Facets of the minimum-adjacency vertex coloring

polytope

Diego Delle Donne∗ Javier Marenco∗

email: {ddelle, jmarenco}@dc.uba.ar

Computer Science Dept., FCEN, University of Buenos AiresPabellon I, Ciudad Universitaria, (1428) Buenos Aires, Argentina

Sciences Institute, National University of General Sarmiento,J. M. Gutierrez 1150, Malvinas Argentinas, (1613) Buenos Aires, Argentina

1 Introduction

In this work we are interested in a combinatorial optimization problem arisingfrom frequency assignment problems in wireless communication networks, thatwas motivated by the types of interference generated in GSM mobile phone net-works [4].

A wireless network employs some portion of the electromagnetic spectrum toestablish communications between the transmitter/receiver network antennas,called TRXs. A certain part of the electromagnetic spectrum is licensed to thecompany operating the network and is divided into discrete channels. EachTRX must operate through one channel, although whenever two TRXs overlaptheir coverage areas co-channel interference occurs if both are using the samechannel, and communications cannot be established within the common area.Moreover, if these conflicting TRXs are assigned to adjacent channels, then theso-called adjacent-channel interference occurs, generating in this case a minorinterference only. In a typical scenario, a good channel assignment must avoidco-channel interference and should avoid adjacent-channel interference.Several other constraints arise in practical settings as, e.g., blocked channelsand separation constraints (see, e.g., [4, 5, 6, 10]) but in this work we focus onthe basic model as stated in the previous paragraph. We are interested in thepolyhedral structure generated by such a combinatorial optimization problem,which includes a graph coloring structure with additional considerations onadjacent channels/colors. Based on these observations, we introduce in thiswork the minimum-adjacency vertex coloring problem and present an initialpolyhedral study for it.

∗Partially supported by Ubacyt Grant X069 and ANPCyT PICT-2007-00518.

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This paper is organized as follows. In Section 2 we formally state the minimum-adjacency vertex coloring problem and provide an integer programming modelfor this problem. Section 3 presents two families of facet-inducing valid inequal-ities.

2 Problem formulation and integer programmingmodel

We first introduce the interference graph G = (V,E) associated with an in-stance, in such a way that V represents the set of TRXs in the network and,whenever the coverage areas of two TRXs overlap, an edge in E joins the cor-responding vertices. Throughout this work we shall use the notation n = |V |and m = |E|. Let C = {1, . . . , t} be a set of consecutive colors representingthe t available channels. A C-coloring of G is a function c : V → C such thatc(i) 6= c(j) for every ij ∈ E. Clearly, a C-coloring of G corresponds to a fea-sible frequency assignment avoiding co-channel interference. Finally, for everyvw ∈ E, we define ψ(vw) ∈ [0, 1] to be the level of interference generated whenthe vertices v and w are assigned adjacent colors/channels.

Minimum-adjacency vertex coloring problem. Given an inteference graphG = (V,E), a set of consecutive colors C = {1, . . . , t}, and an interferencefunction ψ : E → [0, 1], find a C-coloring of G minimizing the total adjacent-channel interference, i.e.,

miny∈C(G,C)

∑{ψ(vw) : vw ∈ E and |y(v)− y(w)| = 1},

where C(G,C) represents the set of all C-colorings of G.

In the minimum-adjacency vertex coloring problem we ask for a frequency as-signment with no co-channel interference (i.e., a C-coloring of G) minimizingthe less critical adjacent-channel interference. This provides a good compro-mise between forbidding both kinds of interference and allowing them with apenalty term in the objective function. This problem is clearly NP-hard sinceit generalizes the classical vertex coloring problem, thus motivating the integerprogramming approach started in this work.

Throughout this work we shall consider ψ(vw) = 1 for every vw ∈ E, so theminimum-adjacency vertex coloring problem reduces to finding a C-coloring ofG which minimizes the number of edges receiving adjacent colors. Note thatthe adjacencies within the color set C are not circular, i.e., the colors 1 and tare not adjacent if t ≥ 3.

In order to state an integer programming formulation for the minimum-adja-cency vertex coloring problem, for every v ∈ V and every c ∈ C we introducethe binary assignment variable xvc representing whether the color c is assigned

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to the vertex v or not. We also introduce, for every vw ∈ E, v < w, the binaryadjacency variable zvw asserting whether the vertices v and w receive adjacentcolors or not. With these definitions, a model for the minimum-adjacency vertexcoloring problem is given by:

min∑

vw∈Eψ(vw) zvw∑

c∈Cxvc = 1 ∀v ∈ V (1)

xvc + xwc ≤ 1 ∀vw ∈ E, v < w, ∀c ∈ C (2)

xvc1 + xwc2 ≤ 1 + zvw ∀vw ∈ E, v < w, ∀c1, c2 ∈ C, |c1 − c2| = 1 (3)

xvc ∈ {0, 1} ∀v ∈ V, ∀c ∈ C (4)

zvw ∈ {0, 1} ∀vw ∈ E, v < w. (5)

Constraints (1) ensure that every vertex is assigned exactly one color from Cand constraints (2) forbid adjacent vertices to be assigned the same color. Con-straints (3) force adjacent vertices to be assigned non-adjacent colors unlessthe corresponding adjacency variable takes the value 1. Finally, constraints (4)and (5) force the variables to be binary. We call this formulation the stable-setmodel, which is a straightforward adaptation of the formulation presented in[12, 13] for the classical vertex coloring problem.

Note that the constraints allow feasible solutions to have active adjacency vari-ables even when the corresponding vertices are not receiving adjacent colors,however, this is not the case in any optimal solution. We say that y = (x, z) isan undominated solution if zvw = 0 for every vw ∈ E such that v are w recievingnon-adjacent colors.

