Table Of ContentGraphs with maximum degree ∆ ≥ 17 and maximum
average degree less than 3 are list 2-distance (∆ + 2)-colorable∗
Marthe Bonamy, BenjaminLévêque,Alexandre Pinlou†
LIRMM,UniversitéMontpellier2,CNRS
3
1 {marthe.bonamy,benjamin.leveque,alexandre.pinlou}@lirmm.fr
0
2
January 31,2013
n
a
J
9
2 Abstract
] For graphs of bounded maximum average degree, we consider the problem of 2-distance
M
coloring. Thisisthe problemof coloringthe verticeswhile ensuringthattwoverticesthatare
D adjacentor haveacommonneighbor receivedifferentcolors. Itisalreadyknown thatplanar
graphsofgirthatleast6andofmaximumdegree∆arelist2-distance(∆+2)-colorablewhen
.
s ∆≥ 24(BorodinandIvanova(2009))and2-distance(∆+2)-colorablewhen∆≥18(Borodin
c
[ and Ivanova (2009)). We prove here that ∆ ≥ 17 suffices in both cases. More generally, we
show that graphs with maximum average degree less than 3 and ∆ ≥ 17 are list 2-distance
1
(∆+2)-colorable. Theproofcanbetransposedtolistinjective(∆+1)-coloring.
v
0
9
0 1 Introduction
7
.
1 In this paper, we consider only simple and finite graphs. A 2-distance k-coloring of a graph G is
0
a coloring of the vertices of G with k colors such that two vertices that are adjacent or have a
3
1 commonneighborreceivedistinctcolors. Wedefineχ2(G)as thesmallestk suchthatGadmits a
: 2-distance k-coloring. This is equivalent to a proper vertex-coloring of the square of G, which is
v
i definedasagraphwiththesamesetofverticesasG,wheretwoverticesareadjacentifandonlyif
X
theyareadjacentorhaveacommonneighborinG. Forexample,thecycleoflength5cannotbe2-
r
a distancecoloredwithlessthan5colorsasanytwoverticesareeitheradjacentorhaveacommon
neighbor: indeed, its square is the clique of size 5. An extension of the 2-distance k-coloring is
thelist2-distance k-coloring, whereinsteadofhaving thesamesetofk colorsforthewholegraph,
everyvertexisassignedsomesetofkcolorsandhastobecoloredfromit. Wedefineχ2(G)asthe
ℓ
smallestk suchthatGadmitsalist2-distancek-coloringofGforanylistassignment. Obviously,
2-distancecoloringisasub-caseoflist2-distancecoloring(wherethesamecolorlistisassignedto
everyvertex),sofor any graphG, χ2(G) ≥ χ2(G). Kostochkaand Woodall[19]evenconjectured
ℓ
thatitisactuallyanequality. Theconjectureisstillopen.
The study of χ2(G) on planar graphs was initiated by Wegner in 1977 [21], and has been ac-
tivelystudiedbecauseoftheconjecturebelow. ThemaximumdegreeofagraphGisdenoted∆(G).
∗ThisworkwaspartiallysupportedbytheANRgrantEGOS12JS0200201
†Secondaffiliation:DépartementMIAp,UniversitéPaul-Valéry,Montpellier3
1
Conjecture1(Wegner[21]). IfGisaplanargraph, then:
• χ2(G) ≤ 7if∆(G) = 3
• χ2(G) ≤ ∆(G)+5if4 ≤ ∆(G) ≤ 7
• χ2(G) ≤ ⌊3∆(G)⌋+1if∆(G) ≥ 8
2
Thisconjectureremainsopen.
NotethatanygraphGsatisfiesχ2(G) ≥ ∆(G)+1. Indeed,ifweconsideravertexofmaximal
degree and its neighbors, they form a set of ∆(G)+1 vertices, any two of which are adjacent or
haveacommonneighbor. Henceatleast∆(G)+1colorsareneededfora2-distancecoloringofG.
