Complexity of trails, paths and circuits in arc-colored

digraphs ∗

Laurent Gourves

Universite Paris-Dauphine, LAMSADE, F-75775, Paris, France

Adria Lyra

Universidade Federal Rural do Rio de Janeiro, UFRRF, Instituto Multidisciplinar, Brazil

Carlos A. Martinhon ∗∗

Fluminense Federal University, Institute of Computation, Niteroi, RJ, Brazil

Jerome Monnot

Universite Paris-Dauphine, LAMSADE, F-75775, Paris, France

Abstract

We deal with different algorithmic questions regarding properly arc-coloreds-t trails, paths and circuits in arc-colored digraphs. Given an arc-coloreddigraph Dc with c ≥ 2 colors, we show that the problem of determiningthe maximum number of arc disjoint properly arc-colored s-t trails can besolved in polynomial time. Surprisingly, we prove that the determinationof a properly arc-colored s-t path is NP-complete even for planar digraphscontaining no properly arc-colored circuits and c = Ω(n), where n denotesthe number of vertices in Dc. If the digraph is an arc-colored tournament, weshow that deciding whether it contains a properly arc-colored circuit passingthrough a given vertex x (resp., properly arc-colored Hamiltonian s-t path)is NP-complete for c ≥ 2. As a consequence, we solve a weak version of an

Email addresses: laurent.gourves@lamsade.dauphine.fr (Laurent Gourves),adrialyra@gmail.com (Adria Lyra), mart@dcc.ic.uff.br (Carlos A. Martinhon ∗∗ ),monnot@lamsade.dauphine.fr (Jerome Monnot)∗ A preliminary version of this paper appeared in 7th Annual Conference on Theory

and Application of Models of Computation, TAMC2010, Czech Republic, [21]∗∗ Partially supported by FAPERJ/Brazil and CNPq/Brazil

Preprint submitted to Discrete Applied Mathematics October 19, 2012

open problem posed in Gutin et. al. [22], whose objective is to determinewhether a 2-arc-colored tournament contains a properly arc-colored circuit.

Key words: Arc-colored digraphs, Properly arc-colored paths/trails andcircuits, Hamiltonian directed path, arc-colored tournaments, Polynomialalgorithms, NP-completeness.

1. Introduction, Notation and Terminology

In the last few years a great number of applications has been modelledas problems in edge-colored graphs [3, 5]. For instance, problems in molec-ular biology correspond to extracting Hamiltonian or Eulerian paths or cy-cles colored in specified pattern [10, 26, 27], transportation and connectiv-ity problems where reload costs are associated to pair of colors at adjacentedges [16, 17, 19], social sciences [9], VLSI optimization [24] among others.In [29, 30], arc-colored multidigraphs are used to model conflict resolution.The Graph Model for Conflict Resolution - gmcr can be viewed as a gametheory-related tool that can assist negotiators with the strategic aspects ofa negotiation. An arc-colored multidigraph of a conflict is a basic structurepermitting the extensive analysis of the possible strategic interactions amongdecision makers (or agents). Starting from a given status quo state (or initialvertex), the objective is to identify the possible compromise resolutions orequilibria (solution vertices) [11, 12, 13].

In this paper, we are specially concerned from an algorithmic perspectivewith different questions regarding properly arc-colored s-t trails, paths andcircuits in arc-colored digraphs. Given a (not necessarily edge-colored) graphG = (V, E), a trail between s and t in G (called s-t trail) is a sequenceρ = (v0, e0, v1, e1, . . . , ek, vk+1) where v0 = s, vk+1 = t and ei = vivi+1 fori = 0, . . . , k and ei 6= ej for i 6= j. A path between s and t in G (called s-tpath) is a trail ρ = (v0, e0, v1, e1, . . . , ek, vk+1) between s and t where vi 6= vj

for i 6= j. To extend the definitions above for digraphs we just change edgesei = vivi+1 by arcs (or oriented edges) ei = ~vivi+1. In this case, s-t trails(resp., s-t paths) are called directed s-t trails (resp., directed s-t paths).

Let Ic = 1, . . . , c be a given set of colors (c ≥ 2). In this work, Dc

denotes a digraph whose arcs have a color in Ic, with no loops and parallel arcslinking the same pair of vertices. The vertex and arc sets of Dc are denotedby V (Dc) and A(Dc), respectively. For a given color i, Ai(Dc) denotes theset of arcs of Dc colored by i. Given Dc and two vertices u, v ∈ V (Dc),

2

we denote by ~uv an arc of A(Dc) and its color by c( ~uv). In addition, wedefine N+

Dc(x) = y ∈ V (Dc) : ~xy ∈ A(Dc) the out-neighborhood of x in Dc

(d+Dc(x) = |N+

Dc(x)| is the out-degree of x in Dc), N−

Dc(x) = y ∈ V (Dc) : ~yx ∈A(Dc) the in-neighborhood of x in Dc (d−

Dc(x) = |N−

Dc(x)| is the in-degree ofx in Dc) and NDc(x) = N+

Dc(x)∪N−

Dc(x) the neighborhood of x ∈ V (Dc). Wesay that, T c defines an arc-colored tournament if it is obtained from a non-oriented complete edge-colored graph Kc by choosing an arbitrary directionfor each colored edge of Kc.

