Complexity of Injective Homomorphisms to Small Tournaments, and of Injective Oriented Colourings ()
1. Introduction
Three natural possible definitions of local injectivity of a homomorphism f from an input oriented graph G to a target oriented graph H are: for every vertex
, the function f is injective when restricted to:
1) The in-neighbourhood
;
2)
and
separately;
3) The union
.
When H is reflexive, that is, has a loop at every vertex, the three definitions are different. When H is irreflexive, that is, has no loops, definitions 2 and 3 coincide. Each of these five situations leads naturally to a notion of locally- injective oriented k-colouring.
Locally-injective homomorphisms (as in possible definition 1) and colourings of oriented graphs were first introduced as an example in monadic second order logic [1]. Consequently, by Courcelle’s Theorem, these problems are all solvable in polynomial time when the input has bounded treewidth. The same holds for the other possible definitions above.
Possible definition 1 has been studied in previous papers for both irreflexive and reflexive targets [2] [3] [4] [5] [6]. A fairly complete theory has been developed. When the target, H, is reflexive there is a dichotomy theorem characterizing the oriented graphs H for which the problem of deciding the existence of a homomorphism to H is Polynomial, and those for which it is NP-complete. When H is irreflexive the complexity has been determined when H has maximum in-degree
or
; when
the situation is as rich as that for all digraph homomorphism problems, and hence all constraint satisfaction problems [5].
Possible definitions 2 and 3 have been studied in [7] [8]. Obstructions to (subgraphs that prevent the existence of) homomorphisms to small tournaments are the focus of [8]. Both definitions are considered. Possible definition 3 is the main focus of [7].
Locally-injective colourings of undirected graphs were first explicitly studied by Hahn, Kratochvil, Siřan and Sotteau [9]. Subsequent papers have considered chordal graphs [10], planar graphs (see [11]) and other graph classes, as well as list versions [12]. The complexity of locally-injective homomorphisms has been extensively studied by Fiala, Kratochvil, and others (e.g. see [13] [14]).
The purpose of this paper is to contribute to the theory of locally-injective homomorphisms and colourings under possible definitions 2 and 3 above. In each of the three cases that arise, the complexity of deciding the existence of a homomorphism to H is determined for the four tournaments on at most three vertices. These results appear in Sections 3, 4, and 5. Later, in Section 6, these results are then used to determine the complexity of the associated locally-injective oriented colouring problems.
We conclude this section by noting that the complexity of deciding whether a given directed graph G has a homomorphism to a tournament H has been studied [15]. There is a dichotomy theorem: the problem is Polynomial when H has at most one directed cycle, and NP-complete when H has at least two directed cycles. The results reported in this paper are first steps towards finding a similar theorem for locally-injective homomorphisms.
2. Notation and Terminology
An oriented graph is a directed graph G with the property that for any two different vertices x and y, at most one of the arcs
belongs to
. An oriented graph G can be viewed as arising from a simple graph H by assigning a direction, or orientation, to each edge. The graph H is called the underlying graph of G, and G is referred to as an orientation of H. The converse of an oriented graph G is the oriented graph Gc with the same vertex set as G, and arc set
.
An oriented graph is reflexive if it has a loop at each vertex, and irreflexive if it has no loops. The superscript “r”, as in
, indicates that the oriented graph under consideration is reflexive. Oriented graphs without this superscript, as in G, are irreflexive.
We use Pn, Tn, and Tn to denote the directed path on n vertices, the directed cycle on n vertices, and the transitive tournament on n vertices, respectively,
. It will be assumed throughout that C3 has vertex set
and arc set
, and that Tn has vertex set
and arc set
.
A homomorphism of an oriented graph G to an oriented graph H is a function
such that
whenever
. When H has a loop, any directed graph has a homomorphism to H: map all vertices of G to a vertex of H with a loop. Thus, when loops are present, the existence of a homomorphism is a non-trivial question only in the presence of some side condition like selecting the image of each vertex from a list of possible images, or local injectivity. The book [16] contains a wealth of information about homomorphism of graphs and digraphs.