Itisthereforenaturaltoaskwhenthislowerboundisreached. Forthatpurpose,wecanstudy,as
suggestedbyWangandLih[20],whatconditionsonthesparsenessofthegraphcanbesufficient
toensuretheequalityholds.
Afirstmeasureofthesparsenessofaplanargraphisitsgirth. ThegirthofagraphG,denoted
g(G), is the length of a shortestcycle. Wang and Lih [20] conjectured that for any integer k ≥ 5,
there exists an integer D(k) such that for every planar graph G verifying g(G) ≥ k and ∆(G) ≥
D(k),χ2(G) = ∆(G)+1. ThiswasprovedbyBorodin,IvanovaandNoestroeva[11,13]tobetrue
fork ≥ 7,eveninthecaseoflist-coloring,andfalsefork ∈ {5,6}. Sofar,inthecaseoflistcoloring,
it is known [3, 18] that we can choose D(7) = 16, D(8) = 10, D(9) = 8, D(10) = 6, D(12) = 5.
Borodin, Ivanova and Neustroeva [12] proved that the case k = 6 is true on a restricted class of
graphs, i.e. for a planar graph G with girth 6 where every edge is incident to a vertex of degree
at most twoand ∆(G) ≥ 179, we have χ2(G) ≤ ∆(G)+1. Dvorˇák etal. [16] proved that thecase
k = 6istrueby allowing onemorecolor, i.e. foraplanar graphGwithgirth6and∆(G) ≥ 8821,
wehave χ2(G) ≤ ∆(G)+2. Theyalso conjecturedthat thesame holdsfor aplanar graph Gwith
girth 5 and sufficiently large ∆(G), but this remains open. Borodin and Ivanova improved [5]
Dvorˇáketal.’sresultandextendedittolist-coloring[6,7]asfollows.
Theorem1(BorodinandIvanova[5]). EveryplanargraphGwith∆(G) ≥ 18andg(G) ≥ 6admitsa
2-distance(∆(G)+2)-coloring.
Theorem2(BorodinandIvanova[7]). EveryplanargraphGwith∆(G) ≥ 24andg(G) ≥ 6admitsa
list2-distance(∆(G)+2)-coloring.
Theorems1and2areoptimalwithregardstothenumberofcolors,asshownbythefamilyof
graphs presented by Borodin et al. [4], which are of increasing maximum degree, of girth 6 and
arenot2-distance(∆+1)-colorable. WeimproveTheorems1and2asfollows.
Theorem3. Every planar graph Gwith∆(G) ≥ 17andg(G) ≥ 6admits alist 2-distance (∆(G)+2)-
coloring.
Another way to measure the sparseness of a graph is through its maximum average degree.
The average degree of a graph G, denoted ad(G), is Pv∈V d(v) = 2|E|. The maximum average degree
|V| |V|
ofa graph G, denotedmad(G), is the maximum ofad(H) over all subgraphsH ofG. Intuitively,
this measuresthe sparsenessof agraph because it stateshow great theconcentration of edgesin
a same area can be. For example, stating that mad(G) has to be smaller than 2 means that G is a
forest. Usingthismeasure,weproveamoregeneraltheoremthanTheorem3.
2
Theorem 4. Every graph G with ∆(G) ≥ 17 and mad(G) < 3 admits a list 2-distance (∆(G) + 2)-
coloring.
Euler’sformulalinksgirthandmaximumaveragedegreeinthecaseofplanargraphs.
Lemma1(Folklore). ForeveryplanargraphG,(mad(G)−2)(g(G)−2) < 4.
ByLemma1,Theorem4impliesTheorem3.
An injective k-coloring [17] of G is a (not necessarily proper)coloring of the vertices of G with
k colors such that two vertices that have a common neighbor receive distinct colors. We define
χ (G) as the smallest k such that G admits an injective k-coloring. A 2-distance k-coloring is an
i
injectivek-coloring,buttheconverseisnottrue. Forexample,thecycleoflength5canbeinjective
colored with 3colors. The list version ofthis coloring is a list injective k-coloring ofG, and χ (G)
i,ℓ
isthesmallestk suchthatGadmitsalistinjectivek-coloring.
