ( 1 ,λ ) in the
boundary of P 1 for λ> 1; in particular , either ( 1 , 2 ) ∈ P 1 or the boundary edge at
w passes through a point in the open interv al (( 1 , 2 ), ( 1 , 1 )) . Also, as ( 0 , − 3 ) is an
interior point, no point in ∂ P 1 is present on the line y = 3 x − 3. W e realize that in this
case P 1 is contained in the triangle con v { p 1 ,w ,( 1 , 3 ) } . Howe ver , this triangle has no
v alid lattice point which could be a verte x of P 1 ; recall that ( 1 , 3 ) has been e xcluded;
see Fig. 7 . This provides the desired contradiction, and thus g ( P  ) ≤ 3.  
The abov e result is sharp as Example 2 sho ws. The following summarizes the kno wn
obstructions to tropical planarity together with our ne w results.
Theorem 3 A t rivalent planar gr aph of genus g ≥ 3 is not tr opically planar if one of
the following holds:
(i) it contains a sprawling node , or
(ii) it contains a sprawling triangle and g ≥ 5 ,o r
(iii) it is cr owded, or
(i v) it is a TIE-fighter , or
(v) it has a heavy cycle with two loops such that the interior lattice points of the
heavy component do not align with the intersection of the two split lines.
123

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Fig. 8 Examples of non realizable graphs of genus 7 and 8
For instance, the preceding result e xcludes the graphs in Fig. 8 .
5 Anti-honeycombs
The purpose of this section is to define a class of lattice polygons each of which
admits a special triangulation. This is moti vated by Theorem 1 , which characterizes
one of these triangulations. W e belie ve that the entire f amily deserves some attention.
Consider three families of parallel lines:
L k ={ y = 2 x + k } , M  ={ 2 y = x −  } , N m ={ y =− x + m } , (3)
where k , , m ∈ Z . By picking a sixtuple π = ( k , k  ; ,   ; m , m  ) , with k < k  , <
 ,
and m < m  we obtain a polygon A π which is defined by three pairs of inequalities,
where each pair comes from one of the parallel families ( 3 ). W e call A π the anti-
hone ycomb polygon of type π ; in general, this is not a lattice polygon. The follo wing
characterizes when the lines from the three families intersect at lattice points; we omit
the proof, which is a direct calculation.
Lemma 7 W e have
(i) L k ∧ M  ∈ Z 2 if and only if k −  is divisible by 3 ;
(ii) L k ∧ N m ∈ Z 2 if and only if k − m is divisible by 3 ;
(iii) M  ∧ N m ∈ Z 2 if and only if  − m is divisible by 3 .
The name comes about from the connection to the “honeycomb polygons” studied
in Brodsky et al. ( 2015 , pp. 10f f). Note that not all of the six inequalities need to be
facet defining, whence A π is a hexagon, a pentagon, a quadrangle or a triangle. For
instance,
A ( k , − 2 k ; k , − 2 k ; k , − 2 k ) = con v { ( − k , − k ), ( 0 , k ), ( k , 0 ) } (4)
is a triangle, and its genus equals
g ( A ( k , − 2 k ; k , − 2 k ; k , − 2 k ) ) = 3 k 2 − 3 k + 2
2 .
123

Beitr Algebra Geom
Fig. 9 Anti-honeycomb triangulation of genus 19 on the left, and the corresponding sk eleton on the right
W e fix a type π = ( k , k  ; ,   ; m , m  ) , and we let V = A π ∩ Z 2 be the set of lattice
points in A π . Intersecting with the shifted lattice
L =  − 1
− 1  + Z  2
1  + Z  1
2 
of index 3 we obtain a subset V  = V ∩ L of the lattice points in A π . The families of
lines ( 3 ) yield weakly compatible splits of the point configuration V  , and they induce
a triangulation Δ 
π of the lattice points in V  . Notice that all the points in V \ V  lie
in the interior int A π . Since none of these lattice points lies on an y of the lines ( 3 )
it follo ws that each of them is contained in the interior of a unique triangle of Δ 
π .
