Page 1
Minimum Cost Homomorphism Dichotomy for
Oriented Cycles
Gregory Gutin1, Arash Rafiey2, and Anders Yeo1
1Department of Computer Science
Royal Holloway, University of London
Egham, Surrey TW20 0EX, UK
gutin(anders)@cs.rhul.ac.uk
2School of Computing Science
Simon Fraser University
Burnaby, B.C., Canada, V5A 1S6
arashr@cs.sfu.ca
Abstract. For digraphs D and H, a mapping f : V (D)→V (H) is a homo-
morphism of D to H if uv ∈ A(D) implies f(u)f(v) ∈ A(H). If, moreover, each
vertex u ∈ V (D) is associated with costs ci(u),i ∈ V (H), then the cost of the
u∈V (D)cf(u)(u). For each fixed digraph H, we have the
minimum cost homomorphism problem for H (abbreviated MinHOM(H)). In
this discrete optimization problem, we are to decide, for an input graph D with
costs ci(u), u ∈ V (D),i ∈ V (H), whether there exists a homomorphism of D
to H and, if one exists, to find one of minimum cost. We obtain a dichotomy
classification for the time complexity of MinHOM(H) when H is an oriented
cycle. We conjecture a dichotomy classification for all digraphs with possible
loops.
homomorphism f is?
1Introduction
For directed (undirected) graphs G and H, a mapping f : V (G)→V (H) is a
homomorphism of G to H if uv is an arc (edge) implies that f(u)f(v) is an
arc (edge). Let H be a fixed directed or undirected graph. The homomorphism
problem for H asks whether a directed or undirected input graph G admits a
homomorphism to H. The list homomorphism problem for H asks whether a
directed or undirected input graph G with lists (sets) Lu⊆ V (H),u ∈ V (G)
admits a homomorphism f to H in which f(u) ∈ Lufor each u ∈ V (G).
Suppose G and H are directed (or undirected) graphs, and ci(u), u ∈ V (G),
i ∈ V (H) are nonnegative costs. The cost of a homomorphism f of G to H
is?
input graph G, together with costs ci(u), u ∈ V (G), i ∈ V (H), we wish to find
a minimum cost homomorphism of G to H, or state that none exists.
The minimum cost homomorphism problem was introduced in [10], where
it was motivated by a real-world problem in defence logistics. We believe it
u∈V (G)cf(u)(u). If H is fixed, the minimum cost homomorphism problem,
MinHOM(H), for H is the following discrete optimization problem. Given an
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2G. Gutin, A. Rafiey, and A. Yeo
offers a practical and natural model for optimization of weighted homomor-
phisms. The problem’s special cases include the homomorphism and list homo-
morphism problems [15,17] and the general optimum cost chromatic partition
problem, which has been intensively studied [13,19,20].
There is an extensive literature on the minimum cost homomorphism prob-
lem, e.g., see [5–10]. These and other papers study the time complexity of
MinHOM(H) for various families of directed and undirected graphs. In par-
ticular, Gutin, Hell, Rafiey and Yeo [6] proved a dichotomy classification for
all undirected graphs (with possible loops): If H is a reflexive proper inter-
val graph or a proper interval bigraph, then MinHOM(H) is polynomial time
solvable; otherwise, MinHOM(H) is NP-hard. It is an open problem whether
there is a dichotomy classification for the complexity of MinHOM(H) when H
is a digraph with possible loops. We conjecture that such a classification exists
and, moreover, the following assertion holds:
Conjecture 1. Let H be a digraph with possible loops. Then MinHOM(H) is
polynomial time solvable if H has either a Min-Max ordering or a k-Min-Max
ordering for some k ≥ 2. Otherwise, MinHOM(H) is NP-hard.
For the definitions of a Min-Max and k-Min-Max ordering see Section 3,
where we give theorems (first proved in [10,9]) showing that if H has one of
the two orderings, then MinHOM(H) is polynomial time solvable. So, it is the
NP-hardness part of Conjecture 1 which is the ‘open’ part of the conjecture.
Very recently Gupta, Hell, Karimi and Rafiey [5] obtained a dichotomy clas-
sification for all reflexive digraphs that confirms this conjecture. They proved
that if a reflexive digraph H has no Min-Max ordering, then MinHOM(H) is
NP-hard. Gutin, Rafiey and Yeo [8,9] proved that if a semicomplete multipar-
tite digraph H has neither Min-Max ordering nor k-Min-Max ordering, then
MinHOM(H) is NP-hard.
