The complexity of tropical graph homomorphisms
Florent Foucaud∗

Ararat Harutyunyan†
Yannis Manoussakis§

Pavol Hell‡

Sylvain Legay§

Reza Naserasr¶

January 31, 2018

arXiv:1607.04777v2 [] 30 Jan 2018

Note to readers. A shorter version of this article appeared in Discrete Applied Mathematics
229 (2017). The present extended version contains all missing proofs and additional figures.
Abstract
A tropical graph (H, c) consists of a graph H and a (not necessarily proper) vertex-colouring c of
H. Given two tropical graphs (G, c1 ) and (H, c), a homomorphism of (G, c1 ) to (H, c) is a standard
graph homomorphism of G to H that also preserves the vertex-colours. We initiate the study of
the computational complexity of tropical graph homomorphism problems. We consider two settings.
First, when the tropical graph (H, c) is fixed; this is a problem called (H, c)-Colouring. Second,
when the colouring of H is part of the input; the associated decision problem is called H-TropicalColouring. Each (H, c)-Colouring problem is a constraint satisfaction problem (CSP), and we
show that a complexity dichotomy for the class of (H, c)-Colouring problems holds if and only if the
Feder–Vardi Dichotomy Conjecture for CSPs is true. This implies that (H, c)-Colouring problems
form a rich class of decision problems. On the other hand, we were successful in classifying the
complexity of at least certain classes of H-Tropical-Colouring problems.

1

Introduction

Unless stated otherwise, the graphs considered in this paper are simple, loopless and finite. A homomorphism h of a graph G to a graph H is a mapping h : V (G) → V (H) such that adjacency is preserved by
h, that is, the images of two adjacent vertices of G must be adjacent in H. If such a mapping exists, we
note G → H. For a fixed graph H, given an input graph G, the decision problem H-Colouring (whose
name is derived from the proximity of the problem to proper vertex-colouring) consists of determining
whether G → H holds. Problems of the form H-Colouring for some fixed graph H, are called homomorphism problems. A classic theorem of Hell and Nešetřil [21] states a dichotomy for this problem: if
H is bipartite, H-Colouring is polynomial-time solvable; otherwise, it is NP-complete.
Tropical graphs. As an extension of graph homomorphisms, homomorphisms of edge-coloured graphs
have been studied, see for example [1, 6, 7, 8, 9]. In this paper, we consider the variant where the vertices
are coloured. We initiate the study of tropical graph homomorphism problems, in which the vertex sets of
the graphs are partitioned into colour classes. Formally, a tropical graph (G, c) is a graph G together with
a (not necessarily proper) vertex-colouring c : V (G) → C of G, where C is a set of colours. If |C| = k,
we say that (G, c) is a k-tropical graph. Given two tropical graphs (G, c1 ) and (H, c2 ) (where the colour
set of c1 is a subset of the colour set of c2 ), a homomorphism h of (G, c1 ) to (H, c2 ) is a homomorphism
of G to H that also preserves the colours, that is, for each vertex v of G, c1 (v) = c2 (h(v)). For a fixed
tropical graph (H, c), problem (H, c)-Colouring asks whether, given an input tropical graph (G, c1 ),
we have (G, c1 ) → (H, c).
The homomorphism factoring problem. Brewster and MacGillivray defined the following related
problem in [10]. For two fixed graphs H and Y and a homomorphism h of H to Y , the (H, h, Y )Factoring problem takes as an input, a graph G together with a homomorphism g of G to Y , and asks
∗ LIMOS,

Université Blaise Pascal, Clermont-Ferrand (France). florent.foucaud@gmail.com
Université Toulouse III (Paul Sabatier), Toulouse (France). aharutyu@math.univ-toulouse.fr
‡ School of Computing Science, Simon Fraser University, Burnaby (Canada). pavol@sfu.ca
§ LRI, Université Paris-Sud, Orsay (France). {legay,yannis}@lri.fr
¶ CNRS - IRIF, Université Paris Diderot, Paris (France). E-mail: reza@irif.fr
† IMT,

1

for the existence of a homomorphism f of G to H such that f = h ◦ g. The (H, c)-Colouring problem
+
+
corresponds to (H, c, K|C|
)-Factoring where K|C|
is the complete graph on |C| vertices with all loops
(and with C the set of colours used by c). (Note that in [10], loops were not considered.)
Constraint satisfaction problems (CSPs). Graph homomorphism problems fall into a more general
class of decision problems, the constraint satisfaction problems, defined for relational structures. A relational structure S over a vocabulary (a vocabulary is a set of pairs (Ri , ai ) of relation names and arities)
consists of a domain V (S) of vertices together with a set of relations corresponding to the vocabulary,
that is, Ri ⊆ V (S)ai for each relation Ri of the vocabulary. Given two relational structures S and T over
the same vocabulary, a homomorphism of S to T is a mapping h : V (S) → V (T ) such that each relation
Ri is preserved, that is, for each subset of V (S)ai of Ri in S, its image set in T also belongs to Ri . For
a fixed relational structure T , T -CSP is the decision problem asking whether a given input relational
structure has a homomorphism to T .
Using this terminology, a graph H is a relational structure over the vocabulary {(A, 2)} consisting of
a single binary relation A (adjacency). Hence, H-Colouring is a CSP. Further, (H, c)-Colouring is
equivalent to the problem C(H, c)-CSP, where C(H, c) is obtained from H by adding a set of k unary
relations to H (one for each colour class of the k-colouring c).
The Dichotomy Conjecture. In their celebrated paper [20], Feder and Vardi posed the following
conjecture.
Conjecture 1.1 (Feder and Vardi [20]). For every fixed relational structure T , T -CSP is polynomial-time
solvable or NP-complete.
Conjecture 1.1 became known as the Dichotomy Conjecture and has given rise to extensive work in
this area, see for example [11, 12, 15, 16, 17, 18]. If the conjecture holds, it would imply a fundamental
distinction between CSP and the whole class NP. Indeed, the latter is known (unless P=NP) to contain
so-called NP-intermediate problems that are neither NP-complete nor polynomial-time solvable [26].
The Dichotomy Conjecture was motivated by several earlier dichotomy theorems for special cases,
such as the one of Schaefer for binary structures [28] or the one of Hell and Nešetřil for undirected
graphs, stated as follows.
Theorem 1.2 (Hell and Nešetřil Dichotomy [21]). Let H be an undirected graph. If H is bipartite, then
H-Colouring is polynomial-time solvable. Otherwise, H-Colouring is NP-complete.
Digraph homomorphisms. Digraph homomorphisms are also well-studied in the context of complexity
dichotomies. We will relate them to tropical graph homomorphisms. For a digraph D, D-Colouring
asks whether an input digraph admits a homomorphism to D, that is, a homomorphism of the underlying
undirected graphs that also preserves the orientation of the arcs.
While in the case of undirected graphs, the H-Colouring problem is only polynomial time for graphs
whose core is either K1 or K2 , in the case of digraphs the problem remains polynomial time for a large
class of digraphs which are cores. The classification of such cores has been one of the difficulties of the
conjecture. Such classifications are given for certain interesting subclasses, see for example [2, 3, 4, 5, 14].
A proof of a conjectured classification of the general case has been announced while this paper was under
review (see [19]). If valid, this would imply the truth of the Dichotomy Conjecture, as Feder and Vardi [20]
showed the following (seemingly weaker) statement to be equivalent to it.
Conjecture 1.3 (Equivalent form of the Dichotomy Conjecture, Feder and Vardi [20]). For every bipartite digraph D, D-Colouring is polynomial-time solvable or NP-complete.
In Section 3, similarly to its above reformulation (Conjecture 1.3), we will show that the Dichotomy
Conjecture has an equivalent formulation as a dichotomy for tropical homomorphisms problems. More
precisely, we will show that the Dichotomy Conjecture is true if and only if its restriction to (H, c)Colouring problems, where (H, c) is a 2-tropical bipartite graph, also holds. In other words, one can
say that the class of 2-tropical bipartite graph homomorphisms is as rich as the whole class of CSPs.
For many digraphs D it is known such that D-Colouring is NP-complete. Such a digraph of order 4
and size 5 is presented in the book by Hell and Nešetřil [22, page 151]. Such oriented trees are also
known, see [23] or [22, page 158]; the smallest such known tree has order 45. A full dichotomy is known
for oriented cycles [14]; the smallest such NP-complete oriented cycle has order between 24 and 36 [13, 14].
2

Using these results, one can easily exhibit some NP-complete (H, c)-Colouring problems. To this end,
given a digraph D, we construct the 3-tropical graph T (D) as follows. Start with the set of vertices V (D)
→ in D, add a path ux x v of length 3 from u to v in T (D),
and colour its vertices Blue. For each arc −
uv
u v
where xu and xv are two new vertices coloured Red and Green, respectively. The following fact is not
difficult to observe.
Proposition 1.4. For any two digraphs D1 and D2 , we have D1 → D2 if and only if T (D1 ) → T (D2 ).
By the above results on NP-complete D-Colouring problems and Proposition 1.4, we obtain a 3tropical graph of order 14, a 3-tropical tree of order 133, and a 3-tropical cycle of order between 72
and 108 whose associated homomorphism problems are NP-complete. Nevertheless, in this paper, we
exhibit (by using other reduction techniques) much smaller tropical graphs, trees and cycles (H, c) with
(H, c)-Colouring NP-complete.
List homomorphisms. Dichotomy theorems have also been obtained for a list-based extension of the
class of homomorphism problems, the list-homomorphism problems. In this setting, introduced by Feder
and Hell in [15], the input consists of a pair (G, L), where G is a graph and L : V (G) → 2V (H) is a
list assignment representing a set of allowed images for each vertex of G. For a fixed graph H, the
decision problem H-List-Colouring asks whether there is a homomorphism h of G to H such that
for each vertex v of G, h(v) ∈ L(v). Problem H-List-Colouring can be seen as a generalization of
H-Colouring. Indeed, restricting H-List-Colouring to the class of inputs where for each vertex v of
G, L(v) = V (H), corresponds precisely to H-Colouring. Therefore, if H-Colouring is NP-complete,
so is H-List-Colouring. For this set of problems, a full complexity dichotomy has been established in
a series of three papers [15, 17, 18]. We state the dichotomy result for simple graphs from [17], that is
related to our work. (A circular arc graphs is an intersection graph of arcs on a cycle.)
Theorem 1.5 (Feder, Hell and Huang [17]). If H is a bipartite graph such that its complement is a circular arc graph, then H-List-Colouring is polynomial-time solvable. Otherwise, H-List-Colouring
is NP-complete.
Given a tropical graph (H, c), the problem (H, c)-Colouring is equivalent to the restriction of HList-Colouring to instances (G, L) where each list is the set of vertices in one of the colour classes of c.
Next, we introduce a less restricted variant of H-List-Colouring that is also based on tropical graph
homomorphisms.
The H-Tropical-Colouring problem. Given a fixed graph H, we introduce the decision problem
H-Tropical-Colouring, whose instances consist of (1) a vertex-colouring c of H and (2) a tropical
graph (G, c2 ). Then, H-Tropical-Colouring consists of deciding whether (G, c1 ) → (H, c).
Alternatively, H-Tropical-Colouring is an instance restriction of H-List-Colouring to instances with laminar lists, that is, lists such that for each pair of distinct vertices v1 , v2 ∈ V (G),
L(v1 ) = L(v2 ) or L(v1 ) ∩ L(v2 ) = ∅. (We remark that H-Tropical-Colouring, as well as H-ListColouring, can also be formulated as a CSP, where certain unary relations encode the list constraints:
so-called full CSPs, see [16] for details.)
Given the difficulty of studying (H, c)-Colouring problems, as will be demonstrated in Section 3,
the study of H-Tropical-Colouring problems will be the focus of the other parts of this paper. This
study is directed by the following question.
Question 1.6. For a given graph H, what is the complexity of H-Tropical-Colouring?
Clearly, (H, c)-Colouring where each vertex receives the same colour, is computationally equivalent to H-Colouring. Therefore, by the Hell-Nešetřil dichotomy of Theorem 1.2, if H is non-bipartite,
H-Tropical-Colouring is NP-complete. Furthermore, by the above formulation of H-TropicalColouring as an instance restriction of H-List-Colouring, whenever H-List-Colouring is polynomialtime solvable, so is H-Tropical-Colouring.
Thus, according to Theorems 1.2 and 1.5, all problems H-Tropical-Colouring where H is not
bipartite are NP-complete, and all problems H-Tropical-Colouring where H is bipartite and its
complement is a circular-arc graph are polynomial-time solvable. Thus, it remains to study H-TropicalColouring when H belongs to the class of bipartite graphs whose complement is not a circular-arc graph.
This class of graphs has been well-studied, and characterized by forbidden induced subgraphs [29]. It is

