beyond the shannon’s bound

8/13/2019 Beyond the Shannons Bound

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arXiv:1309.6069v

1

[cs.DS]24Sep

2013

Beyond the Shannons Bound

Michal Farnik Lukasz Kowalik Arkadiusz Socala

Abstract

Let G= (V, E) be a multigraph of maximum degree . The edges ofG can becolored with at most 32 colors by Shannons theorem. We study lower bounds on

the size of subgraphs ofG that can be colored with colors.Shannons Theorem gives a bound of 3

2 |E|. However, for = 3, Kaminski and

Kowalik [7,8] showed that there is a 3-edge-colorable subgraph of size at least 79 |E|,unless G has a connected component isomorphic to K3+e (a K3 with an arbitraryedge doubled). Here we extend this line of research by showing that G has a -edgecolorable subgraph with at least 3

21 |E| edges, unless is even and G contains

2K3or is odd andGcontains

12 K3+e. Moreover, the subgraph and its coloring

can be found in polynomial time.Our results have applications in approximation algorithms for the Maximum k-

Edge-Colorable Subgraph problem, where given a graphG(without any bound on its

maximum degree or other restrictions) one has to find a k-edge-colorable subgraphwith maximum number of edges. In particular, for every even k 4 we obtain a2k+23k+2 -approximation and for every odd k 5 we get a 2k+13k -approximation. When4k13 this improves over earlier algorithms due to Feige et al. [5].

1 Introduction

In this paper we consider undirected multigraphs (though for simplicity we will callthem graphs). A graph is k-edge-colorable if there exists an assignment ofk colorsto the edges of the graph, such that every two incident edges receive different colors.By Shannons theorem [13], 3

2

colors suffice to color any multigraph, where denotes the maximum degree. This bound is tight, e.g. for every even consider thegraph (/2)K3, and for odd consider the graph/2 K3+ e (see Fig. 1 and 2and Section1.1 for definitions).

It is natural to ask how manyedges of a graph of maximum degree can becolored with less than

32

colors. Themaximum k-edge-colorable subgraph of G

Theoretical Computer Science Department, Faculty of Mathematics and Computer Science, Jagiel-lonian University, Krakow, Poland; email: [email protected]

Institute of Informatics, University of Warsaw, Poland; email: [email protected] of Mathematics, Informatics and Mechanics, University of Warsaw, Poland; email:

[email protected].

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edges when is odd ( 12 K3 is one of them). It is natural to ask whether these areagain the only bottlenecks and how does the next bottleneck look like. We partially

answer this question, at least for large value of , with the following theorem.

Theorem 2. LetG be a multigraph of maximum degree and lett be an integer suchthat

32

t 12

22 1 1.34. Assume that G does not contain a three-vertex subgraph with more than t edges. Then G has a -edge-colorable subgraphwith at least t|E| edges. Moreover, the subgraph and its coloring can be found inpolynomial time.

By Vizings theorem for multigraphs, every multigraph of maximum degree and maximal edge multiplicity has a -edge colorable subgraph with at least +edges. Below we state an immediate corollary from Theorem2 which improves this

bound for

[2226 , /2).

Corollary 3. LetG be a multigraph of maximum degree and maximal edge mul-tiplicity 6

22 20.45. ThenG has a-edge-colorable subgraph with at

least 3 |E| edges. Moreover, the subgraph and its coloring can be found in polynomialtime.

Below we state a (not immediate) corollary from Theorem 1. It will be useful inapplications in approximation algorithms.

Theorem 4. LetG be a connected multigraph of maximum degree3. ThenGhas a-edge-colorable subgraph with at least

1.

2

32 |E| edges when is even andG=

2K3,2. 2+13 |E| edges when is odd andG= 12 K3+ e.

Moreover, the subgraph and its coloring can be found in polynomial time.

Again, the bounds in Theorem4 are tight. The smallest examples are 2K3 withone edge removed when is even and a graph consisting of two copies of 12 K3 + ewith the two vertices of degree 1 joined by an edge when is odd.

1.2 Approximation Algorithms

One may ask why we study (G) and not, say +1(G). Our main motivation

is that finding large -edge-colorable subgraphs has applications in approximationalgorithms for the Maximum k-Edge-Colorable Subgraphproblem (aka Max-imum Edge k-coloring [5]). In this problem, we are given a graph G (without anyrestriction on its maximum degree) and the goal is to compute a maximum k-edge col-orable subgraph ofG. It is known to be APX-hard when k2 [3,6,4]. The researchon approximation algorithms for max k-ECS problem was initiated by Feige, Ofekand Wieder [5]. (In the discussion below we consider only multigraphs, consult [8]for an overview for simple graphs.)

Feige et al. [5] suggested the following simple strategy. Begin with finding amaximumk-matching Fof the input graph, i.e. a subgraph of maximum degree k

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In this paper we also apply the constructive versions of our combinatorial boundswith the algorithmic technique from [7, 8] and we obtain new approximation algo-

rithms which improve the previously known approximation ratios for 4k13.Thecurrent state of art in approximating Maximum k-Edge-Colorable Subgraphfor multigraphs is given in Table1.

1.3 Preliminaries

Our notation is mostly consistent with the one used in [8], which we recall below.Let G = (V, E) be a graph. For a vertex x V by N(x) we denote the set of

neighbors of x and N[x] = N(x) {x}. For a set of verticesSwe denote N(S) =xSN(x) \ S and N[S] =

xSN[x]. We also denote byI[S] the subgraph whose

set of vertices is N[S] and set of edges is the set of edges ofG incident with S. For

a subgraphH ofG we denote N[H] = N[V(H)] and I[H] =I[V(H)].Apartialk-coloringof a graph G = (V, E) is a function : E {1, . . . , k}{}such that if two edges e1, e2Eare incident then (e1)=(e2), or (e1) =(e2) =. From now on by a coloringof a graph we will mean a partial (G)-coloring. Wesay that an edge e is uncolored if(e) =; otherwise, we say that e is colored. Fora vertex v, (v) is the set of colors of edges incident with v , i.e. (v) ={(e) : eI[v]}\{}, while (v) ={1, . . . , k} \ (v) is the set of free colors at v .