In this paper, we show that the same result (as for semicomplete multi-
partite digraphs) holds for oriented cycles. This provides a further support for
Conjecture 1. In fact, we prove a graph-theoretical dichotomy for the com-
plexity of MinHOM(H) when H is an oriented cycle. The fact that Conjecture
1 holds for oriented cycles follows from the proof of the graph-theoretical di-
chotomy. In the proof, we use a new concept of a (k,l)-Min-Max ordering
introduced in Section 3. Our motivation for Conjecture 1 partially stems from
the fact that we initially proved polynomial time solvability of MinHOM(H)
when V (H) has a (k,l)-Min-Max ordering by reducing it to the minimum cut
problem. However, we later proved that (k,l)-Min-Max orderings can simply
be reduced to p-Min-Max orderings for p ≥ 1 (see Section 3).
Homomorphisms to oriented cycles have been investigated in a number of
papers. Partial results for the homomorphism problem to oriented cycles were
obtained in [11] and [18]. A full dichotomy was proved by Feder [3]. Feder,
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Minimum Cost Homomorphism Dichotomy for Oriented Cycles3
Hell and Rafiey [4] obtained a dichotomy for the list homomorphism problem
for oriented cycles. Notice that our dichotomy is different from the ones in [3]
and [4].
Bulatov [2] proved that there exists a dichotomy classification for the list
homomorphism problem for digraphs, but no such dichotomy has been ob-
tained and even conjectured to the best of our knowledge. For the homomor-
phism problem for digraphs, we do not even know whether a dichotomy exists
and there is no conjecture of such a classification for the general case.
The rest of this paper is organized as follows. In the next section we consider
so-called levels of vertices in oriented paths and cycles. The concepts of Min-
Max ordering, k-Min-Max ordering and (k,l)-Min-Max ordering are considered
in Section 3. In Section 4 we obtain a dichotomy classification for MinHOM(H)
when H is a balanced oriented cycle. For all oriented cycles H, a dichotomy is
proved in Section 5.
2Levels of Vertices in Oriented Paths and Cycles
In this paper [p] denotes the set {1,2,...,p}. Let D be a digraph. We will use
V (D) (A(D)) to denote the vertex (arc) set of D. We say that xy (x,y ∈ V (D))
is an edge of D if either xy or yx is an arc of D. A sequence b1b2...bpof distinct
vertices of D is an oriented path if bibi+1is an edge for every i ∈ [p − 1]. If
b1b2...bpis an oriented path, we call C = b1b2...bpb1an oriented cycle if bpb1
is an edge. An edge bibi+1(here bpbp+1= bpb1) of an oriented path P or cycle
C is called forward (backward) if bibi+1∈ A(D) (bi+1bi∈ A(D)).
Let P = b1b2...bp be an oriented path. We assign levels to the vertices
of P as follows: we set levelP(b1) = 0, and levelP(bt+1) = levelP(bt) + 1, if
btbt+1is forward and and levelP(bt+1) = levelP(bt) − 1, if btbt+1is backward.
We say that P is of type r if r = max{levelP(bi) : i ∈ [p]} = levelP(bp) and
0 ≤ levelP(bt) ≤ r for each t ∈ [p].
An oriented cycle C is balanced if the number of forward edges equals the
number of backward edges; if C is not balanced, it is called unbalanced. Note
that the fact whether C is balanced or unbalanced does not depend on the
choice of the vertex b1or the direction of C.
Let C = b1b2...bpb1be an oriented cycle. It has two directions: b1b2...bpb1
and b1bpbp−1...b1. In what follows, we will always consider the direction in
which the number of forward arcs is no smaller than the number of backward
arcs. We can assign levels to the vertices of C as follows: level(b1) = k, where
k is a non-negative integer, and level(bt+1) = level(bt) + 1, if btbt+1is forward
and and level(bt+1) = level(bt) − 1, if btbt+1is backward. Clearly, the value of
each level(bi), i ∈ [p], depends on both k and the choose of the initial vertex
b1. Feder [3] proved the following useful result.
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4G. Gutin, A. Rafiey, and A. Yeo
Proposition 1. The integer k and initial vertex b1 in an oriented cycle C
can be chosen such that level(b1) = 0 and level(bi) ≥ 0 for every i ∈ [p]. If
C is unbalanced, then k and b1 can be chosen such that level(b1) = 0 and
level(bi) > 0 for every i ∈ [p] \ {1}.