3

a rich class of graphs that includes all cycles of length at least 6, all trees with at least one vertex from
which there are three branches of length at least 3, and an many other graphs [29].
Observe that for any induced subgraph H 0 of a graph H, one can reduce H 0 -Tropical-Colouring
to H-Tropical-Colouring by assigning, in the input colouring of H, a dummy colour to all the
vertices of H − H 0 . Hence, if H-Tropical-Colouring is polynomial-time solvable, then H 0 -TropicalColouring is also polynomial-time solvable. Conversely, if H 0 -Tropical-Colouring is NP-complete,
so is H-Tropical-Colouring. Therefore, to answer Question 1.6, it is enough to consider minimal
graphs H such that H-Tropical-Colouring is NP-complete.
A first question is to study the case of minimal graphs H for which H-List-Colouring is NPcomplete; such a list is known and it follows from Theorem 1.5. In particular, it contains all even
cycles of length at least 6. In Section 4, we show that for every even cycle C2k of length at least 48,
C2k -Tropical-Colouring is NP-complete. On the other hand,for every even cycle C2k of length at
most 12, C2k -Tropical-Colouring is polynomial-time solvable. Unfortunately, for each graph H in
the above-mentioned list that is not a cycle, H-Tropical-Colouring is polynomial-time solvable, and
thus larger graphs will be needed in the quest of a similar characterization of NP-complete H-TropicalColouring problems.
In Section 5, we show that for every bipartite graph H of order at most 8, H-Tropical-Colouring
is polynomial-time solvable, but there is a bipartite graph H9 of order 9 such that H9 -TropicalColouring is NP-complete.
Finally, in Section 6, we study the case of trees. We prove that for every tree T of order at most 11,
T -Tropical-Colouring is polynomial-time solvable, but there is a tree T23 of order 23 such that
T23 -Tropical-Colouring is NP-complete.
We remark that our NP-completeness results are finer than those that can be obtained from Proposition 1.4, in the sense that the orders of the obtained target graphs are much smaller. Similarly, we note
that the results in [10] imply the existence of NP-complete H-Tropical-Colouring problems, and H
can be chosen to be a tree or a cycle. However, similarly as in Proposition 1.4, these results are also based
on reductions from NP-complete D-Colouring problems, where H is obtained from the digraph D by
replacing each arc by a path (its length depends on D, but it is always at least 3). Thus, the NP-complete
tropical targets obtained in [10] are trees of order at least 133 and cycles of order at least 72, which is
much more than the ones exhibited in the present paper.

2

Preliminaries and tools

In this section we gather some necessary preliminary definitions and results.

2.1

Isomorphisms, cores

For tropical graph homomorphisms, we have the same basic notions and properties as in the theory of
graph homomorphisms. A homomorphism of tropical graph (G, c1 ) to (H, c2 ) is an isomorphism if it is
a bijection and it acts bijectively on the set of edges.
Definition 2.1. The core of a tropical graph (G, c) is the smallest (in terms of the order) induced tropical
subgraph (G0 , c|G0 ) admitting a homomorphism of (G, c) to (G0 , c|G0 ).
In the same way as for simple graphs, it can be proved that the core of a tropical graph is unique. A
tropical graph (G, c) is called a core if its core is isomorphic to (G, c) itself. Moreover, we can restrict
ourselves to studying only cores. Indeed it is not difficult to check that (G, c1 ) admits a homomorphism
to (H, c2 ) if and only if the core of (G, c1 ) admits a homomorphism to the core of (H, c2 ).

2.2

Formal definitions of the used computational problems

We now formally define all the decision problems used in this paper.
H-Colouring
Input: A (di)graph G.
Question: Does there exist a homomorphism of G to H?

4

H-List-Colouring
Input: A graph G and a list function L : V (G) → 2V (H) .
Question: Is there a homomorphism f of G to H such that for every vertex x of G, f (x) ∈ L(x)?
(H, c)-Colouring
Input: A tropical graph (G, c1 ).
Question: Does (G, c1 ) admit a homomorphism to (H, c)?
H-Tropical-Colouring
Input: A vertex-colouring c of H, and a tropical graph (G, c1 ).
Question: Does (G, c1 ) admit a homomorphism to (H, c)?
T -CSP
Input: A relational structure S over the same vocabulary as T .
Question: Does S admit a homomorphism to T ?
k-SAT
Input: A pair (X, C) where X is a set of Boolean variables and C is a set of k-tuples of literals of
X, that is, variables of X or their negation.
Question: Is there a truth assignment A : X → {0, 1} such that each clause of C contains at least
one true literal?
NAE k-SAT
Input: A pair (X, C) where X is a set variables and C is a set of k-tuples of variables of X.
Question: Is there a partition of X into two classes such that each clause of C contains at least one
variable in each class?
It is a folklore result that 2-SAT is polynomial-time solvable, a fact for example observed in [25]. On
the other hand, 3-SAT is NP-complete [24], and NAE 3-SAT is NP-complete as well [27] (even if the
input formula contains no negated variables).

2.3

Bipartite graphs

We now give several facts that are useful when working with homomorphisms of bipartite graphs.
Observation 2.2. Let H be a bipartite graph with parts A, B. If φ : G → H is a homomorphism of G
to H, then G must be bipartite. Moreover, if G and H are connected, then φ−1 (A) and φ−1 (B) are the
two parts of G.
The next proposition shows that for bipartite target graphs, we may assume (at the cost of doubling
the number of colours) that no two vertices from two different parts of the bipartition are coloured with
the same colour.
Proposition 2.3. Let (H, c) be a connected tropical bipartite graph with parts A, B, and assume that
vertices in A and B are coloured by c with colours in set CA and CB , respectively. Let c0 be the colouring
with colour set (CA × 0) ∪ (CB × 1) obtained from c with c0 (x) = (c(x), 0) if x ∈ A and c0 (x) = (c(x), 1)
if x ∈ B. If (H, c0 )-Colouring is polynomial-time solvable, then (H, c)-Colouring is polynomial-time
solvable.
Proof. Let (G, c1 ) be a bipartite tropical graph. We may assume G is connected since the complexity of
(H, c)-Colouring and (H, c0 )-Colouring stays the same for connected inputs. Let c01 and c001 be the
colourings obtained from c1 by performing a similar modification as for c0 : c01 (x) = (c1 (x), 0) if x ∈ A
and c01 (x) = (c1 (x), 1) if x ∈ B, and c001 (x) = (c1 (x), 1) if x ∈ A and c001 (x) = (c1 (x), 0) if x ∈ B. Now it is
clear, by Observation 2.2, that (G, c1 ) → (H, c) if and only if either (G, c01 ) → (H, c0 ) or (G, c001 ) → (H, c0 ).
Since the latter condition can be checked in polynomial time, the proof is complete.

5

2.4

Generic lemmas for polynomiality

We now prove several generic lemmas that will be useful to prove that a specific (H, c)-Colouring
problem is polynomial-time solvable.
Definition 2.4. Let (H, c) be a tropical graph. A vertex of (H, c) is a forcing vertex if all its neighbours
are coloured with distinct colours.
This is a useful concept since in any mapping of a tropical graph (G, c0 ) to a target containing a
forcing vertex x, if a vertex of G is mapped to x, then the mapping of all its neighbours is forced. We
have the following immediate application:
Lemma 2.5. Let (H, c) be a tropical graph. If all vertices of H are forcing vertices, then (H, c)Colouring is polynomial-time solvable.
Proof. Choose any vertex x of the instance (G, c1 ), and map it to any vertex of (H, c) with the same
colour. Once this choice is made, the mapping for the whole connected component of x is forced. Hence,
try all O(|V (H)|) possibilities to map x, and repeat this for every connected component of G. The
tropical graph (G, c1 ) is a YES-instance if and only if every connected component admits a mapping.
Lemma 2.6 (2-SAT). Let (H, c) be a tropical graph and let {S1 , . . . , Sk } be a collection of independent
sets of H, each of size at most 2. Assume that for every tropical graph (G, c1 ) admitting a homomorphism
to (H, c), there exists a partition P = P1 , . . . , P` of V (G) into ` ≤ k sets and a homomorphism f :
(G, c1 ) → (H, c) such that for every i ∈ {1, . . . , `}, there is a j = j(i) ∈ {1, . . . , k} such that all vertices
of Pi map to vertices of Sj . Then (H, c)-Colouring is polynomial-time solvable.
Proof. We reduce (H, c)-Colouring to 2-SAT. For every set Si , if Si contains only one vertex s, s
represents TRUE. If Si contains two vertices s, s0 , one of them represents TRUE, the other FALSE (note
that if some vertex belongs to two distinct sets Si and Sj , it is allowed to represent, say, FALSE with
respect to Si and TRUE with respect to Sj ). Now, given an instance (G, c1 ) of (H, c)-Colouring, we
build a 2-SAT formula over variable set V (G) that is satisfiable if and only if (G, c1 ) → (H, c), as follows.
For every edge xy of G, assume that in f , x is mapped to a vertex of Si and y is mapped to a vertex
of Sj (necessarily if (G, c1 ) → (H, c) we have i 6= j since Si , Sj induce independent sets). Let Fxy be a
disjunction of conjunctive 2-clauses over variables x, y. For every edge uv between a vertex u in Si and a
vertex v in Sj , depending on the truth value assigned to u and v, add to Fxy the conjunctive clause that
would be true if x is assigned the truth value of u and y is assigned the truth value of v. For example:
if u = F ALSE and v = T RU E add the clause (x ∧ y). When Fxy is constructed, transform it into an
equivalent conjunction of disjunctive clauses and add it to the constructed 2-SAT formula. Now, by the
construction, if the formula is satisfiable we construct a homomorphism by mapping every vertex x to
the vertex of the corresponding set Si that has been assigned the same truth value as x in the satisfying
assignment. By construction it is clear that this is a valid mapping. On the other hand, if the formula is
not satisfiable, there is no homomorphism of (G, c1 ) to (H, c) satisfying the conditions, and hence there
is no homomorphism at all.
As a corollary of Lemma 2.6 and Proposition 2.3, we obtain the following lemma:
Lemma 2.7. If (H, c) is a bipartite tropical graph where each colour is used at most twice, then (H, c)Colouring is polynomial-time solvable.
Given a set S of vertices, the boundary B(S) is the set of vertices in S that have a neighbour out of
S.
Lemma 2.8. Let (H, c) be a tropical graph containing a connected subgraph S of forcing vertices such
that:
(a) every vertex in B(S) is coloured with a distinct colour (let C(S) be the set of colours given to vertices
in B(S)), and (b) no colour of C(S) is present in V (H) \ S.
If (H − S)-List-Colouring is polynomial-time solvable, then (H, c)-Colouring is polynomial-time
solvable.

6

Proof. Let S = V (H) \ S. Let (G, c1 ) be an instance of (H, c)-Colouring. Consider an arbitrary vertex
v of G with c1 (v) = i. Then, v must be mapped to a vertex coloured i. For every possible choice of
mapping v, we will construct one instance of (H − S)-List-Colouring. To construct an instance from
such a choice, we first partition V (G) into two sets: the set VS containing the vertices that must map to
vertices in S (and their images are determined), and the set VS containing the vertices that must map to
vertices of S. We now distinguish two basic cases, that will be repeatedly applied during the construction.
Case 1: vertex v is mapped to a vertex in S. If v has been mapped to a vertex x of S, since x is
a forcing vertex, the mapping of all neighbours of v is determined (anytime there is a conflict we return
NO for the specific instance under construction). We continue to propagate the forced mapping as much
as possible (i.e. as long as the forced images belong to S) within a connected set of G containing v.
This yields a connected set Cv of vertices of G whose mapping is determined, and whose neighbourhood
Nv = N (Cv ) \ Cv consists of vertices each of which must be mapped to a determined vertex of S. We
add Cv to VS . We now remove the set Cv from G and repeat the procedure for all vertices of Nv using
Case 2.
Case 2: vertex v is mapped to a vertex in S. We perform a BFS search on the remaining vertices
in G, until we have computed a maximal connected set Cv of vertices containing v in which no vertex
is coloured with a colour in C(S). Then, for every vertex x of Cv with a neighbour y that is coloured i
(i ∈ C(S)), by Property (a) we know that y must be mapped to a vertex in B(S), and moreover the
image of y is determined by colour i. Hence the neighbourhood Nv = N (Cv ) \ Cv has only vertices whose
mapping is determined. We add Cv to set VS and apply Case 1 to every vertex in Nv .
End of the procedure. Once V (G) has been partitioned into VS and VS (where the mapping of all
vertices in VS ∪ N (VS ) is fixed), we can reduce this instance to a corresponding instance of (H − S)-ListColouring.
In total, (G, c1 ) is a YES-instance if and only if at least one of the O(|V (G)|) constructed instances
of (H − S)-List-Colouring is a YES-instance.
The next lemma is similar to Lemma 2.8 but now the boundary is distinguished using edges.
Lemma 2.9. Let (H, c) be a tropical graph containing a connected subgraph S of forcing vertices with
boundary B = B(S) and N = N (B) \ S. Assume that the following properties hold:
(a) for every pairs xy, x0 y 0 of distinct edges of B × N , we have (c(x), c(y)) 6= (c(x0 ), c(y 0 )), and
(b) for every edge xy of B × N , there is no edge in (H − S) × (H − S) whose endpoints are coloured
c(x) and c(y). If (H − S)-List-Colouring is polynomial-time solvable, then (H, c)-Colouring is
polynomial-time solvable.
Proof. The proof is almost the same as the one of Lemma 2.8, except that now, while computing an
instance of (H − S)-List-Colouring, the distinction between VS and VS is determined by the edges of
B × N.
The next lemma identify some unique features of a tropical graph to simplify the problem into a
list-homomorphism problem.
Definition 2.10. A Unique Tropical Feature in a tropical graph (H, c) is a vertex or an edge of H that
satisfies one of the following conditions.
Type 1. A vertex u of H whose colour class is {u}.
Type 2. An edge uv of H such that there is no other edge in H whose vertices are coloured c(u) and c(v),
respectively.
Type 3. A vertex u of H such that N (u) is monochromatic in (H, c) with colour s, and every vertex coloured
s that does not belong to N (u) has no neighbour coloured with c(u).
Type 4. A forcing vertex u of H such that for each pair v, w of distinct vertices in N (u), there is no path
v 0 u0 w0 in H − u with c(v) = c(v 0 ), c(u) = c(u0 ) and c(w) = c(w0 ).
Definition 2.11. Let (H, c) be a tropical graph and S a set of Unique Tropical Features of (H, c). S is
partitioned into four sets as S = S1 ∪ S2 ∪ S3 ∪ S4 , where Si is the set of unique tropical features of type
i in S. We define H(S) as follows : V (H(S)) = (V (H) ∪ {uv |u ∈ S4 , v ∈ N (u)}) \ (S1 ∪ S3 ∪ S4 ) and
E(H(S)) = (E(H[V (H(S))]) \ S2 ) ∪ {uv v|u ∈ S4 , v ∈ N (u)}.
7

