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Discrete Applied Mathematics 145 (2005) 326–331

www.elsevier.com/locate/dam

An exact algorithm for the channel assignment problem

Daniel Král’aaDepartment of Applied Mathematics and Institute for Theoretical Computer Science, Charles University, Malostranské námˇestí 25,

118 00 Prague, Czech Republic

Received 11 June 2002; received in revised form 9 October 2002; accepted 16 January 2004Available online 25 September 2004

Abstract

A channel assignment problem is a triple(V , E, w) whereV is a vertex set,E is an edge set andw is a function assigningedges positive integer weights.An assignmentcof integers between 1 andK to the vertices is proper if|c(u)− c(v)|�w(uv) foreachuv ∈ E; the smallestK for which there is a proper assignment is called the span. The input problem is set to bel-boundedif the values ofw do not exceedl. We present an algorithm running in time O(n(l + 2)n) which outputs the span forl-boundedchannel assignment problems withn vertices.An algorithm running in time O(nl(l +2)n) for computing the number of differentproper assignments of span at mostK is further presented.© 2004 Elsevier B.V. All rights reserved.

1. Introduction

The channel assignment problem naturally generalizes graph coloring. Achannel assignment problemis a triple(V , E, w)

whereV is a vertex set, E is a set consisting of pairs of elements ofV (callededges) andw is a function assigning edgesof E positive integer weights. Ife is an edge ofE, the numberw(e) is called theweightof the edgee. The value ofw(uv)

is considered to be zero ifuv /∈ E. Hence we further consider channel assignment problems to be just pairs(V , w) where

w :(

V2

)→ {0,1,2, . . .} implicitly assuming that the edges areuvwith w(uv) >0. Anassignmentc of positive integers to the

vertices, i.e., the elements ofV , isproper, if |c(u) − c(v)|�w(uv) for each edgeu, v ∈ V . Thespan of a proper assignmentisthe maximum integer assigned to a vertex; thespan of a channel assignment problemis the smallest integerk for which there isa proper assignment of the spank. A channel assignment problem and its special case called distance graph labelling[6,7,9]canbe used to model real-world problems: The vertices can be thought as radio transmitters, integers assigned to them as frequenciesand the weights of the edges as minimal differences of frequencies which can be assigned to the transmitters corresponding totheir end-vertices without interference.

Determining the span of a channel assignment problem is very hard: the decision version of the problem for the span is NP-complete even if the underlying graph has tree-width 3 (cf.[14]).A channel assignment problem isl-boundedif theweight of eachedge is atmostl. Since the decision version of the graph coloring problem iswell-known to beNP-complete, the decision problemfor the span of 1-bounded channel assignment problem is NP-complete, too. In Section 3, we find an algorithm for computingan optimal assignment of anl-bounded channel assignment running in time O(n(l + 2)n) and in space O((l + 2)n) wheren = |V | (Corollary 7) in a similar computation model which we assume in this paper. This improves a previous algorithm due toMcDiarmid running in time O(n2(2l +1)n) from [13], see also[12]. Our algorithm is based on a concept of separated sequences

E-mail address:kral@kam.mff.cuni.cz(D. Král’).1 Institute for Theoretical Computer Science is supported by Ministry of Education of Czech Republic as project LN00A056.

0166-218X/$ - see front matter © 2004 Elsevier B.V. All rights reserved.doi:10.1016/j.dam.2004.01.020

D. Král’/ Discrete Applied Mathematics 145 (2005) 326–331 327

introduced in Section 2. Next, wemodify the algorithm to an algorithm running in timeO(nl(l+2)n)which computes the numberof proper assignments of span at mostK (Theorem 8); note that the running time of the counting algorithm is independent onK. The computational models considered in this paper are the following: in case of the first algorithm, we assume that we arecapable to store O(n log l)-bit pointers and each operation with O(log n)-bit numbers requires a unit of time. In case of thesecond algorithm, since the resulting numbers might be�(log n)-bit numbers, we assume that each operation with�(b)-bitnumbers requires a unit of time, if the results are�(b)-bit numbers. The just presentedmodels are implicitly assumed throughoutthe paper.