Since the proposition was proved in [3], we will not give its complete proof.
Instead, we will outline a procedure for finding appropriate k and b1 and
remark on how the procedure can be used in showing the proposition.
Let C = b1b2...bpb1be an oriented cycle. We may assume that b1is chosen
in such a way that if C has a backward edge, then bpb1is a backward edge.
Compute mi, the number of the forward arcs minus the number of backward
arcs in the oriented path b1b2...bi, for each i ∈ [p]. Set k = |min{mi: i ∈ [p]}|.
Assign the level to each vertex of C using the level definition and starting from
assigning level k to b1. By the definition of k, the level of each vertex bj is
non-negative and there are vertices biof level zero. Choose such a vertex bi
with maximum index i and reassign the levels to the vertices of C as follows.
Consider C?= bibi+1...bpb1b2...biand set level(bi) = 0 and the rest of the
levels according to the order of vertices given in C?.
This procedure can be turned into a proof of the proposition by observing
that if C?is unbalanced, then the level of b1 in C?will be greater than the
level of b1in C. Thus, the levels of all vertices vj, j ∈ [i − 1] will be greater
than their levels in C, implying that the only level zero vertex in C?is bi.
Thus, in the rest of the paper, we may assume that the ‘first’ vertex of b1
of an oriented cycle C = b1b2...bpb1is chosen in such a way that the levels of
all vertices of C satisfy Proposition 1.
We will extensively use the following notation: V L(C) = {bt: level(bt) =
0,t ∈ [p]}, h(C) = max{level(bj) : j ∈ [p]}, and V H(C) = {bt: level(bt) =
h(C),t ∈ [p]}. Note that for unbalanced cycles C we have |V L(C)| = 1.
The concepts of this section are illustrated on Figure 1. In particular, Z
is balanced with forward edges b1b2,b3b4,... and backward edges b2b3,b4b5,....
We have level(b1) = level(b3) = level(b5) = 0, level(b2) = level(b4) = level(b6) =
level(b8) = level(b14) = 1, level(b7) = level(b9) = level(b11) = level(b13) = 2
and level(b10) = level(b12) = 3. Thus, h(Z) = 3, V L(Z) = {b1,b3,b5}, V H(Z) =
{b10,b12}.
3k-Min-Max and (k,l)-Min-Max Orderings
All known polynomial cases of MinHOM(H) can be formulated in terms of
certain vertex orderings. In fact, all known polynomial cases can be partitioned
into two classes: digraphs H admitting a Min-Max ordering of their vertices
and digraphs H having a k-Min-Max ordering of their vertices (k ≥ 2). Both
types of orderings are defined in this section, where we also introduce a new
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Minimum Cost Homomorphism Dichotomy for Oriented Cycles5
b
b2
b
b
b
b
b
b
b
b
b
b
b
13
4
5
b6
7
8
9
10
11
12
13
14
Fig.1.
b1b2,b3b2,b3b4,b5b4,b5b6,b6b7,b8b7b8b9,b9b10,b11b10,b11b12,b13b12,b14b13,b1b14 are arcs.
A leveldiagramof oriented cycleZ=b1b2...b14b1,where
type of ordering, a (k,l)-Min-Max ordering. It may be surprising, but we prove
that the new type of ordering can be reduced to the two known orderings.
Let H be a digraph and let (v1,v2,...,vp) be an ordering of the vertices
of H. Let e = vivrand f = vjvsbe two arcs in H. The pair vmin{i,j}vmin{s,r}
(vmax{i,j}vmax{s,r}) is called the minimum (maximum) of the pair e,f. (The
minimum (maximum) of two arcs is not necessarily an arc.) An ordering
(v1,v2,...,vp) is a Min-Max ordering of V (H) if both minimum and maxi-
mum of every two arcs in H are in A(H). Two arcs e,f ∈ A(H) are called a
crossing pair if {e,f} ?= {g?,g??}, where g?(g??) is the minimum (maximum) of
e,f. Clearly, to check that an ordering is Min-Max, it suffices to verify that
the minimum and maximum of every crossing pair of arcs are arcs, too. The
concept of Min-Max ordering is of interest due to the following:
Theorem 1. [10] If a digraph H has a Min-Max ordering of V (H), then
MinHOM(H) is polynomial-time solvable.