In other words, H(S) is the graph obtained from H by removing unique tropical features of type 1,
2, and 3, and for each unique tropical feature u of type 4, replacing N [u] by d(u) pending edges.
Lemma 2.12. Let (H, c) be a tropical graph and S a set of unique tropical features of (H, c). If (H(S))List-Colouring is polynomial-time solvable, then (H, c)-Colouring is polynomial-time solvable.
Proof. Let (G, c0 ) be an instance of (H, c)-Colouring. We are going to construct a graph G0 and
associate to each vertex of G0 a list of vertices of H(S) such that there is a list-homomorphism from G0
to H(S) (with respect to these lists) if and only if there is a tropical homomorphism of (G, c0 ) to (H, c).
We proceed with sequential modifications, by considering the unique tropical features of S one by one.
First, we can see the instance (G, c0 ) of (H, c)-Colouring as an instance of H-List-Colouring by
giving to each vertex u in G the list L(u) of vertex in H coloured c0 (u). If at any point in the following,
we update the list of a vertex to be empty, we can conclude that there is no tropical homomorphism
between (G, c0 ) and (H, c).
For each unique tropical feature u of type 1 in S, there is a colour s such that only the vertex u is
coloured s in (H, c). Every vertex in (G, c0 ) coloured s must be mapped to u and has a list of size at most
one. For each vertex v in (G, c0 ) coloured s, we update the list of each of its neighbours w such that L(w)
becomes L(w) ∩ N (u). We can then delete v from (G, c0 ) and forget L(v) without affecting the existence
of a list-homomorphism. Indeed, if a homomorphism exists, then it must map each neighbour of v to a
neighbour of u. Moreover, there is no other vertex of (G, c0 ) that can be mapped to u.
For each unique tropical feature uv of type 2 in S, there is no other edge than uv in H such that the
colour of its vertices are c(u) and c(v). Every edge in (G, c0 ) whose vertices are coloured c(u) and c(v)
must be mapped to uv. For each edge xy in (G, c0 ) such that c0 (x) = c(u) and c0 (y) = c(v), we update the
list of x and y such that L(x) becomes L(x) ∩ {u} and L(y) becomes L(y) ∩ {v}. We can then delete the
edge uv from (G, c0 ) without changing the existence of a list-homomorphism. Indeed, if a homomorphism
exists, it must map x to u and y to v. Again, there is no other edge of (G, c0 ) that can be mapped to uv.
For each unique tropical feature u of type 3 in S, N (u) is monochromatic in (H, c) of colour s and any
vertex coloured s with a neighour coloured c(u) must belong to N (u). Let v be a vertex of G such that
c(v) = c(u) and N (v) is monochromatic in (G, c0 ) of colour s. Then, we can assume that v is mapped to
u. Indeed, in every tropical homomorphism of (G, c0 ) to (H, c), if v is not mapped to u, it is mapped to a
vertex at distance 2 from u, and one obtains another valid tropical homomorphism by only changing the
mapping of v to u. For each such vertex v, we update the list of its neighbours w such that L(w) becomes
L(w) ∩ N (u). We can then delete v from (G, c0 ) without affecting the existence of a list-homomorphism.
Indeed, if a homomorphism exists, it maps every neighbour of v to a neighbour of u. Moreover, there no
other vertex of (G, c0 ) can be mapped to u.
Finally, let u be a vertex of type 4 in S. Thus, by the definition of type 4, for each v, w ∈ N (u), there
is no other path v 0 u0 w0 in H such that c(v) = c(v 0 ), c(u) = c(u0 ) and c(w) = c(w0 ). Furthermore, since u
is a forcing vertex, we have c(v) 6= c(w) for any two neighbours v and w of u.
Let x be a vertex of G such that c0 (x) = c(u) and such that at least two neighbours of x are of colours
c(v) or c(w), one of each. Then, as x is of type 4, any homomorphism of (G, c0 ) to (H, c) must map all
such vertices x to u. Remove all such vertices from G and let (G0 , c0 ) be the remaining tropical graph.
For any vertex y of G0 if it is of colour c(u), it may then either map to another vertex of this colour, or
all its neighbours must map a same neighbour of u. Let (H1 , c) be a tropical graph obtained from (H, c)
by removing the vertex u, and then adding one new vertex for each vertex in NH (u) and assigning the
colour c(u) to it. It follows that (G0 , c0 ) admits a homomorphism to (H 0 , c) if and only (G, c0 ) admits a
homomorphism to (H, c), proving our claim.
In conclusion, we have built an instance (G0 , L) of H(S)-List-Colouring that maps to H(S) if and
only if (G, c0 ) maps to (H, c), thus proving our claim. We remark, furthermore, that these changes used
to introduced (G0 , L) and H(S) are compatible even between different types of vertices, thus we may
allow S to contain a combination of such vertices. However, in this work we will only consider sets S
whose elements are all of a same type.

3

(H, c)-Colouring and the Dichotomy Conjecture

Since each (H, c)-Colouring problem is a CSP, the Feder–Vardi Dichotomy Conjecture (Conjecture 1.1)
would imply a complexity dichotomy for the class of (H, c)-Colouring problems. As we mentioned
8

before a proof of the conjecture has been recently announced, thus every (H, c)-Colouring is either
polynomial time solvable or it is an NP-complete problem. Here we point out that an independent proof
even on a very restricted set of (H, c) would also prove the original conjecture.
Following the construction of Feder and Vardi ([20, Theorem 10]) and based on its exposition in the
book by Hell and Nešetřil [22, Theorem 5.14], one can modify their gadgets to prove a similar statement
for the class of 2-tropical bipartite graph homomorphism problems.
Theorem 3.1. For each CSP template T there is a 2-coloured graph (H, c) such that (H, c)-Colouring
and T -CSP are polynomially equivalent. Moreover, (H, c) can be chosen to be bipartite and homomorphic
to a 2-coloured forcing path.
Proof. We follow the proof of Theorem 5.14 in the book [22] proving a similar statement for digraph
homomorphism problems. The structure of the proof in [22] is as follows. First, one shows that for each
CSP template T , there is a bipartite graph H such that the T -CSP problem and the H-Retraction
problem are polynomially equivalent. Next, it is shown that for each bipartite graph H there is a digraph
H 0 such that H-Retraction and H 0 -Retraction are polynomially equivalent. Finally it is observed
that H 0 is a core and thus H 0 -Retraction and (H 0 , c)-Colouring are polynomially equivalent. We
adapt this proof to the case of 2-tropical graph homomorphism problems.
The construction of H 0 from H in [22] is through the use of so-called zig-zag paths. In our case,
we replace these zig-zag paths by specific 2-coloured graphs that play the same role. This will allow
us to construct a 2-coloured graph H 0 from a bipartite graph H such that H-Retraction and H 0 Retraction are polynomially equivalent. Our paths will have black vertices denoted by B and white
vertices denoted by W . Hence the path W B 4 W 4 B consists of one white vertex, four black vertices, four
white vertices and a black vertex. The maximal monochromatic subpaths are called runs. Thus the
above path is the concatenation of four runs: the first and last of length 1, the middle two of length 4.
Given an odd integer `, we construct a path P consisting of ` runs. The first and the last run each
consist of a single white vertex. The interior runs are of length four. We denote that last (rightmost)
vertex of P by 0. From P we construct ` − 2 paths P1 , . . . , P`−2 . Path Pi (i = 1, 2, . . . , ` − 2) is obtained
from P by replacing the ith run of length four with a run of length 2. We denote the rightmost vertex of
Pi by i.
Similarly, for an even integer k, we construct a second family of paths Q and Qj , (j = 1, 2, . . . , k − 2).
The leftmost vertex of Q is 1 and the leftmost vertex of Qj is j. The paths are described below:
P

:= W |B 4 W 4 ·{z
· · W 4 B }4 W

Pi

4
:= W |B 4 ·{z
· · W }4 B 2 W
· · · B }4 W
| {z

(i odd)

:= W |B 4 ·{z
· · B }4 W 2 |B ·{z
· · B} W

(i even)

Q

`−2

i−1

Pi

i−1

`−i−2
4
4

:= W |B 4 W 4 ·{z
· · B 4 W }4 B
k−2

Qj

4
:= W |B 4 ·{z
· · W }4 B 2 W
· · W }4 B
| ·{z

(j odd)

:= W |B 4 ·{z
· · B }4 W 2 B
··W}B
| ·{z

(j even)

j−1

Qj

j−1

`−i−2

k−j−2
4
4
k−j−2

We observe the following (c.f. page 156 of [22]):
1. The paths P and Pi (i = 1, 2, . . . ` − 2) each admit a homomorphism onto a 2-colour forcing path
of length 2` − 1, (that is, a path consisting of one run of length 1, ` − 2 runs each of length 2 and
a final run of length 1: W BBW W B · · · W ).
2. The paths Q and Qj (j = 1, 2, . . . k − 2) each admit a homomorphism onto a 2-colour forcing path
of length 2k − 1.
3. Pi → Pi0 implies i = i0 .
4. Qj → Qj 0 implies j = j 0 .
5. P → Pi for all i.
6. Q → Qj for all j.
7. if X is a 2-tropical graph and x is a vertex of X such that f : X → Pi and f 0 : X → Pi0 for i 6= i0
with f (x) = i and f 0 (x) = i0 , then there is a homomorphism F : X → P with F (x) = 0.
9