From now on, we write pac instead of properly arc-colored. A pac path(resp., pac trail) is a directed path (resp., trail) such that any two successivearcs have different colors. A pac path or trail in Dc is closed if its end-vertices coincide and its first and last arcs differ in color. They are alsorefereed, respectively, as pac circuits and directed pac closed trails. We saythat a directed pac s-t path (resp., circuit) is Hamiltonian if it visits eachvertex of Dc exactly once. The length of a directed trail, path, closed trail orcircuit is the number of its arcs. Here, we only deal with pac paths of lengthgreater or equal than 2.

1.1. Some Related work

Problems regarding properly edge-colored paths, trails and cycles (or pec

paths, trails and cycles, for short) in c-edge-colored (undirected) graphs havebeen widely studied from a graph theory and algorithmic point of views (see[1, 3, 31], the book [5] and the recent survey [23]). For instance, in [28],the author gives polynomial algorithms for several problems, including thedetermination of a pec s-t path (if one exists). More recently, the authors in[1] introduced the notion of trail-path graph. Using this concept, they extendSzeider’s Algorithm to deal with pec s-t trails and they propose a polynomialalgorithm for the determination of a pec s-t trail. A characterization of c-edge-colored graphs containing pec cycles was first presented by Yeo [31]and generalized in [1] for pec closed trails. In both cases, the proposedresults are used to construct polynomial time algorithms to check whetheran edge-colored graph contains a pec cycle or a pec closed trail.

When dealing with pec paths or trails with additional constraints inc-edge-colored graphs, the results are less optimistic. For example, it iswell known that deciding whether a general 2-edge-colored graph (coloredin blue and red) contains a pec Hamiltonian cycle, a pec Hamiltonian s-tpath, or a pec cycle passing through a prescribed pair of vertices are NP-complete problems [5]. Basically, the idea is to start from the proof of NP-

3

completeness of these problems in uncolored digraphs and to use the directedversion of the Haggkvist’s transformation which consists in replacing each arce = ~xy by a path of length 2, ~xve and ~vey (where ve is a new vertex) withcolors blue and red, respectively. Moreover, it is proved in [9] that decidingwhether a 2-edge-colored graph contains a pec s-t path passing through avertex z is NP-complete. On the other side, many problems of this kind be-come polynomial in 2-edge-colored complete graphs. For instance, in [9] theauthors proved that finding a pec s-t path passing through a vertex z (if any)can be solved in polynomial time. The authors of [14] produced a nice charac-terization of c-edge-colored complete graphs which admit a pec Hamiltonianpath (with a non specified source and destination), and then they deduce anew polynomial algorithm for finding it (if one exists). In [20], the authorsconsider a number of path and trail problems restricted to c-edge-coloredgraphs with no pec cycles or pec closed trails. These classes are interestingsince they constitute, in a certain sense, the edge-colored counterpart of non-colored acyclic digraphs. Finally, in [25] a characterization of c-edge-coloredmultigraphs which contain a pec Eulerian trail is given. A O(n2logn) algo-rithm for finding a pec Eulerian trail in c-edge-colored multigraphs (if oneexists) is described in [7].

In [2, 16, 19], the authors consider edge-colored s-t paths, trails andwalks with minimum reload costs. In this case, we are given a c-edge-colored graph and a c × c matrix R = [ri,j] (for i, j ∈ Ic) whose entriesdefine the reload cost when changing color i for color j. Given a trail (path)ρ = (v1, e1, v2, e2, . . . , ek, vk+1) between vertices s and t, we define the reloadcost of ρ as r(ρ) =

∑k−1j=1 rc(ej),c(ej+1). In [2], the authors deal with the deter-

mination of a minimum directed s-t trail with reload costs in edge-coloreddigraphs whose total cost is given by the sum of reload costs (between succes-sive colors in a trail) and positive costs associated to each arc. As discussedin [19], pec trails (or paths) in edge-colored graphs may be converted intoreload s-t trails (or paths) by conveniently choosing reload costs between eachpair of colors (for instance, by setting ri,i = 1 and ri,j = rj,i = 0, ∀i, j ∈ Ic,i 6= j and by setting arc costs equal to 0. In this case, we seek a s-t trail (ors-t path) with total reload cost 0).

Finally, if we deal with c-arc-colored digraphs, the existing results con-cerning the complexity of finding pac s-t paths and circuits are rather rareand less optimistic. To our best knowledge, there is only one result whichsays that deciding whether a pac circuit exists in 2-arc-colored digraphs, isNP-complete [22].

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1.2. Contributions

In Section 2, we deal with the problem of finding pac s-t trails in c-arc-colored digraphs Dc with c ≥ 2. We prove that decide whether Dc containsa directed pac closed trail can be done in polynomial time. Further, wepresent a polynomial time algorithm for determining the maximum numberof arc disjoint pac s-t trails. In Section 3, we restrict our attention to pathproblems over c-arc-colored digraphs. We show that the determination ofone pac s-t path is NP-complete even if Dc is a planar c-arc-colored digraphcontaining no pac circuits and c = Ω(|V (Dc)|). Finally, in Section 4 wefocus on c-arc-colored tournaments. We prove that deciding whether a c-arc-colored tournament T c (for c = Ω(|V (T c)|2)) contains a pac circuit passingthrough a given vertex x is NP-complete. This solves a weak version of anopen problem initially posed by Gutin, Sudako and Yeo [22], whose objectiveis to determine whether T c (for c = 2) contains a pac circuit. In addition, weprove that deciding whether T c has a pac s-t path or a pac Hamiltonian s-tpath is NP-complete. Notice that there is no evident link between decidingwhether a c-arc-colored tournament possesses a pac s-t path and a pac

Hamiltonian s-t path, although finding a pac s-t path seems to be an easierproblem than finding a pac Hamiltonian s-t path.