2. Separated sequences

In this section, definitions and notation related to separated sequences are introduced.Let (V , w) be a fixedl-bounded channel assignment problem in this paragraph. A sequenceV = V0, . . . , Vk−1 of k subsets

of V is non-decreasingif Vi−1 ⊆ Vi for 1� i �k − 1. A sequenceV is separatedif it is non-decreasing andw(vivj )�j − i

for anyvi ∈ Vi\Vi−1, vj ∈ Vj \Vj−1, 1� i �j �k − 1. In particular, ifV is separated, thenVi\Vi−1 is an independent set forany 1� i �k − 1. Note that there is no condition on the weights of the edges joining the vertices ofV0 to the remaining vertices.A consecutive subsequence of a separated sequence is separated, too. Unless stated otherwise, a (non-decreasing, separated)sequence is a sequence formed by subsets of the vertex set in the rest.A non-decreasing sequenceV′ = V ′

0, . . . , V′k′−1 extendsa non-decreasing sequenceV = V0, . . . , Vk−1 if there existsi0,

max{0, k − k′ + 1}� i0�k − 1 such that the following holds:

Vi0+j = V ′j for 0�j �k − 1− i0.

The sequenceV′′ = V0, . . . , Vi0 = V ′0, Vi0+1 = V ′

1, . . . , Vk−1 = V ′k−1−i0

, . . . , V ′k′−1 is called anextensionof V with V′

and writeV ⊕ V′ for it; it will always be clear from the context which extension we have in mind if there are more possibleextensions. The extension of a separated sequence with another separated sequence is not necessarily a separated sequence ingeneral. However, the following holds.

Lemma 1. Let (V , w) be anl-bounded channel assignment problem, V′ a separated sequence of length at leastl + 1 whichextends a separated sequenceV. If the lengthV ⊕ V′ is the length ofV increased by one, thenV ⊕ V′ is separated.

Proof. LetV⊕V′ =V0, . . . , Vk , 1� i, j �k, vi ∈ Vi\Vi−1 andvj ∈ Vj \Vj−1. If both i < k andj < k, thenw(vivj )� |i −j |becauseV is separated. Ifi = k andj �k − l, thenw(vivj )� l� |i − j | = |k − j | because the problem isl-bounded. Ifi = k

andj > k − l, thenw(vivj )� |i − j | becauseV′ is separated. �

The span of a channel assignment problem is related to separated sequences in the next lemma.

Lemma 2. Let (V , w) be a channel assignment problem and letK �0 be an integer. The set of proper assignments of(V , w)

with the span at mostK is in one-to-one correspondence to the set of separated sequences of lengthK + 1, V0, . . . , VK , withV0 = ∅ andVK = V .

Proof. Letcbe a proper assignment. LetVi ={v|c(v)� i} for 0� i �K. The just defined sequence is separated sinceVi\Vi−1={v|c(v) = i} and|c(u) − c(v)|�w(uv); the conditionsV0 = ∅ andVK = V are also satisfied. On the other hand, we can definefor a separated sequenceV0=∅, V1, . . . , VK−1, VK =V a proper assignmentc(v)=min{i|v ∈ Vi}. SinceV0=∅ andVK =V ,the assignmentc is defined for all the vertices and its span is at mostK. Because the sequence is separated,c is proper. It isstraightforward to verify that the just defined correspondence is really one-to-one.�

The immediate corollary of Lemma 2 is the following:

Lemma 3. The span of a channel assignment problem(V , w) is the smallestK such that there exists a separated sequenceV0, . . . , VK with V0 = ∅ andVK = V .

3. The algorithms

We describe our algorithms forl-bounded channel assignment problems in this section. Let(V , w) be a givenl-boundedchannel assignment problem. We writen for |V | in the rest of this paper.Gl (V , w) is the directed graph whose vertices are all

328 D. Král’/ Discrete Applied Mathematics 145 (2005) 326–331

the separated sequences of lengthl and a sequenceV is joined by an arc to a sequenceV′, if V′ extendsV andV ⊕ V′ is aseparated sequence of lengthl +1. LetV∅

lbe the (separated) sequence∅, . . . , ∅ of lengthl andVV

lbe the (separated) sequence

V, . . . , V of lengthl. Note thatGl (V , w) might contain loops, in particular, there are loops incident withV∅landVV

l.V∅

lis

called theinitial separated sequenceandVVlthefinal separated sequence. We first show thatGl (V , w) can be constructed in

time O(n(l + 2)n).