3 Polyhedral study

In this section we introduce the polytope associated with the stable-set formu-lation, we characterize the dimension of this polytope for |C| > χ(G), and wepresent some families of facet-inducing valid inequalities.

Definition 1 Given a graph G = (V,E) and a set of colors C, we definePS(G,C) to be the convex hull of the incidence vectors (x, z) ∈ Rnt+m of feasiblesolutions to the stable-set model (1)-(5).

Proposition 1 If |C| > χ(G) and E 6= ∅, then dim(PS(G,C)) = n(t− 1) +m,and a minimal equation system is defined by (1).

Proof. Let λ ∈ Rnt+m and λ0 ∈ R such that λy = λ0 for every feasible solutiony ∈ PS(G,C). It suffices to show that (λ, λ0) is a linear combination of themodel constraints (1).

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Let vw ∈ E, and consider a feasible solution y = (x, z) ∈ PS(G,C) such thatthe colors assigned to v and w are nonadjacent and, furthermore, zvw = 0. Sucha solution can always be constructed since |C| > χ(G) ≥ 2 (as E 6= ∅). Lety′ = (x, z′) ∈ PS(G,C) be the feasible solution obtained from y by settingz′vw = 1 and leaving the remaining variables unchanged. The solution y′ isclearly feasible and only differs from y in the zvw-variable, hence λzvw = 0.Let v ∈ V . Since |C| > χ(G), there exists a feasible solution y = (x, z) ∈PS(G,C) such that at least one color from C is not used by any vertex. Letc ∈ C be the color assigned to the vertex v by y, and let c′ ∈ C be a color notused in y. Define y′ = (x′, z′) ∈ PS(G,C) to be the feasible solution obtainedfrom y by setting x′vc = 0, x′vc′ = 1, and leaving the remaining x-variablesunchanged. The solution y′ is feasible as the color c′ is not used in y and, fur-thermore, x only differs from x′ in the x′vc- and x′vc′ -variables, and possibly somez-variables. Since λz = 0, we conclude λxvc

= λxvc′ . By arbitrarily renamingthe colors, we conclude that λxvc

= λxvc′ for any c, c′ ∈ C.

By combining these observations, we conclude that λ is a linear combination ofthe coefficient vectors of the model constraints (1). Since the coefficient vectorsof these constraints are linearly independent, we conclude that (1) is a minimalequation system for PS(G,C), so dim(PS(G,C)) = n(t− 1) +m. 2

3.1 Consecutive colors inner clique inequalities

In this and the following section we introduce two classes of facet-inducing in-equalities for PS(G,C) arising from a clique K ⊆ V and a distinguished vertexk ∈ K. In these constructions, the clique structure is crucial both for validityand facetness, suggesting the importance of the cliques of the interference graphfor the structure of PS(G,C).

Let K ⊆ V be a clique, k ∈ K be some vertex of K, and {c1, c2, c3} ⊆ C be aset of three consecutive colors. If some vertex v ∈ K \ {k} receives the color c2then an adjacency with k is generated if the latter receives either c1 or c3. Thisidea can be generalizaed for a set Q = {c1, . . . , cq} ⊆ C of an odd number ofconsecutive colors; if vertices from K \ {k} are using every color ci ∈ Q with anodd index i, then any color cj ∈ Q with an even index j generates adjacencieswhen assigned to k. The following lifted family of valid inequalities arises fromthis observation.

Definition 2 (Consecutive colors inner clique inequality) Let K ⊆ V bea clique of G, and fix a vertex k ∈ K. Let Q = {c1, ..., cq} ⊆ C with an odd q,be a set of consecutive colors such that ci+1 = ci + 1 for i = 1, ..., q − 1. LetI = {c1, c3, ..., cq} and P = {c2, c4, ..., cq−1}. We define the consecutive colorsinner clique inequality to be(xkc1+xkcq+

∑ci∈I\{c1,cq}

2xkci+∑cj∈P

xkcj)+

∑v∈K\{k}

∑cj∈P

xvcj ≤q − 1

2+

∑v∈K\{k}

zvk.

(6)

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Note that if Q = {c1, c2, c3}, K = {v, w} and k = v then the resulting consec-utive colors inner clique inequality dominates the constraint (3) correspondingto the edge vw and the colors c1 and c2 (resp. the edge vw and the colors c3and c2).

Proposition 2 The consecutive colors inner clique inequalities (6) are valid forPS(G,C).

Proof. Let y = (x, z) ∈ PS(G,C) be a feasible solution. Define δc ∈ {0, 1} as

δc =∑

v∈K\{k}

xvc,

i.e., δc = 1 iff c is assigned to a vertex from K \ {k}. For any color cj ∈ P ,if δcj (xkcj−1 + xkcj+1) = 1, then a color adjacency exists between k and somevertex from K \ {k}. So, we can bound the RHS of (6) as

q − 1

2+∑cj∈P

δcj (xkcj−1+ xkcj+1

) ≤ q − 1

2+

∑v∈K\{k}

zkv. (7)

Note that no color adjacency is counted more than once since∑

c∈Q xkc ≤ 1.On the other hand, we can write the LHS of (6) as∑

cj∈P(xkcj−1

+ xkcj + xkcj+1) +

∑cj∈P

δcj . (8)

Hence we can prove validity of (6) by showing that (8) is not greater than theLHS of (7). To this end, we can show, for every cj ∈ P , that

(xkcj−1 + xkcj + xkcj+1) + δcj ≤ 1 + δcj (xkcj−1 + xkcj+1) (9)

and then, sum over P .

If δcj = 0, then (9) holds as (xkcj−1+ xkcj + xkcj+1

) ≤ 1. Finally, if δcj = 1 theinequality holds only if xkcj = 0, and this is true as δcj = 1.