2. pac trails and closed trails in arc-colored digraphs

Here, we are interested in the complexity of finding pac s-t trails andclosed trails in arc-colored digraphs. These problems turn out to be polyno-mial using minimum cost flow computation.

Theorem 1. Given an arbitrary c-arc-colored digraph Dc, finding a pac s-ttrail in Dc (if any) can be done within polynomial time.

Proof: A more general version of this problem was polynomially solvedin the Proposition 1 of [2] (remark that s-t paths in [2] are not necessar-ily simple). Here, given reload costs ri,j associated to each pair of col-ors i, j ∈ Ic (see Subsection 1.1 for the definition of reload costs), andcosts w(e) associated to each arc e = ~xy, the objective is minimize f(ρ) =∑k

i=1 w(ei) +∑k−1

j=1 rc(ej),c(ej+1), where ρ = (v1, e1, . . . , ek, vk+1) with v1 = s,vk+1 = t and ei 6= ej for i 6= j, is a sequence of arcs in a directed s-t trail(here, let us call this problem the Minimum Reload+Weight Directed s-t Trailproblem).

5

Hence, in order to find whether there exists a pac s-t trail in Dc, it sufficesto set w(e) = 0 for every arc e = ~xy of Dc, assign reload costs ri,i = 1 andri,j = rj,i = 0 for i, j ∈ Ic with i 6= j. Thus, from Proposition 1 of [2] thereexists a reload+weight directed s-t trail ρ with total cost f(ρ) = 0, if andonly if, Dc has a pac s-t trail. Therefore, we can find a pac s-t trail withinpolynomial time (provided that one exists).

Corollary 2. Let Dc be a c-arc-colored digraph with c ≥ 2. The problem offinding a directed pac closed trail in Dc (if any) can be solved in polynomialtime.

Proof: Our construction is done in various steps. For each vertexx ∈ V (Dc) (one at a time) we build a new arc-colored digraph, say Dc

x, byreplacing x by two new vertices x1, x2 with N+

Dcx(x1) = N+

Dc(x) and N−

Dcx(x2) =

N−

Dc(x) (all incoming and outgoing arcs are colored as in Dc) and find, if oneexists, a pac trail from x1 to x2 in Dc

x. If the associated directed closedtrail passing by x in Dc, say τ , is not pac (since that both incoming andoutgoing arcs of τ passing by x may have the same color) we proceed withthe following step: for each color i with N i

Dc(x) 6= ∅, we delete all outgoingarcs at x1 with colors j 6= i and we delete all incoming arcs at x2 colored i.Now, we try to find a pac trail between x1 and x2 (in any). We repeat theprocess for another color (other than i) until a directed pac trail passing byx in Dc is obtained. Obviously, these steps are polynomially bounded.

Corollary 3. The problem of maximizing the number of arc disjoint pac s-ttrails in Dc can be solved in polynomial time.

Proof: Analogously to [2] for the minimum reload+weight directed s-ttrail problem, the idea is to apply a splitting procedure to all vertices v ofV (Dc) \ s, t (with k1(v) incoming arcs and k2(v) outgoing arcs) and con-struct a non-colored digraph H(v) with unitary arc capacities as illustratedin the Figure 1. Let H be the new uncolored digraph obtained after repeatingthis process for each v ∈ V (Dc) \ s, t.

In the sequel, after constructing digraph H (associated to Dc) we assignthe same reload and arc costs as in Theorem 1 and we define unitary arccapacities. Then it suffices to solve a sequence of minimum cost flow problemsfrom s to t in H . The algorithm proceeds as follows: (1) Set θ ← n − 2;(2) Solve the minimum cost flow problem between s and t in H by sending

6

(b)(a)

1

2

1

2

H(v)

k2(v)

k1(v)

v

Gc

Figure 1: Splitting at vertex v ∈ V (Dc) with k1(v) incoming arcs and k2(v) outgoing arcs.

θ units of flow and obtain ρ (if one exists); (3) If H contains a feasible flowρ with f(ρ) = 0 then we are done (return ρ, θ and stop). Otherwise, setθ ← θ − 1 and go to step 2. We clearly get a polynomial time procedureto maximize the number of pac trails from s to t since the minimum costflow problem is polynomial time solvable. Note in the minimum cost flowproblem that, since all arcs have unitary arc capacities and the mass balanceconstraints must be satisfied at all vertices of V (Dc) \ s, t, some of thesecan be visited several times for one or more pac s-t trails.

3. pac paths in arc-colored digraphs with no pac circuits

Finding pec paths, pec trails and pec cycles in edge-colored graphs arepolynomial problems [1, 28]. However finding pac paths or pac circuits inarc-colored digraphs seems harder. For example, the authors of [22] provedthat deciding whether a 2-arc-colored digraph contains a pac circuit is NP-complete. However, the pac s-t path problem is polynomial in the followingsimple case:

Theorem 4. Deciding whether a 2-arc-colored digraph Dc with no circuitsat all contains a pac path from s to t can be solved in polynomial time.