We call a homomorphism f of an oriented graph G to an oriented graph H:
• ios-injective if, for every vertex x of G, the restriction of f to
is injective, as is the restriction of f to
; and
• iot-injective if, for every vertex x of G, the restriction of f to
is injective.
These two concepts are the same when H is an irreflexive oriented graph, and different when H is a reflexive oriented graph.
The designations “ios” and “iot” arise from the local injectivity being on in- neighbourhoods and out-neighbourhoods separately, and on in-neighbourhoods and out-neighbourhoods together. In introducing the designations “ios” and “iot”, the qualifier “locally” has been dropped as it is part of the definition.
It is easy to see that the composition of two ios-injective homomorphisms is an ios-injective homomorphism, and similarly for iot-injective homomorphisms.
The following structure and its converse will be particularly useful. We define the hat H3 to be the oriented graph with vertex set
and edge set
. See Figure 1. The vertices v0 and v2 will be referred to as the ends of H3 or
. Whether or not H is reflexive, in an ios-injective or iot-injective homomorphism of H3 or
to H, the vertices v0 and v2 must have different images.
3. Irreflexive Targets
In this section we show that, if T is an irreflexive tournament on at most 3 vertices, then the problem of deciding whether a given oriented graph has an ios-injective (and hence also iot-injective) homomorphism to T is Polynomial. A given oriented graph has an ios-injective homomorphism to T1 if and only if it has no edges, and has an ios-injective homomorphism to T2 if and only if it is a disjoint union of copies of T1 and T2. A given oriented graph, G, has an ios-injective homomorphism to C3 if and only if it has maximum in-degree 1, maximum out-degree 1, and has a homomorphism to C3. It follows that G has an ios-injective homomorphism to C3 if and only if it is a disjoint union of directed paths, and directed cycles of length a multiple of 3. These conditions are easy to check in polynomial time. It remains to consider ios-injective homomorphisms to the transitive triple.
Proposition 3.1. The problem of deciding whether a given oriented graph has an ios-injective homomorphism to T3 is Polynomial.
Proof. Let G be a given digraph. If the underlying graph of G has a vertex of degree 3 or more, then G has no ios-injective homomorphism to T3. Hence assume that G is an orientation of a graph with maximum degree at most 2. Therefore the underlying graph of G is a disjoint union of paths and cycles, and hence has treewidth at most 2. Since ios-injective homomorphism is expressible in monadic second-order logic, the statement now follows from Courcelle’s Theorem.
4. Ios-Injective Homomorphisms to Small Reflexive Targets
In this section, we determine the complexity of deciding whether there exists an ios-injective homomorphism from a given oriented graph G to the fixed oriented graph H when H is one of the four reflexive tournaments on at most three vertices.
It is clear that an oriented graph G has an ios-injective homomorphism to
if and only if it has maximum in-degree at most one and maximum out-degree at most one, that is, if and only if neither H3 nor
is a subgraph of G. Consequently, the only oriented graphs which have an ios-injective homomorphism to
are disjoint unions of directed paths and directed cycles.
Proposition 4.1. The problem of deciding whether a given oriented graph has an ios-injective homomorphism to
is Polynomial.
Proof. We describe a reduction to 2-SAT. Associate the vertices t0 and t1 of
with false and true, respectively. Given an oriented graph G, the corresponding instance of 2-SAT has the set of variables
. Since no oriented graph with a vertex of in-degree at least 3, or a vertex of out-degree at least 3, has an ios-injective homomorphism to
, we can assume that
and
.
The set of clauses is constructed as follows.
1) If
, then
is a clause.
2) If
, then
is a clause.
3) If
, then
is a clause.
4) If v and w are the ends of a copy of H3 or
, then
and
are clauses.
All clauses in groups (1) and (2) are satisfied if and only if the image of any vertex of out-degree 2 is t0 and the image of any vertex of in-degree 2 is t1. All clauses in group (3) are satisfied if and only if the mapping corresponding to the truth assignment preserves arcs. And finally, all clauses in group (4) are satisfied if and only if the ends of a copy of H3 or
are assigned different images. It follows that there is an ios-injective homomorphism of G to
if and only all clauses are satisfied.