Since the solution y is arbitrary, the inequality (6) is valid for PS(G,C). 2

Theorem 1 If |C| > χ(G) and |C| ≥ |K|+5, then the consecutive colors innerclique inequalities (6) are facet-defining for PS(G,C).

Proof. Let π ∈ Rnt+m be the coefficient vector of (6) and π0 ∈ R the independentterm of (6). The face of PS(G,C) defined by this inequality is

F = {y ∈ Rnt+m : πy = π0} ∩ PS(G,C).

Let λ ∈ Rnt+m and λ0 ∈ R such that λy = λ0,∀y ∈ F . We will prove that λ isa linear combination of π and the coefficient vectors of the equations (1) fromthe model, implying that F is a facet of PS(G,C).

We first characterize the integer solutions y = (x, z) ∈ F in four types:

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Type (I) If the vertex k is not assigned to any color from Q then the only casein which (6) is satisfied by equality is when q−1

2 vertices from the cliqueare using colors from P , i.e., every color from P , and no other vertex fromK is assigned to a color adjacent to the one assigned to k. That is,

• xkc = 0, for all c ∈ Q,

• for all cj ∈ P , there exists v ∈ K \ {k} such that xvcj = 1, and

• zkv = 0 for all v ∈ K \ {k}.

Type (II) The vertex k uses a color cj ∈ P . In order to satisfy (6) with equalityin this case, every other color from P must be assigned to vertices fromK and the colors cj−1 and cj+1 cannot be assigned to any vertex from theclique, as it will generate color adjacencies with k without increasing theLHS of (6). Obviously, cj cannot be assigned to vertices from K, since itis being used by k. This is,

• xkcj = 1 with cj ∈ P ,

• for every ci ∈ P \ {cj}, there exists v ∈ K \ {k} such that xvci = 1,and

• xvcj−1= xvcj = xvcj+1

= 0 for all v ∈ K \ {k}.

Type (III) The vertex k uses c1 (resp. cq). In order to satisfy (6) with equalityin this case, every color from P \{c2} (resp. in P \{cq−1}) must be assignedto vertices from the clique and the color preceding c1 (resp. following cq)cannot be assigned to any other vertex from the clique, as it will generatecolor adjacencies with k without increasing the LHS of (6). The color c2(resp. cq−1) may or may not be assigned to a vertex from K, but for everyv ∈ K \ {k}, if v is not assigned to c2 (resp. cq−1), then zkv = 0. Wecall type (IIIa) and (IIIb) to the solutions from this type using c1 or cq,respectively. This is,Type (IIIa):

• xkc1 = 1,

• for every cj ∈ P \ {c2}, there exists v ∈ K \ {k} such that xvcj = 1,

• xv,c1−1 = 0 for all v ∈ K \ {k}, and

• for every v ∈ K \ {k}, zkv = 1⇔ xvc2 = 1.

Type (IIIb):

• xkcq = 1,

• for every cj ∈ P \{cq−1}, there exists v ∈ K \{k} such that xvcj = 1,

• xv,cq+1 = 0 for all v ∈ K \ {k}, and

• for every v ∈ K \ {k}, zkv = 1⇔ xvcq−1= 1.

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Type (IV) The vertex k uses a color ci ∈ I \ {c1, cq}. In order to satisfy theequality in this case, every color from P \ {ci−1, ci+1} must be assignedto vertices from the clique. The colors ci−1 and ci+1 may or may not beassigned to vertices from the clique, but for every v ∈ K \ {k}, if v is notusing any of these colors then zkv = 0.

• xkci = 1 with ci ∈ I \ {c1, cq},• for all cj ∈ P \ {ci−1, ci+1}, there exists v ∈ K \ {k} such thatxvcj = 1, and

• for every v ∈ K \ {k}, zkv = 1⇔ (xvci−1= 1 ∨ xvci+1

= 1).

The following illustrates the types of solutions described above. The whitespaces in the table denote that there are no restrictions for the vertices receivingthe corresponding colors and the symbol ∅K denotes that the color cannot beassigned to a vertex from the clique. The vertices v represent vertices from theclique.

c c1 c2 c3 c4 c5 c6 . . . cq−1 cq(I)→ ∅K k ∅K v v v v

(II)→ v ∅K k ∅K v v(IIIa)→ ∅K k v v v(IIIb)→ v v v k ∅K

(IV )→ v k v

To prove that λ is a linear combination of π and the coefficient vectors of theequations (1) from the model we will state several claims on the coefficientsfrom λ.

Let vw ∈ E such that vw /∈ {kj : j ∈ K}, we will see that λzvw= 0. Let

y = (x, z) ∈ PS(G,C) be a solution of type (II) in which vertices v and w receivenonadjacent colors and zvw = 0. This solution can be obtained by permutingthe color classes from any valid solution, in order to obtain the non-adjacencybetween v and w. We now build the solution y′ = (x′, z′) ∈ PS(G,C) from ybut setting z′vw = 1. This solution is still valid and since zvw is not involved in(6), the equality holds. As both y and y′ belong to F , we know that λy = λ0and λy′ = λ0, so λy = λy′. We also know that y equals y′ in every coefficientbut zvw, and so λzvwzvw = λzvwz

′vw. Hence,

[Claim 1] λzvw = 0 ∀vw ∈ E such that vw /∈ {kj : j ∈ K}. (10)

Take now a vertex v /∈ K, we will see that λxvc= λxvc′ for every pair of colors

c, c′ ∈ C. As |C| > χ(G), it is possible to build a solution where the color c′ isnot assigned to any vertex. Assume first that c′ /∈ {ci : i ∈ P}, we will showlater what happens in that case. Consider a solution y = (x, z) ∈ PS(G,C) oftype (I) in which c′ is not assigned to any vertex and in which v receives the

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color c (this can be achieved by permuting the color class of v in any solution,assuming that the color assigned to v does not belong to {ci : i ∈ P}). We nowbuild the solution y′ = (x′, z′) from y, in which the vertex v receives the colorc′. We can visualize the difference between these solutions in the next diagram:

c c′

y → v ∅y′ → v

The solution y′ is feasible as c is not being used by any vertex in y. Moreover,this solution belongs to the face F since it also represents a type (I) solution.As both y and y′ belong to F , we can deduce again that λy = λy′. In this case yequals y′ in every coefficient except for xvc, xvc′ and eventually some adjacencyvariables incidents to v, however (10) asserts that λzvw

= 0,∀vw ∈ E, and sowe obtain

λxvcxvc + λxvc′xvc′ = λxvcx′vc + λxvc′x

′vc′ ,

and using the corresponding variable values we get λxvc= λxvc′ .