Someresultson 2-distance coloring have their counterparton injective coloring with oneless
color. ThisisthecaseofTheorems1and2[8,9]. TheproofofTheorem4alsoworkswithcloseto
no alteration for list injective coloring, thus yielding a proof that every graph G with ∆(G) ≥ 17
andmad(G) < 3admitsalistinjective(∆(G)+1)-coloring.
In Sections2 and 3, weintroduce themethodand terminology. In Sections4 and 6, we prove
Theorem4anditscounterpartoninjectivecoloringbyadischargingmethod.
2 Method
Thedischargingmethodwas introducedin thebeginningofthe20th century. Ithasbeenusedto
provethecelebratedFourColor Theoremin[1,2]. Adischargingmethodissaidtobelocal when
the weight cannot travel arbitrarily far. Borodin, Ivanova and Kostochka introduced in [10] the
notionofglobaldischargingmethod,wheretheweightcantravelarbitrarilyfaralongthegraph.
We prove for induction purposes a slightly stronger version of Theorem 4 by relaxing the
constraint on the maximum degree. Namely, we relax it into “For any k ≥ 17, every graph G
with ∆(G) ≤ k and mad(G) < 3 verifies χ2(G) ≤ k + 2” so that the property is closed under
ℓ
vertex-oredge-deletion. Agraphisminimalforapropertyifitsatisfiesthispropertybutnoneof
itssubgraphsdoes.
The first step is to consider a minimal counter-example G, and prove it cannot contain some
configurations. To do so, we assume by contradiction that G contains one of the configurations.
We consider a particular subgraph H of G, and color it by minimality (the maximum average
degreeofanysubgraphofGisboundedbythemaximumaveragedegreeofG). Weshowhowto
extendthecoloringofH toG,acontradiction.
Thesecondstepis toprovethatagraphthatdoesnotcontainanyoftheseconfigurationshas
amaximumaveragedegreeofatleast3. Tothatpurpose,weassigntoeachvertexitsdegreeasa
weight. Weapplydischarging rulestoredistributeweightsalongthegraphwith conservationof
the total weight. As some configurations are forbidden, we can then prove that after application
ofthedischargingrules,everyvertexhasafinalweightofatleast3. Thisimpliesthattheaverage
degree of the graph is at least 3, hence the maximum average degree is at least 3. So a minimal
counter-examplecannotexist.
Wefinallyexplainhowthesameproofholdsalsoforlistinjective(∆+1)-coloring.
3
3 Terminology
In the figures, we draw in black a vertex that has no other neighbor than the ones already rep-
resented, in white a vertex that might have other neighbors than the ones represented. White
verticesmaycoincidewithotherverticesofthefigure. Whenthereisalabelinsideawhitevertex,
itisanindicationonthenumberofneighborsithas. Thelabel’i’means"exactlyineighbors",the
label’i+’(resp. ’i−’)meansthatithasatleast(resp. atmost)ineighbors.
Let u be a vertex. The neighborhood N(u) of u is the set of vertices that are adjacent to u. Let
d(u) = |N(u)| be the degree of u. A p-link x−a −...−a −y, p ≥ 0, between x and y is a path
1 p
between x and y such that d(a ) = ... = d(a ) = 2. When a p-link exists between two vertices x
1 p
and y, we say they are p-linked. If there is a p-link x−a −...−a −y between x and y, we say
1 p
x is p-linked through a to y. A partial 2-distance list coloring of G is a 2-distance list-coloring of a
1
subgraphH ofG.
A vertexis weak whenit is ofdegree3 and is 1-linked totwovertices ofdegreeat most14, or
twice1-linkedtoavertexofdegreeatmost14(seeFigure1). Aweakvertexisrepresentedwitha
wlabelinside(w ifitisnotweak).
14− 14−
x
Figure1: Aweakvertexx.
Avertexissupportwhenitiseither(seeFigure2):
Type(S ): avertexofdegree2adjacenttoanothervertexofdegree2;
1
Type(S ): avertexofdegree2thatisadjacenttoavertexofdegree3whichisadjacenttoavertex
2
ofdegree2andtoavertexofdegreeatmost7;
Type(S ): aweakvertex1-linkedtoanotherweakvertex.