Lemma 4. Let (V , w) be a channel assignment problem. There is an algorithm which outputs all the separated sequences oflengthk and which runs in timeO(n(k + 1)n) and in spaceO(n2). The number of separated sequences of lengthk is at most(k + 1)n.

Proof. It is possible to design a simple backtracking algorithm based on the following one-to-one correspondence betweensequencesa1, . . . , an of integers between 0 andk of lengthn and non-decreasing sequencesV0, . . . , Vk−1 of lengthk:

ai = min{j |vi ∈ Vj } ∪ {k}.

The algorithm simply generates all such sequences of integers and one can check that the edgesvivj , 1�j < i, do not violatethe condition thatV is separated, when generatingai . This straightforwardly leads to an O(n(k + 1)n)-time algorithm. �

Lemma 5. Let(V , w)be anl-bounded channel assignment problem.The graphGl (V , w) can be constructed in timeO(n(l+2)n)

and in spaceO((l + 2)n). The graphGl (V , w) contains at most(l + 2)n arcs.

Proof. We create a complete(l +1)-ary treeT of depthn (the depth of a tree containing only a root is zero). The paths from theroot to the leaves inT correspond to sequences of integers between 0 andl and hence to non-decreasing sequences of subsets ofV of lengthl as described in the proof of Lemma 4.We use twice the algorithm from Lemma 4. First, all the separated sequencesof lengthl are generated and it is stored in the leaves ofT whether the corresponding sequence is separated. Next, we will storea list of arcs leading from the corresponding vertex ofGl (V , w) in each leaf ofT: all the separated sequencesV of lengthl +1are generated and split to two sequencesV′ andV′′ of lengthl each such thatV=V′ ⊕V′′ (V can be uniquely decomposedtoV′ andV′′) and we add to the list of arcs leading fromV′ stored inT an arc fromV′ toV′′.

The data structuresT looks as follows: each internal node containsl+1 pointers and each leaf contains a single bit determiningwhether it corresponds to a separated sequence and a list of pointers to leaves corresponding to its successors inGl (V , w). Thesum of the lengths of these lists is O((l + 2)n) which strictly dominates the space O((l + 1)(l + 1)n) needed by the rest ofthe algorithm. The graphGl (V , w) can be immediately obtained from the leaves ofT. The running time of the algorithm isdominated by the step when the algorithm from Lemma 4 is used for the second time and it is O(n(l + 2)n).

Since each arc ofGl (V , w) corresponds to a separated sequence of lengthl + 1, the number of arcs ofGl (V , w) is at most(l + 2)n due to Lemma 4. From this, it is clear that space O((l + 2)n) is enough to storeGl (V , w) and hence it covers all spacedemands of the algorithm.�

We are now ready to relateGl (V , w) to proper assignments.

Theorem 6. Let(V , w) be anl-bounded channel assignment problem. The length of the shortest directed path betweenV∅l

and

VVl

in Gl (V , w) is equal to the span of(V , w) increased byl − 1.

Proof. Consider a shortest directed pathV0, . . . ,VL fromV∅ltoVV

lin Gl (V , w). It follows from the definition ofGl (V , w)

thatVi−1⊕Vi is a separated sequence of lengthl +1 (1� i �L). Since(V , w) is l-bounded, it straightforwardly follows fromLemma 1 thatV0 ⊕ · · · ⊕ VL is a separated sequence of lengthL + l. We can construct a separated sequenceV of lengthL + l − 2(l − 1) by cutting firstl − 1 and lastl − 1 elements ofV0 ⊕ · · · ⊕ VL. Since the first element ofV is ∅ and the lastone isV , the span of(V , w) is at mostL − l + 1 due to Lemma 3.We prove the opposite inequality of the statement of the theorem in this paragraph. LetK be the span of(V , w) andV a

separated sequence of lengthK + 1 starting with∅ and ending withV (it exists due to Lemma 3).V is extended toV′ byaddingl − 1 empty sets to the beginning ofV and by addingl − 1 setsV to the end. The length ofV′ isK + 2l − 1.V′ can besplit to sequencesV0, . . . ,VL of lengthl formed by consecutive subsequences ofV′. Note thatV0 = V∅

l,VL = VV

land

L = K + l − 1. This establishes the theorem.�

Theorem 6 implies together with Lemmas 2 and 5 the following:

D. Král’/ Discrete Applied Mathematics 145 (2005) 326–331 329

Corollary 7. There is an algorithm for computing the span of anl-bounded channel assignment problem and constructing anoptimal assignment which runs in timeO(n(l + 2)n) and spaceO((l + 2)n).

Proof. The proof of Theorem 6 actually describes the way how to find an optimal assignment of the problem. The shortest pathbetween two given vertices in a directed graph can be found in time linear in the number of arcs if the graph is represented byadjacency lists by a simple breadth-first search algorithm, see, e.g.,[3]. �

The first algorithm can be modified to count proper assignments.

Theorem 8. There is an algorithm for computing the number of proper assignments of span at mostK for a givenl-boundedchannel assignment problem and givenK; the algorithm runs in timeO(nl(l + 2)n) and in spaceO((l + 2)n).

Proof. Proper assignments of span at mostK one-to-one correspond to separated sequences of lengthK + 1 starting with anempty set and ending with the whole vertex set due to Lemma 2. It follows from the proof of Theorem 6 that separated sequencesof lengthK + 1 one-to-one correspond to directed walks (i.e., they may contain the same vertex more times) betweenV∅

land

VVlof lengthK + l − 1 inGl (V , w). We reduced the problem to computing the number of directed walks inGl (V , w).

Observe thatGl (V , w) is acyclic. First, the vertices ofGl (V , w) are topologically sorted, i.e., a sequence of verticesV1, . . . ,VN of Gl (V , w) is found which satisfies that ifViVj is an arc, theni �j . A topological sorting algorithm runsin time O(m) wherem is the number of arcs of the digraph (see, e.g.,[3]); this gives the time bound O((l + 2)n). Note thatV1 =V∅

landVN =VV

l. For each vertexGl (V , w), a triple(A, p, k0) is computed whereA is an array of sizek0 indexed by

numbers from 0 tok0 − 1,p is a polynomial andk0 is an integer. The polynomials are represented as sequences of their coeffi-cients; later we show that the degrees of all the polynomials which appear in the algorithm are at mostn. We write(Ai, pi , ki

0)

for a triple assigned to the vertexVi . It is demanded that the triple(Ai, pi , ki0) assigned toVi satisfies the following:

The number of directed walks from the vertexV1 = V∅

lto the vertexViof lengthk is

{Ai [k] if 0�k < k0.

pi(k) if k0�k.

The triples can be computed in the order determined by the topological ordering. The triple assigned toV1 consists of an emptyarrayA1, a polynomialp1(k) := 1 and an integerk0 := 0. Let us assume that the triples for allVi , i < i0 have been computed.We first compute a triple(A′, p′, k′

0) which has the same meaning as(Ai0, pi0, ki0) except that it counts the directed walks

fromV1 toVi0 which end with exactly oneVi0. Let I be the set of thosei for which there is an arcViVi0 in Gl (V , w). Thefollowing holds due to the definition of the triples:

k′0 := 1+ maxi∈I ki ,

A′[0] := 0,

A′[k] :=∑

i∈I,k−1<ki0

Ai [k − 1] +∑

i∈I,k−1�ki0

pi(k − 1) for 1�k < k′0,

p′(k) :=∑i∈I

pi(k − 1).

If there is not a loopVi0Vi0 inGl (V , w), then any walk fromV1 toVi0 ends with exactly oneVi0 and hence(Ai0, pi0, ki0)=(A′, p′, k′

0). If there is a loopVi0Vi0, then the triple(Ai0, pi0, ki0) is determined as follows:

ki00 := k′

0,

Ai0[k] :=∑

0�k′ �k

A′[k′] for 0�k < k0,

pi0(k) :=∑

0�k′<k′0

A′[k′] +∑

k′0�k′ �k

p′(k′).