We will sometimes call a Min-Max ordering also a 1-Min-Max ordering.
The reason for this will become apparent in the rest of this section.
A collection V1,V2,...Vkof subsets of a set V is called a k-partition of V
if V = V1∪ V2∪ ··· ∪ Vk, Vi∩ Vj= ∅ provided i ?= j.
Let H = (V,A) be a digraph and let k ≥ 2 be an integer. We say that H
has a k-Min-Max ordering of V (H) if there is a k-partition of V into subsets
V1,V2,...Vkand there is an ordering Vi= (vi
such that
1,vi
2,...,vi
?(i)) of Vifor each i
(i) Every arc of H is an arc from Vito Vi+1for some i ∈ [k] and
(ii) (Vi,Vi+1) = (vi
of the subdigraph of H induced by Vi∪ Vi+1for each i ∈ [k]. (All indices
are taken modulo k.)
1,vi
2,...,vi
?(i)vi+1
1
vi+1
2
,...,vi+1
?(i+1)) is a Min-Max ordering
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6G. Gutin, A. Rafiey, and A. Yeo
In such a case, (V1,V2,...,Vk) is a k-Min-Max ordering of V (H); k-Min-Max
orderings are of interest due to the following:
Theorem 2. [9] If a digraph H has a k-Min-Max ordering of V (H), then
MinHOM(H) is polynomial-time solvable.
Our study of MinHOM(H) for oriented cycles H has led us to the following
new concept.
Definition 1. Let H = (V,A) be a digraph and let k ≥ 2 and l be integers.
For l < k we say that H has a (k,l)-Min-Max ordering if there is a (k+l−2)-
partition of V into subsets V1,V2,...,Vk,U2,U3,...,Ul−1(set U1= V1, Ul=
Vk) and there is an ordering Vi= (vi
and there is an ordering Ui= (ui
that
1,vi
2,...,ui
2,...,vi
?u(i)) of Uifor each 1 ≤ i ≤ l such
?v(i)) of Vifor each 1 ≤ i ≤ k
1,ui
(i) Every arc of H is an arc from Vito Vi+1) for some i ∈ [k − 1], or is an
arc from Ujto Uj+1for some j ∈ [l − 1].
(ii) (Vi,Vi+1) is a Min-Max ordering of the subdigraph H[Vi∪ Vi+1] for all
i ∈ [k − 1].
(iii) (Ui,Ui+1) is a Min-Max ordering of the subdigraph H[Ui∪ Ui+1] for all
i ∈ [l − 1].
(iv) (V1,V2,U2) is a Min-Max ordering of the subdigraph H[V1∪ V2∪ U2].
(v) (Ul−1,Vk−1,Vk) is a Min-Max ordering of the subdigraph H[Vk−1∪Ul−1∪
Vk].
It turns out that (k,l)-Min-Max orderings can be reduced to p-Min-Max
orderings as follows from the next assertion:
Theorem 3. If a digraph H has a (k,l)-Min-Max ordering, then MinHOM(H)
is polynomial-time solvable.
Proof. Let H have a (k,l)-Min-Max ordering as described in Definition 1. Let
d = k − l. We will show that H has a d-Min-Max ordering, which will be
sufficient because of Theorems 1 and 2. (Recall that a 1-Min-Max ordering is
simply a Min-Max ordering.) Let us consider two cases.
Case 1: d = 1. It is not difficult to show that the ordering
(V1,V2,U2,V3,U3,...,Vk−2,Uk−2,Vk−1,Vk)
is a Min-Max ordering. Indeed, all crossing pairs of arcs are only in the sub-
graphs given in (ii)-(v) of Definition 1. According to the definition, the maxi-
mum and minimum of every crossing pair is in H.
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Minimum Cost Homomorphism Dichotomy for Oriented Cycles7
Case 2: d ≥ 2. Let si= max{p : i+pd ≤ l} for each i ∈ [d]. Consider the
following orderings for each 2 ≤ i ≤ d
Wi= (Vi,Ui,Vi+d,Ui+d,Vi+2d,Ui+2d,...,Vi+sid,Ui+sid,Vi+(si+1)d)
and the ordering
W1= (V1,V1+d,U1+d,V1+2d,U1+2d,...,V1+s1d,U1+s1d,V1+(s1+1)d).