Proof: We use an algorithm which maintains a set of colors C(v) foreach vertex v. At the beginning of the algorithm, C(v) = ∅ for all v. SinceDc has no circuits, the level of a vertex v, denoted by ℓ(v), is the length (i.e.number of arcs) of the longest path between s and v. Therefore ℓ(s) = 0,ℓ(v) ≤ n − 1 for all v and ℓ(u) < ℓ(v) for all arc ~uv. For digraphs with no

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circuits, a longest path can be found in time O(n + m) [5]. Let us considerthe following algorithm.

For j = 1 to n− 1 Do

For i = 0 to j − 1 Do

For all arc ~uv such that ℓ(u) = i and ℓ(v) = j Do

If u = s or C(u)\c( ~uv) 6= ∅Then C(v)← C(v)∪c( ~uv)

End For

End For

End For

By construction, a color in C(v) indicates the color of the last arc of apac path from s to v. Note that, if C(v) = ∅ it follows that u 6= s andC(u) \ c( ~uv) = ∅. So, for all arcs ~wu of Dc with ℓ(w) ≤ ℓ(u) we have thatc( ~wu) = c( ~uv), thus avoiding the presence of a pac path between s and v.Therefore, Dc admits a pac s-t path, if and only if, C(t) 6= ∅.

Unfortunately, this result does not hold in 2-arc-colored digraphs with nopac circuits (note that non pac circuits are allowed in this case).

Theorem 5. Deciding whether a 2-arc-colored digraph with no pac circuitscontains a pac path from s to t is NP-complete.

Proof: We use a reduction from the Path with Forbidden Pairs Problem(pfpp, in short). In pfpp, we are given a (non-colored) digraph D = (V, A),two vertices v, w ∈ V and a collection C = a1, b1, . . . , aq, bq of pairs ofvertices (ai 6= bi) from V \v, w. The objective is to determine whether thereexists a directed path connecting v to w and passing through at most onevertex from each pair. This problem was shown NP-complete [15] even if D isacyclic and all pairs of C are required to be disjoint, i.e., ai, bi∩aj , bj = ∅for i 6= j (see problem [GT54] page 203 in [18]).

Let D = (V, A) be an acyclic digraph containing v, w ∈ V and a subset Cof disjoint pairs of vertices. W.l.o.g., assume that d−

D(v) = d+D(w) = 0. The

construction of Dc is done in two steps. We build a digraph D′ at first andthen we build Dc from D′. The digraph D′ = (V ′, A′) is such that V ′ = V ∪s, A′ = A∪A′

1∪A′

2 and vertices v and w are replaced by u and t respectively.

8

(a) (b)

1

u

5

1

Color 1

Color 2

5

3 3

sv w

442

2

t

Figure 2: Reduction from the pfpp with C = 1, 2, 3, 4 to the pac s-t path problem.Color 1 (resp., 2) corresponds to red (resp., blue). No pac circuits are allowed

.

Let A′

1 := ~sa1, ~sb1, ~aqu, ~bqu and A′

2 := ~aiai+1, ~aibi+1, ~biai+1, ~bibi+1 : i =1, . . . , q−1. For the moment, notice that two arcs connecting the same pairof vertices may exist.

We build Dc as follows: for arcs in A′

1, ~sa1 and ~sb1 are colored in blue

(color 2), while arcs ~aqu and ~bqu are colored in red (color 1). Next, we applya directed version of Haggkvist’s transformation (see Subsection 1.1): eacharc e = ~xy of A ∪ A′

2 is replaced by a directed path of length two, that is~xve, ~vey, except for arcs incident to t. If e = ~xy ∈ A, then ~xve is colored

in blue and ~vey is colored in red. If e = ~xt, then e is colored in blue. Byextension, arcs ~xve, ~vey are in A. If e = ~xy ∈ A′

2 then ~xve is colored in redand ~vey is colored in blue. By extension, arcs ~xve, ~vey are in this case in A′

2.The construction is completed (an example is given in Figure 2).

This construction is clearly done within polynomial time and Dc is a 2-arc-colored digraph. Further, we can directly establish the following intermediateproperty.

Property 1. Any pac path of Dc cannot use two consecutive arcs ~xy and~yz such that ~xy ∈ A (resp., ~xy ∈ A′

1 ∪A′

2) and ~yz ∈ A′

1 ∪A′

2 (resp., ~xy ∈ A)except if y = u.

From Property 1, we deduce that any pac path of Dc from s to t ispartitioned into two parts and uses some arcs in A′

1 ∪ A′

2 at first and then,

9

once vertex u has been visited, it uses some arcs in A.Let us show that Dc does not contain any pac circuit. Since (V ′, A) has

no circuits (by hypothesis) and (V ′, A′

1∪A′

2) has no circuits (by construction),any circuit of Dc must contain two consecutive arcs such that the first arcis in A (resp., A′

1 ∪ A′

2) and the second arc is in A′

1 ∪ A′

2 (resp., A). UsingProperty 1, the circuit is not pac.