We now show that the problem of deciding the existence of an ios-injective homomorphism to
is NP-complete. Some “gadget” oriented graphs which map to
only in special ways will be used in the NP-completeness proof. For an integer
, the oriented graph Dd is constructed from a directed cycle
by adding the vertices
and arcs
,
.
Lemma 4.2. In an ios-injective homomorphism of Dd to
the vertices
all have the same image.
Proof. Let f be an ios-injective homomorphism of Dd to
. Without loss of generality, suppose
. Then
is either c1 or c2.
Suppose first that
. Then, observing that an ios-injective homomorphism of an irreflexive directed 3-cycle to
either assigns every vertex the same image, or assigns no two vertices the same image, it must be that
. By injectivity
, so
the only other out-neighbour of c1. It follows that
. Similarly,
, so that
. Continuing in this way, the vertices
map to
, respectively, and the vertices
map to
, respectively.
Now suppose that
. By our observation regarding homomorphisms of irreflexive directed 3-cycles, it must be that
. Arguing as in the previous paragraph, ios-injectivity implies
and
which in turn implies
. Continuing in this way, the vertices
map to
, respectively, and the vertices
map to
, respectively.
For
, let Xd be the oriented graph constructed from Dd by adding d new vertices
and the arcs belonging to
, where addition is modulo 3d. The following is a consequence of Lemma 4.2.
Corollary 4.3. In an ios-injective homomorphism of Xd to
, the vertices of the directed cycle
must all be assigned the same image. Futher, any partial mapping in which these vertices are all assigned the same image can be extended to an ios-injective homomorphism of Xd to
.
Theorem 4.4. The problem of deciding if a given oriented graph G has an ios-injective homomorphism to
is NP-complete.
Proof. The transformation is from 3-colouring of graphs with minimum degree at least 3. Suppose a graph G is given. For each vertex
, regard the edges incident with x as being in 1-1 correspondence with the integers
so that it is meaningful to talk about the ith edge incident with x. Construct a digraph G' as follows. For each vertex
there is a copy of
. (Note that
.) Each edge of G is replaced by an oriented path on three vertices. Suppose
is the ith edge incident with w and the jth edge incident with z. Add a new vertex
and arcs from vertex
of the copy of
corresponding to w, and from vertex
of the copy of
corresponding to z, to
. The transformation can be accomplished in polynomial time. We will show that G is 3-colourable if and only if there is an ios-injective homomorphism of G' to
.
Suppose that G is 3-colourable, and fix a 3-colouring using the colours
. If the colour of x is ci, then map vertices
of the copy of
corresponding to x to ci and extend this to an ios-injective homomorphism to
. The ends of each oriented path that replaced an edge of G are now assigned different images, and the mapping so far can be extended to the remaining vertex of each oriented path that replaced an edge of G.
Suppose G' has an ios-injective homomorphism to
. Then, in each copy of
, all vertices of the directed cycle
are assigned the same image. Assign this colour to x. By the construction of G' and ios-injectivity, adjacent vertices of G are assigned different colours.
We conclude this section by showing that the problem of deciding whether a given oriented graph G has an ios-injective homomorphism to
is NP-com- plete. A useful technical lemma is established first.
Lemma 4.5. Let F be the oriented graph in Figure 2. Then for
, there exists an ios-injective homomorphism of F to
that maps u to x, and any such homomorphism also maps v to x.
Proof. We sketch the proof that in an ios-injective homomorphism of F to
that maps u to t1, the vertex v also maps to t1.
Referring to Figure 2, it is straightforward to check that in any ios-injective homomorphism of F to
, the vertices labelled
must map to
, respectively. It is also easy to check that the vertices labelled a must have the same image, and similarly for the vertices labelled b, e and f. It will follow from the argument below that the vertices labelled c must have the same image, and similarly for the vertices labelled d.
We show that the vertices labelled t1 must map to t1. Suppose u maps to t1. Then by injectivity its out-neighbour labelled c maps to t0 or t2. Since c has in-degree 2, it must map to t2. Therefore d maps to t0. The in-neighbour of c labelled t1 must map to t1 or t2. But its in-neighbour labelled b has an out- neigh-bour labelled t2, so the in-neighbour of c labelled t1 must map to t1. By
Figure 2. The oriented graph F in Lemma 4.5.
injectivity, the out-neighbour labelled a of this vertex must map to t1, so the symmetrically located vertex labelled a must also map to t1, and its in-neighbour labelled t1 must map to t1. A similar argument shows that the other vertices labelled t1 must map to t1.