3
Avertexispositive whenitis ofdegreeatleast4andis adjacenttoasupportvertex. Avertex
u is locked if it has two neighbors v and v , where v and v are both 1-linked to the same two
1 2 1 2
b 14− 14− g
b
7− a f
u x a b u x a c d u x c d e
Type(S1) Type(S2) Type(S3)
Figure2: Supportverticesx.
4
vertices w and w that have a common neighbor, and d(v ) = d(v ) = d(w ) = d(w ) = 3 (see
1 2 1 2 1 2
Figure3). Thisconfigurationiscalledalock.
v w
1 1
u x
v w
2 2
Figure3: Alockedvertexu.
4 Forbidden Configurations
Inallthepaper,kisaconstantintegergreaterthan17andGisaminimalgraphsuchthat∆(G) ≤ k
andGadmitsno2-distance(k+2)-list-coloring.
Wedefineconfigurations(C )to(C )(seeFigures4,5and6). Notethatconfigurationssimilar
1 11
toConfigurations(C ),(C )and(C )alreadyexistedinthelitterature,forexamplein[16].
1 2 4
• (C )isavertexuwithd(u) ≤ 1
1
• (C )isavertexuwithd(u) = 2thathas twoneighborsv,w anduis 1-linkedthroughv toa
2
vertexofdegreeatmostk−1.
• (C )isavertexuwithd(u) = 3thathasthreeneighborsv,w,xwithd(w)+d(x) ≤ k−1,and
3
uis1-linkedthroughv toavertexofdegreeatmostk−1.
• (C )isavertexuwithd(u) = 3thathasthreeneighborsv,w,xwithd(w)+d(x) ≤ k−1,and
4
v hasexactlythreeneighborsu,y,z withd(z) ≤ 7andd(y) = 2.
• (C ) is a vertex u with d(u) = 3 that has three neighbors v,w,x with d(x) ≤ k −1 and u is
5
1-linked through v (resp. through w) to a vertex of degreeat most 14. (Notethat u is weak
vertex.)
• (C ) is a vertex u with d(u) = 4 that has four neighbors v,w,x,y with d(w) ≤ 7, d(x) ≤ 3,
6
d(y) ≤ 3,anduis1-linkedthroughv toavertexofdegreeatmost14.
• (C ) is a vertex u with d(u) = 4 that has four neighbors v,w,x,y with d(x)+d(y) ≤ k −1
7
anduis1-linkedthroughv (resp. throughw)toavertexofdegreeatmost14.
• (C ) is a vertexu with d(u) = 5 that has five neighborsv,w,x,y,z with d(w) ≤ 7, d(x) ≤ 3,
8
d(y) ≤ 3,d(z) = 2,anduis1-linkedthroughv toavertexofdegreeatmost7.
• (C )isavertexuwithd(u) = 6thathassixneighborsv,w,x,y,z,t withd(w) ≤ 7, d(x) ≤ 3,
9
d(y) ≤ 3,d(z) = 2,d(t) = 2,anduis1-linkedthroughv toavertexofdegreeatmost7.
• (C )is a vertexuwithd(u) = 7that has sevenneighborsv,w ,...,w with d(v) ≤ 7and u
10 1 6
is1-linkedthroughw ,1 ≤ i≤ 6,toavertexofdegreeatmost3.
i
5
x
x
−
(k−1) x
w u 7− z x
v v (k−1)−
u
w d(x)+d(w) u
u v ≤k−1 y w v
d(x)+d(w) (k−1)−
1− u w ≤k−1 14− 14−
(C1) (C2) (C3) (C4) (C5)
Figure4: Forbiddenconfigurations(C )to(C ).
1 5
• (C ) is a vertex uwith d(u) = k that has threeneighbors v,w,x with x is a supportvertex,
11
v,wareboth1-linkedtoasamevertexyofdegree3,andv (resp. w)is1-linkedtoavertexof
degreeatmost14distinctfromy. (Notethatv,w areweakvertices.)