LetV={vV : (v)=}. In what follows,(G, ) = (V, 1()) is calledthe graph of free edges. Every connected component of the graph(G, ) is called a

free component. If a free component has only one vertex, it is called trivial.Below we state a few lemmas proved in [8] which will be useful in the present

paper. Although the lemmas were formulated for simple graphs one can easily checkthat the proofs apply to multigraphs as well.

Lemma 5 ([8], Lemma 7). Let(G, ) be a colored graph that maximizes the numberof colored edges. For any free component Q of (G, ) and for every two distinctverticesv, wV(Q)

(a) (v) (w) =,(b) for everya(v), b(w) there is an(ab,vw)-path.

For a free component Q, by(Q) we denote the set of free colors at the verticesofQ, i.e. (Q) =

vV(Q) (v).

Corollary 6([8], Lemma 8). Let(G, )be a colored graph that maximizes the numberof colored edges. For any free component Q of(G, ) we have|(Q)| 2|E(Q)|. InparticularQ has at most

2

edges.

LetQ1, Q2be two distinct free components of (G, ). Assume that for some pair ofvertices xV(Q1) andyV(Q2), there is an edgexyEsuch that(xy)(Q1).Then we say that Q1 seesQ2 withxy, or shortly Q1 seesQ2.

Lemma 7 ([8], Lemma 10). Let(G, )be a colored graph that maximizes the numberof colored edges. If Q1, Q2 are two distinct free components of (G, ) such that Q1seesQ2 then(Q1) (Q2) =.

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We use the notion of the potential function introduced in [8]:

(G, ) = (c, n/2

, n/21

, . . . , n1

),

where c is the number of colored edges, i.e. c =|1({1, . . . , })| and ni is thenumber of free components with i edges for every i = 1, . . . , /2.

1.4 Our Approach and Organization of the Paper

Informally, our plan for proving the main results is to consider a coloring that maxi-mizes the potential and injectively assign many colored edges to every free compo-nent in the coloring. To this end we introduce edges controlled by a component (eachof them will be assigned to the component which controls it) and edges influencedby a component (as we will see every edge is influenced by at most two components;

if it is influenced by exactly two components, we will assign halfof the edge to eachof the components).

In Section2 we develop structural results on colorings maximizing . Informally,these results state that in such a coloring every free component influences/controlsmany edges. Then, in Subsection2.3we prove lower bounds for the number of edgesassigned to various types of components, using a convenient formalism of sendingcharge.

In Section3we develop a method of collapsing subgraphs. We use it to reducesome special graphs, for which maximizing does not give a sufficiently good result,to graphs better suited to our needs. The paper concludes with Section 4 containingthe proofs of the main results.

2 The structure of colorings maximizing

2.1 Moving free components

Note that ifP and Qare distinct free components of a coloring (G, ) then E(P) E(I[Q]) =.

Definition 8. Let (G, ) be a colored graph and let Pbe a nontrivial free com-ponent of (G, ). An elementary moveofP in is a coloring such that:

1. can be obtained from by uncoloringk edges ofI[P] \ E(P) and coloring kedges ofP for some k0,

2. |I[P] has exactly one nontrivial free component, denote it P.If the above holds we say that and P have been obtained respectively from andPby an elementary move. Sometimes we will write shortly is an elementary moveof, meaning that is an elementary move of a free component of .

Note that in particular is the trivial elementary move of any of its components.Note also that and have the same number of uncolored edges. Furthermore thecomponent P is either replaced with a component P or merged with a component

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Q into P Q. Either way an elementary move does not decrease the potential .Furthermore we have the following:

Remark 9. If maximizes the potential then an elementary move of a com-ponent P cannot cause a merge of nontrivial components and hence P is a freecomponent of.

Thus if maximizes the potential and is an elementary move of then thereis a one-to-one correspondence between free components of and free componentsof . For a free component P of and a free component P of we write thatP= P() if either P =P or is the elementary move ofP toP.

Lemma 10. Let(G, ) be a coloring that maximizes the potential and letP andQ be two distinct nontrivial free components. Let be an elementary move ofP in. Then,

|E(I[Q])=

|E(I[Q]).

Proof. Recall that is obtained from by uncoloring some edgesE1E(I[P]), andcoloring some edges E2E(P). Then E2 E(I[Q]) =since P and Q are distinctfree components. Moreover, E1 E(I[Q]) =for otherwise (G, )> (G, ).Lemma 11. Let(G, ) be a coloring that maximizes the potential and letP andQbe two distinct nontrivial free components. Suppose thatP andQ can be obtainedrespectively by elementary movesP andQ ofP andQ. Then:

(i) Q is a free component ofP,

(ii) P is a free component ofQ,

(iii) P|E(I[P])

E(I[Q])= Q

|E(I[P])

E(I[Q]),

(iv) let us define: E(G) {1, . . . , }{}as follows|E(G)\(E(I[P])E(I[Q])) = |E(G)\(E(I[P])E(I[Q])),

|E(I[P]) = P|E(I[P]),|E(I[Q]) = Q|E(I[Q]).

Then, is an elementary move ofQ inP such thatQ() =Q and is anelementary move ofP inQ such thatP(

) = P.

Proof. (i) and (ii) follow directly from Lemma10. For (iii), note that ifeE(I[P])E(I[Q]) then e

E(Q), since P and Q are distinct free components. Hence, by

Lemma10, P(e) =(e). By symmetry, also Q(e) =(e), which proves the claim.Note that (iii) implies that is well-defined in (iv). If we compare to we seethat only edges inE(P) andE(Q) get new colors. Since the former are incident onlyto edges in I(P) and the latter only to edges in I(Q), and P and Q are properpartial edge-colorings (i.e. incident edges get different colors), so is also a properpartial edge-coloring. The rest of the claim follows from Lemma 10.