We now show that v maps to t1. By the above argument and injectivity, the vertex labelled c on the right of the figure maps to t2. A symmetric argument shows that the vertex labelled d on the bottom of the figure must map to t0. Now, by injectivity, v maps to t1, as wanted.
Similar arguments show that if u maps to t0 then so does v, and if u maps to t2 then so does v.
Theorem 4.6. The problem of deciding if a given oriented graph has an ios-injective homomorphism to
is NP-complete.
Proof. The transformation is from 3-edge colouring of cubic graphs [17]. Suppose such a graph G is given. Construct a graph G' as follows. For each
, regard the edges incident with x as being in 1-1 correspondence with the integers 1, 2, 3 so that it is meaningful to talk about the ith edge incident with x.
Let H4 denote the orientation of
in which there is a vertex of in-degree 3. In the sequel we refer to H4 as an in-star. Start with a collection of
disjoint copies of H4. Let Sx denote the copy of H4 corresponding to vertex x. Regard the leaves of each oriented graph Sx to be in 1-1 correspondence with
. Suppose
is the ith edge incident with x and the jth edge incident with y. Add a new copy of the oriented graph F shown in Figure 2 and identify the vertices labelled u and v having in-degree one with the ith leaf of Sx and the jth leaf of Sy. The transformation may be accomplished in polynomial time. We claim that G is 3-edge-colourable if and only if G' has an ios-injective homomorphism to
.
Suppose G has a 3-edge-colouring
(the colours are the vertices of
). For any edge xy of G, map the vertices labelled u and v in the corresponding copy of F to
. Finally, map the centre of each in-star of G' to its only possible image, t2.
Conversely, suppose G' has an ios-injective homomorphism to T3. For each edge xy of G, the vertices labelled u and v in the corresponding copy of F in G' must have the same image. Use this for the colour of xy. The resulting assignment is a 3-edge-colouring because the leaves of each in-star Sx in G' must have different images.
5. Iot-Injective Homomorphisms to Small Reflexive Targets
In this section we consider the complexity of deciding whether there exists an iot-injective homomorphism from a given oriented graph G to the fixed oriented graph H, when H is one of the four reflexive tournaments on at most three vertices.
It is clear that an oriented graph has an iot-injective homomorphism to
if and only if it contains no oriented path on three vertices, that is, if and only if it is a disjoint union of copies of T1 and T2.
We now turn our attention to
. No orientation of a graph with a vertex of degree three has an iot-injective homomorphism to
. Thus, if G admits an iot-injective homomorphism to
, then the underlying graph of G is a disjoint union of paths and cycles. The following proposition can be proved using a reduction to 2-SAT, or by an appeal to Courcelle’s Theorem.
Proposition 5.1. The problem of deciding whether a given oriented graph has an iot-injective homomorphism to
is Polynomial.
We next consider iot-injective homomorphism to
. Consider the family of oriented cycles
such that each
is comprised of two disjoint perfect matchings oriented in opposite directions; that is,
and
.
Lemma 5.2. Let
have order n. Then (1) B has an iot-injective homomorphism to
if and only if
(mod 6), and (2) B has an iot- injective homomorphism to
if and only if
(mod 4).
Proof. Let
.
We first consider iot-injective homomorphism of B to
. Suppose B has n vertices. Let x be a vertex of out-degree two. Without loss of generality x maps to c1. Then its out-neighbours map to c1 and c2. Let y be the out-neighbour that maps to c1. Its out-neighbour must map to c2. Continuing in this way, starting from x, the images of consecutive vertices are
. Therefore an iot-injective homomorphism exists if and only if
(mod 6).
We now consider iot-injective homomorphism of B to
. Suppose B has n vertices. Let x be a vertex of out-degree two. Then x maps to t0 or t1.