8. if Y is a 2-tropical graph and y is a vertex of Y such that f : Y → Qj and f 0 : Y → Qj 0 for j 6= j 0
with f (y) = j and f 0 (y) = j 0 , then there is a homomorphism F : Y → Q with F (y) = 1.
We note that 2-colour forcing paths in 2-tropical graphs can be used to define height analogously
to height in directed acyclic graphs. More precisely, suppose G is a connected 2-tropical graph that
admits a homomorphism onto a 2-colour forcing path, say F P , of even length. Let the vertices of F P be
h0 , h1 , . . . , h2t . Observe that each vertex in the path has at most one white neighbour and at most one
black neighbour. Thus once a single vertex u in G is mapped to F P , the image of each neighbour of u is
uniquely determined and by connectivity, the image of all vertices is uniquely determined. In particular,
as G maps onto F P , there is exactly one homomorphism of G to the path. (More precisely, if the path
has length congruent to 0 modulo 4, there is an automorphism that reverses the path. In this case there
onto
are two homomorphisms that are equivalent up to the reversing.) We then observe that if g : G → F P ,
h : H → F P , and f : G → H, then for all vertices u ∈ V (G), g(u) = h(f (u)). This allows us to define
the height of u ∈ V (G) to be hi when g(u) = hi . Specifically, vertices at height hi in G must map to
vertices at height hi in H.
For each problem T in CSP there is a bipartite graph H such that T -CSP and H-Retraction
are equivalent [20, 22]. Let H be a bipartite graph with parts (A, B), with A = {a1 , . . . , a|A| } and
B = {b1 , . . . , b|B| }. Let ` (respectively k) be the smallest odd (respectively even) integer greater than or
equal to |A| (respectively |B|). To each vertex ai ∈ A attach a copy of Pi identifying i in Pi with ai in
A. Colour all original vertex of H white. To each vertex bj ∈ B attach a copy of Qj identifying j in Qj
with bj in B. Call the resulting 2-tropical graph (H 0 , c). See Figure 1 for an illustration.
Let G be an instance of H-Retraction. In particular, we may assume without loss of generality
that H is a subgraph of G, G is connected, and G is bipartite. We colour the original vertices of G white.
Let (A0 , B 0 ) be the partite classes of G where A ⊆ A0 and B ⊆ B 0 . To each vertex v of A0 \A, we attach
a copy of P , identifying v and 0. To the vertices of A ∪ B, we attach paths Pi and Qj as described above
to create a copy of H 0 . Call the resulting 2-tropical graph (G0 , c0 ). In particular, note that (G0 , c0 ) and
(H 0 , c0 ) both map onto a 2-colour forcing path of length 2` + 2k − 1. The (original) vertices of G and
H are at height 2` − 1 and 2` for colour classes A and B respectively. In particular, by the eight above
properties, under any homomorphism f : G0 → H 0 the restriction of f to G must map onto H with
vertices in A0 mapping to A and vertices in B 0 mapping to B.
Using the eight properties of the paths above and following the proof of Theorem 5.14 in [22], we
conclude that G is a YES instance of H-Retraction if and only if (G0 , c0 ) is a YES instance of (H 0 , c)Retraction.
On the other hand, let (G0 , c0 ) be an instance of (H 0 , c)-Retraction. We sketch the proof from [22].
We observe that (G0 , c0 ) must map to a 2-colour forcing path of length 2` + 2k − 1. The two levels of G0
corresponding to H induce a bipartite graph (with white vertices) which we call G. The components of
G0 − E(G) fall into two types: those which map to lower levels and those that map to higher levels than
G. Let Ct be a component that maps to a lower level. After required identifications we may assume Ct
contains only one vertex from G (say v) and Ct must map to some Pi . If Pi is the unique Pi path to
which Ct maps, then we modify G0 by identifying v and i. Otherwise, Ct maps to two paths and (by the
properties 5–8) hence to all paths. The resulting graph (G0 , c0 ) retracts to (H 0 , c) if and only if G retracts
to H.
H

···

···

···

..
.
..
.

..
.

···

···

Figure 1: Construction of a 2-tropical target H 0 from a H-Retraction problem.

10

4

Minimal graphs H for NP-complete H-List-Colouring

Recall the dichotomy theorem for list homomorphism problems of Feder, Hell and Huang (Theorem 1.5):
H-List-Colouring is polynomial-time solvable if H is bipartite and its complement is a circular arc
graph, otherwise NP-complete. Alternatively, the latter class of graphs was characterized by Trotter and
Moore [29] in terms of seven families of forbidden induced subgraphs: six infinite ones and a finite one.
See their descriptions in Table 1, as reproduced from [17]. To concisely describe these seven families, they
employ the following notation: Let F = {Si : 1 ≤ i ≤ k} be a family of subsets of {1, 2, . . . , `}. Define
HF to be the bipartite graph (X, Y ) with X = {x1 , x2 , . . . , x` } and Y = {y1 , y2 , . . . , yk } such that xi yj
is an edge if and only if i ∈ Sj . The families C, T , W, D, M, N and G in Table 1 are defined in this way.
Note that the graph Ci in C is the cycle of length i. See Figure 2 for an illustration of the other families
from Table 1. Also note that G1 , which is a claw where each edge is subdivided twice, is the only tree in
the table.
Given the above characterization, we can reformulate Theorem 1.5 as follows.
Theorem 4.1 (Restatement of Theorem 1.5, Feder, Hell and Huang [17]). If H contains one of the
graphs defined in Table 1 as an induced subgraph, then H-List-Colouring is NP-complete. Otherwise,
H-List-Colouring is polynomial-time solvable.
C6 = {{1, 2}, {2, 3}, {3, 1}}
C8 = {{1, 2}, {2, 3}, {3, 4}, {4, 1}}
C10 = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 1}}
...
T1 = {{1, 2}, {2, 3}, {3, 4}, {2, 3, 5}, {5}}
T2 = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {2, 3, 4, 6}{6}}
T3 = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {2, 3, 4, 5, 7}, {7}}
...
W1 = {{1, 2}, {2, 3}, {1, 2, 4}, {2, 3, 4}, {4}}
W2 = {{1, 2}, {2, 3}, {3, 4}, {1, 2, 3, 5}, {2, 3, 4, 5}, {5}}
W3 = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {1, 2, 3, 4, 6}, {2, 3, 4, 5, 6}, {6}}
...
D1 = {{1, 2, 5}, {2, 3, 5}, {3}, {4, 5}, {2, 3, 4, 5}}
D2 = {{1, 2, 6}, {2, 3, 6}, {3, 4, 6}, {4}, {5, 6}, {2, 3, 4, 5, 6}}
D3 = {{1, 2, 7}, {2, 3, 7}, {3, 4, 7}, {4, 5, 7}, {5}, {6, 7}, {2, 3, 4, 5, 6, 7}}
...
M1 = {{1, 2, 3, 4, 5}, {1, 2, 3}, {1}, {1, 2, 4, 6}, {2, 4}, {2, 5}}
M2 = {{1, 2, 3, 4, 5, 6, 7}, {1, 2, 3, 4, 5}, {1, 2, 3}, {1}, {1, 2, 3, 4, 6, 8}, {1, 2, 4, 6}, {2, 4}, {2, 7}}
M3 = {{1, 2, 3, 4, 5, 6, 7, 8, 9}, {1, 2, 3, 4, 5, 6, 7}, {1, 2, 3, 4, 5}, {1, 2, 3}, {1}, {1, 2, 3, 4, 5, 6, 8, 10},
{1, 2, 3, 4, 6, 8}, {1, 2, 4, 6}, {2, 4}, {2, 9}}
...
N1 = {{1, 2, 3}, {1}, {1, 2, 4, 6}, {2, 4}, {2, 5}, {6}}
N2 = {{1, 2, 3, 4, 5}, {1, 2, 3}, {1}, {1, 2, 3, 4, 6, 8}, {1, 2, 4, 6}, {2, 4}, {2, 7}, {8}}
N3 = {{1, 2, 3, 4, 5, 6, 7}, {1, 2, 3, 4, 5}, {1, 2, 3}, {1}, {1, 2, 3, 4, 5, 6, 8, 10}, {1, 2, 3, 4, 6, 8},
{1, 2, 4, 6}, {2, 4}, {2, 9}, {10}}
...
G1 = {{1, 3, 5}, {1, 2}, {3, 4}, {5, 6}}
G2 = {{1}, {1, 2, 3, 4}, {2, 4, 5}, {2, 3, 6}}
G3 = {{1, 2}, {3, 4}, {5}, {1, 2, 3}, {1, 3, 5}}
Table 1: Six infinite families C, T , W, D, M, N and family G of size 3 of forbidden induced subgraphs
for polynomial-time H-List-Colouring problems.
In this section, we first turn our attention to the family of even cycles of length at least 6. We show
that C2k -Tropical-Colouring is polynomial-time solvable for any k ≤ 6. On the other hand, for any
k ≥ 24, C2k -Tropical-Colouring is NP-complete. We then prove that for all other minimal graphs
H described in Table 1, H-Tropical-Colouring is polynomial-time solvable.

11

yi+4

xi+4
yi+3
yi+3
yi+1

xi+2

yi+2

x2

···

xi

xi+3

x1
x3

y1

x1

xi+1

···

xi+2 yi+2

x2

yi+1

y2

y1

xi+1

yi

(a) Ti

(b) Wi

yi+4

x1

x2

y1

x2i+4 yi+3

yi+3

x1

y2i+3 y2i+4
···

xi+1 xi+2
yi+2

···

y2

xi+3

yi+2

yi+1

···

x2

x3

yi+1

xi+4

x4

···

x2i

y2

y1

x2i+1 x
2i+2 x2i+3

···

(c) Di

(d) Mi

x2i+4 yi+2

y2i+2 y2i+3
···

y2i+4
x1

x2

yi+1

x3

x4

x2i

···

x2i+1 x
2i+2 x2i+3

···
yi

y1
(e) Ni

x2

x4

y2

y3
y1

x1

x3
x5

x6
y4

x2
y2

x3

x5

x4

y3

y3

y2

x5

x4

x3

y5

y4

x1

y4

x1

x6

y1

x2

y1

(f) G1

(g) G2

(h) G3

Figure 2: Illustration of the families defined in Table 1 (except the cycles in C).

4.1

Polynomial-time cases for even cycles

We now prove that the tropical homomorphism problems for small even cycles are polynomial-time
solvable.
Theorem 4.2. For each integer k with 2 ≤ k ≤ 6, C2k -Tropical-Colouring is polynomial-time
solvable.
Proof. Since C4 -List-Colouring is polynomial-time solvable, C4 -Tropical-Colouring is polynomial12

time solvable.
We will consider all cases k ∈ {3, 4, 5, 6} separately. But in each of those cases we note that if (C2k , c)
is not a core, then the core is path, and since the Pk -List-Colouring is polynomial-time solvable for
any k ≥ 1, C2k -Tropical-Colouring would also be polynomial-time solvable. Hence in the rest of
the proof we always assume (C2k , c) is a core. Furthermore, by Proposition 2.3, we can assume that the
colour sets of c in X and Y are disjoint.
First, assume k = 3. There are three vertices in each part of the bipartition of C6 . If one vertex
is coloured with a colour not present anywhere else in the part, Lemma 2.12 implies again that (C6 , c)Colouring is polynomial-time solvable. Hence, we can assume that each part of the bipartition is
monochromatic. But then (C6 , c) is not a core, a contradiction with our assumption.
Suppose k = 4. There are four vertices in each part of the bipartition (X, Y ) of C8 . If there is a
vertex that, in c, is the only one coloured with its colour, since Pk -List-Colouring is polynomial-time
solvable for any k ≥ 1, by Lemma 2.12 (C8 , c)-Colouring is polynomial-time solvable. Hence we may
assume that each colour appears at least twice, in particular each part of the bipartition is coloured with
either one or two colours. If some part, say X, is coloured with only one colour (say Blue) then (C8 , c) is
not a core which again contradicts our assumption. Hence, in each part, there are exactly two vertices of
each colour. In this case we can use Lemma 2.6 with S1 , S2 , S3 and S4 being the four sets of two vertices
with the same colour. It follows that (C8 , c)-Colouring is polynomial-time solvable.
Assume that k = 5, and let c be a vertex-colouring of C10 . By similar arguments as in the proof of
Theorems 5.1 and 6.1, using Lemma 2.12 and the fact that (H, c) should not be homomorphic to a P2 or P3 -subgraph, each part of the bipartition (X, Y ) contains exactly two vertices of one colour and three
vertices of another colour, say X has three vertices coloured 1 and two vertices coloured 2, and Y has
three vertices coloured a and two vertices coloured b.
The cyclic order of the colours of X can be either 1−1−1−2−2 or 1−1−2−1−2 (up to permutation
of colours and other symmetries). If this order is 1 − 1 − 1 − 2 − 2, then the vertex of Y adjacent to
the two vertices coloured 2 satisfies the hypothesis of Lemma 2.12 and hence (C10 , c)-Colouring is
polynomial-time solvable. The same argument can be applied to Y , hence the cyclic order of the colours
of Y is a − a − b − a − b.
Hence, there is a unique vertex y of Y whose two neighbours are coloured 1. If c(y) = b, then the
second vertex of Y coloured b is in the centre of a 3-vertex path coloured 1 − b − 2 that satisfies the
hypothesis of Lemma 2.12, hence (C10 , c)-Colouring is polynomial-time solvable. Therefore, we have
c(y) = a. By the same argument, the unique vertex of X adjacent to two vertices of Y coloured a must be
coloured 1. Therefore, up to symmetries c is one of the three colourings 1 − a − 1 − a − 2 − b − 1 − a − 2 − b,
1 − a − 1 − b − 2 − a − 1 − a − 2 − b and 1 − a − 1 − b − 2 − a − 1 − b − 2 − a (in the cyclic order).
We are going to use the Lemma 2.6 to conclude the case k = 5. In a homomorphism to (C10 , c), a
vertex coloured 2 or b can only be mapped to the two vertices in (C10 , c) of the corresponding colour.
A vertex v coloured 1 adjacent to at least one vertex coloured b or a vertex coloured a adjacent to at
least one vertex coloured 2 also can only be mapped to two vertices of (C10 , c) (the ones having the
same properties as v). However, a vertex coloured 1 all whose neighbours are coloured a can be mapped
to three different vertices in (C10 , c) (say x1 , x2 , x3 , the vertices coloured 1, that all have a neighbour
coloured a). But at least one of x1 , x2 , x3 , say x1 , has a common neighbour coloured a with one of the
two other vertices (say x2 ). Therefore, if there is a homomorphism h of some tropical graph (G, c1 ) to
(C10 , c) mapping a vertex v of G coloured 1 all whose neighbours are coloured a to x1 , we can modify
h so that v is mapped to x2 instead. In other words, there is a homomorphism of (G, c1 ) to (C10 , c)
where none of the vertices coloured 1 all whose neighbours are coloured a is mapped to x1 . Therefore
such vertices have two possible targets: x2 and x3 . The same is true for vertices coloured a all whose
neighbours are coloured 1. Thus, (C10 , c) satisfies the hypothesis of Lemma 2.6 and (C10 , c)-Colouring
is polynomial-time solvable.
Finally, assume now that k = 6. Again, using Lemma 2.12, we can assume than each part of the
bipartition has at most three colours, and each colour appears at least twice. Furthermore, if there are
exactly three colours in each part, each colour appears exactly twice and hence (C12 , c)-Colouring is
polynomial-time solvable by Lemma 2.6. If one part of the bipartition has one colour and the other has
at most two colours, then (C12 , c) would not be a core. Therefore, the numbers of colours of the parts in
the bipartition are either one and three, two and three, or two and two.