Observe that W1,W2,...,Wdform a partition of V (H) and that every arc is
from Wito Wi+1for some i ∈ [d], where Wd+1= W1. As in Case 1, it is not
difficult to see that (W1,W2,...,Wd) is a d-Min-Max ordering of H.
? ?
Remark 1. Notice that not always a p-Min-Max ordering can be reduced to a
(k,l)-Min-Max ordering. As an example, consider the directed cycle C5.
4Balanced Oriented Cycles
We say that a balanced oriented cycle C = b1b2...bpb1is of the form (l+h+)q
with q ≥ 1 if P = C − bpb1can be written as P = P1R1P2R2...PqRq, where
V (Pi) ∩ V L(C) ?= ∅, V (Pi) ∩ V H(C) = ∅,
V (Ri) ∩ V L(C) = ∅, V (Ri) ∩ V H(C) ?= ∅
for each i ∈ [q]. We write l+h+instead of (l+h+)1. For example, the cycle in
Figure 1 is of the form l+h+. Balanced oriented cycles C of the form l+h+are
considered in the following:
Theorem 4. Let C = b1b2...bpb1 be a balanced oriented cycle of the form
l+h+. Then MinHOM(C) is polynomial time solvable.
Proof. Let q = min{j : bj ∈ V H(C)}, let m = h(C) and let V = V (C).
Consider the following ordering V = (bp,bp−1,...,bq+1,b1,b2...,bq) of V . We
can define the following natural (m+1)-partition of V : V1,V2,...,Vm+1, where
Vj= {bs∈ V (C) : level(bs) = j − 1}. Note that every arc of C is an arc from
Vjto Vj+1for some j ∈ [m].
Let Vjbe the ordering of Vjobtained from V by deleting all vertices not
in Vj and let Vj = (sj
has no crossing pair of arcs for any j ∈ [m] since, for every pair sj
of arcs in the digraph, we have that either α ≤ γ and β ≤ δ, or α ≥ γ and
β ≥ δ. Thus, C has an (m+1)-Min-Max ordering of vertices and, by Theorem
2, MinHOM(C) is polynomial-time solvable.
1,sj
2,...,sj
b(j)). Observe that the digraph C[Vj∪ Vj+1]
αsj+1
β
, sj
γsj+1
δ
? ?
The following lemma was first proved in [12]; see also [3,21] and Lemma
2.36 in [17].
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8G. Gutin, A. Rafiey, and A. Yeo
Lemma 1. Let P1and P2be two oriented paths of type r. Then there is an
oriented path P of type r that maps homomorphically to P1and P2such that
the initial vertex of P maps to the initial vertices of P1and P2and the terminal
vertex of P maps to the terminal vertices of P1 and P2. The length of P is
polynomial in the lengths of P1and P2.
We need a modified version of Lemma 1, Lemma 2. We say that an oriented
path b1b2...bpof type r is of the form (l+h+)kif the balanced oriented cycle
b1b2...bpar−1ar−2...a2a1b1
is of the form (l+h+)k, where b1a1a2...ar−2ar−1bpis a directed path.
Lemma 2. Let P1 and P2 be two oriented paths of type r. Let P1 be of the
form h+l+and let P2be of the form (l+h+)k, k ≥ 1. Then there is an oriented
path P of type r that maps homomorphically to P1and P2such that the initial
vertex of P maps to the initial vertices of P1and P2and the terminal vertex of
P maps to the terminal vertices of P1and P2. The length of P is polynomial
in the lengths of P1and P2, and P is of the form (l+h+)k.
Proof. We will show that our construction implies that |V (P)| ≤ |V (P1)| ×
|V (P2)|.
We first prove the lemma for the case when k = 1. The proof is by induction
on r ≥ 0. If 0 ≤ r ≤ 1, the claim is trivial. Assume that r ≥ 2. Let P1 =
a1a2...ap, let P2 = b1b2...bq, let s1 = min{i :
s2 = min{i :
β2= min{levelP2(bi) : s2≤ i}. Without loss of generality assume that β1≤ β2
and let t1 = min{i :
levelP2(bi) = β1and i ≤ s2}. Note that β1> 1 as P1is of form h+l+.
By the induction hypothesis, there is an appropriate oriented path P?that
can be mapped homomorphically to a1a2...as1−1and b1b2...bs2−1. There is
also an oriented path P??that can be mapped homomorphically to as1as1+1...at1
and bs2bs2−1...bt2(by reversing the two paths and then reversing the path we
get by the induction hypothesis). Furthermore there is an oriented path P???
that can be mapped homomorphically to at1+1at1+2...apand bt2+1bt2+2...bp.