Finally, using Property 1, we claim that we have a directed path fromv to w in D visiting at most one vertex from each pair of C, if and onlyif, we have a pac path from s to t in Dc. To see that, let Γ with |Γ| ≤ qbe a subset of vertices of C belonging to a directed path from v to w in D.As a consequence, we can construct a directed pac sub-path from u to t inDc, say α, visiting the same set Γ of vertices and only containing arcs of A.Therefore, we can determine a pac path from s to t in Dc by concatenatinga pac sub-path from s to u and containing no vertices of Γ (only with arcsof A′

1 ∪ A′

2), which always exist in this case, with the pac sub-path α fromu to t. Conversely, consider a pac path from s to t in Dc (note that all pac

s-t paths in Dc contain vertex u). Thus, by Property 1, it follows that theassociated pac sub-path from s to u only contains arcs of A′

1 ∪ A′

2 and thepac sub-path from u to t only contains arcs of A (each of them containing atmost one vertex of ai, bi for i = 1, .., q). After deleting all arcs of A′

1 ∪ A′

2

in Dc, the resulting path from u to t will be directly associated to a pathfrom v to w in D passing through at most one vertex from each pair of C.

Now, we show that our previous theorem can be extended to include anynumber of colors. Formally, we have the following result:

Corollary 6. Deciding whether a c-arc-colored digraph Dc with no pac cir-cuits contains a pac s-t path is NP-complete, even if c = Ω(|V (Dc)|2).

Proof: We extend Theorem 5 to construct digraphs with 2n vertices,c = Ω(n2) colors and with no pac circuits. To do so, we first construct a2-arc-colored digraph Dc′

ϕ with no pac circuits, c′ = 2 and with n verticesas done in Theorem 5. Next, we build a tournament T c

n with n vertices con-taining no circuits and additional colors Ic with Ic ⊇ Ic′. For that, given anon edge-colored complete graph Kn, it suffices to choose arbitrary verticesof Kn, one at a time, and change all adjacent (non-oriented) edges by incom-ing arcs with non-coincident colors of Ic. Note that, T c

n contains no circuits.Next, to connect T c

n and Dc′

ϕ , we choose two arbitrary vertices v1 ∈ V (Dc′

ϕ )

10

and v2 ∈ V (T cn) and add arc ~v1v2 with an arbitrary color of Ic′. The resulting

connected digraph Dc has 2n vertices and Ω(n2) different colors (the colorsinside T c

n). Therefore, pac s-t paths in 2-arc-colored digraphs (with no pac

circuits) correspond to pac s-t paths in digraphs with c = Ω(n2) colors (withno pac circuits) and vice versa.

Theorem 5 can also be extended to planar c-arc-colored digraphs.

Corollary 7. Deciding whether a planar c-arc-colored digraph DcP with no

pac circuits contains a pac s-t path is NP-complete even for c = Ω(|V (DcP )|).

Proof: We initially prove our result for c′ = 2 colors (red and blue). Ba-sically, given an arbitrary Dc′ with n vertices and containing no pac circuits,the idea is to conveniently change all intersections by new vertices in orderto make it planar. Note that the number of intersections is polynomiallybounded on the size of Dc′.

Thus, whenever we have an intersection between 2 arcs ~ab and ~cd, saycolored j, we add 3 new vertices f1, f2 and f3 and replace arcs ~ab, ~cd by 2

sets of arcs ~af1, ~cf2, ~f2b, ~f3d and ~f1f2, ~f2f3, respectively colored j and i.

However, if ~ab and ~cd have different colors (say i and j), we add the vertices

f1, f2, f3 and change ~ab and ~cd by arcs ~cf2, ~f2f1, ~f3d all colored j, and

arcs ~af2, ~f2f3, ~f1b all colored i (see Figure 3). Denote by Dc′

P1the resulting

planar 2-arc-colored digraph containing no pac circuits. Therefore, if wehave some path passing by ~ab (resp., ~cd) in Dc′, we have a path passing byvertices a and b (resp., c and d) in Dc′

P1.

Now, we define a new planar (non-oriented) graph GP with n verticesand maximum number of edges given by 3n−6 (see Bondy and Murthy [8]).Choose an arbitrary vertex of GP , one at a time, and change all adjacent(non-oriented) edges by incoming arcs with non-coincident colors of Ic (withIc ⊇ Ic′). Repeat this process for all the remaining vertices of GP . Obviously,the resulting planar digraph, say Dc

P2, contains no pac circuits. Finally,

to connect DcP1

and DcP2

, choose two arbitrary vertices v1 ∈ V (Dc′

P1) and

v2 ∈ V (DcP2

) and add arc ~v1v2 with an arbitrary color of Ic′ and such that

~v1v2 has no intersection with any other arc of Dc′

P1and Dc

P2, respectively. Now,

since Ic′ ⊆ Ic and Ic contains Ω(n) colors we can prove that our result alsoholds for digraphs Dc

P with 2n vertices (since V (DcP ) = V (Dc′

P1) ∪ V (Dc

P2))

and c = Ω(|V (DcP )|) colors.

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color i

color j

f1

b

f2

c

f3

d

aa

b

d

c

c

da

b

f1 f3

f2

a

b

d

c

Figure 3: (a)Intersection of arcs with the same color (above). (b) Intersection of arcs withdifferent colors (below).