13

Assume that one part, say X, is monochromatic (say Red) and the other, Y , has three colours (thus
two vertices of each colour). For the graph to be a core and not satisfy Lemma 2.12, the three colours
of Y must form the cyclic pattern x − y − z − x − y − z. In this case, considering any vertex v of colour
Red in an input tropical graph (G, c1 ), in any homomorphism (G, c1 ) → (C12 , c), all the neighbours of
v with the same colour must be identified. Furthermore, no Red vertex in (G, c1 ) can have neighbours
of three distinct colours. Therefore, the mapping of each connected component is forced after making a
choice for one vertex. Since there are two choices per vertex, we have a polynomial-time algorithm for
(C12 , c)-Colouring.
Assume now that one part, say X, contains two colours (a and b) and the other, Y , contains three
colours (x, y and z). Note that there are exactly two vertices of each colour in Y . We are going to
use Lemma 2.6 to conclude this case. A vertex of some input graph (G, c1 ) coloured x, y or z can only
be mapped to two possible vertices in (C12 , c). A vertex of (G, c1 ) coloured a or b (say a) and having
all its neighbours of the same colour, say x, might be mapped to more than two vertices of (C12 , c).
However, once again, there are always two of these vertices that, together, are adjacent to all the vertices
of colour x (indeed, there are only two vertices of colour x). These two vertices are the designated targets
for Lemma 2.6. A vertex coloured a (or b) with two different colours in its neighbourhood can only be
mapped to two possible vertices if there is no pattern x − a − y − a − x − a − y in the graph (up to
permutation of colours). Hence, if there is no such pattern in the graph (up to permutation of colours),
(C12 , c) satisfies the hypothesis of Lemma 2.6 and (C12 , c)-Colouring is polynomial-time solvable. On
the other hand, if there is a pattern x − a − y − a − x − a − y in the graph, then there is a unique path
coloured a − x − b or a − y − b in the graph and, by Lemma 2.12, (C12 , c)-Colouring is polynomial-time
solvable as well.
Therefore, we are left to consider the cases where there are exactly two colours in each part. We
assume first that there are two vertices coloured a and four vertices coloured b in one part, say X. If the
neighbours of vertices of colour a all have the same colour, say x, then (C12 , c) is not a core because it
can be mapped to its sub-path coloured a − x − b − y. We suppose without loss of generality that the
coloured cycle contains a path coloured y − a − x − b. Then, if there is no other path coloured y − a − x − b,
by Lemma 2.12 (C12 , c)-Colouring is polynomial-time solvable. Therefore, there is another such path
in (C12 , c). If this other path is part of a path x − a − y − a − x, then the problem is polynomial-time
solvable by applying Lemma 2.12 to the star a − y − a. Up to symmetry, we are left with two cases:
y − a − x − b − . − b − . − a − . − b − . − b or y − a − x − b − . − b − . − b − . − a − . − b (where a dot could
be colour x or y). The first case must be y − a − x − b − . − b − y − a − x − b − . − b, because otherwise,
(C12 , c) is not a core. Any placement of the remaining x’s and y’s yields a polynomial-time solvable case
using Lemma 2.8. Similarly, the second case must be y − a − x − b − . − b − . − b − y − a − x − b. Then,
(C12 , c)-Colouring is polynomial-time solvable because of Lemma 2.9, with a − x − b − y − a as forcing
set and x − b − . − b − . − b − y, which contains no vertex coloured a, as the other set.
Finally, we can assume, without loss of generality, that there are exactly three vertices for each of
the two colours in each part. There are three possible configurations in each part: a − a − a − b − b − b,
a − a − b − b − a − b or a − b − a − b − a − b, up to permutations of colours. If one part of the bipartition is in
the first configuration, then, either we have the pattern a − x − b or a − y − b that satisfies the hypothesis
of Lemma 2.12, or we have two paths a − x − b, in which case, there is a unique path a − y − a or b − y − b
which satisfies the hypothesis of Lemma 2.12. Suppose some part of the bipartition is in the second case.
Then, if we have the pattern a−x−a−.−b−x−b−.−a−.−b−., there is a unique path a−x−b satisfying
the hypothesis of Lemma 2.12. Otherwise, we have the pattern a − x − a − . − b − y − b − . − a − . − b − ., in
which case we can apply Lemma 2.6 in a similar way as for C10 . Therefore, both parts of the bipartition
must be in the third configuration. But then every vertex is a forcing vertex and we can apply Lemma 2.5.
This completes the proof.

4.2

NP-completeness results for even cycles

We now show that C2k -Tropical-Colouring is NP-complete whenever k ≥ 24. We present a proof
using a specific 4-tropical 48-cycle. The proof holds similarly for any larger even cycle. It also works
similarly for some 3-tropical cycles C2k for k ≥ 24 and for 2-tropical cycles C2k for k ≥ 27.
We use the colour set {G, B, R, Y } (for Green, Blue, Red and Yellow).
We define Px,y to be a tropical path of length 8, with vertices x = x0 , x1 , . . . , x7 , x8 = y where
{c(x), c(y)} = {G, B}, c(x5 ) = R and all others are coloured Yellow. Thus, Px,y represents one of the two

14

non-isomorphic tropical graphs from Figure 3. The distance of the only vertex of colour R from the two
ends defines an orientation from one end to another. Thus, in our figures, an arc between two vertices u
and v is a Puv path.
G

Y

Y

Y

Y

R

Y

Y

x

x1

x2

x3

x4

x5

x6

x7

B

Y

Y

Y

Y

R

Y

Y

x

x1

x2

x3

x4

x5

x6

x7

B
y
G
y

Figure 3: The two non-isomorphic graphs of type Pxy .
Similarly, Qz,t is defined to be a tropical path of length 10 with vertices z = z0 , z1 , . . . , z9 , z10 = t where
{c(z), c(t)} = {G, B}, c(z5 ) = R and all others are coloured Yellow. See Figure 4 for an illustration. In
this case, as the only vertex of colour R is at the same distance from both ends, the two possible colourings
of the end-vertices correspond to isomorphic graphs. Hence, in our figures, a dotted edge will be used to
represent a Q-type path between two vertices.
G

Y

Y

Y

Y

R

Y

Y

Y

Y

B

z

z1

z2

z3

z4

z5

z6

z7

z8

z9

t

Figure 4: The Q-type path Qz,t .
The following lemma is easy to observe.
Lemma 4.3. The following is true.
1. Px,y admits a tropical homomorphism to Pu,v if and only if c(x) = c(u) and c(y) = c(v).
2. Qz,t admits a tropical homomorphism to Pu,v both in the case where c(z) = c(u) and c(t) = c(v),
and in the case where c(z) = c(v) and c(t) = c(u).
By Lemma 4.3, in our abbreviated notation of arcs and dotted edges, a dotted edge can map to a
dotted edge or to an arc as long as the colours of the end-vertices are preserved. However, to map an arc
to another arc, not only the colours of the end-vertices must be preserved, but also the direction of the
arc.
With our notation, the tropical directed 6-cycle of Figure 5 corresponds to a 4-tropical 48-cycle,
(C48 , c).
b1
B

g2

b2

g1

G

G

B

B
G
g0

b0

Figure 5: A short representation of the 4-tropical 48-cycle (C48 , c).
Our aim is to show that NAE 3-SAT reduces (in polynomial time) to (C48 , c)-Colouring.
Theorem 4.4. For any k ≥ 24, C2k -Tropical-Colouring is NP-complete.
Proof. We prove the statement when k = 24 and observe that the same reduction holds for any k ≥ 24.
Indeed, one can make Px,y and Qz,t longer while still satisfying Lemma 4.3.
(C48 , c)-Colouring is clearly in NP. To show NP-hardness, we show that NAE 3-SAT can be reduced
in polynomial-time to (C48 , c)-Colouring.
15

Let (X, C) be an instance of NAE 3-SAT. To partition X into two parts, it is enough to decide, for
each pair of elements of X, whether they are in a same part or not. Thus, we are expected to define a
binary relation among variables which satisfies the following conditions.
1. Xp ∼ Xq ∧ Xq ∼ Xr ⇒ Xq ∼ Xr (Partition)
2. Xp  Xq ∧ Xq  Xr ⇒ Xp ∼ Xr (Partition into two parts)
To build our gadget, we start with a partial gadget associated to each pair of variables of X. To each
pair xi , xj ∈ X, we associate the 4-tropical 6-cycle (Cxi xj , c) of Figure 6. Here, UG (coloured Green) is a
common vertex of all such cycles, but all other vertices are distinct.
b1xi xj
gx2i xj

gx1i xj

B
G

G

B

B

b2xi xj

b0xi xj

G
UG

Figure 6: (Cxi xj , c)
We are interested in possible mappings of this partial gadget into our tropical 48-cycle, (C48 , c) of
Figure 5. By the symmetries of (C48 , c), we assume, without loss of generality, that UG maps to g0 .
Having this assumed, we observe the following crucial fact.
Claim 4.5. There are exactly two possible homomorphisms of (Cxi xj , c) to (C48 , c).
1. A mapping σ given by σ(UG ) = g0 , σ(b0xi xj ) = b0 , σ(gx1i xj ) = g1 , σ(b1xi xj ) = b1 , σ(gx2i xj ) = g2 and
σ(b2xi xj ) = b2
2. A mapping ρ give by ρ(UG ) = g0 , ρ(b0xi xj ) = b0 , ρ(gx1i xj ) = g1 , ρ(b1xi xj ) = b0 , ρ(gx2i xj ) = g1 and
ρ(b2xi xj ) = b0
The main idea of our reduction lies in Claim 4.5. After completing the description of our gadgets,
we will have a 4-tropical graph containing a copy of Cxi xj for each pair xi , xj of variables. If we find a
homomorphism of this graph to (C48 , c), then its restriction to Cxi xj is either a mapping of type σ, or of
type ρ. A σ-mapping would correspond to assigning xi and xj to two different parts, and a ρ-mapping
would correspond to assigning them to a same part of a partition of X.
Observation 4.6. It is never possible to map b2xi xj to b1 or to map b1xi xj to b2 .
To enforce the two conditions, partitioning X into two parts by a binary relation, we add more
structures. Consider the three partial gadgets (Cxp xq , c), (Cxq xr , c) and (Cxp xr , c). Considering b1xp xq
of (Cxp xq , c), we choose vertices b2xp xr and b2xq xr from (Cxp xr , c) and (Cxq xr , c) respectively, and connect
them by a tree as in Figure 7. The internal vertices of these trees are all new and distinct.
1
B bxp xq

G

B

G

B
b2xq xr

2
B bxp xr

Figure 7: Tree connecting b1xp xq , b2xp xr and b2xq xr
We build similar structures on (b1xp xr , b2xq xr , b2xp xq ) and on (b1xq xr , b2xp xq , b2xp xr ), where the order corresponds to the structure. Let (Cxp xq xr , c) be the resulting partial gadget (see Figure 8).
16

b1xp xq
B
G
b2xp xq

G

B

B

G
B

G

G
G
UG

G

B

b2xp xr

B

B

G

G

G

B

B
b1xq xr

B
G

B

B

G

b2xq xr

G

b1xp xr

Figure 8: Cxp xq xr
Claim 4.7. In any mapping of (Cxp xq xr , c) to (C48 , c), an odd number of (Cxi xj , c) is mapped to (C48 , c)
by a ρ-mapping. Furthermore, for any choice of an odd number of (Cxi xj , c) (that is either one or all
three of them), there exists a mapping of (Cxp xq xr , c) to (C48 , c) which induces a ρ-mapping exactly on
our choice.
Proof of claim Indeed, each (Cxi xj , c) can be mapped to (C48 , c) only by σ or ρ, which implies that
there are eight ways to map the union of (Cxp xq , c), (Cxp xr , c) and (Cxq xr , c) to (C48 , c). Of these eight
ways, four map an odd number of (Cxi xj , c) to (C48 , c) by a ρ-mapping. The four remaining ways are
to map all (Cxi xj , c) to (C48 , c) by a σ-mapping, or to choose one of them to map by a σ-mapping and
to map the two others by a ρ-mapping. One can check easily that the union of (Cxp xq , c), (Cxp xr , c),
(Cxq xr , c) and the tree of Figure 7 has six ways to be mapped to (C48 , c). Indeed, it is no longer possible
to map all (Cxi xj , c) by σ nor to map (Cxp xr , c) by σ and (Cxp xq , c) and (Cxq xr , c) by ρ. By symmetry,
this implies Claim 4.7. ()
Finally, to complete the gadget, what remains is to forbid the possibility of a ρ-mapping for all three
of (Cxp xq , c), (Cxp xr , c) and (Cxq xr , c) in the case where (xp xq xr ) is a clause in C. This is done by adding
a b1xp xq b2xq xr -path shown in Figure 9.
B