Suppose first that x maps to t0. Let v be an out-neighbour of x. If v were mapped to t0, then its other in-neighbour must also map to t0, in violation of injectivity. Therefore, the out-neighbours of x map to t1 and t2. Let y be the out-neighbour that maps to t2. Then y’s other in-neighbour, z, must map to t1 and z’s other out-neighbour, a, must also map to t1. The vertex a has another in-neighbour, b. By injectivity, b maps to t0. Continuing in this way, starting from x, the images of consecutive vertices are
. Therefore
(mod 4).
Now suppose x maps to t1. As above, the out-neighbours of x map to t1 and t2. Let y be the out-neighbour that maps to t1. The vertex y has another in- neigh-bour, z, which by injectivity maps to t0. Now, following the same argument as in the previous paragraph we have that, starting from x, the images of consecutive vertices are
. Again,
(mod 4).
It now follows that an iot-injective homomorphism exists if and only if
(mod 4).
Corollary 5.3. For
, let
have 6t vertices. In any iot-injective homomorphism f of
to
we have
, when
(mod 6).
Proof. This follows from the argument in Lemma 5.2.
Theorem 5.4. The problem of deciding whether an oriented graph has an iot-injective homomorphism to
is NP-complete.
Proof. The transformation is from 3-colouring of connected graphs [18]. Suppose such a graph G is given. Construct a graph G' as follows. For each
, regard the edges incident with x as being in 1-1 correspondence with the integers
so that it is meaningful to talk about the ith edge incident with x. Replace every vertex
with a copy Rx of
where, without loss of generality, the vertices
, have in-degree 2. Suppose xy is the ith edge incident with x and the jth edge incident with y. Construct an oriented path Pxy by adding a new vertex txy and joining each of
and
to it by adding a directed path of length two (the midpoint of each such directed path is a new vertex). The transformation can be carried out in polynomial time. We claim that G is 3-colourable if and only if G' has an iot-injective homomorphism to
.
Suppose G has a 3-colouring
. For each vertex x, map the vertices
of Rx to
. By Corollary 5.3, this partial mapping extends to an iot-injective homomorphism of Rx to
. We claim that this mapping of the oriented cycles Rx extends to the oriented paths Pxy, where
. Since adjacent vertices in G must receive different colours, this mapping of the copies of
assigns the vertices
of Rx a different image than it assigns the corresponding vertices of Ry. Suppose, without loss of generality, that the vertices
of Rx are mapped to c1 and the corresponding vertices of Ry are mapped to c2. The in-neighbours of the vertices in Rx are mapped to c1 and c3, while the neighbours of the corresponding vertices in Ry are mapped to c2 and c1. The vertex txy can be mapped to c3 and the assignment extended to an iot-injective homomorphism of Pxy to
. This proves the claim, and completes the proof of the implication.
On the other hand, suppose G' has an iot-injective homomorphism to
. Fix such a mapping. Then, for each
, the vertices
all have the same image; assign this to be the colour of vertex v of G.
We claim that vertices x and y that are adjacent in G are assigned different colours. Suppose not. By symmetry of
, assume both are assigned c1. Suppose also that xy is the i-th edge incident with x and the j-th edge incident with y. Then, the vertices
and
both map to c1. By construction,
has two in-neighbours in Rx and one out-neighbour on the directed path to txy, and similarly for
. Since both
and
map to c1, in each case their in-neighbours must map to c1 and c3. By injectivity, in each case their out-neighbour on the directed path to txy must map to c2. Therefore txy has 2 in-neighbours that map to c2, which violates injectivity. This proves the claim, and completes the proof.
Finally, we consider iot-injective homomorphism to
. The following lemma can be proved similarly to Lemma 4.5. The proof of Lemma 4.5 relies only on injectivity on in-neighbourhoods or out-neighbourhoods, and never both at the same vertex.
Lemma 5.5. Let F be the oriented graph in Figure 2. Then For
, there exists an iot-injective homomorphism of F to
that maps u to x, and any such homomorphism also maps v to x.
The proof of the following theorem is similar to that of Theorem 4.6 and is omitted. For details, see [7].
Theorem 5.6. The problem of deciding whether an oriented graph has an iot-injective homomorphism to
is NP-complete.