If c′ ∈ {ci : i ∈ P}, there are no type (I) solutions in F with c′ free of vertices.However, we can show that λxvc

= λxvc′ by resorting to solutions of type (IV).Hence,

[Claim 2] λxvc= λxvc′ ∀v /∈ K ∀c, c′ ∈ C. (11)

Next, consider a vertex v ∈ K \ {k}, we will see that for every cj ∈ P , λxvcj=

λxvc − λzkvfor all c ∈ C \ P . Let y = (x, z) ∈ PS(G,C) be a solution of type

(IV) where v is assigned to cj and k to cj−1 (if j = 2, then the solution wastype (IIIa)). Assume there is a color c ∈ C \(P ∪{cj−2}) free from vertices (thiscan be achieved since |C| > χ(G)). We now build another solution y′ = (x′, z′)from y but assigning c to v.

c cj−2 cj−1 cjy → ∅ k vy′ → v k

Note that y′ is feasible as y assingns no vertex to color c. Moreover, y′ representsa type (IV) solution (or a type (IIIa), if j = 2) so it belongs to F . Again, as inthe previous cases, λy = λy′. Both solutions are identical for every coefficientbut for xvcj , xvc, zkv and other adjacency variables for v. The latter, however,are void since (10) asserts that their coefficients in λ are null. Hence,

λxvcj+ λzkv

= λxvc.

It is worth mentioning that if cj−2 /∈ P , then in order to prove the equality forc = cj−2 we can use a type (IV) solution in which k is assigned to cj+1 instead

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of cj−1 (it will be a type (IIIb) solution if j = q − 1), following the same line.So, as c is an arbitrary color from C \ P , we have

[Claim 3] λxvcj+ λzkv

= λxvc ∀v ∈ K \ {k}, ∀cj ∈ P, ∀c ∈ C \ P. (12)

Note also that (12) implies

[Claim 4] λxvc= λxvc′ ∀v ∈ K \ {k} ∀c, c

′ ∈ C \ P. (13)

We will see next that λxkc1= λxkcq

= λxkc− λzkv

for all c ∈ C \ Q and

v ∈ K \ {k}. First we will prove it for λxkc1.

Let y = (x, z) ∈ PS(G,C) be a type (IIIa) solution in which v is assigned thecolor c2 and there is a color c ∈ C \ Q free from vertices (again this can beachieved as |C| > χ(G)). We also assume that no vertex from K is assigned tothe colors adjacent to c; this is possible if |C \ {c}| ≥ |K|+ 2, which is impliedby the hypothesis. We now build a solution y′ = (x′, z′) from y by assigning cto k.

c c1 c2 c3 c4 . . .y → ∅K ∅ ∅K k v vy′ → ∅K k ∅K v v

Note that y′ is feasible as y assingns no vertex to color c. Moreover, y′ belongsto F as it represents a type (I) solution and therefore, as in the previous cases,λy = λy′. Here, y equals y′ in every variable but xkc1 , xkc, zkv and eventuallysome adjacency variables that can be ignored based on (10). We thus obtain

λxkc1+ λzkv

= λxkc. (14)

Analogously, by taking a type (IIIb) solution instead of a type (IIIa) solution,we get

λxkcq+ λzkv

= λxkc. (15)

Finally, from (14) and (15) we have

[Claim 5] λxkc1= λxkcq

= λxkc− λzkv

, ∀v ∈ K \ {k},∀c ∈ C \Q. (16)

Note that v is an arbitrary vertex from K \ {k}, therefore (16) also implies

[Claim 6] λzkv= λzkw

∀v, w ∈ K \ {k}. (17)

Next we will see that for every ci ∈ I\{c1, cq}, we have λxkci= λxkc

−λzkv−λzkw

,for all c ∈ C \Q and v, w ∈ K \ {k}. Let y = (x, z) be a type (IV) solution inwhich k is assigned to ci, v to ci−1 and w to ci+1. Assume also that c is freefrom vertices in y and that the colors adjacent to c are free from vertices fromK. We now build a solution y′ = (x′, z′) from y by assigning c to k.

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c c1 c2 . . . ci−1 ci ci+1 . . . cq−1 cqy → ∅K ∅ ∅K v v k w vy′ → ∅K k ∅K v v w v

The solution y′ is feasible and satisfy (6) with equality as it is a type (I) solution.Again, we get λy = λy′ and in this case the only difference between y and y′ isgiven by xkci , xkc, zkv, zkw and eventually some adjacency variables. Hence,

[Claim 7] λxkci= λxkc

− λzkv− λzkw

, (18)

∀v, w ∈ K \ {k}, ∀i ∈ I \ {1, q},∀c ∈ C \Q.