G

B

b1xp xq

G

B
b2xq xr

Figure 9: Partial clause gadget.
Let f (X, C) the final gadget we have just built. Assuming that there are v variables and c clauses,
the 4-tropical graph f (X, C) has 1 + 53 × v 2 + 132 × v 3 + 33 × c vertices. To complete our proof we want
to prove the following.
(X, C) is a YES instance of NAE 3-SAT if and only if the 4-tropical graph f (X, C) admits a homomorphism to (C48 , c).
It follows directly form our construction that if f (X, C) → (C48 , c), then (X, C) is a YES instance of

17

NAE 3-SAT. We need to show that if (X, C) is a YES instance, then there exists a homomorphism of
f (X, C) to (C48 , c).
Let (X, C) be a YES instance of NAE 3-SAT. There exists a partition p : X → {A, B} such that
every clause in C is not fully included in A or B. We build a homomorphism of f (X, C) to (C48 , c) in the
following way. UG is mapped to g0 . For each pair of variables xi , xj ∈ X, we map Cxi xj by a ρ-mapping if
and only if p(xi ) = p(xj ), and by a σ-mapping otherwise. For every triple of variable xp , xq , xr ∈ X, there
is an odd number of pairs xi , xj of variables in {xp , xq , xr } such that p(xi ) = p(xj ). It follows from Claim
4.7 that one can extend the mapping to any Cxp xq xr . Moreover, as two such structures only intersect on
Cxi xj , we can extend the mapping to every Cxp xq xr . It only remains to map the b1xp xq b2xq xr -path added
for the clause, shown in Figure 9. If (xp , xq , xr ) is a clause in C, then p(xp ) 6= p(xq ) or p(xq ) 6= p(xr ).
It follows that Cxp xq or Cxq xr is mapped by a σ-mapping, in which case the b1xp xq b2xq xr -path shown in
Figure 9 can also be mapped. We have shown that there is a homomorphism of f (X, C) to (C48 , c). This
concludes the proof.
We observe that the proof could be slightly modified to obtain variations of Theorem 4.4.
Remark 4.8.
1. In the reduction from Theorem 4.4, Red vertices are never in the same part of the bipartition as
Blue and Green vertices. It follows that one could colour every Red vertex Blue, and Theorem 4.4
would still hold, for 3-tropical cycles.
2. The idea of this proof can also be extended for a 2-tropical 54-cycle. To do this we first insert a Red
vertex between x5 and x6 in Pxy and a Red vertex between z5 and z6 in Qzt . We observe that the
proof follows similarly. However, in this case, all blue vertices are in one part and all green vertices
are on the other part of the bipartition. Thus, as in the previous claim, we can remove two colours
now and use the natural bipartition to distinguish two sets of colours for each colour class.

4.3

Other families of minimal graphs

Next, we show that for each of the minimal graphs H from Table 1 (other than even cycles) that make
H-List-Colouring NP-complete, H-Tropical-Colouring is polynomial-time solvable.
Theorem 4.9. f For every graph H belonging to one of the six families T , W, D, M, N and G described
in Table 1, H-Tropical-Colouring is polynomial-time solvable.
Proof. We assume for contradiction, that for some integer i and a family F among T , W, D, M, N and
G, there is a problem (Fi , c)-Colouring that is not polynomial-time solvable.
Family T . Suppose xi+4 is coloured m. Suppose yi+3 is coloured a, a 6= m by Proposition 2.3. Then, yi+4
cannot be coloured a (otherwise it can be folded onto yi+3 ), so it is coloured b. Because of Lemma 2.12,
there must be another P3 coloured amb on the graph, but for the graph to be a core, the vertex coloured
m of this P3 must not be adjacent to yi+3 . However, note that yi+3 is adjacent to every vertex of X
except for x1 and xi+3 , both of which have degree 1 and cannot create a 3-vertex path coloured a-m-b.
Family W. Now, we consider Wi . We try to find a colouring c of Wi such that (Wi , c)-Colouring
is not polynomial-time solvable. Suppose yi+2 is coloured with colour a. Suppose xi+3 is coloured m.
yi+4 cannot be coloured a, otherwise it can be folded onto yi+2 , so we may assume it is coloured b. yi+3
cannot be coloured b, for otherwise yi+4 can be folded onto it, so it is coloured a or d. Suppose first that
it is coloured d. By Lemma 2.12, there is another vertex coloured a, and the only one which could not
be folded onto yi+2 is yi+1 , so it must be coloured a. Similarly, y1 is coloured d. By Lemma 2.12, there
is another edge besides xi+3 , yi+4 with endpoints coloured m and b, but for the graph to be a core, the
edge xi+3 yi+4 must not be able to fold onto it. However, it is easily verified that this is impossible. So,
we must assume that yi+3 is coloured a. There is no other vertex in Y coloured a, otherwise it can be
folded onto yi+2 or yi+3 . We can assume, without loss of generality, that a connected subgraph of the
source graph, coloured only with m and b, and with only vertices of colour a at distance 1, will be sent
to xi+3 and yi+4 . Knowing this, we can contract each such subgraph to a single vertex, coloured with
a new colour ω, and similarly replace xi+3 and yi+4 by a single vertex coloured ω, adjacent to yi+2 and
yi+3 . There will be a homomorphism between the source graph and (Wi , c) if and only if there is one

18

after such transformation. However, the graph obtained after such transformation will not contain any
induced subgraph from the table above, which yields a contradiction.
Family D. Now, consider Di and a colouring c such that (Di , c)-Colouring is not polynomial-time
solvable. Suppose xi+4 is coloured m and yi+4 is coloured a. By Lemma 2.12, there is another vertex
coloured a. We may assume that y1 is such a vertex because it is the only one that cannot be folded on
yi+4 . Then, x1 cannot have colour m, for otherwise it can be folded onto xi+4 , so it is coloured l. By
Lemma 2.12, there is another vertex in X coloured l, say v. y1 x1 can be folded onto yi+4 v, which yields
a contradiction.
Family M. Now, consider Mi and a colouring c such that (Mi , c)-Colouring is not polynomial-time
solvable. Suppose x2 is coloured m and yi+2 is coloured a. By Lemma 2.12, there is another vertex in
X coloured m. The only vertex which can be coloured m without being able to be folded onto x2 is x1 .
This is because yi+2 is adjacent only to x1 and x2 is adjacent to every vertex in Y except for yi+2 . So we
may assume x1 is coloured m. By Lemma 2.12, there is another vertex in Y coloured a, say v. x1 yi+2
can be folded onto x2 v since x2 is adjacent to every vertex in Y except yi+2 , which yields a contradiction.
Family N . Now, consider Ni and a colouring c such that (Ni , c)-Colouring is not polynomial-time
solvable. Suppose x2 is coloured m. For 3 ≤ j ≤ 2i + 3, xj cannot be coloured m, for otherwise it can
be folded onto x2 since N (xj ) ⊂ N (X2 ). By Lemma 2.12, x1 or x2i+4 must be coloured m. Both x1 and
x2i+4 have a neighbour of degree 1 (namely, yi+1 and y2i+4 , respectively), which are the two only vertices
in Y not adjacent to x2 . By Lemma 2.12, neither x1 yi+1 nor x2i+4 y2i+4 can be an edge of unique colour.
Either exactly one of them is coloured ma and a neighbour v of x2 is coloured a, in which case the graph
is not a core because the edge can be folded on x2 v (since N (x1 ) \ {yi+1 } and N (x2i+4 ) \ {y2i+4 } are both
subsets of N (x2 )), or both x1 yi+1 and x2i+4 y2i+4 are coloured ma and the graph is not a core because
x2i+4 y2i+4 can be folded on x1 yi+1 since N (x2i+4 ) \ {y2i+4 } ⊂ N (x1 ), yielding a contradiction.
Family G. We try to find a colouring c of G1 such that (G1 , c)-Colouring is not polynomial-time
solvable. The colour of y1 is, say, a. By Lemma 2.12, colour a must be present somewhere else in Y . By
symmetry, we can assume y2 is coloured a. The two neighbours of y2 cannot be coloured with the same
colour, for otherwise we can fold x2 on x1 , implying that (G1 , c) is not a core, a contradiction. Without
loss of generality, x1 and x2 are coloured 1 and 2 respectively. By Lemma 2.12 applied to edge y2 x2 ,
there must be another edge coloured a2. However, if a neighbour of y1 is coloured 2, we can fold y2 x2
onto y1 and the graph is not a core, a contradiction. It follows that the other edge coloured a2 is either
y3 x4 or y4 x6 . By symmetry, we can assume that x4 is coloured 2 and y3 is coloured a. x3 cannot be
coloured 1 or 2, for otherwise (G1 , c) is not a core. Therefore, x3 is coloured with a third colour, say 3.
At this point, y1 x1 y2 x2 is coloured a1a2 and y1 x3 y3 x4 is coloured a3a2. Consider the colour of y4 . It
must be a by Lemma 2.12. There are only two uncoloured vertices, x5 and x6 , which must be coloured 1
and 3 by Lemma 2.12. The graph is not a core in both cases as we can either fold x6 y4 onto x1 y1 or the
edge x6 y4 onto x3 y1 , a contradiction.
Now, let c be a colouring of G2 such that (G2 , c)-Colouring is not polynomial-time solvable. Suppose
the vertex y2 is coloured with a. Then, y1 cannot be coloured a, for otherwise it can be folded onto y2 ,
which yields a contradiction. Therefore, y1 is coloured b. Because of Lemma 2.12, y3 and y4 must be
coloured a and b. By symmetry, we may assume y3 is coloured a and y4 is coloured b. Suppose x5 is
coloured m. Then x1 , x2 , x3 and x4 cannot be coloured m, for otherwise y3 x5 can be folded on y2 . Thus,
y3 x5 is the only edge coloured am. Lemma 2.12 yields a contradiction.
Now, let c be a colouring of G3 such that (G3 , c)-Colouring is not polynomial-time solvable. By
Lemma 2.12, there are at most two colours in each part of the bipartition. If x1 and x2 have the same
colour, x2 can be folded onto x1 , a contradiction. Similarly, if y1 and y4 have the same colour, y1 can be
folded onto y4 . Then x1 , y1 , x2 and y4 induce a complete bipartite graph with every colour of c, implying
that (G3 , c) is not a core, a contradiction.

5

Bipartite graphs of small order

In this section, we show that for each graph H of order at most 8, H-Tropical-Colouring is polynomialtime solvable. On the other hand, there is a graph H9 of order 9 such that H9 -Tropical-Colouring
is NP-complete.