4. pac circuits, paths and Hamiltonian paths in c-arc-colored tour-

naments

The problems of finding pac s-t paths and pac circuits in c-arc-coloredtournaments are challenging. For example, the complexity of determininga pac circuit in a 2-arc-colored tournament was posed in [5, 22] and it isstill open. Here, we propose and solve a weaker version of this problem.We show that deciding whether a c-arc-colored tournament contains a pac

circuit passing through a given vertex x is NP-complete. As a consequence,we prove that finding pac s-t paths in tournaments is also NP-complete.

We also deal with the determination of pac Hamiltonian s-t paths inarc-colored tournaments. When restricted to uncolored tournaments, one ofthe earliest results is Redei’s theorem, which states that every tournamenthas an Hamiltonian directed path (the endpoints are not specified). More re-cently, in [6] the authors gave a polynomial algorithm to find an Hamiltoniandirected s-t path (if one exists) in an uncolored tournament. Recently in [14](using a nice characterization) the authors show that the problem of findingpec Hamiltonian s-t path is polynomial in c-edge-colored complete graphsfor c ≥ 3, solving a conjecture posed in [4] (the case c = 2 was previouslysolved in [4]). Unfortunately, unless P=NP, we prove that these resultscannot be extended to the directed case.

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Thus, we begin with the following result:

Theorem 8. Deciding whether a c-arc-colored tournament contains a pac

circuit visiting a given vertex x is NP-complete even for c = Ω(|V (Dc)|2).

Proof: We initially prove our result for c′ = 2 and then we extend itfor c = Ω(|V (Dc)|2). Thus, we start from the 2-arc-colored digraph Dc′ =(V ′, A′) built in Theorem 5 and we complete it in order to get a 2-arc-coloredtournament T c′. The idea is to get a tournament whose pac circuits passingthrough x = s (if one exists) also visit vertex t. Then, directed paths fromv to w in D (visiting at most one vertex from each pair of C), instance ofthe Path with Forbidden Pairs Problem, correspond to pac circuits passingthrough s in T c′ and vice-versa.

Recall that in the construction of Dc′ (see the proof of Theorem 5), wereplace each arc e ∈ A (resp., e from A′

2), except those which are incident tot, by a directed path of length two in A (resp., in A′

2) where the added vertexis denoted by ve. If e ∈ A (resp. e ∈ A′

2) then we suppose that ve ∈ V (A)(resp., ve ∈ V (A′

2)).Now, we show how to build the tournament T c′. The construction is done

in four steps:

(1) Build a set of arcs A′

3 as follows. Add a red arc ~ts and a blue arc ~us.Then, add a blue arc ~tx for each x /∈ NDc′ (t), a blue arc ~xu for eachx /∈ NDc′ (u) and a blue arc ~xs for each x /∈ NDc′ (s). Do A(Dc′) ←A(Dc′) ∪A′

3.

(2) Build a set of arcs A′

4 as follows. Choose an arbitrary vertex ve ofV (A) (resp., V (A′

2)) with an incoming blue (resp., red) arc ~yve (resp.,

~aive or ~bive), and add a blue (resp., red) arc ~vex for every x /∈ NDc′ (ve).Repeat the process for the remaining vertices ve of V (A) (resp., V (A′

2))by following an arbitrary order. Let A′

4 be this new set of arcs and doA(Dc′)← A(Dc′) ∪ A′

4.

(3) Build a set of blue arcs A′

5 = ~aqx : ∀x /∈ NDc′ (aq) ∪ ~bqy : ∀y /∈(NDc′ (bq) ∪ aq). Recall that aq, bq is the last pair of C. SetA(Dc′)← A(Dc′) ∪ A′

5.

(4) Build a set A′

6 of blue arcs with endpoints in V (Dc′) \ (s, u, t, aq, bq∪ve : ve ∈ V (A) ∪ V (A′

2)) and arbitrary directions. Set A(Dc′) ←A(Dc′) ∪A′

6.

13

The construction is completed. It is clearly done within polynomial time,and T c′ is a 2-arc-colored tournament. We now give some useful properties:

Property 2. The following properties hold:

(i) Any pac circuit passing through s (resp., u) in T c′ uses ~ts and one arc

among ~sa1, ~sb1 (resp., uses exactly one arc among ~aqu, ~bqu and onearc ~uve ∈ A).

(ii) No pac circuit passing through s in T c′ uses an arc of A′

4.

(iii) No pac circuit passing through s in T c′ uses an arc of A′

5 ∪ A′

6.