We now show λxkcj= λxkc

− λzkvfor every cj ∈ P for all c ∈ C \ Q and

v ∈ K \ {k}. Let y = (x, z) be a type (II) solution in which k is assigned tocolor cj . Assume for a moment that cj 6= c2 and call v the vertex of the cliqueassigned to cj−2. Assume also that the color cj−1 is free from vertices in y. Wenow build the solution y′ = (x′, z′) from y in which we move k to the color cj−1.

c1 c2 . . . cj−2 cj−1 cj cj+1 . . . cq−1 cqy → v v ∅ k ∅K vy′ → v v k ∅K v

The resulting solution is valid and it satisfies (6) with equality as it is a type(IV) solution. Once again, λy = λy′ and the differences between y and y′ aregiven by xkcj , xkcj−1 , zkv and eventually some adjacency variables. So we have

λxkcj= λxkcj−1

+ λzkv.

If cj = c2, a similar argument can be given by moving k to the color cj+1,defining v to be the vertex of K using cj+2, and analogously deducing

λxkcj= λxkcj+1

+ λzkv.

By (17) and (18) we conclude that

[Claim 8] λxkcj= λxkc

− λzkv, ∀v ∈ K \ {k}, ∀j ∈ P,∀c ∈ C \Q, (19)

as both cj−1 and cj+1 belong to I.

Finally we show that λxkc= λxkc′ for all c, c′ ∈ C \Q. Let y = (x, z) be a type

(I) solution in which k is assigned to the color c and no vertex gets the color c′.In addition, assume that the colors adjacent to c and to c′ are free from verticesfrom K; such a solution exists if |C \ {c}| ≥ |K| + 4, which is implied by thehypothesis. We now build a solution y′ = (x′, z′) from y by assigning c′ to k.

c c′

y → ∅K k ∅K ∅K ∅ ∅Ky′ → ∅K ∅K ∅K k ∅K

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The solution y′ is feasible as y assigns no vertex to c′. Moreover, y′ belongs toF as it is a type (I) solution. In this case the difference between both solutionsis given by xkc, xkc′ and eventually some adjacency variables, so we obtain

[Claim 9] λxkc= λxkc′ ∀c, c

′ ∈ C \Q. (20)

To conclude the proof, we now show that λ is a linear combination of π and thecoefficient vectors of (1).

Claim 2 asserts that given a vertex v /∈ K, the coefficient in λ for the variablesxvc is the same for any color c. Call βv this value, so for v /∈ K

λxvc= βv ∀c ∈ C.

Claim 4 asserts that for a vertex v ∈ K \ {k}, the coefficient in λ for xvc is thesame for any color c ∈ C \ P . Call βv this value, then for v ∈ K \ {k} we have

λxvc= βv ∀c ∈ C \ P.

Similarly, Claim 9 states that λxkcis the same for any color c outside Q. By

defining βk to be this value we have

λxkc= βk ∀c ∈ C \Q.

On the other hand, Claim 6 shows that λzkv= λzkw

for every pair of vertexv, w ∈ K \ {k}. Call α this value, so

λzkv= α ∀v ∈ K \ {k}.

We can now compute the remaining coefficients of λ. Claim 3 implies

λxvcj= βv − α ∀j ∈ P, ∀v ∈ K \ {k},

and from Claim 5 we have

λxkc1= λxkcq

= βk − α.

Finally, from Claim 7 we deduce

λxkci= βk − 2 α, ∀i ∈ I \ {1, q},

and from Claim 8 we obtain

λxkcj= βk − α, ∀j ∈ P.

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In this manner, using also Claim 1, we verify that λ can be obtained as thelinear combination

λ =∑v∈V

βve(v) − απ,

where e(v) represents the coefficient vector of the constraint (1) correspondingto the vertex v. Hence, the consecutive colors inner clique inequalities definefacets of PS(G,C). 2

It is interesting to note that we can define the set Q = {c1, . . . , cq} even when cqrepresents a color outside the limits of C, i.e., when cq−1 is the last color fromC. The inequalities obtained by omitting the variables associated to cq are stillvalid, and they also define facets of PS(G,C). The same applies to c1, when c2represents the first color of C.

To conclude this section, note that for any undominated integer solution y =(x, z) ∈ PS(G,C) there are several consecutive colors inner clique inequalitiessatisfied with equality by y. To this end, take a vertex k ∈ V and call c the colorassigned to it. If c 6= 1, t, then by defining Q = {c− 2, c− 1, c, c+ 1, c+ 2} andtaking any clique K such that k ∈ K, the associated valid inequality is satisfiedwith equality by y. If c is the first color resp. the last color of C, by definingQ = {c, c + 1, c + 2} resp. Q = {c − 2, c − 1, c} we obtain a consecutive colorsinner clique inequality satisfied with equality by y.

3.2 Consecutive colors subset clique inequalities

The class of valid inequalities introduced in this section arises from similarconsiderations as for the inner clique inequalities.

Definition 3 (Consecutive colors subset clique inequality) Let K ⊆ Vbe a clique of G, and fix a vertex k ∈ K. Let Q = {c1, ..., cq} ⊆ C be a set ofconsecutive colors such that ci+1 = ci + 1 for i = 1, ..., q − 1. We define theconsecutive colors subset clique inequality to be

(xkc1+2xkc2+

q−2∑i=3

3xkci+2xkcq−1+xkcq)+

∑v∈K\{k}

q−1∑i=2

xvci ≤ (q−2)+∑

v∈K\{k}

zvk.

(21)

Proposition 3 The consecutive colors subset clique inequalities (21) are validfor PS(G,C).

Proof. Let y = (x, z) ∈ PS(G,C) be a feasible solution. Define δc ∈ {0, 1} as

δc =∑

v∈K\{k}

xvc,

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i.e., δc = 1 iff c is assigned to a vertex from K \ {k}. For every i = 2, . . . , q − 1,if δci(xkci−1 + xkci+1) = 1, then a color adjacency exists between k and somevertex from K \ {k}. So, we can bound the RHS of (21) as

(q − 2) +

q−1∑i=2

δci(xkci−1 + xkci+1) ≤ (q − 2) +∑

v∈K\{k}

zkv. (22)

Note that no color adjacency is counted more than once since∑c∈Q

xkc ≤ 1.