Employing stellar subdi visions at the points in V \ V  this yields a triangulation Δ π
of V , which we call the anti-hone ycomb triangulation of type π . Figure 2 , Example 1 ,
and Theorem 1 are concerned with the case π = ( − 2 , 4 ;− 2 , 4 ;− 2 , 4 ) of genus 4.
Figure 9 sho ws a genus 19 anti-honeycomb triangulation along with its corresponding
skeleton for π = ( 4 , − 8 ; 4 , − 8 ; 4 , − 8 ) .
The honeycomb curv es yield moduli cones of maximal dimension 2 g + 1, where
g is the genus, cf. (Brodsky et al. 2015 , Theorem 1). In contrast the anti-hone ycomb
curves form a lar ge family whose moduli cones are much smaller . F or instance, a direct
polymake (Gawrilo w and polymake 2000 ) computation sho ws that the moduli cone
of Δ ( 4 , − 8 ; 4 , − 8 ; 4 , − 8 ) is only 28-dimensional, whereas the upper bound 2 g + 1 equals 39.
Example 3 T wo interesting classes of anti-honeycomb quadrangles are:
A ( k , − 2 k ; k , k − 3 ; k , − 2 k ) = con v { ( − k , − k ), ( k , 0 ), ( k − 1 , 1 ), ( 1 − k , 2 − k ) } ,
A ( k , 3 − 2 k ; k , k − 3 ; k , − 2 k ) = con v { ( − k , − k ), ( k − 2 , − 1 ), ( k − 1 , 1 ), ( 1 − k , 2 − k ) } .
They arise from the triangle A ( k , − 2 k ; k , − 2 k ; k , − 2 k ) in ( 4 ) by imposing M k − 3 and L 3 − 2 k ,
respecti vely , as additional facets. The quadrangle A ( k , − 2 k ; k , k − 3 ; k , − 2 k ) is a lattice trape-
zoid of genus 2 k − 1, while A ( k , 3 − 2 k ; k , k − 3 ; k , − 2 k ) is a lattice parallelogram of genus
2 k − 2. These examples pro vide anti-honeycomb polygons of arbitrary genus. Addi-
tionally , their skeleta are hyperelliptic, despite the fact that the interior lattice points
do not lie on a line. The first fe w cases are shown in Fig. 10 . Because of their shapes
we call them anti-hone ycomb quadrangles of zigzag type .
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Fig. 10 Anti-honeycomb quadrangles of zigzag type and their sk eleta; genus 3, 4 and 5
The anti-honeycomb polygons of lo w genus can be found by directly inspecting
the possibilities of stacking copies of the elliptic “b uilding block” A ( 1 , − 2 ; 1 , − 2 , 1 , − 2 ) .
Proposition 1 The anti-hone ycomb polygons of genus g ≤ 6 ar e unimodularly equiv-
alent to quadrangles of zigza g type or to the triangle A ( 2 , − 4 ; 2 , − 4 ; 2 , − 4 ) .
Up to an af fine transformation, which is not a lattice transformation, the three fami-
lies of lines in ( 3 ) form a Coxeter hyperplane arrangement of type ˜
A 2 . This generalizes
to arbitrary dimensions, and so does the construction of the anti-honeycomb triangu-
lations. The resulting anti-honeycomb polytopes are af fine images of the “alcov ed
polytopes” of Lam and Postniko v ( 2007 ). Further details will be explored else where.
6 Conclusion
The classification of the tropically planar graphs of genus g ≤ 5 was obtained in
Brodsky et al. ( 2015 ). Theorem 3 no w allows for a combinatorial characterization:
Corollary 1 A trivalent planar graph of genus g ≤ 5 is tr opically planar if and only if
none of the obstructions in Theorem 3 occurs.
Proof The tri valent graphs of lo w genus ha ve been classified in Balaban ( 1976 ). F or
g = 3 there are five such graphs, one of which has a spra wling node; the other four are
tropically planar (Brodsky et al. 2015 , Theorem 5.1.) F or g = 4 there are 17 graphs:
one is non-planar , three hav e a sprawling node, the remaining 13 are tropically planar
(Brodsky et al. 2015 , Theorem 7.1). This w as known before.