6. Colourings
Recall that a (proper) oriented k-colouring of an oriented graph G is a homomorphism to a tournament on k vertices. We therefore make the following definitions:
1) A proper ios-injective oriented k-colouring of an oriented graph G is an ios-injective homomorphism to an irreflexive tournament on k vertices.
2) An improper ios-injective oriented k-colouring of an oriented graph G is an ios-injective homomorphism to a reflexive tournament on k vertices.
3) An improper iot-injective oriented k-colouring of an oriented graph G is an iot-injective homomorphism to a reflexive tournament on k vertices.
A proper iot-injective oriented k-colouring of a graph G would be an iot- injective homomorphism to an irreflexive tournament on k vertices. Since tournaments have no directed 2-cycles, these are the same as proper ios-injective oriented k-colourings.
For each fixed integer k and each injective colouring problem defined above, we will determine the complexity of deciding whether a given oriented graph G has an injective colouring with k colours. The approach to proving NP- completeness is similar to that for oriented colourings that are injective on in- neighbourhoods [2] [6]: prove that it is NP-complete to decide the existence of an injective homomorphism of the given type to the tournament
, that consists of a directed three cycle dominated by every vertex of a transitive tournament of size
, and then obtain the desired result as a corollary. We consider the three situations in turn after establishing a useful lemma.
Lemma 6.1. Let G be an oriented graph such that Um is a subgraph of G. For
, if G has a
-injective homomorphism to a tournament T (respectively, reflexive tournament
), then Um (respectively,
) is a subgraph of T.
Proof. The tournament Um has the property that every two different vertices have a common in-neighbour or a common out-neighbour. Hence no two of its vertices can be assigned the same image by a
-injective homomorphism. Consequently, the image of G must contain Um.
6.1. Proper Ios-Injective Colourings
Theorem 6.2. For each fixed
, the problem of deciding if a given oriented graph has an ios-injective homomorphism to Um is NP-complete.
Proof. We first show that the problem of deciding whether a given oriented graph G has an ios-injective homomorphism to U4 is NP-complete. The transformation is from the problem of deciding if a given cubic graph is 3-edge- colourable [17]. Let G be a given cubic graph. Construct an oriented graph G' by replacing each edge xy of G by an oriented path Pxy with vertices
and arcs
. The transformation can be accomplished in polynomial time. We claim that G is 3-edge-colourable if and only if there is an ios-injective homomorphism of G' to U4.
Suppose that G is 3-edge colourable. Then, for each vertex x of G, each of the colours 1, 2 and 3 appears on an edge incident with x. An ios-injective homomorphism of G' to U4 is obtained by mapping all vertices of G to the vertex of out-degree 3 in U4, assigning the colour of the edge xy to the vertices v1 and v4 of Pxy, and extending this pre-colouring to the vertices v2 and v3 of Pxy.
Suppose that there is an ios-injective homomorphism of G' to U4. Every vertex of G has out-degree 3 in G', so an ios-injective homomorphism of G' to U4 must map it to the unique vertex of out-degree 3 in U4. Similarly, the vertices v1, v2, v3, and v4 in each oriented path Pxy have positive in-degree in G', so an ios-injective homomorphism of G' to U4 must map each of them to a vertex of the directed 3-cycle. In any such mapping, v1 and v4 map to the same vertex, and the three out-neighbours of each vertex of G (in G') map to different vertices of the 3-cycle. Assigning each edge xy of G the image of the vertex v1 (and v4) in Pxy gives a 3-edge-colouring of G.
NP-completeness of ios-injective homomorphism to Um follows from NP- completeness of ios-injective homomorphism to U4. Given an instance G of ios-injective homomorphism to U4, construct G' by adding the new vertices belonging to
and the arcs
. Since m is a constant, the transformation can be accomplished in polynomial time. Each vertex of G in G' has in-degree
and therefore cannot map to the
vertices of Um with in-degree less than
. An ios-injective homomorphism of G to U4 can be extended to an ios-injective homomorphism of G' to Um.
Corollary 6.3. Let k be a fixed positive integer. If
, the problem of deciding if a given oriented graph G has a proper ios-injective oriented k- colouring is Polynomial. If
, the problem of deciding if a given oriented graph G has a proper ios-injective oriented k-colouring is NP-complete.