Lemma2. GdoesnotcontainConfigurations(C )to(C ).
1 11
Proof. Given apartial 2-distancelist-coloring ofG, aconstraintofavertexuiscolor appearingon
avertexatdistanceatmost2fromuinG.
NotationreferstoFigures4,5and6.
Claim1. Gdoesnotcontain(C ).
1
Proof. SupposebycontradictionthatGcontains(C ). UsingtheminimalityofG,wecolorG\{u}.
1
Since∆(G) ≤ k,andd(u) ≤ 1,vertexuhasatmostk constraints(oneforitsneighborandatmost
k −1 for the vertices at distance 2 from u). There are k+2 colors available in the list ofu, so the
coloringofG\{u}canbeextendedtoG,acontradiction.
Claim2. Gdoesnotcontain(C ).
2
Proof. Suppose by contradiction that G contains (C ). Using the minimality of G, we color G \
2
{u,v}. Vertexuhasatmostk+1constraints. Hencewecancoloru. Thenvhasatmostk−1+2=
k+1constraints. Hencewecancolorv. SowecanextendthecoloringtoG,acontradiction.
Claim3. Gdoesnotcontain(C ).
3
Proof. SupposebycontradictionthatGcontains(C ). UsingtheminimalityofG,wecolorG\{v}.
3
Becauseofu,verticeswandxhavedifferentcolors. Wediscoloru. Vertexvhasatmostk−1+2 =
k+1constraints. Hencewecancolorv. Vertexuhasatmostd(w)+d(x)+2 ≤ k+1constraints.
Hencewecancoloru. SowecanextendthecoloringtoG,acontradiction.
Claim4. Gdoesnotcontain(C ).
4
Proof. Supposebycontradiction thatGcontains (C ). Letebetheedgeuv. Usingtheminimality
4
ofG,wecolorG\{e}. Wediscoloruandv. Vertexuhasatmostd(w)+d(x)+2≤ k+1constraints.
Hencewecancoloru. Vertexv hasatmost7+3+2 ≤ k+1constraints. Hencewecancolorv. So
wecanextendthecoloringtoG,acontradiction.
6
x x 7−
3− 7−
z v
y z v t
w 7− 3− y 14−
u u
w u u
v d(x)+d(y) v y 3− 7− w y 3− 7− w
≤k−1 3− 3−
14− 14− x x
(C6) (C7) (C8) (C9)
Figure5: Forbiddenconfigurations(C )to(C ).
6 9
Claim5. Gdoesnotcontain(C ).
5
Proof. Suppose by contradiction that G contains (C ). Using the minimality of G, we color G \
5
{u,v,w}. Vertexuhasatmostk−1+2 = k+1constraints. Hencewecancoloru. Verticesv and
w have at most 14+3 ≤ k +1 constraints respectively. Hence we can color v and w. So we can
extendthecoloringtoG,acontradiction.
Claim6. Gdoesnotcontain(C ).
6
Proof. SupposebycontradictionthatGcontains(C ). UsingtheminimalityofG,wecolorG\{v}.
6
Wediscoloru. Vertexv hasatmost14+3≤ k+1constraints. Hencewecancolorv. Vertexuhas
atmost2+3+3+7 ≤ k+1constraints. Hencewecancoloru. Sowecanextendthecoloringto
G,acontradiction.
Claim7. Gdoesnotcontain(C ).
7
Proof. Suppose by contradiction that G contains (C ). Using the minimality of G, we color G \
7
{v,w}. Wediscoloru. Vertexuhasatmostd(x)+d(y)+2 ≤ k+1constraints. Hencewecancolor
u. Verticesv andwhaveatmost14+4≤ k+1constraintsrespectively. Hencewecancolorv and
w. SowecanextendthecoloringtoG,acontradiction.
Claim8. Gdoesnotcontain(C ).
8
Proof. SupposebycontradictionthatGcontains(C ). UsingtheminimalityofG,wecolorG\{v}.