Lemma 12. Let(G, )be a coloring that maximizes the potential. Let0, 1, . . . , kbe a sequence of colorings such that0 =k = and for i= 1, . . . , k coloring i isan elementary move of a free component ini1. LetP0 be a free component of 0and letPi = Pi1(i). ThenPk = P0.

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Proof. Suppose that for some index j it holds thatj+1 is an elementary move ofPjin j and j is an elementary move of some free component R ofj1 distinct from

Pj1. Then Pj =Pj1. Define a coloring j as follows: j |E(I[Pj1])= j+1|E(I[Pj1])and j |E(G)\E(I[Pj1]) = j1|E(G)\E(I[Pj1]). Then by Lemma 10, j is an ele-mentary move of Pj1 in j1 and j (Pj1) = Pj+1. Moreover, by Lemma 10,j+1|E(G)\(E(I[R])E(I[Pj1])) = j1|E(G)\(E(I[R])E(I[Pj1])), j+1|E(I[R]) = j |E(I[R]),and j+1|E(I[Pj1]) = j |E(I[Pj1]). Hence, by Lemma11(iv), j+1 is an elementarymove ofR in j . Thus we may replace j with

j in the sequence0, 1, . . . , k and

redefine Pj accordingly. Note that by replacing j with

j we have decreased thesum of indexes j such thatj+1 is an elementary move ofPj inj. We continue thisprocess as long as possible and obtain a sequence of colorings and an index 0i0ksuch that for every 1ii0 i is an elementary move ofPi1 ini1 and for everyi0 < i k i is an elementary move of some free component ofi1 distinct fromPi1. Note that 0 and k were never changed.Suppose thatP0 and Pk are two distinct free components of . Let Q0= Pk andQi= Qi1(i). By induction we obtain that Pi and Qi are distinct free componentsofi. Note that for i i0 we have Qi = Qi1 and for i > i0 we have Pi = Pi1.Thus Qi0 =Q0= Pk and Pi0 =Pk, a contradiction.

We say that a coloring is amoveof if maximizes the potential and thereis a sequence of colorings 0= , 1, . . . , k =

such that i is an elementary moveof a free component in i1. We say that a free componentP of is obtained froma free component P in if for every i = 0, . . . , k there is a free component Pi ofisuch that P0= P, P

= Pk and for i > 0 we have Pi = Pi1(i). We denote the freecomponent of a coloringobtained from a free componentPbyP(). This notationextends the notation introduced earlier for elementary moves. Note that P() doesnot depend on the sequence 1, . . . , k1 since given a different sequence 1, . . . ,

t1

we may apply Lemma12 to the sequence , k1, . . . , 1, , 1, . . . , t1,

.

Theorem 13. Let(G, ) be a coloring that maximizes the potential. Then:

(i) if is a move of then is a move of;

(ii) LetPbe a nontrivial free component of and let be a move of. Then thereis a sequence of elementary moves{i}i=1,...,k, such that for everyi= 1, . . . , k,i is a move of a component Pi1 in i1, where P0 = P, 0 = , for everyi= 1, . . . , k we havePi = P(i), Pk = P(

) andk|E(I[Pk])= |E(I[Pk]);(iii) if P1, . . . , P k are distinct nontrivial free components of and 1, . . . , k are

moves of then there is a moveof such thatPi() =Pi(i)and|E(I[Pi()])=i|E(I[Pi(i)]) for i= 1, . . . , k.

Proof.

(i) By Remark9 if is an elementary move of a free component P in then isan elementary move ofP() in . Since a move is a sequence of elementarymoves the claim follows.

(ii) Let 0, . . . , t be any sequence of elementary moves such that 0 = andt=

. We use the process described in the proof of Lemma 12to redefine the

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Definition 16. For a free component Q of a coloring (G, ) we define a set ofvertices con1(Q) ={v V(G) : v V(Q()) for some elementary move ofQsuch that|E(Q()) \ E(Q)| 1}.Lemma 17. Let (G, ) be a coloring that maximizes the potential and letQ be a

free component with|E(Q)| 2. Ifu, vcon1(Q) andu=v then(u) (v) =.Proof. Assume that we have two distinct u, v con1(Q) such that there exists acolora(u)(v). By Lemma5the verticesu and v cannot both belong toV(Q).By symmetry we can assume uV(Q).

Since u con1(Q), one can uncolor an edge ux, x V(Q), and color an edgeeE(Q), obtaining a proper coloring . Note thate is incident withx for otherwisewe can color e without uncoloring ux, and increase the potential . Hencee = xyfor some y V(Q) and (ux) (y). It follows that v V(Q) for otherwise Qsees the (possibly trivial) component containing u, a contradiction with Lemma 7.Let vp be the edge such that one can uncolor vp and color an edge ofQ with color(vp), which exists because v con1(Q). By Corollary6,|(Q)| 4, so there is acolor b (Q) \ {(ux), (vp)}. Let z be the vertex ofQ such that b(z). Notethat a (z) for otherwise Q sees the (possibly trivial) component containing v, acontradiction with Lemma7. Consider a maximal path Pwhich starts at z and hasedges colored in a and b alternately. Swap the colors a and b on the path. As aresult, a becomes free in Q. Also, Ptouches at most one of the vertices u, v , soa isstill free in at least one of them, by symmetry say in u. Since b=(ux), the vertexu is still in con1(Q). Hence we arrive at the first case again.

2.3 Sending charges

In this section we consider a connected colored graph (G, ) which maximizes thepotential .

We put one unit of charge on each colored edge ofG. Every edge divides its chargeequally between the nontrivial components that control its endpoints (by Lemma 14there are at most two such components). For every nontrivial free component P, letch(P) denote the amount of charge sent to P. Then the number of colored edges

in G is at least

Pch(P) |E| minP ch(P)ch(P)+|E(P)| , where the summation is over allnontrivial free components in (G, ). In what follows we give lower bounds for ch(P)for various types of free components.