Proof: For (i). Due to step (1) of the above procedure, there is aunique red arc incident to s (resp., t) which is ~ts. Thus, any pac circuitpassing through s also visits t. Moreover, vertex s only has two outgoingarcs ~xa1 and ~xb1 which are colored in blue.Concerning vertex u, ~aqu and ~bq are the only red arcs incident to u. Thus,if a pac circuit visits u then it contains one of these two arcs as incomingarc and one arc ~uve ∈ A as outgoing arc. Actually, vertex u has only arcs~uve ∈ A and ~us as outgoing arcs and no pac circuit can use the blue arc ~us

since all arcs going out of s are blue.For (ii). By contradiction, assume that there is a pac circuit passing

through s, ρ = (v1, e1, . . . , ek, vk+1) with v1 = vk+1 = s and containingsome arcs of A′

4. Consider the first arc ep ∈ A′

4 used by ρ (i.e., eq /∈ A′

4

for q = 1, . . . , p − 1). By construction ep = ~vex and from (i), we deducek > p > 1 (i.e., x /∈ s, t). Since ep−1 /∈ A′

4 and ep−1 /∈ A′

3 from (i),arc ep−1 = ~yve ∈ A ∪ A′

2. Thus, ep−1 has the same color as ep, which is acontradiction.

For (iii). By contradiction. Firstly assume that there is a pac circuitpassing through s, ρ = (v1, e1, . . . , ek, vk+1) with v1 = vk+1 = s and contain-ing some arcs of A′

5. Like previously, consider the first arc ep ∈ A′

5 of ρ (i.e.,eq /∈ A′

5 for q = 1, . . . , p − 1). W.l.o.g., suppose that ep = ~aqx (the same

result holds for ep = ~bqx); we get x 6= u from (i). Then, ep−1 = ~veaq ∈ A iscolored in red and from (ii) we deduce that ep−2 = ~yve ∈ A and is coloredin blue. Since all arcs in A′

6 are blue like ep−2, by induction we deduce thateq ∈ A for q = 1, . . . , p− 1. We obtain a contradiction since from (i) e1 ∈ A′

1

(i.e., e1 ∈ ~sa1, ~sb1).

14

Now, suppose that a pac circuit passing through s, ρ = (v1, e1, . . . , ek, vk+1)with v1 = vk+1 = s contains some arcs in A′

6. Consider the last arc ep =~xy ∈ A′

6 used by ρ (i.e., eq /∈ A′

6 for q = p + 1, . . . , k + 1). Since ep is coloredin blue and y 6= t (from (i)), we deduce that ep+1 is colored in red. Then,we get y = ai or y = bi and ep+1 = ~yve ∈ A′

2 since ep+1 /∈ A′

6. Moreover,from (ii), ep+2 = ~vez ∈ A′

2 is colored in blue. Now, since ek ∈ A (the arc ofρ incoming in vertex t) is also colored in blue, the pac sub-path of ρ from x

to t = vk must contain arc ~aqu or ~bqu (using Property 1 of Theorem 5, it isthe only way to flip arcs of A′

2 to arcs of A). Thus, this pac circuit ρ canbe decomposed into three pac paths: ρ1 from y to u, ρ2 from u to s (andcontaining arc ek+1 = ~ts) and ρ3 from s to y. In particular, the pac path ρ3

begins with a blue arc (by (i)), only uses arcs in A′

2 and ends by a blue arc,which is impossible since ρ3 does not contain u. Actually, path ρ3 cannotuse some arcs of A′

6. We have e2 = ~x1ve ∈ A′

2 with x1 ∈ a1, b1 (since thearc must be colored in red) and using (ii), arc e3 = ~vex2 with x2 ∈ a2, b2 iscolored in blue. Thus, e4 /∈ A′

5 ∪ A′

6. Then, the result follows by induction.Notice that it may exist a pac circuit containing one arc e = ~xy ∈ A′

6 (butnot passing through s). In this case, this pac circuit is composed of two pac

paths ρ1 from y to u and ρ2 from u to y: ρ1 only uses arcs of A′

2 from y to

aq (or bq) and uses arc ~aqu ∈ A′

1 (or ~bqu ∈ A′

1) while ρ2 only uses arcs of Afrom u to x and uses arc e = ~xy ∈ A′

6. •Using Properties 1 and 2, we can easily see that we have a directed path

from u to w in D and visiting at most one vertex from each pair of C, if andonly if, we have a pac circuit passing through s in T c′.

Finally, to extend our result for tournaments with c = Ω(|V (Dc)|2) weperform the same steps as described in Corollary 6. In this case, we constructtwo (non connected) tournaments T c′ and T c

n (each with n vertices) suchthat T c′ for c′ = 2 is defined as above and T c

n contains no circuits and at

least |Ic| = n(n−1)2

different colors with Ic ⊇ Ic′ (see Corollary 6). Now to

connect both tournaments, we introduce an arc ~xy for every x ∈ V (T c′) andy ∈ V (T c

n) all colored with arbitrary colors of Ic. The resulting connected

tournament, say T c2n, has 2n vertices and at least n(n−1)

2different colors (the

colors inside T cn). Obviously, we have a pac circuit passing by x in T c

2n, ifand only, we have a pac circuit passing by x in T c′.

Corollary 9. Deciding whether a c-arc-colored tournament T c contains apac s-t path is NP-complete even for c = Ω(|V (T c)|2).

15

Proof: In the proof of Theorem 8, we have a pac circuit passing throughs if and only if we have a pac s-t path in T c.

In the sequel, we explain why the construction of Theorem 8 cannot beused to solve the open problem raised in [22] (regarding the existence of pac

circuits in tournaments). Actually, the main difficulty is to decide how toconnect the vertices inside subset C (See step 4 – subset A′

6 in Theorem 8).To see that, suppose a subset C in the non-colored digraph D (with source vand destination w) containing 3 pairs (ai, bi) for i = 1, 2, 3. In addition, alsoconsider (v, a1) in A(D). Now, in the construction of T c (step 4), supposethat arc (a1, a2) is blue (that is, (a1, a2) in A′

6). Thus, after constructing thetournament through the steps 1-4, the circuit (a1, a2); (a2, ve); (ve, a3); (a3, u);(u, v′

e); (v′

e, a1) is a pac circuit not passing by s since (a1, a2); (ve, a3); (u, v′

e)are blue and (a2, ve); (a3, u); (v′

e, a1) are red. In this way, our constructiondoes not hold for the former open problem.