On the other hand, we can write the LHS of (21) as

q−1∑i=2

(xkci−1 + xkci + xkci+1) +

q−1∑i=2

δci . (23)

Hence, we can prove validity of (21) by showing that (23) is not greater thanthe LHS of (22). To this end, we can show, for every i = 2, . . . , q − 1, that

(xkci−1+ xkci + xkci+1

) + δci ≤ 1 + δci(xkci−1+ xkci+1

) (24)

and then, sum over {2, . . . , q − 1}.

If δci = 0, then (24) holds as (xkci−1+ xkci + xkci+1

) ≤ 1. Finally, if δci = 1 theinequality holds only if xkci = 0, and this is true as δci = 1.

Since the solution y is arbitrary, the inequality (21) is valid for PS(G,C). 2

Theorem 2 If |C| > χ(G) and |C| ≥ |K| + 5, the consecutive colors subsetclique inequalities (21) are facet-defining for PS(G,C).

Proof. Let Q′ = Q \ {c1, cq} and Q′′ = Q′ \ {c2, cq−1}. Let π ∈ Rnt+m be thecoefficient vector from (21) and let π0 ∈ R the independent term of (21). Theface of PS(G,C) defined by this inequality is

F = {y ∈ Rnt+m : πy = π0} ∩ PS(G,C).

Let λ ∈ Rnt+m and λ0 ∈ R such that λy = λ0,∀y ∈ F . We will prove that λ isa linear combination of π and the coefficient vectors of the equations (1) fromde model, implying that F is a facet of PS(G,C).

We first characterize the integer solutions y = (x, z) ∈ F in four types:

Type (I) If the vertex k is not assigned to any color from Q, then the onlyway to satisfy (21) with equality is by assigning q − 2 colors from Q′ tovertices from K, i.e., every color from Q′. In addition, no vertex from Kcan use a color adjacent to the color assigned to k. This is,

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• xkc = 0, for all c ∈ Q,

• for every c ∈ Q′, there exists v ∈ K \ {k} such that xvc = 1, and

• zkv = 0 for all v ∈ K \ {k}.

Type (II) The vertex k is assigned to c1 (resp. cq). In this case (21) holdswith equality only if every color from Q′ \ {c2} (resp. Q′ \ {cq−1}) isassigned to a vertex from K. Moreover, no vertex from the clique can usethe color preceding c1 (resp. following cq) as that would generate a coloradjacency with k without increasing the LHS of the inequality. The colorc2 (resp. cq−1) may or may not be assigned to vertices from K but forevery v ∈ K \ {k}, if that color is not assigned to v then zkv = 0. Wecall type (IIa) and type (IIb) to the solutions of this type in which k isassigned to c1 and cq, respectively. This is,Type (IIa):

• xkc1 = 1,

• for every c ∈ Q′ \ {c2}, there exists v ∈ K \ {k} such that xvc = 1,

• xv,c1−1 = 0 for all v ∈ K \ {k}, and

• for every v ∈ K \ {k}, zkv = 1⇔ xvc2 = 1.

Type (IIb):

• xkcq = 1,

• for every c ∈ Q′ \ {cq−1}, there exists v ∈ K \ {k} such that xvc = 1,

• xv,cq+1 = 0 for all v ∈ K \ {k}, and

• for every v ∈ K \ {k}, zkv = 1⇔ xvc2 = 1.

Type (III) The vertex k is assigned to c2 resp. cq−1. In this case (21) holdswith equality only if every color from Q′′ \ {c3} (resp. Q′′ \ {cq−2}) isassigned to a vertex from K. Moreover, no vertex from the clique canuse c1 (resp. cq) as that would generate a color adjacency with k withoutincreasing the LHS of the inequality. The color c3 (resp. cq−2) may ormay not be assigned to vertices from K but for every v ∈ K \ {k}, ifthat color is not assigned to v then zkv = 0. We call type (IIIa) and type(IIIb) to the solutions of this type in which k is assigned to c2 and cq−1,respectively. This is,Type (IIIa):

• xkc2 = 1,

• for every c ∈ Q′′ \ {c3}, there exists v ∈ K \ {k} such that xvc = 1,

• xv,c1 = 0 for all v ∈ K \ {k}, and

• for every v ∈ K \ {k}, zkv = 1⇔ xvc3 = 1.

Type (IIIb):

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• xkcq−1 = 1,

• for every c ∈ Q′′ \ {cq−2}, there exists v ∈ K \ {k} such that xvc = 1,

• xv,cq = 0 for all v ∈ K \ {k}, and

• for every v ∈ K \ {k}, zkv = 1⇔ xvcq−2= 1.

Type (IV) The vertex k is assigned to a color c ∈ Q′′. The inequality (21)holds with equality only if every color from Q′ \{c−1, c, c+ 1} is assignedto a vertex from K. The colors c−1 and c+1 may or may not be assignedto vertices from K but for every v ∈ K \ {k}, if v does not receive eitherc− 1 or c+ 1, then zkv = 0. This is,

• xkc = 1 for some c ∈ Q′′,• for every c ∈ Q′ \ {c− 1, c, c+ 1}, there exists v ∈ K \ {k} such thatxvc = 1, and

• for every v ∈ K \ {k}, zkv = 1⇔ (xv,c−1 = 1 ∨ xv,c+1 = 1).