Observation 18. Let P be a free component of (G, ) and let be a move of .Then every edge of G sends the same amount of charge to P() in the coloring

as it does to P in.

Proof. By Lemma15for an arbitrary free component Q of we have con(Q()) =con(Q). Thus P controls an endpoint of an edge iff P() controls this endpoint.FurthermorePis the only free component of controlling an endpoint of an edge iffP() is the only free component of controlling an endpoint of that edge.

For a subsetSEby degSv we denote the degree ofv in the graph (V, S) andV(S) denotes the set of endpoints of all edges in S. The following observation follows

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immediately from Lemma 14 (we state explicitly also the weaker bound 12 |V(S)con(P)| because it will be sufficient in many cases).Observation 19. LetPbe a nontrivial free component and letSbe an arbitrary setof colored edges. Then edges in Ssend the charge of at least 12

vcon(P)degSv

12 |V(S) con(P)| to P.

Let Q be a free component in a colored graph (G, 1). If a colored edge e isincident with Q,|1(Q)| (G) 1, and 1(e)1(Q), then we say that edge e isdominatedbyQ in the coloring 1.

Lemma 20. LetP be a nontrivial free component in(G, ) and let1 be a move of. LetP = P(1). Every edge dominated byP in1 sends its whole charge to P.

Proof. Let e = xy and x

V(P). (Then x

con(P).) By Lemma 14 it suffices

to show that there is no nontrivial component Q such that ycon(Q). Assume fora contradiction that such a component exists. Since y con(Q), there is a move2M(Q) such thatyV(Q(2)). By Theorem13(iii) there is a move 3of suchthat P(3) = P

and Q(3) = Q(2) and 3|E(I[P(3)]) = 1|E(I[P]). In particular3(e) = 1(e), so P(3) sees Q(3) through e in the colored graph (G, 3) and byLemma 7 they have disjoint sets of free colors. However, since Q is nontrivial, byCorollary6we have|(Q)| 2, a contradiction.Lemma 21. Every free component P of (G, ) with|E(P)| = 1 receives at least(G) of charge.

Proof. We will show that for each color a {1, . . . , } the componentP receives atleast one unit of charge from edges colored with a. Suppose that a / (P). Theneach of the two vertices of P is incident with an edge colored with a, hence theseedges send at least 1 by Observation 19. Now suppose that a (P). Denote thevertices of P by x and y so that a (x). By Lemma 5, a (y); let z be theneighbor ofy such that (yz) = a. Uncoloringyz and coloring xy is an elementarymove sozcon(P) and hence z y sends 1 to Pby Observation19.Lemma 22. Let P be a free component of (G, ) and let U con(P) be a set ofvertices such that (v)(w) = for every two distinct vertices v, w U. Thench(P)(|U| 1)|E(P)|.

Proof. LetSadenote the set of edges incident to Uand colored with color a. Sinceais free in at most one vertex ofU, we have|V(Sa) U| |U| 1. SinceUcon(P),by Observation19we infer that Sa sends at least

|U|12 of charge to P. Taking the

sum over all colors we obtain that P receives at least (|U|1)2 of charge. Moreoverby Lemma6 we have|E(P)| 2, hence the claim.

In what follows, a single edgeis an edge of multiplicity 1.

Lemma 23. LetPbe a free component of(G, ) isomorphic to the2-path. Assumethat|(P)| (G)1. If both edges ofPare single edges inG, thench(P)2(G).

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Proof. By Lemma17 and Lemma22, if|con1(P)| 5, we are done. In what followswe assume| con1(P)| 4. Let Sa denote the set of edges incident to con1(P) andcolored with color a. We will show that for every color a the edges of Sa send atleast 2 to P; then the claim clearly follows.

Let P = xyz and let a (x). Then, by Lemma5we have a (y), so thereis an yv edge such that (yv) = a. Since P has single edges we have v / V(P).Furthermore we may move P by uncoloring yv and coloring xy, thus v con1(P)and hence con1(P) ={x,y ,z ,v}.

Consider any edge e colored with c(P) (v) and incident to con1(P). Ifeis incident to V(P) and c (P) then e sends 1 to P by Lemma20. Assumee isincident to v but not to V(P). Then c (P). Color c is free in at most one ofx, z; by symmetry we can assume c(x). Pick any color cx(x). By Lemma5,cx (y) and hence there is an edge vy colored with cx. Consider the elementarymove obtained by uncoloringvyand coloringxy withcx. We obtainv con1(P).SincePhas single edges it follows that v = v. Furthermorec(P()) and hencee is dominated by P(), so e sends 1 to P by Lemma20. Finally, assume c(v).Then e is incident with V(P), if both endpoints ofe are in V(P) then e sends 1 toP. Assume that only one endpoint of e is in V(P); by symmetry we can assume eis not incident with x. Then, again e is dominated by P(), so e sends 1 to P byLemma20. Hence, for everyc(P)(v) every edge inScsends 1 toP. Moreover,forc(P)(v) we have|Sc| (| con1(P)| 1)/2 = 3/2, so|Sc| 2 and the edgesin Sc send 2 to P, as required.

Finally consider c (P) (v). Since every vertex in con1(P) is incident withan edge in Sc, by Observation19we infer that Sc sends at least

12 | con1(P)| = 2 of

charge to P, as required.

Lemma 24. LetP be a free component of (G, ) consisting of ak-foldxy edge fork 2. Assume that|(P)| = (G). Then eitherch(P)2(G) orG has exactlythree vertices.

Proof. First assume|N(P)| = 1, i.e. N(P) ={z}, for some vertex z. Since (x)(y) ={1, . . . , }and (x)(y) =by Lemma5, so(x) =(y) and(y) =(x).It follows that the number of edges between z and P is|(x)|+|(y)| = . Thismeans that z has no neighbors outside P and G has exactly three vertices x, y andz.