Using the above formulas, we can compute all the triples (there are explicit formulas for sums of type∑k

i=k′0x� for a fixed�

which can be used to calculatepi0(k); observe that∑k

i=k′0x� is a polynomial inx of degree� + 1) and we can determine the

number of directed walks fromV1 = V∅ltoVN = VV

lhaving computed(AN , pN , kN

0 ).

330 D. Král’/ Discrete Applied Mathematics 145 (2005) 326–331

It remains to establish the time and space demands of the algorithm. We first realize that the length of any directed path inGl (V , w) is at mostnl: for any pathV1, . . . ,Vk the sequenceV1 ⊕ · · · ⊕ Vk is a separated (and hence non-decreasing)sequence of lengthk + l − 1 and it does not contain a subsequence consisting of at leastl + 1 identical sets (otherwise thesequenceV1, . . . ,Vk contained the same vertex twice). This immediately gives thatk + l −1�nl and hencek�nl. Sinceki

0 is

the length of the longest directed path fromV1 toVi (this can be easily proved by induction from the formulae definingki0 for

1� i �N ), the above implies thatki0�nl for all 1� i �N . The degree of a polynomial is increased only at a vertex with a loop;

hence the degree ofpi is at most the maximum number of vertices with loops on a path fromV1 toVi . On any directed pathV1, . . . ,Vk there are at mostn vertices with loops because the verticesVi with loops are precisely the vertices correspondingto sequences consisting ofl identical subsets ofV , and the sequenceV1 ⊕ · · · ⊕ Vk is separated (and hence non-decreasing).We may conclude that the degree of anypi is at mostn. The time and space needed by the algorithm can be now estimated: thegraphGl (V , w) can be constructed in time O(n(l +2)n) and in space O((l +2)n). Computation of(A′, p′, k′) requires additionof at mostnl values for each arc leading to a vertex and hence it takes total time O(nl(l + 2)n)for all the vertices ofGl (V , w)

becauseGl (V , w) contains at most(l + 2)n arcs. In addition, at every vertex with a loop, we need extra time to compute theactual triple from(A′, p′, k′). This extra time is at most O(q(l, n)) for some polynomialq(l, n). SinceGl (V , w) contains atmost(l +1)n vertices, the time bound O(nl(l +2)n) dominates the extra time O((l +1)nq(l, n)) spent at the vertices with loops.The space needed by the algorithm consists of major parts: the space needed to storeGl (V , w) which is O((l + 2)n) and thespace needed to store the triples which is O(nl(l + 1)n) (each triple requires a space O(nl)—O(nl) for the arrayA and O(n) forthe coefficients ofp). The bound O((l + 2)n) dominates the bound O(nl(l + 1)n). �

Note that we actually proved that the number of proper satisfying assignments of spanK is a polynomial inK for sufficientlylargeK. This result has been previously proved using other techniques in[11,16].

4. Conclusion

For the case of ordinary graph coloring (a 1-bounded channel assignment problem), our algorithm runs in time O(3n)—weomit polynomial factors in time bounds in the whole concluding section. There are better algorithms for this special case: thefirst one with running time O(2.4422n) was obtained by Lawler in[10] and an improvement to O(2.4150n) from [5] is due toEppstein.All these algorithms are based onmaximal independent set (MIS) technique. There are a series of papers on algorithmsfor 3-coloring of graphs[1,2,4,15]and only one of them,[1], is not MIS-based. There are instances of the 2-bounded channelassignment problem[8] for which any MIS-based algorithm outputs a span arbitrarily larger than the actual span is, so MIS-technique does not seem to be applicable in this case. Anyway, there could be faster exact algorithms for both the bounded andunbounded channel assignment problems based on other approaches. In particular, the following problem is open:

• Find an algorithm running in timeO(Kn) for a fixed constantK for the (unbounded) channel assignment problemonnverticesor prove that there is no such algorithm under reasonable complexity theory assumptions.

Acknowledgements

The author would like to thank Jan Kára and Jan Kratochvíl for carefully reading the preliminary version of this manuscript.The comments of the anonymous referees which helped to improve the presentation are also appreciated.

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