The following illustrates the types of solutions described above. The whitespaces in the table denote that there are no restrictions for the vertices receivingthe corresponding colors and the symbol ∅K denotes that the color cannot beassigned to a vertex from K. The vertices v represent vertices from K.

c c1 c2 c3 c4 c5 c6 . . . cq−2 cq−1 cq(I) → ∅K k ∅K v v v v v . . . v v

(IIa) → ∅K k v v v v . . . v v(IIb) → v v v v v . . . v k ∅K

(IIIa) → ∅K k v v v . . . v v(IIIb) → v v v v v . . . k ∅K(IV ) → v k v . . . v v

To show that λ is a linear combination of π and the coefficient vectors of themodel constraints (1) we will prove several claims on the coefficients of λ.

Let vw ∈ E such that vw /∈ {kj : j ∈ K}. Let y = (x, z) ∈ PS(G,C) be asolution of type (I) in which the vertices v and w receive nonadjacent colors andzvw = 0. This solution can be obtained by permuting the color classes from anyvalid solution, in order to obtain the non-adjacency between v and w. We nowbuild the solution y′ = (x′, z′) ∈ PS(G,C) from y by setting z′vw = 1. Thissolution is still valid and since zvw is not involved in (21), equality holds. As bothy and y′ belong to F , we know that λy = λ0 and λy′ = λ0, so λy = λy′. We alsoknow that y equals y′ in every coefficient but zvw, and so λzvw

zvw = λzvwz′vw.

Hence,

[Claim 1] λzvw= 0 ∀vw ∈ E such that vw /∈ {kj : j ∈ K}. (25)

Take now a vertex v /∈ K, we will see that λxvc= λxvc′ for every pair of colors

c, c′ ∈ C. As |C| > χ(G), it is possible to build a solution where the color c′

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is not assigned to any vertex. Consider a solution y = (x, z) ∈ PS(G,C) oftype (I) in which c′ is not assigned to any vertex and in which v receives thecolor c (this can be achieved by permuting the color class corresponding to v inany solution). we now build the solution y′ = (x′, z′) from y, in which vertex vreceives the color c′. We can visualize the difference between these solutions inthe next diagram:

c c′

y → v ∅y′ → v

The solution y′ is feasible xvc = 0 for every v ∈ V . Moreover, y′ ∈ F since it is atype (I) solution. As both y and y′ belong to F , we get λy = λy′. In this case yequals y′ in every coefficient except for xvc, xvc′ and eventually some adjacencyvariables, however (25) states that λzvw

= 0,∀vw ∈ E, and so obtain

λxvcxvc + λxvc′xvc′ = λxvc

x′vc + λxvc′x′vc′ ,

and using the corresponding variable values we get λxvc= λxvc′ . Since c and c′

are arbitrary colors, we conclude

[Claim 2] λxvc= λxvc′ ∀v /∈ K ∀c, c′ ∈ C. (26)

Next, consider a vertex v ∈ K \ {k} and a color c ∈ Q′, we will see thatλxvc

= λxvc′ − λzkvfor all c′ ∈ C \Q′. Let y = (x, z) ∈ PS(G,C) be a solution

of type (IV) in which v is assigned to the color c and k is assigned to a coloradjacent to c. Assume that c′ is free from vertices (such a solution exists as|C| > χ(G)). We now build another solution y′ = (x′, z′) from y by assigning c′

to v.

c′ c1 . . . c c+ 1 . . . cqy → ∅ . . . v k . . .y′ → v . . . k . . .

The solution y′ is feasible xvc = 0 for every v ∈ V . Moreover, y′ ∈ F since it isa type (IV) solution, so it belongs to F and, therefore, λy = λy′. Both solutionsare identical for every coefficient but xvc, xvc′ , zkv and some adjacency variablesfor v. The latter are void since (25) asserts that their coefficients in λ are null.Hence,

λxvc + λzkv= λxvc′ .

If there are no solutions of type (IV) in PS(G,C), if |Q| < 5, the same conclu-sion can be achieved by resorting to type (IIa) or type (IIb) solutions. So, weconclude:

[Claim 3] λxvc= λxvc′ − λzkv

∀v ∈ K \ {k}, ∀c ∈ Q′,∀c′ ∈ C \Q′. (27)

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Note also that (27) implies

[Claim 4] λxvc = λxvc′ ∀v ∈ K \ {k} ∀c, c′ ∈ C \Q′. (28)

We show next that λxkc1= λxkcq

= λxkc− λzkv

for all c ∈ C \ Q and v ∈K \{k}. Let y = (x, z) ∈ PS(G,C) be a solution of type (IIa) in which a vertexv ∈ K \ {k} uses the color c2 and a color c ∈ C \Q is free from vertices (againthis can be achieved as |C| > χ(G)). We also assume that no vertex from K isassigned to a color adjacent to c; this is possible if |C \ {c}| ≥ |K| + 2, whichis implied by the hypothesis. We now build a solution y′ = (x′, z′) from y butassigning c to k.

c c1 c2 . . .y → ∅K ∅ ∅K ∅K k v . . .y′ → ∅K k ∅K ∅K v . . .

The solution y′ is feasible xvc = 0 for every v ∈ V . Moreover, y′ ∈ F since it isa type (I) solution and, therefore, λy = λy′. Here, y equals y′ in every variablebut xkc1 , xkc, zkv and eventually some adjacency variables that can be ignoredbased on (25). So we deduce

λxkc1+ λzkv

= λxkc. (29)

Analogously, by taking a type (IIb) solution instead of a type (IIa) solution weget

λxkcq+ λzkv

= λxkc. (30)

Finally, from (29) and (30) we have

[Claim 5] λxkc1= λxkcq

= λxkc− λzkv

, ∀v ∈ K \ {k},∀c ∈ C \Q. (31)

Note that v is an arbitrary vertex from K \ {k}, therefore (31) also implies

[Claim 6] λzkv= λzkw

∀v, w ∈ K \ {k}. (32)

Similarly as c is any color from C \Q, we also obtain

[Claim 7] λxkc= λxkc′ ∀c, c

′ ∈ C \Q. (33)

Let y = (x, z) ∈ PS(G,C) be a type (IIIa) solution in which a vertex v ∈ K\{k}is assigned to c3 and the color c1 is free from vertices (such a solution existsas |C| > χ(G)). We also assume that no vertex from K is assigned to a colorpreceding c1; this is possible if |C \ {c}| ≥ |K| + 2, which is implied by thehypothesis. We now build a solution y′ = (x′, z′) from y but assigning c1 to k.

c1 c2 c3 . . .y → ∅K ∅ k v . . .y′ → ∅K k v . . .