Let P = P?P??P???(where the arc between the last vertex of P?to the first ver-
tex of P??is oriented from P?to P??and similarly the arc between P??and P???
is oriented in that direction). Note that P is of type r and form h+l+and
maps homomorphically to P1and P2such that the initial vertex of P maps
to the initial vertices of P1and P2and the terminal vertex of P maps to the
terminal vertices of P1and P2. Furthermore |V (P)| ≤ (s1− 1)(s2− 1) + (t1−
s1+1)(s2−t2+1)+(p−s1)(q−t2). As (s1−1)+(t1−s1+1)+(p−s1) = p and
(s2− 1),(s2− t2+ 1),(q − t2) ≤ q we have |V (P)| ≤ pq ≤ |V (P1)| × |V (P2)|.
levelP1(ai) = r} and let
s1 ≤ i} andlevelP2(bi) = r}. Let β1 = min{levelP1(ai) :
levelP1(ai) = β1and i ≥ s1} and let s2 = max{i :
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Minimum Cost Homomorphism Dichotomy for Oriented Cycles9
Now we proceed by induction on k ≥ 1. The base case has already been
proved. Assume that k ≥ 2 and let P1= a1a2...apand P2= b1b2...bq. Let
t = max{i : levelP2(bi) = 0} and let s = max{i : i < t, levelP2(bi) = r}. By
the induction hypothesis, there is an appropriate oriented path P?that can be
mapped homomorphically to P1and b1b2...bs. Also, there is an appropriate
oriented path P??(P???) that can be mapped homomorphically to apap−1...a1
and bsbs+1...bt(P1and btbt+1...bq). Now obtain a new oriented path P by
identifying the terminal vertex of P?with the initial vertex of P??and the
terminal vertex of P??with the initial vertex of P???. Observe that P satisfies
the required properties.
? ?
Consider the oriented cycle C0
that 1,2,3,4 is a Min-Max ordering of V (C0
polynomial-time solvable.
4= 12341 with arcs 12,32,14,34. Observe
4) and, thus, MinHOM(C0
4) is
Theorem 5. Let C = b1b2...bpb1 be a balanced oriented cycle of the form
(l+h+)k, k ≥ 2, and let C ?= C0
Proof. Let C ?= C0
V L(C)}, t = min{j : j > q,bj∈ V H(C)} and m = h(C). Let P1= b1b2...bs,
P2= bqbq−1...bs, P3= bqbq+1...btand P4= b1bpbp−1...bt. Note that each
Pjis of type m. By Lemma 2 there is a path Q1of type m which is mapped
homomorphically to P4, P2 and P3. There is also a path Q2 of type m and
which is mapped homomorphically to P1and P3. Since C ?= C0
vertices vertices of Q1are mapped to the end-vertices vertices of P4, the path
Q1contains more than two vertices. Furthermore, by Lemma 2 we may assume
that Q1is of the form (l+h+)k−1and Q2is of the form l+h+.
Let x (y) be the terminal vertex of Q1 (Q2). Form a new oriented path
Q = q1q2...qlby identifying x with y and let 1 ≤ r ≤ l be defined such that
Q1= q1q2...qrand Q2= qlql−1...qr. As Q1contains more than two vertices
we have that r ≥ 3.
Let D be an arbitrary digraph. We will now reduce the problem of find-
ing a maximum independent set in D (i.e. in the underlying graph of D) to
MinHOM(C). Replace every arc ab of D by a copy of Q identifying q1with
a and qlwith b, and denote the obtained digraph by D?. For every path Q in
D?which we added in the construction of D?we define the cost function c as
follows, where M is a number greater than |V (D)|:
(i) cb1(q1) = cb1(ql) = 0 and cbq(q1) = cbq(ql) = 1;
(ii) cbs(qr) = cbt(qr) = 0 and cb(qr) = M for all b ∈ V (C) − {bs,bt};
(iii) cb2(q) = M for all q ∈ {q2,q3,...,qr−1} (?= ∅, as r ≥ 3);
(iv) if s = 2 then cb1(qr−1) = M;
(v) all other costs of mapping vertices of D?to H are zero.