Now assume that z1 and z2 are two distinct vertices in N(P). For everyi = 1, 2

we can uncolor any edge between zi and Pand color one of the edges xy obtainingan elementary move i. Thusz1, z2 con1(P) and by Lemma17we have (z1) =(z2) =. Every colored edge incident to x or y is dominated byPand it sends 1 toPby Lemma20. Similarly, fori = 1, 2 every colored edge incident to zi is dominatedbyP(i) and it sends 1 to Pby Lemma20. So each colored edge in I[{x,y,z1, z2}]sends its charge to P. Notice that for each color there are at least two distinct edgesin I[{x,y ,z1, z2}] colored with this color: one incident with Pand one incident withthe zi not connected to P by the first edge. Thus P receives at least two units ofcharge from edges of a given color and hence the claim.

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we are done, so in what follows we assume 7. On the other hand, by Corollary6we infer that |(Q)| 2|E(Q)| = 6. Hence, {6, 7} and|(Q)| 1.We will show that for every color a, edges in Sa send 2 units of charge to Q; thench(Q) 2(G) 12 = 4|E(Q)|, as required. First assumea (Q). Then everyedge in Sa is dominated by Q and by Lemma 20 sends 1 to Q. It follows that Sasends to Q at least|Sa| (|V(Q)| 1)/2= 2 units of charge. Finally, ifa(Q)then|V(Sa) V(Q)|= 4, so by Observation19 edges in Sa send at least 42 = 2 unitsof charge to Q, as required.CASE 2. |E(Q)| 4. Note that Q contains a cycle (possibly of length 2, i.e. amultiple edge). Since|V(Q)|= 4, all cycles in Q have length at most 4. Our strategyis to show that con(Q) contains a subset Uof cardinality at least 5 such that everypair of vertices in Uhas distinct free colors. Then the claim follows from Lemma22.CASE 2.1. Qcontains a cyclevpqv of length 3 or a cycle uvpquof length 4. In the

prior case let u be the fourth vertex in V(Q) and assume w.l.o.g. uv E(Q). Leta(u). By Lemma5each of the vertices v, p, q is incident with an edge coloredwitha. At least one of these edges has the other endpointxoutsideQ. By symmetrythere are two cases: (i) there is an edge vxcolored with a or (ii)vqis colored withaand there is an edge px colored with a. In both cases xcon(Q): in case (i) we canmove Q by uncoloring vx and coloring vu with a, and in case (ii) we can move Qby uncoloringvqand px and coloring uv and pqwith a. Moreover, (x) (Q) =,for otherwiseQ sees the trivial free component{x}, a contradiction with Lemma 7.Vertices ofQ have disjoint free colors by Lemma5, so we can putU={u,v,p,q,x}.CASE 2.2. Qdoes not contain cycles, but it contains a multiple edge. Then, afterignoring multiplicity of edges, Q is a treeTQ. First assume thatQ contains a double

edge uv such that v is a leaf of TQ. Let a (u). By Lemma 5 there is an edgevx colored with a. Consider the elementary move obtained by uncoloring vx andcoloring uv with a. IfxV(Q), then xcon1(Q), hence by Lemma17 we can putU= con1(Q). Otherwise,Q(

) contains a cycle of length 3 or 4, so the claim followsby Case 2.1 and Observation18.

We are left with the case when every multiple edge of Q is not incident with aleaf ofTQ. Since|V(Q)| = 4 it means that TQ is a path uvpq, and the edge vp hasmultiplicity at least two in Q. Leta(p) and consider the edgevxcolored with a,which exists due to Lemma5. We may moveQ by uncoloringvx and coloringvp witha. Ifx /V(Q) then again xcon1(Q) and by Lemma17 we can putU= con1(Q).Ifx V(Q) then either x = u and we have obtained the case where the multipleedge is incident to a leaf of TQ or x = qand we have obtained the case where Qcontains a cycle of length 3; in both cases we get the claim by Observation 18.

3 Collapsing subgraphs

Definition 28. Let G be a multigraph with maximum degree and H an in-duced subgraph of G such that|V(H)| = 3. We say that H is k-collapsible if|E(V(H), V(G) \ V(H))| k and|E(H\ x)| |E(x, V(G) \ V(H))| for every vertexxV(H)

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IfH is ak-collapsible subgraph ofG then we may obtain a graph G by removingfrom G all edges in E(H) and identifying the three vertices in V(H) into a single

vertex h. Note that G has the maximum degree at most max{k, (G)}. We saythat G is obtained by collapsing H to h in G.Lemma 29. Let G be obtained by collapsing a k-collapsible subgraph H to h in amultigraphG with maximum degree.

(i) If G does not contain an induced subgraph on three vertices with more than + k2 edges then G does not contain an induced subgraph on three verticeswith more than + k2 edges.

(ii) Given a partial coloring of G which colors at least p|E(G)| edges we canconstruct a partial coloring of G which colors at least min{p, |E(H)|}|E(G)|edges.

Proof. 1. Every induced subgraph of G that does not contain h is isomorphicto an induced subgraph ofG hence we only need to consider subgraphs ofG

containing h. However a subgraph of G containing h and two other verticescan have at most +

degG(h)

2

+ k2 edges.

2. The edges inG correspond to edges in Gthat are not in E(H). Thus we mayuse the partial coloring to obtain a partial coloring of E(G)\E(H). Theedges inG incident with h correspond to edges in E(V(H), V(G) \ V(H)). Fora vertex x V(H) we use the colors of edges in E(x, V(G)\V(H)) to coloredges in E(H\x). We use the remaining colors to color arbitrary edges of

E(H). We obtained a partial coloring such that min{, |E(H)|} edges ofE(H) are colored. Thus the claim follows.

Lemma 30. Let G be a connected multigraph of maximum degree . Let be acoloring maximizing.

(i) If = 4 andG= 2K3 then colors at least 45 |E| edges.(ii) If = 5 andG does not contain a3-collapsible subgraph then colors at least

56 |E| edges.