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The solution y′ is valid and it satisfies (21) with equality as it is a type (IIa)solution. Once again, λy = λy′ and the differences between y and y′ are givenby xkc2 , xkc1 , zkv and eventually some adjacency variables. So we have

λxkc2+ λzkv

= λxkc1. (34)

Analogously, with a type (IIIb) instead of a type (IIIa) solution, we can show

λxkcq−1+ λzkv

= λxkcq. (35)

Finally, by combining (34), (35) and (31) we have

[Claim 8] λxkc2= λxkcq−1

= λxkc− 2λzkv

, ∀v ∈ K \ {k},∀c ∈ C \Q. (36)

Let y = (x, z) ∈ PS(G,C) be a type (IV) solution in which k is assigned to acolor c ∈ Q′′, a vertex v1 ∈ K \ {k} is assigned to c + 1 and the color c − 1 isfree from vertices (again, this can be achieved as |C| > χ(G)). Assume for nowthat c > c3, and call v2 to the vertex from K assigned to c − 2 (such vertexexists as y is a type (IV) solution). We now build a solution y′ = (x′, z′) fromy by moving k to the color c− 1.

c1 . . . c− 2 c− 1 c c+ 1 . . . cqy → . . . v2 ∅ k v1 . . .y′ → . . . v2 k v1 . . .

The solution y′ is valid as c − 1 is free from vertices in y. Moreover, y′ ∈F as y′ represents a type (IV) solution and, therefore, λy = λy′. The onlydifference between y and y′ is given by xkc, xk,c−1, zkv1

, zkv2and eventually

some adjacency variables that can be ignored, so we obtain

λxkc+ λzkv1

= λxk,c−1+ λzkv2

and using (32) we conclude

λxkc= λxk,c−1

, ∀c ∈ Q′′ \ {c3}. (37)

Suppose now that c = c3, then the color c − 2 = c1 may be free from verticesfrom the clique. We keep in y the color c− 1 = c2 free from vertices and buildy′ the same way than before, moving k to color c− 1, namely to c2.

c1 c2 c3 c4 . . . cqy → ∅K ∅ k v1 . . .y′ → ∅K k v1 . . .

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This solution is valid and it belongs to F as it is a type (IIIa) solution, henceλy = λy′. The difference between y and y′ is given by xkc3 , xkc2 , zkv1 andeventually some adjacency variables that can be ignored, so we obtain

λxkc3+ λzkv1

= λxkc2.

Combining this result with (37) and then using (36) we conclude that

[Claim 9] λxkc= λxkc′ − 3λzkv

, ∀c ∈ Q′′,∀v ∈ K \ {k},∀c′ ∈ C \Q. (38)

We are now in position of proving that λ is in fact a linear combination of πand the coefficient vectors of (1).

Claim 2 asserts that given a vertex v /∈ K, the coefficient in λ for the variablesxvc is the same for any color c. Define βv to be this value, so for v /∈ K

λxvc = βv ∀c ∈ C.

Claim 4 asserts that for a vertex v ∈ K \ {k}, the coefficient in λ for xvc is thesame for any color c ∈ C \Q′. Define βv to be this value, then for v ∈ K \ {k}we have

λxvc = βv ∀c ∈ C \Q′.

Similarly, Claim 7 asserts that λxkcis the same for any color c outside Q. By

defining βk to be this value, we have

λxkc= βk ∀c ∈ C \Q.

On the other hand, Claim 6 shows that λzkv= λzkw

for every pair of verticesv, w ∈ K \ {k}. By defining α to be this value, we get

λzkv= α ∀v ∈ K \ {k}.

We can now compute the remaining coefficients of λ. Claim 3 implies

λxvc = βv − α ∀c ∈ Q′, ∀v ∈ K \ {k},

and from Claim 5 we have

λxkc1= λxkcq

= βk − α.

Finally, from Claim 8 we conclude

λxkc2= λxkcq−1

= βk − 2 α

and from Claim 9 we obtain

λxkc= βk − 3 α, ∀c ∈ Q′′.

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Finally, by applying Claim 1, we can see that λ can be obtained as the linearcombination

λ =∑v∈V

βve(v) − απ,

where e(v) represents the coefficient vector of (1) corresponding to vertex v.Hence, the consecutive colors subset clique inequalitites define facets of PS(G,C).2

As in the consecutive colors inner clique inequalities, the set Q = {c1, . . . , cq}can be defined even when c1, resp. cq, represents a color outside the limits of C.The inequalities obtained by omitting the variables associated to these colorsare still valid, and they also define facets of PS(G,C).

Again, we conclude this section noting that for any undominated integer solutiony = (x, z) ∈ PS(G,C) there are several consecutive colors subset clique inequal-ities satisfied with equality by y. To this end, take a vertex k ∈ V and call c thecolor assigned to it. If c 6= 1, t, then by defining Q = {c− 2, c− 1, c, c+ 1, c+ 2}and taking any clique K such that k ∈ K, the associated valid inequality issatisfied with equality by y. If c is the first color resp. the last color of C, bydefining Q = {c, c+ 1, c+ 2} resp. Q = {c− 2, c− 1, c} we obtain a consecutivecolors subset clique inequality satisfied with equality by y (these are actuallythe same inequalities as in the inner clique).

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