We finish the paper by considering the pac Hamiltonian s-t path problem.

Theorem 10. Deciding whether a 2-arc-colored tournament T c contains apac Hamiltonian s-t path is NP-complete.

Proof: We use a reduction from the directed Hamiltonian s′-t′ pathproblem in general uncolored digraphs (DHPP in short). Given a digraphD = (V, A) and two vertices s′, t′, DHPP asks whether a directed Hamiltonians′-t′ path exists. DHPP is NP-complete (see problem [GT39] page 199 in[18]).

Let D = (V, A) be a digraph where V = v1, . . . , vn and v1 = s′, vn =t′, instance of DHPP. W.l.o.g., assume that d−

D(v1) = d+D(vn) = 0. The

construction of the 2-arc-colored tournament T c is done in two steps: wefirst build a 2-arc-colored digraph Dc and then we complete Dc into T c.

The 2-arc-colored digraph Dc = (V ′, A′) is built in the following way:

V ′ = viin, v

iout : i = 1, . . . , n and A′ = A′

1 ∪ A′

2 where A′

1 = ~vioutv

jin : ~vivj ∈

A and A′

2 = ~viinv

iout : i = 1, . . . , n. Arcs in A′

1 are colored in red while arcsin A′

2 are colored in blue. See Figure 4 for an illustration of Dc.Next we build the tournament T c from Dc as follows. For every missing

arc in Dc, we apply the following procedure where 1 ≤ i < j ≤ n is assumed.If the endpoints of the missing arc are vi

in and vjin (resp., vi

in and vjout), add

a blue arc ~vjinvi

in (resp., ~vjoutv

iin). If the endpoints of the missing arc are vi

out

16

v1

v2

v3

v4

D

v1in v1

out

v2in v2

out

v3in v3

out

v4in v4

out

Dc

Figure 4: A digraph D and the 2-arc-colored digraph Dc. Dotted arcs are colored in blue

and rigid arcs are colored in red.

v1

v2

v3

v4

D

v1in v1

out

v2in v2

out

v3in v3

out

v4in v4

out

T c

Figure 5: A digraph D and the 2-arc-colored tournament T c. Dotted arcs are colored inblue and rigid arcs are colored in red.

and vjin (resp., vi

out and vjout), add a red arc ~vj

inviout (resp., ~vj

outviout). These

new blue (resp., red) arcs define a set denoted by A′′

2 (resp., A′′

1).The construction is completed (see Figure 5 for an illustration). It is

clearly done within polynomial time. The resulting tournament is 2-arc-colored. Its blue arcs belong to A′

2 ∪A′′

2 while its red arcs belong to A′

1 ∪A′′

1.Let us give an intermediate property.

Property 3. No pac path from v1in to vn

out in T c can use an arc of A′′

1 ∪A′′

2.

Proof: By contradiction suppose that a pac path ρ = (v0, e0, v1, e1, . . . ,ek, vk+1) linking v0 = v1

in to vk+1 = vnout uses some arcs of A′′

1 ∪A′′

2. Considerthe last arc ep ∈ A′′

1∪A′′

2 used by ρ (that is eq /∈ A′′

1∪A′′

2 for q = p+1, . . . , k+1).

If ep = ~vjinv

iin or ep = ~vj

outviin (i < j) then it belongs to A′′

2 and it is blue.

17

We have viin 6= vn

out so the path must contain an arc going out of viin which

does not belong to A′′

1 ∪ A′′

2. This arc ep+1 = ~viinv

iout is blue, contradiction.

Otherwise, ep = ~vjinv

iout (i 6= j) or ep = ~vj

outviout. Therefore ep ∈ A′′

1 and itis red. We have vi

out 6= vnout since ~vn

invnout is the unique arc coming into vn

out.Then, the path must contain an arc ep+1 /∈ A′′

1 ∪A′′

2 going out of viout but all

arcs of A′

1∪A′

2 going out of viout are red since they belong to A′

1, contradiction.

We deduce from Property 3 that any pac path from v1in to vn

out in T c

only uses arcs of A′

1∪A′

2. Thus, D admits a directed Hamiltonian path froms′ = v1 to vn = t′, if and only if, T c has a pac Hamiltonian path from s = v1

in

to t = vnout.

5. Conclusions and final remarks

As a future direction, one related question is the problem is to polyno-mially decide whether a 2-arc-colored tournament contains a directed pac

Hamiltonian path with non-fixed extremities. Finally, note that our resultsover arc-colored digraphs can be easily extended to handle arc-colored mul-tidigraphs. To do that, given an arc-colored multidigraph we can easily con-struct a new arc-colored digraph by conveniently adding new vertices andcolors between vertices with multiple arcs.

6. Acknowledgements

We are very grateful to the anonymous referees for their insightful com-ments. We also thanks CNPq\Brazil and FAPERJ\Brazil for their partialfinancial support.

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