Proof. We will show that in both cases each free component P of receives at least

|E(P)| units of charge, which gives the claim.If|E(P)| = 1 then we are done by Lemma 21. Thus by Corollary 6 we mayassume that|E(P)|= 2. We will first consider two special cases and then show thatall other cases can be reduced via Observation 18 to those two cases.

IfP is a path of length two and its edges are single edges in G then we are doneby Lemma23.

IfPconsists of a double edge then we are done either by Lemma24or by Lemma25.

Assume that P is a path xyz and there are at least two xy edges in G. Let adenote the color of a colored xy edge. Ifa(z) then we uncolor the a-colored xyedge and color theyz edge witha. We obtain a component consisting of a double edge

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and we are done by Observation18. So assume thata /(z). Sincea(x) (y),we have a(P). By Lemma5 this means that = 5 and (x) ={b}, (z) ={c}for some distinct colorsb and c. Consider the edges ywb andywccolored respectivelywith b and c. Ifwb =z or wc =x then we can uncolor this edge, color respectivelyxy or y z and obtain a component consisting of a double edge; again we are done byObservation18. Otherwise,{x, y}{wb, wc}= and we may movePto a componentwith edges ywb and ywc. Note that if wb= wc then both edges ywb and ywc aresingle in G, because deg(y)5. We obtain a new component consisting of a doubleedge or two single edges in G and again we are done Observation18.

In the following lemma we will use the function . Although the definition mayseem artificial it will be justified in the proof.

Let(, k , t) = min({72} {32e | e,, Z, e2, 2e, 0, + ,e +

t, 2+

k + 1}

).

Lemma 31. Let (G, ) be a coloring maximizing . Let 6 be the maximaldegree of G and lett 32

and0k be integers such that G does not

contain ak-collapsible subgraph andG does not contain an induced subgraph on threevertices with more than t edges. LetH be a set of three vertices ofG and letQ bea free component of (G, ) such that|E(Q)| 2 and V(Q) H con1(Q). If foreveryv H, for every color a (v) the two vertices ofH\ {v} are connected byan edge colored witha, thench(Q)(, k , t)|E(Q)|.Proof. Let =|(H)|, by Corollary6we have |(Q)| 2|E(Q)|. By Lemma17,for every color a (H) there are exactly two vertices ofH incident with an edge

colored with a, so by Observation19 Q receives the charge of at least from edgescolored with colors in (H). Clearly, for every color a (H) there are exactlythree vertices ofH incident with an edge colored with a, and hence Q receives thecharge of at least 32( ) charge from edges colored with colors not in (H). Thusch(Q) 12(3). In what follows, we show that 12(3)(, k , t)|E(Q)|, thatis we will define asuch that for e =|E(Q)| either ch(Q) 72 or all the inequalitiesin definition of are satisfied.

Note that for any color b / (H) there are either two or three distinct edgescolored withb and incident with a vertex ofH. LetC2and C3be the sets of colors in{1, . . . , }\(H) such that there are respectively two or three distinct edges coloredthat color and incident with a vertex ofH. Let=|C3|, then|C2|= . We

will show that if 2+ k and |E(Q)| then the subgraph ofG induced byH is k-collapsible. Indeed, for b C3 all three edges in I[H] colored with b belongto E(H, V(G)\H). For b C2 one edge belongs to E(H, V(G)\H) and one hasboth endpoints in H. Note that in E(H, V(G) \ H) there are no edges colored withcolors in(H) so|E(H, V(G) \ H)|= 3+ ( )k. For every vertexxHand color b C2 if there is an edge in E(x, V(G)\ H) colored with b then there isalso an edge in E(H\x, H\x) colored with b. Recall that no edges colored witha color from (H) leave H. Hence, to show that|E(G[H]\ x)| |E(x, V(G)\ H)|it suffices to prove that each pair of vertices in H is connected by at least edgesin 1((H) {}). However for each x V(Q) we have degQ(x) |(x)| so wemay assign to each edge ofQ two colors from (Q) using one free color from each of

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its endpoints. By Lemma5 we may assign the colors so that every color is assignedat most once. Consider an edge eE(Q), say e= uv , and let p = H\ {u, v}. Letc1 (u) and c2 (v) be the colors assigned to e, then by the assumption of thelemma there is an up edge colored with c2 and anvpedge colored with c1. It followsthat every edgeeE(Q) forms aK3together with the edges in E(G[H]) colored withthe assigned colors. Thus each pair of vertices in H is connected by at least|E(Q)|edges in1((Q){}). Hence if 2+k and |E(Q)|then the subgraphofG induced onH isk-collapsible, contradicting the assumption. If >|E(Q)|thench(Q) 12(3 ) 12(2+ 3)> 12(4|E(Q)| + 3|E(Q)|) = 72 |E(Q)| and the claimfollows. Suppose 2+ > k. Note that G[H] contains edges colored withcolors in (H), edges colored with colors in C2 and e uncolored edges.Thus t + +e= e + , so by definition ch(Q)(, k , t)|E(Q)|and the claim follows.

Lemma 32. Let (G, ) be a coloring maximizing . Let Q be a free componentof (G, ) such that|V(Q)| {2, 3}. Let 6 be the maximal degree of G and lett 32

and0k be integers such thatG does not contain ak-collapsible

subgraph andG does not contain an induced subgraph on three vertices with more thant edges. Thench(Q)(, k , t)|E(Q)|.Proof. CASE 1. |V(Q)| = 3 and|E(Q)| 3. Let V(Q) ={u,v,p}. Note thatQcontains a cycle or a multiple edge.CASE 1.1. Q contains a cycle of length 3, i.e. the cycle uvpu. Pick any a(Q),by symmetry assume thata(u). By Lemma5there is anvxedge colored with a.Ifx