There are exactly 71 tri valent graphs of genus 5. Among them only 52 are planar
without a sprawling node (Brodsk y et al. 2015 ). Of these 14 were ruled out by explicit
computations (Brodsky et al. 2015 ), which lea ves 38 tropically planar graphs of genus
5. One of the ke y contributions of Coles et al. ( 2019 ), Figure 8 is to obtain obstructions
to tropical planarity , which rules out another ten, which are crowded or TIE-fighters.
As our ne w contribution we can no w discuss the remaining four graphs, which
are sho wn in Fig. 11 . Firstly , we observ e that all of these exhibit a hea vy cycle. The
graph labeled “a” has a heavy c ycle with two loops, b ut the component aw ay from the
two loops is not hyperelliptic; i.e., it is ruled out by Theorem 2 . The second graph,
labeled “b” also has a hea vy cycle with tw o loops, the component (of genus 3) away
from the two loops is e ven h yperelliptic. Ho wev er , we see that the interior point z
dual to the hea vy cycle w ould lie in between the other two interior lattice points in the
123

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a b c d
Fig. 11 The four genus 5 graphs that are not ruled out by any prior kno wn criteria
a b c
d e f
g h
Fig. 12 The eight tri valent planar graphs of genus 6, which are not tropically planar (Coles et al. 2019 ), b ut
which are not cov ered by Theorem 3
hyperelliptic polygon dual to the genus 3 component; the latter contradicts Theorem
2 , which says that the interior lattice points in the genus 3 component should lie belo w
z on the line spanned by z and w , where w is the point of intersection of the two
splits. Thus “b” is ruled out by Theorem 2 , too. The graphs labeled “c” and “d” feature
sprawling triangles, whence the y are taken care of by Theorem 1 . This completes our
combinatorial characterization of the tropically planar graphs of genus at most five.  
For genus 6, there are 388 tri valent graphs altogether , 354 of which are planar
(Balaban 1976 ). In Coles et al. ( 2019 ) it was sho wn that 152 tropically planar graphs
of genus 6 remain. There are 28 graphs which are non-realizable and could not be ruled
out using any prior kno wn criteria; cf. (Coles et al. 2019 , Figure 17). Out of these 28
graphs, 19 ha ve a hea vy cycle with two loops and can be ruled out using Theorem 2
because the genus is too high. One of the remaining graphs has a spra wling triangle,
and thus excluded by Theorem 1 . W e are left with eight graphs of genus 6, which are
sho wn in Fig. 12 ; for these we are not aware of an y a priori obstruction. There are 672
troplanar graphs of genus 7 according to Coles et al. ( 2019 , T able 1); ho we ver , the full
list does not seem to be av ailable.
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It would be interesting to kno w how the tropically plane curv es of a fixed genus fit
into the moduli space of all tropical curves. F or genus 3 this was recently answered in
terms of modifications by Hahn et al. ( 2019 ).
Acknowledgements Open Access funding provided by Projekt DEAL. Open Access funding pro vided by
Projekt DEAL. W e are grateful to Dominic Bunnett and Marta Panizzut for helpful comments on an early
draft of this article and for requiring a minor correction. Moreov er , we thank two anonymous referees, whose
suggestions and requests helped to improv e the paper further . M. Joswig has been supported by Deutsche
Forschungsgemeinschaft (EXC 2046: “MA TH + ”, SFB-TRR 195: “Symbolic T ools in Mathematics and
their Application”, and GRK 2434: “Facets of Comple xity”). A. K. T e wari has been supported by Deutsche
Forschungsgemeinschaft (SFB-TRR 195: “Symbolic T ools in Mathematics and their Application”).
Open Access This article is licensed under a Creative Commons Attrib ution 4.0 International License, which
permits use, sharing, adaptation, distribution and reproduction i n any medium or format, as long as you gi ve
appropriate credit to the original author(s) and the source, provide a link to the Creati ve Commons licence,
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