4. Then MinHOM(C) is NP-hard.
4. Let s = min{j : bj∈ V H(C)}, q = min{j : j > s,bj∈
4and the end-
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10G. Gutin, A. Rafiey, and A. Yeo
Consider a mapping g from V (Q) to V (C), where g(q1) = bq, g(ql) = bq,
and Q1 and Q2 are both homomorphically mapped to P3 (i.e., g(qr) = bt).
Observe that g is a homomorphism from Q to C of cost 2. This implies, in
particular, that there is a homomorphism from D?to H of cost less than M.
We now consider three other homomorphisms from Q to C:
(a) f(q1) = b1and f(ql) = bq, and Q1is mapped to P4and Q2is mapped
to P3homomorphically (i.e., f(qr) = bt). The cost of f is 1.
(b) f?(q1) = bq and f?(ql) = b1, and f?maps Q1 to P2 and Q2 to P1
homomorphically (i.e., f?(qr) = bs). The cost of f?is 1.
(c) f??(q1) = b1 and f??(ql) = b1. We will show that the cost of f??is at
least M. If this is not the case then f??(qr) ∈ {bs,bt} by (ii). First assume that
f??(qr) = bs. By (iii) no vertex of V (Q1) − {qr} is mapped to b2and if s = 2
then by (iv) qr−1is not mapped to b1. However as Q1is of the form (l+h+)k−1
and the path b1bpbp−1...bsis of the form (l+h+)kwe get a contradiction to
f??mapping Q1to C and f??(q1) = b1and f??(qr) = bs. So now assume that
f??(qr) = bt. However Q2is of the form (l+h+) and there is no path in C from
b1to btof the form (l+h+), a contradiction. Therefore the cost of f??is at least
M.
By the above a minimum cost homomorphism h : D?→ C maps all vertices
from a maximum independent set in D to b1and all other vertices from D to
bq. As finding a maximum independent set in a digraph is NP-hard we see that
MinHOM(C) is NP-hard.
? ?
5Dichotomy and Unbalanced Oriented Cycles
We are ready to prove the following main result:
Theorem 6. Let C be an oriented cycle. If C is unbalanced or C is balanced
of the form l+h+or C = C0
Otherwise, MinHOM(C) is NP-hard.
4, then MinHOM(C) is polynomial-time solvable.
By Theorems 4 and 5, to show Theorem 6 it suffices to prove the following:
Theorem 7. Let C = b1b2...bpb1 be an unbalanced oriented cycle. Then
MinHOM(C) is polynomial-time solvable.
Proof. It is well-known that the minimum cost homomorphism problem to a
directed cycle is polynomial-time solvable (see, e.g., [7]). Thus, we may as-
sume that C is not a directed cycle. By Proposition 1, we may assume that
level(b1) = 0 and level(bi) > 0 for all i ∈ [p] \ {1}. Let q = max{j : bj ∈
V H(C)}.
Consider the oriented path P = b1b2...bq. Let Vi+1 = {bj ∈ V (P) :
level(bj) = i} for all i ∈ [k] ∪ {0}, where k = level(bq). Now consider the
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Minimum Cost Homomorphism Dichotomy for Oriented Cycles 11
oriented path Q = b1bpbp−1...bq. Assign levels to the vertices of Q stating
from levelQ(b1) = 0 and continuing as described in Section 1. Observe that all
vertices of Q get non-negative levels. Let Ui+1= {bj∈ V (Q) : levelQ(bj) = i},
i ∈ [l] ∪ {0}, where l = levelQ(bq). Clearly, V1= U1= {b1}; set Ul+1= Vk+1.
Consider the ordering U = (b1,bp,bp−1,...,bq) of the vertices of Q. For
i ∈ [l + 1], the ordering Ui is obtained from U by deleting all vertices not
in Ui. Consider the ordering V = (b1,b2,...,bq) of the vertices of P. For
i ∈ [k + 1], the ordering Viis obtained from V by deleting all vertices not in
Vi. Observe that the ordering (V1,V2,U2) of the vertices of C[V1∪ V2∪ U2]
has no crossing arcs. Similarly, the ordering (Ul,Vk,Vk+1) of the vertices of
C[Ul∪Vk∪Vk+1] has no crossing arcs, and orderings (Vi,Vi+1) and (Uj,Uj+1)
(i ∈ [k],j ∈ [l]) have no crossing arcs. Thus, C has a (k + 1,l + 1)-ordering of
vertices. Now we are done by Theorem 3.
? ?
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