=p then we can move Q by uncoloring vx and coloring uv with a. We obtain a

component with four vertices and the claim follows by Lemma27and Observation18.Thus we can assume that for every color in (Q) there is an edge incident with twovertices in V(Q) colored with that color. The claim follows from Lemma 31 appliedfor H=V(Q).CASE 1.2. Qcontains a multiple edge. By symmetry assume thatuv is a multipleedge ofQ and vp is an edge ofQ. Let a (v), by Lemma 5there is an ux edgecolored witha. We may move Q by uncoloring ux and coloring uv with a. Ifx=pthen we obtain a component with four vertices and the claim follows by Lemma27and Observation18. Ifx = p then we obtain a component with three vertices and acycle of length 3. Hence the claim follows by Observation18and Case 1.1.CASE 2.

|V(Q)

|= 3 and

|E(Q)

| = 2. If

7 then by Lemma 5 and Observa-

tion19we have ch(Q) |V(Q)|12 7 = 72 |E(Q)|. So we can assume that = 6.Note that if there is a moveofsuch that con1(Q())4 then by Observation18,Lemma 17 and Observation 19 we have ch(Q) = ch(Q()) | con1(Q())|12 =6412 =

92 |E(Q)|. Thus we assume that con1(Q())3 for every move of. Let

E(Q) ={uv,vp}and cu(u). By Lemma5there is anvx edge colored withcu. Byuncoloringvxand coloringuv with cu we obtain an elementary move 1 of. Hencex con1(Q), so x = p. Thus for every color cu (u) there is an vp edge coloredwith cu and therefore for every cu1(u) there is an vp edge colored with cu in 1.Letcv1(v) and consider the py edge colored in 1 with cv. We may moveQ(1)by uncoloring py and coloring vp with cv. Hence y con1(Q(1)) ={u,v,p}, so

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y= u. Similarly for every cp1(p) there is an uv edge colored in 1 withcp. Thuswe have obtained that for every color in 1({u,v,p}) there is an edge incident withtwo vertices in{u,v,p} colored with that color. The claim follows from Observation18and Lemma31 applied for (G, 1) and H={u,v,p}.CASE 3. |V(Q)| = 2. If |E(Q)| = 1 then by Lemma 21 we have ch(Q) 4|E(Q)|. Suppose|E(Q)| 2, let V(Q) ={u, v} and a (u). We may move Qby uncoloring the vx edge colored with a and coloring a uv edge with a. We obtaina free component Q with three vertices and are done by Case 2 and Observation18.

4 Proof of the main results

Now we are ready to describe the algorithm used to find the colorings from Theorem 4,Theorem2 and Theorem1in polynomial time.

Algorithm 1. LetG be a multigraph with maximal degree 4. Let t be aninteger such that

32

t0 and G does not contain a subgraph with 3 verticesand more than t edges.

1. Let k = min{, 2(t ) + 1}, i:= 0 and G0 := G.2. WhileGi contains ak-collapsible subgraphH leti := i + 1 andGi be the graph

obtained by collapsing H in Gi1. By Lemma 29 Gi has maximal degree atmost and does not contain a subgraph with 3 vertices and more then t edges.The resulting graph Gi does not contain a k-collapsible subgraph.

3. Let be the empty partial coloring ofGi, i.e. (E(Gi)) ={}.4. As long as (Gi, ) contains a free component with more than/2 edges, use

the procedure described in the proof of Lemma 5 to find a new partial coloring, with increased number of colored edges and replace by .

5. For every free component P determine O(1) colored edges which send chargeto P so that Pgets the required amount; the edges are specified in proofs ofrelevant lemmas in Subsection2.3.

6. If in Step 5charge is claimed from an uncolored edge or more than 1 unit ofcharge is claimed from a colored edge proceed as follows. The proofs of lemmasfrom Subsection2.3provide a coloring with ()> (). Replace with and repeat step5.

7. For j =i, . . . , 1 use Lemma29to obtain a coloring ofGi1 from a coloring ofGi.

Now let us argue that the algorithm above takes only polynomial time. Let n, mdenote the number of vertices and edges of the input graph G. Finding ak-collapsiblesubgraph can be easily done in O(n3) time. Since after collapsing such a subgraphthe number of vertices decreases, Step2takesO(n4) time. One can easily check thatthe procedure from the proof of Lemma 5used to find the coloring in Step4 takesO(nm) time. Since after each such operation the number of colored edges increases,

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Step4 takes O(nm2) time. Now we focus on Step5. Note that although in principlea component Pcan receive charge from many edges, which may be hard to find, in

the proofs in Section 2.3 we needed only charge from edges incident to P or to acomponent obtained fromPby at most two elementary moves, and the moves werealways specified. Since every such edge sends at least 1/2 unit of charge and everycomponent needs onlyO(1) units of charge per edge, to each componentPwe assignO(|E(P)|) = O() edges and it is easy to find these edges in O(m) time. Henceone execution of Step5takes onlyO(mn) time. There are O(mn/2) possible valuesof the potential , so the total time spent on Step 5 is O(m2n/2+1). The laststep takes time linear in the size of all the k-collapsible subgraphs found, which isbounded by O(m). We conclude that the algorithm takes O(m2n/2+1) time, whichis polynomial for every fixed value of .

Now we will prove that if G satisfies certain conditions then Algorithm 1 con-

structs a coloring which colors sufficiently many edges. It will be convenient to beginwith the proof of Theorem2.

Proof of Theorem2. Let G be a multigraph of maximum degree . Fix an integert such that

32

t 12

22 1 and G does not contain a subgraph with 3vertices and more than t edges. Let be the coloring constructed for G and t byAlgorithm 1. Let k = min{, 2(t) + 1}. Assume G contains a k-collapsiblesubgraph H. Let G be the graph obtain by collapsing H. Since +k/2 = t,Lemma 29(i) tells us that G does not contain a three vertex subgraph with morethan t edges. Moreover, Lemma 29(ii) states that if we show G has a -edge-colorable subgraph with at least t|E(G)|edges then G contains a -edge-colorablesubgraph with at least

t|E(G)

|edges. Hence in what follows we can assume that G

does not contain k-collapsible subgraphs.If 5 then 32

1 < 12

22 1 so t = 32

. Then the claim followsfrom Shannons theorem.