19

Theorem 5.1. For any bipartite graph H of order at most 8, H-Tropical-Colouring is polynomialtime solvable.
Proof. It suffices to prove that for each bipartite graph H of order at most 8 and each colouring c of
H, (H, c)-Colouring is polynomial-time solvable. In fact, by Proposition 2.3 it suffices to show the
statement for colourings of H such that the colour sets in the two parts of the bipartition are disjoint. To
prove that (H, c)-Colouring is polynomial-time solvable it is enough to prove it for the core of S(H, c),
it is also enough to prove it for each connected component of (H, c). Thus in the rest of the proof we
always assume that (H, c) is connected core. Let (X, Y ) be the bipartition of H.
Since the only graphs of order at most 8 in the characterization of minimal NP-complete graphs H
with H-List-Colouring NP-complete are the cycles C6 and C8 [17] (see Table 1), by Theorem 1.5, if H
does not contain an induced 6-cycle or an induced 8-cycle, then H-List-Colouring is polynomial-time
solvable and therefore H-Tropical-Colouring is polynomial-time solvable. Therefore H contains an
induced 6-cycle or an induced 8-cycle.
If H contains an induced copy of C8 , then H is isomorphic to C8 itself and hence we are done by
Theorem 4.2. Therefore, we can assume that H contains an induced copy of C6 . Again by Theorem 4.2,
if H is isomorphic to C6 , we are done.
Now, assume that H is a bipartite graph of order 7 or 8 with an induced copy of C6 . If one part, say
X, is of order 3, then all its vertices belong to each 6-cycle of H. Hence, for each x ∈ X, (H − x)-ListColouring is polynomial-time solvable. Thus, if X is not monochromatic, we can apply Lemma 2.12
and (H, c)-Colouring is polynomial-time solvable. Therefore we may assume X is monochromatic, say
Blue. If Y contains at most two colours, then (H, c) contains as a subgraph the path on three vertices
where the central vertex is Blue and the other vertices are coloured with the colours of Y .But then
(H, c) maps to this subgraph and, therefore, it is not a core, a contradiction. Hence, Y contains at least
three colours. If |Y | = 4, then Y contains two colours that are the unique ones coloured with their
colour. Moreover, (H − {x, y})-List-Colouring contains no 6-cycle and, therefore, by Lemma 2.12
(H, c)-Colouring, is polynomial-time solvable. Hence we can assume that |Y | = 5. If Y contains at
least four colours, by the same argument we are done, therefore, we assume that Y contains exactly three
colours. If (H, c) contains a star with a Blue centre and a three leaves of different colours, then (H, c)
is not a core. Therefore the neighbourhood of each vertex of X contains at most two colours. Assume
that the three vertices y1 , y2 , y3 of Y in the 6-cycle have three different colours. Let y4 , another element
of Y be of the same colour as yi . By the previous observation, y4 can only be adjacent to neighbours of
yi . But then mapping y4 to y1 is a homomorphism which means (H, c) is not a core. Therefore, we can
assume that c(y1 ) = c(y2 ) = 1 and c(y3 ) = 2. Then, the vertex coloured 3 has degree 1 and is adjacent
to the common neighbour of y1 and y2 . But then again, (H, c) is not a core.
Therefore, H is a bipartite graph of order 8 and |X| = |Y | = 4. If there are at least three colours
in one part of the bipartition (say X), then two vertices x1 , x2 in X form two colour classes of size 1.
Moreover, H − {x1 , x2 } has no 6-cycle and therefore, by Lemma 2.12, (H, c)-Colouring is polynomialtime solvable. We may then assume that each part of the bipartition contains at most two colours. If
one part, say X, contains exactly one colour (say Blue), then (H, c) contains a path on three vertices
with every colour of c (the central vertex is Blue) and is not a core, a contradiction. Therefore each part
of the bipartition contains exactly two colours. If in each part, each colour has exactly two vertices, we
can apply Lemma 2.6 to show that (H, c)-Colouring is polynomial-time solvable. Therefore, we can
assume that there is a colour, say Blue, where exactly three vertices of one part, say x1 , x2 , x3 from part
X, coloured Blue (x4 is coloured Green). If H − x4 contains no induced 6-cycle (it cannot contain an
8-cycle since it has order 7), then (H − x4 )-List-Colouring is polynomial-time solvable and we can use
Lemma 2.12 and (H, c)-Colouring is polynomial-time solvable. Hence we may assume H − x4 contains
an induced 6-cycle C. Note that C must contain three vertices of X and therefore contains all three of
x1 , x2 , x3 . If the three other vertices y1 , y2 an y3 of C are coloured with the same colour, then (H, c)
is not a core, a contradiction. Therefore assume, without loss of generality, that c(y1 ) = c(y2 ) = 1 and
c(y3 ) = 2. Then, in order for (H, c) to be a core, we cannot have both x1 and y1 (respectively, y2 and
x3 ) of degree 3. More precisely, either d(y1 ) = d(x3 ) = 2 and d(x1 ) = d(y2 ) = 3, or d(y1 ) = d(x3 ) = 3
and d(x1 ) = d(y2 ) = 2. In both cases, we have d(y3 ) = 2, for otherwise (H, c) contains a 4-cycle with all
four colours, and (H, c) is not a core. If c(y4 ) = 1, then (H, c) contains a path on four vertices coloured
2-Blue-1-Green; moreover there is no edge in (H, c) whose endpoints are coloured Green and 2, therefore
(H, c) is homomorphic to the above path and is not a core. If c(y4 ) = 2, then (H, c) contains a 4-coloured

20

4-cycle and again (H, c) is not a core, a contradiction. As no such tropical graph exists, we have shown
that for all possible cases the (H, c)-colouring problem is polynomial-time solvable.
Denote by H9 the graph obtained from a 6-cycle by adding a pendant degree 1-vertex to three
independent vertices (see Figure 10).
B
R

G
B

B

B

B
B

Y

Figure 10: The 4-tropical graph H9 .
Theorem 5.2. H9 -Tropical-Colouring is NP-complete.
Proof. We show that (H9 , c)-Colouring is NP-complete, where c is the 4-colouring of H9 illustrated in
Figure 10. We describe a reduction from C6 -List-Colouring, which is NP-complete [17]. We label the
vertices in C6 from 1 to 6 sequentially. We also do that in the C6 included in H9 . We assume without
loss of generality that the vertex adjacent to the Red vertex is labelled 1, and the one adjacent to the
Green one is labelled 3. It follows that the vertex adjacent to the Yellow vertex is labelled 5.
Let (G, L) be an instance of C6 -List-Colouring, where L is the list-assignment function. If G is not
bipartite, then G has no homomorphism to C6 , so we can assume that G is bipartite. Since G and C6 are
bipartite, we may assume that ∀u ∈ V (G), either L(u) ⊆ {1, 3, 5}, or L(u) ⊆ {2, 4, 6}. Thus |L(u)| ≤ 3.
From (G, L), we build an instance f (G, L) of (H9 , c)-Colouring as follows. First, we consider a
copy G0 of G, we let G0 ⊂ f (G, L) and colour every vertex of G0 Black. We call u0 the copy of vertex u
in G0 . Then, for each vertex u of G, we add a gadget Hu to f (G, L) that is attached to u0 . The gadget
is described below and depends only on L(u).
• If L(u) = {1} (respectively, {3} or {5}), then Hu is a single Red (respectively, Green or Yellow)
vertex of degree 1 adjacent only to u0 .
• If L(u) = {2} (respectively, {4} or {6}), then Hu consists of two 2-vertex path: a Red–Black path
and a Green–Black path (respectively, a Green–Black path and a Yellow–Black path or a Yellow–
Black path and a Red–Black path) whose Black vertex is of degree 2 and is adjacent to u0 (the
other vertex is of degree 1).
• If L(u) = {2, 4} (respectively, {4, 6} or {2, 6}), then Hu is a 2-vertex Green–Black (respectively,
Yellow–Black or Red–Black) path whose Black vertex is of degree 2 and adjacent to u0 (the other
vertex is of degree 1).
• If L(u) = {1, 3} (respectively, {3, 5} or {1, 5}), then Hu is a 5-vertex Red–Black–Black–Black–
Green path (respectively, Green–Black–Black–Black–Yellow or Yellow–Black–Black–Black–Red)
whose middle Black vertex is of degree 3 and adjacent to u0 (the endpoints of the path are of
degree 1 and the other two vertices have degree 2).
• If L(u) = {1, 3, 5}, then Hu is a 3-vertex Black–Black–Red path with the black leaf adjacent to u0 .
• If L(u) = {2, 4, 6}, then Hu is a 4-vertex Black–Black–Black–Red path with the black leaf adjacent
to u0 .

21

Let us prove that G has a homomorphism to C6 that fulfils the constraints of list L, if and only if
f (G, L) → (H9 , c).
For the first direction, consider a list homomorphism h of G to C6 with the list function L. We build
a homomorphism h0 of f (G, L) to (H9 , c) as follows. First of all, each copy v 0 of a vertex v of G with
h(v) = i is mapped to i in (H9 , c). It is clear that this defines a homomorphism of the subgraph G0 of
f (G, L) to the Black 6-cycle in (H9 , c). It is now easy to complete h0 into a homomorphism of f (G, L)
to (H9 , c) by considering each gadget Hu independently.
For the converse, let hT be a homomorphism of f (G, L) to (H9 , c). Then, we claim that the restriction
of hT to the vertices of the subgraph G0 of f (G, L) is a list homomorphism of G to C6 with list function
L. Indeed, let u0 be a vertex of G0 . If Hu has one vertex (say a Red vertex), then L(u) = {1}. Then
necessarily u0 is sent to a neighbour of a vertex coloured Red in (H9 , c). Since the only such neighbour
is vertex 1, u0 ∈ hT (u). All the other cases follow from similar considerations.

6

Trees

We now consider the complexity of tropical homomorphism problems when the target tropical graph is
a tropical tree.
It follows from the results in Section 4 that for every tree T of order at most 10, T -TropicalColouring is polynomial-time solvable. Indeed, such a tree needs to contain a minimal tree T of order
at most 10 for which T -List-Colouring is NP-complete, and the only such tree is G1 , which has
order 10 [17]. (See Table 1.) We proved in Theorem 4.9 that G1 -Tropical-Colouring is polynomialtime solvable. With some efforts, one can extend this to trees of order at most 11.
Theorem 6.1. For every tree T of order at most 11, T -Tropical-Colouring is polynomial-time
solvable.
Proof. Let G1 be the smallest tree such that G1 -List-Colouring is NP-hard, as defined in Table 1
of Section 4 (G1 has order 10 and is obtained from a claw by subdividing each edge twice). We let
V (G1 ) = {c, x1 , y1 , z1 , x2 , y2 , z2 , x3 , y3 , z3 }, with edges cxi , xi yi , yi zi for i = 1, 2, 3.
Assume for a contradiction that there is a tree T0 of order 11 such that T -Tropical-Colouring
is not polynomial-time solvable. Then, T0 is a connected core. Once again, by Proposition 2.3, we may
assume that the colour sets of the two parts in the bipartition of T0 are disjoint. By Theorem 1.5, for
any tree T which does not contain G1 as an induced subgraph, T -List-Colouring is polynomial-time
solvable, and therefore T -Tropical-Colouring is polynomial-time solvable. Hence G1 is a subtree of
T0 .
There are four non-isomorphic trees of order 11 which contain G1 , depending on where we attach
the additional vertex a. If in T0 , a is adjacent to c, then the same arguments as in the proof of Theorem 4.9 showing that G1 -Tropical-Colouring is polynomial-time solvable show that T0 -TropicalColouring is polynomial-time solvable, a contradiction.
Let (A, B) be the bipartition of T0 with {c, y1 , y2 , y3 } ⊆ A and {x1 , x2 , x3 , z1 , z2 , z3 } ⊆ B. For the
remainder, we may assume that no vertex (except a) is the only one with its colour, for otherwise,
by Lemma 2.12, T0 -Tropical-Colouring would be polynomial-time solvable. In particular, A − a is
coloured with at most two colours and B − a is coloured with at most three colours.
Assume first that a is adjacent to a vertex xi of G1 , say x1 . The colours of x1 and z1 must be distinct,
otherwise (T0 , c0 ) is not a core. Without loss of generality, assume that c0 (x1 ) = 1 and c0 (z1 ) = 2.
Without loss of generality the central vertex c is Black. The supplementary vertex a must be coloured
with a different colour than c and y1 (say with colour Red), otherwise (T0 , c0 ) is not a core. Hence
y1 is not Red. Assume first that y1 is Green. Then (without loss of generality), y2 is Black and y3
is Green, otherwise we could apply Lemma 2.12. But by Lemma 2.12, there must be two edges with
endpoints 1 and Green, and one with endpoints 2 and Green. Hence c0 (x3 ) = 1 and c0 (z3 ) = 2 (if
c0 (x3 ) = 2 and c0 (z3 ) = 1 then (T0 , c0 ) is not a core). But again by Lemma 2.12 we need another edge
with endpoints Black and 1, and one with endpoints Black and 2. But in both cases (T0 , c0 ) is not a core,
a contradiction. This shows that vertex y1 must be Black. Then, since (T0 , c0 ) is a core, vertex c has no
neighbour coloured 2. But if there is no second edge with endpoints coloured 2 and Black, then we could
apply Lemma 2.12. Hence one of y2 and y3 , say y2 , must be Black, and c0 (z2 ) = 2. If c0 (x2 ) = 1, (T0 , c0 )
is not a core, therefore c0 (x2 ) = 3, and c0 (x3 ) ∈ {1, 3}. If y3 is Black, then (T0 , c0 ) is not a core, hence