Proof. An oriented graph G has a proper ios-injective oriented k-colouring if and only if
has an ios-injective homomorphism to Uk.
6.2. Improper Ios-Injective Colourings
Theorem 6.4. For each fixed
, the problem of deciding if a given oriented graph has an ios-injective homomorphism to
is NP-complete.
Proof. The transformation is from the problem of deciding whether there exists an ios-injective homomorphism of a given oriented graph G to
, which is NP-complete by Theorem 4.4. Suppose the oriented graph G is given. We may assume that
and
, otherwise G cannot have an ios-in-jective homomorphism to
.
Construct G' from G as follows. For each
, if x has in-degree at most one in G, add a set of
new vertices and arcs joining each of them to x. If x has in-degree two in G, do the same using a set of
new vertices. The transformation can be accomplished in polynomial time. We claim that G has an ios-injective homomorphism to
if and only if G' has an ios-injective homomorphism to
.
An ios-injective homomorphism of G to
can clearly be extended to an ios-injective homomorphism of G' to
.
Suppose f is an ios-injective homomorphism of G' to
. Since each vertex
has in-degree at least
in G' and every vertex of
not belonging to the copy of
has in-degree at most
, the vertex x must map to a vertex of the directed 3-cycle in
. The restriction of f to
is the desired mapping.
Corollary 6.5. Let k be a fixed integer. If
, the problem of deciding if a given oriented graph G has an improper ios-injective oriented k-colouring is Polynomial. If
, the problem of deciding if a given oriented graph G has an improper ios-injective oriented k-colouring is NP-complete.
Proof. When
the transformation is from the problem of deciding whether there exists an ios-injective homomorphism of a given oriented graph G to
, which is NP-complete by Theorem 4.4. Since there is no ios-injective homomorphism of Dd (from Lemma 4.2) to
, an oriented graph G has an improper ios-injective oriented 3-colouring if and only if
has an ios-injective homomorphism to
.
When
, the transformation is from the problem of deciding whether there exists an ios-injective homomorphism of a given oriented graph G to
. Given an oriented graph G, the transformed instance of improper ios-injective oriented k-colouring is the oriented graph
. The claim that this graph has an improper ios-injective oriented k-colouring if and only if G has an ios-injective homomorphism to
follows from Lemma 6.1.
6.3. Improper Iot-Injective Colourings
Theorem 6.6. For each fixed
, the problem of deciding if a given oriented graph has an improper iot-injective homomorphism to
is NP-complete.
Proof. The transformation is from the problem of deciding whether there exists an iot-injective homomorphism of a given oriented graph G to
, which is NP-complete by Theorem 4.4. Given an oriented graph G, the transformed instance G' is constructed by starting with G and proceeding as follows. For each vertex
, add a copy of
and arcs from each of its vertices to x. Then for every vertex t of each copy of
that was added, add three vertices,
, and arcs from t to each of them. The oriented graph G has an iot-injective
-colouring if and only if G' has an iot-injective
-colouring.
The proof of the following is identical to that of Corollary 6.5, except for replacing “ios” by “iot”.
Corollary 6.7. Let k be a fixed integer. If
, the problem of deciding whether an oriented graph has an improper iot-injective k-colouring is Polynomial. If
, the problem of deciding whether an oriented graph has an improper iot-injective k-colouring is NP-complete.
For a given oriented graph G, we denote by
and
, the smallest number of colours in a proper ios-injective oriented colouring, an improper ios-injective oriented colouring, and an improper iot-injective oriented colouring of G, respectively. The superscript “r” is used to designate the improper colourings because the target graph being reflexive is what allows adjacent vertices to be assigned the same colour. A project for future research is to find tight bounds for these parameters. The upper bounds should be exponential in the in-degree and out-degree consider the disjoint union of all tournaments on a fixed number of vertices. Weak upper bounds can be obtained using the methods in [2] [4] [6] [7]. Tight bounds and efficient algorithms for trees can be obtained as in [2] [4] [7].
Supported
Research of the 2nd and 3rd authors supported by NSERC.