Assume that 6. We will show that every free component P of (G, ) receivesat least t |E(P)| charge. It follows that colors at least t |E(G)|/( t + 1) =t|E(G)|edges, as required. Note thatt

12

22 1 implies t

22+43 2.9.

Let Pbe a free component of (G, ). If |V(P)| 4 then by Corollary 26 andLemma 27 we have ch(Q) 4|E(Q)|. If|V(P)| 3 then by Lemma 32 we havech(P)(, k , t)|E(P)|. Thus it remains to show that(, k , t) t .

Let , and e be as in definition of(, k , t). Ifk = then 2 > so 66+ 6 >9. Consequently

3

2e = 6

2

4e > 9

2

2 = 7

2 and thus (, k , t)

tas required. So assume k= 2(t)+1. Then 2+ >2t2 so 2+3> 2t+.It follows that 93= 73 + 27 3 + 2 + 2= 4 + (2+3)>4+2t+= 4+2t. In particular, this gives 3


20/22

edges. If 8 then 32 112

22 1 and the claim follows by Theorem2.

If = 4 then the claim follows by Lemma 30. Hence we are left with the case

= 6. Then t = 8; let k = min{, 2(t) + 1} = 5. By the argument fromthe beginning of the proof of Theorem2 we obtain that it is sufficient to show that(6, 5, 8) t = 3. Let , and e be as in definition of (, k , t). We have2+ 6 6 and 6 + and thus 12 2+ 2 3. Hence 4 and32e 1844 = 72 . So (6, 5, 8) = 72 >3 and the claim follows.

Now assume is odd, 5 and G does not contain 12 K3+ e as a subgraph.This means that G does not contain a three vertex subgraph with more than t =32( 1) edges. We will show that Gcontains a -edge-colorable subgraph with atleast t = /(32 1) edges. Similarly as for even , if 11 then 32( 1)12

22 1 and the claim follows by Theorem 2. If = 5 then by Lemma29 we

can assume thatG does not contain a 3-collapsible subgraph so the claim follows by

Lemma 30. Thus we only need to verify the claim for = 7 and = 9 and, asargued in the beginning of the proof of Theorem 2, this can be achieved by showingrespectively(7, 5, 9) t = 72and (9, 7, 12) t = 3. If = 7 then 2+76 and 7 + and thus 142+ 2 3 1. Hence5 and, since 2eand e is integer, e2. It follows that 32e 2154 = 4, so(7, 5, 9) = 72 . If = 9then 2+ 9 8 and 9 +and thus 182+ 2 3 1. Hence6and 32e 2766 = 72 . We get (9, 7, 12) = 72 , as required.Proof of Theorem4. Let G be a connected multigraph of maximum degree . For = 3 the claim was proved by Kamiski and Kowalik [7, 8], so in what follows weassume 4. Suppose is even. Since G= 2K3and Gis connected it follows thatG does not contain

2K

3as a subgraph. Hence the claim follows from Theorem 1.

Assume is odd. If G consists of two copies of 12 K3+ e joined by an edgethen we can easily color 2 + 1 of the 3 edges of G and the claim follows. Soassume that G does not consist of two copies of 12 K3+e joined by an edge. LetH be the set of subgraphs of G isomorphic to 12 K3 + e. Then each H H isa 2-edge-connected component of G joined with G\H by a single edge eH. LetG = G

V(G) \HH V(H)

. We can color at least 233 |E(G)| edges of G by

Theorem1. The partial coloring ofG gives a partial coloring ofG. For everyH Hwe can additionally color eHand of the

312 edges ofH. Thus we have colored at

least 233 |E(G)| +

HH( + 1) = 233 |E(G)| +

HH

+1|E(H)|+1(|E(H)| + 1) =

233 |E(G)|+

HH

2+23+1(|E(H)| + 1) > 2+13 |E(G)| edges of G and the claim

follows.

5 Approximation Algorithms

Following [1], let ck(G) be the maximum number of edges of a k-edge-colorable sub-graph ofG. We use the following result of Kaminski and Kowalik.

Theorem 33 ([7, 8]). Let G be a family of graphs and letF be a k-normal familyof graphs. Assume there is a polynomial-time algorithm which for everyk-matchingH of a graph inG, such thatH F finds itsk-edge colorable subgraph with at least

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|E(H)| edges. Moreover, let

= minA,BFA is notk-regular

ck(A) + ck(B) + 1

|E(A)| + |E(B)| + 1 and = minAFck(A) + 1

|E(A)| + 1 .

Then, there is an approximation algorithm for the maximum k-ECS problem forgraphs inG with approximation ratio min{,,}.

Since the definition ofk-normal family is very technical, we refer the reader to [8]for its definition. As a direct consequence of Theorem 4 and Theorem33 we get thefollowing two results.

Theorem 34. Let k 4 be even. The maximum k-ECS problem has a 2k+23k+2-approximation algorithm for multigraphs.

Proof. Let F ={ k2K3}. It is easy to check that F is k-normal. Now we give thevalues of parameters , and from Theorem 33. By Theorem 4, = 2k3k2 . Wehave ck(

k2K3) =k and|E( k2K3)|= 3k2. Hence, = and = 2k+23k+2 .

Theorem 35. Let k 5 be odd. The maximum k-ECS problem has a 2k+13k -approximation algorithm for multigraphs.

Proof. Let F={k12 K3+ e}. It is easy to check that F is k-normal. Now we givethe values of parameters , , and for Theorem33. By Theorem4, = 2k3(k1) .We have ck(

k12 K3 +e) = k and|E( k12 K3 + e)| = 3k12 . Hence, = 2k+13k and

=

2k+2

3k+1 .

Acknowledgments

The second and third author are partially supported by National Science Centre ofPoland (grant N206 567140). The work has been done while the first author wasstaying at University of Warsaw.

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[3] G. Cornuejols and W. Pulleyblank. A matching problem with side conditions.Discrete Mathematics, 29:135159, 1980.

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