22

y3 is Red. But both neighbours of y3 must have distinct colours, which means we can apply Lemma 2.12
to one of the edges incident with y3 , a contradiction.
Assume now that a is adjacent to a vertex yi of G1 , say y1 . Then, the colours of a, x1 and z1 must be
distinct, say c0 (x1 ) = 1, c0 (z1 ) = 2, c0 (a) = 3. Without loss of generality the central vertex c is Black.
By Lemma 2.12, there is another vertex coloured Black. If y1 is Black, then by Lemma 2.12 we have
two further edges with endpoints Black-2 and Black-3. But these edges cannot be both incident with c
(otherwise (T0 , c0 ) is not a core), hence there is another Black vertex. Then in fact, Lemma 2.12 implies
that both y2 and y3 are Black. But then, any way to complete c0 implies that (T0 , c0 ) is not a core, a
contradiction. Therefore, y1 is not Black (say it is Red) and we can assume that y2 is Black, and since
we need a second Red vertex, y3 is Red. But one of the type of edges among Red-1, Red-2 and Red-3
will appear only once, and we can apply Lemma 2.12, a contradiction.
We assume finally that a is adjacent to a vertex zi of G1 , say z1 . Without loss of generality, vertex a
is Black, vertex z1 is coloured 1, and vertex y1 is Red (otherwise, (T0 , c0 ) is not a core). By Lemma 2.12,
there must be another 3-vertex path coloured Black-1-Red. This path must be cxi yi with c Black, for
otherwise (T0 , c0 ) is not a core. We can assume that c0 (x2 ) = 1 and y1 is Red. Then c0 (x1 ) 6= 1, assume
c0 (x1 ) = 2. Then again by Lemma 2.12 there is another 3-vertex path coloured Black-2-Red. The only
possibility is that c0 (x3 ) = 2 and y3 is Red. Then c0 (z3 ) ∈
/ {1, 2}, otherwise (T0 , c0 ) is not a core. Hence
we assume c0 (z3 ) = 3, which by Lemma 2.12 implies c0 (z2 ) = 3. But then there is a unique 3-vertex path
coloured 1-Red-3, and by Lemma 2.12, (T0 , c0 )-Colouring is polynomial-time solvable, a contradiction.
This completes the proof.
Let T23 be the tree of order 23 shown in Figure 11.
R B Rc

F1

F2 Bc

B

G

B

c

R

Bc
B

G

Gc

T

B

Bc
B
Rc

B

G
B
B

B
R

Gc

Figure 11: The 7-tropical tree (T23 , c)
Theorem 6.2. T23 -Tropical-Colouring is NP-complete.
Proof. We give a reduction from 3-SAT to (T23 , c)-Colouring, where c is the colouring of Figure 11.
Given an instance (X, C) of 3-SAT, we construct an instance f (X, C) = (GX,C , cX,C ) of (T23 , c)Colouring.
To construct the graph GX,C , we first define the following building blocks. See Figure 12 for illustrations.
• The block S1,2 is a graph built from a 7-vertex black-coloured path with vertex set {x1 , . . . , x7 }
where a BlackCross leaf is attached to vertices x1 and x7 , a RedDot leaf is attached to vertices x2
and x6 , and a GreenDot leaf is attached to vertex x4 .
• The block S1,T is a graph built from a 7-vertex black-coloured path with vertex set {x1 , . . . , x7 }
where a BlackCross leaf is attached to vertices x1 and x7 , a RedDot leaf is attached to vertices x2
and x6 , and a RedCross leaf is attached to vertex x4 .
• The block S1,T is a graph built from a 7-vertex black-coloured path with vertex set {x1 , . . . , x7 }
where a BlackCross leaf is attached to vertices x1 and x7 , a GreenDot leaf is attached to vertices
x2 and x6 , and a GreenCross leaf is attached to vertex x4 .
• The NOT-block is depicted in Figure 12(b).
23

• The A-block is depicted in Figure 12(c).
Illustrations of these blocks can be found in Figure 12.
Bc R

G

Bc R

R Bc

Bc G

R Bc

Rc

Gc

G Bc

B B B B B B B

B B B B B B B

B B B B B B B

S1,2

S1,T

S2,T

(a) The blocks S1,2 , S1,T and S2,T .
Rc Bc

Bc R
Rc B R

Bc

S2,T

B B B B B

B
Bc R
1

G
G Bc Rc

S1,2

Bc

A
B

B
R Bc G

Bc

Gc Bc

x
x0 B B B B B B B B B B B B B

x
0
x
B

S1,2

B

S1,2

1

B B B B B
Bc Gc

(b) The variable gadget of x, essentially a NOT-block.

S1,T

G Bc

(c) The A-block and its representation as an
arrow.

Figure 12: The building blocks of GX,C .
We now define gadgets for each variable of X and each clause of C. The graph GX,C is formed by
the set of all variable and clause gadgets.
• For a variable x ∈ X, the variable gadget of x consists of the four vertices x0 , x1 , x̄0 and x̄1 ,
coloured respectively BlackDot, BlackCross, BlackDot and BlackCross, joined by a NOT-block as
described in Figure 12(b). The image of x0 and x1 in (T23 , c) correspond to the truth-value of the
litteral x. Similarly, the image of x̄0 and x̄1 correspond to the truth-value of the litteral x̄. For a
litteral l, we use the notation l0 (resp. l1 ) to describe either x0 (resp. x1 ) when l = x with x ∈ X,
or x̄0 (resp. x̄1 ) when l = x̄ with x ∈ X.
• For each clause c = (l1 , l2 , l3 ) ∈ C, there is a clause gadget of c (as drawn in Figure 13) connecting
vertices l10 , l20 and l30 .
l11 Bc
l10 B
S1,2
Bc

l21 Bc
l20 B
S1,2

B A

l31 Bc
l30 B
S1,2

B A

B

Bc

Bc

A

Figure 13: Example of a clause gadget of clause (l1 , l2 , l3 ). The full details of the A-blocks and S1,2 -blocks
are represented in Figure 12.
We now show that GX,C → (T23 , c) if and only if (X, C) is satisfiable.
Assume first that there is a homomorphism h of GX,C to (T23 , c). We first prove some properties of
h.
Claim 6.3. The homomorphism h satisfies the following properties.
(1) For each literal l of a variable of X, vertices l0 and l1 are mapped to the two vertices of one of the
pairs T , F1 or F2 . The same holds for the extremities of the blocks S1,2 , S1,T , S2,T and A.
(2) The two extremities of each block S1,2 are both mapped either to the vertices of T , or to vertices of
F1 ∪ F2 .

24

(3) The two extremities of each block S1,T are both mapped either to the vertices of F2 , or to vertices
of F1 ∪ T .
(4) The two extremities of each block S2,T are both mapped either to the vertices of F1 , or to vertices
of F2 ∪ T .
(5) For each variable x of X, exactly one of x0 and x̄0 is mapped to a vertex of T , and the other is
mapped to a vertex of F1 or F2 .
(6) In any A-block, either some extremity is mapped to T (then the other extremity can be mapped to
any of F1 , F2 or T ), or the left extremity is mapped to F2 and the right extremity, to F1 .
Proof of claim. (1) This is immediate since the only pairs in (T23 , c) consisting of two adjacent BlackDot
and BlackCross vertices are the ones of T , F1 and F2 .
(2)–(4) We only prove (2), since the three proofs are not difficult and similar. By (1), the extremities
of S1,2 are mapped to vertices of T ∪ F1 ∪ F2 . If one extremity is mapped to T , the remainder of the
mapping is forced and the claim follows. If one extremity is mapped to F1 ∪ F2 , one can easily complete
it to a mapping where the other extremity is mapped to either F1 or F2 .
(5) By (1), x0 and x̄0 must be mapped to a vertex of T ∪ F1 ∪ F2 . Without loss of generality, we
can assume that x0 corresponds to the left extremity of the NOT-block Nx connecting x0 and x̄0 . First
assume that x0 and x̄0 are mapped to the vertex of T coloured BlackDot. Then, considering the vertices
of Nx from left to right, the mapping is forced and the degree 3-vertex of Nx at distance 2 both of a
RedDot and a RedCross vertex must be mapped to the vertex c of T23 . But then, continuing towards
the right of Nx , x̄0 cannot be mapped to a vertex of T . Therefore, we may assume that both x0 and
x̄0 are mapped to the BlackCross vertices of F1 ∪ F2 . If x0 is mapped to the BlackCross vertex in F1 ,
then again going through Nx from left to right the mapping is forced; the central vertex of Nx must be
mapped to a vertex of F2 , and x̄0 must be mapped to a vertex of T , a contradiction. The same applied
when x0 is mapped to the BlackCross vertex in F2 , completing the proof of (5).
(6) An A-block is composed of two parts: the upper part and the lower part. Observe that if the
left extremity of an A-block is mapped to F1 , then using (2) and (4), the mapping of the upper part of
the A-block is forced and the right extremity has to be mapped to T . Similarly, if the left extremity is
mapped to F2 , by (2) and (3) the right extremity cannot be mapped to F2 . On the other hand, for all
other combinations of mapping the extremities to T , F1 or F2 the mapping can be extended. ()
We are ready to show how to construct the truth assignment A(h). If h(l0 ) ∈ T for some literal l,
we let l be True and if h(l0 ) ∈ F1 ∪ F2 , we let l be False. By Claim 6.3(5), this is a consistent truth
assignment for X. For any clause c = (l1 , l2 , l3 ), in the clause gadget of c, we have three A-blocks forming
a directed triangle. Hence, by Claim 6.3(6), there must be one of the three extremities of this triangle
mapped to a vertex of T . Therefore, by Claim 6.3(2), at least one of the vertices l10 , l20 and l30 is mapped
to T . This shows that A(h) satisfies the formula (X, C).
Reciprocally, if there is a solution for (X, C), one can build a homomorphism of GX,C to (T23 , c)
by mapping, for each literal l, the vertices l0 and l1 to one of the vertex pairs F1 , F2 and T of (T23 , c)
corresponding to the truth value of l (if l is False, we may choose one of F1 and F2 arbitrarily). Then,
using Claim 6.3, one can easily complete this to a valid mapping.

7

Conclusion

We have shown that the class of (H, c)-Colouring problems has a very rich structure, since they fall into
the classes of CSPs for which a dichotomy theorem would imply the truth of the Feder–Vardi Dichotomy
Conjecture. Hence, we turned our attention to the class of H-Tropical-Colouring problems, for
which a dichotomy theorem might exist. Despite some initial results in this direction, we have not been
able to exhibit such a dichotomy, and leave this as the major open problem in this paper.
Towards a solution to this problem, we propose a simpler question. All bipartite graphs H that
we know with problem H-Tropical-Colouring being NP-complete contain, as an induced subgraph,
either an even cycle of length at least 6 (for example cycles themselves or H9 ), or the graph G1 from
Table 1, that is, a claw with each edge subdivided twice (this is the case for T23 ). Hence, we ask the
following. (A bipartite graph is chordal if it contains no induced cycle of length at least 6.)
25

Question 7.1. Is it true that for any chordal bipartite graph H with no induced copy of G1 , H-TropicalColouring is polynomial-time solvable?
Note that Question 7.1 is not an attempt at giving an exact classification, since G1 -TropicalColouring and C2k -Tropical-Colouring for k ≤ 6 are polynomial-time solvable.
Another interesting question would be to consider the restriction of H-Tropical-Colouring to 2tropical graphs. Recall that by Remark 4.8(2), one can slightly modify the gadgets from Theorem 4.4 and
the colouring of the cycle, to obtain a 2-colouring c of C54 such that (C54 , c)-Colouring is NP-complete.
Finally, we relate our work to the (H, h, Y )-Factoring problem studied in [10] and mentioned in
+
+
the introduction. Recall that (H, c)-Colouring corresponds to (H, c, K|C|
)-Factoring where K|C|
is
the complete graph on |C| vertices with all loops, and with C the set of colours used by c. In [10],
the authors studied (H, h, Y )-Factoring when Y has no loops. Using reductions from NP-complete
D-Colouring problems where D is an oriented even cycle or an oriented tree, they proved that for any
fixed graph Y which is not a path on at most four vertices, there is an even cycle C and a tree T such that
(C, hC , Y )-Factoring and (T, hT , Y )-Factoring are NP-complete (for some suitable homomorphisms
hC and hT ). Note that C and T here are fairly large. We can strengthen these results as follows. Consider
our reduction of Theorem 4.4 showing in particular that C48 -Tropical-Colouring is NP-complete. As
noted in Remark 4.8(1), the given colouring c of C48 can easily be made a proper colouring by separating
the red vertices into two classes, according to which part of the bipartition of C48 they belong to. Then,
one can observe that c is in fact a homomorphism to a tree T1 obtained from a claw where one edge is
subdivided once (the three vertices of degree 1 are coloured Blue, Black and Green, and the two other
vertices are the two kinds of Red). Thus, for any graph Y containing this subdivided claw as a subgraph,
we deduce that (C48 , c1|T1 , Y )-Factoring is NP-complete. We can use a similar approach for our result
of Theorem 6.2, that T23 -Tropical-Colouring is NP-complete. Note that the colouring c2 we give is
in fact a homomorphism to a tree T2 which is obtained from a star with five branches by subdividing
one edge once. Thus, for any graph Y containing T2 as a subgraph, (T23 , c2|T2 , Y )-Factoring is NPcomplete. Of course we can apply this argument by replacing T1 and T2 by the underlying graph of any
loop-free homomorphic image of (C48 , c1 ) and (T23 , c2 ), respectively.
Acknowledgements. We thank Petru Valicov for initial discussions on the topic of this paper.

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