8
Wediscolor u. Vertexuhas at most7+3+3+2+1 ≤ k+1 constraints. Hencewe can color u.
Vertexvhasatmost7+5 ≤ k+1constraints. Hencewecancolorv. Sowecanextendthecoloring
toG,acontradiction.
Claim9. Gdoesnotcontain(C ).
9
Proof. SupposebycontradictionthatGcontains(C ). UsingtheminimalityofG,wecolorG\{v}.
9
Wediscoloru. Vertexuhasatmost7+3+3+2+2+1 ≤ k+1constraints. Hencewecancolor
u. Vertex v has at most 7+6 ≤ k +1 constraints. Hence we can color v. So we can extend the
coloringtoG,acontradiction.
7
v
3− w6 7− w1 3− v z1 y1
−
14
w w x u z2 y
5 2
− u − k
3 3
w w support z3
4 3
−
14
3− 3− w z4 y2
(C10) (C11)
Figure6: Forbiddenconfigurations(C )and(C ).
10 11
Claim10. Gdoesnotcontain(C ).
10
Proof. Suppose by contradiction that G contains (C ). Using the minimality of G, we color G \
10
{u,w ,...,w }. Vertex u has at most 7 +6 ≤ k + 1 constraints. Hence we can color v. Vertices
1 6
w have at most 3+7 ≤ k +1 constraints. Hence we can color w ,...,w . So we can extend the
i 1 6
coloringtoG,acontradiction.
Claim11. Gdoesnotcontain(C ).
11
Proof. Suppose by contradiction that G contains (C ). Since x is a support vertex, and u is of
11
degreek,itisoftype(S ),(S )or(S )ofsupportverticeswiththenotationofFigure2. Notethat
1 2 3
someverticesmaycoincidebetweenFigure2andFigure6.
WedefineasetofverticesAasfollows:
{a} ifxisofType(S )
1
A= {a,c} ifxisofType(S )
2
{a,c} ifxisofType(S )
3
Usingtheminimality ofG, wecolor G\({v,w,x,y, z ,...,z }∪A). Ifxis ofType(S )(resp.
1 4 1
(S )),a(resp. c)hasatmostk+1constraints,hencewecancolorit. Forthethreetypes(S ),xhas
2 i
atmostk−3+1+2 = k constraints,thusithasatleast2available colors. Vertexy hasatmostk
constraints,thusithasatleast2available colors. Bothv andwhaveatmostk−3+1+1≤ k−1
constraints,sotheyhaveatleast3available colorsintheirlist.
Wenowexplainhowtocolorv,w,x,y(otheruncoloredverticeswillbecoloredafter). Suppose
xandycanbeassignedthesamecolor,thenbothv andwhaveatleast2available colorsandthus
canbecolored.
Supposethelistsofavailable colorsofxandyaredisjoint. Wecolorv withacolornotappear-
inginthelistofx. Thenwecolory thathask+1constraints. (Vertexxhasstillatleast2available
colors.) Thenwecolorw thathask+1constraintsandfinallyx.
Nowweassumethatwecannotassignthesamecolortoxandyandthattheirlistsofavailable
colors are not disjoint. This means that x and y are either adjacent or have a common neighbor.
So some vertices coincide between Figure 2 and Figure 6. The different cases where x and y are
eitheradjacentorhaveacommonneighborarethefollowing:
(S ) – b =y
1
8
(S ) – b =y
2
– a = y andw.l.o.gb = z ,c = z andd= w.
2 3
(S ) – b =y
3
– d = y,andw.l.o.g. f = z ,g = v ande = z .
2 3
In all thesecases, y has at most 1 contraint. So we can color x,v,w,y, in this orderas they all
haveatmostk+1constraintswhentheyarecolored.
IfxisofType(S )(resp. (S )),vertexa(respverticesa,c)hasatmost11constraints(resp. 18,
2 3
6), sowecan colorthem. Theverticesz haveatmost17 ≤ k+1, sowecan colorthem. Thusthe
i
coloringhavebeenextendedtoG,acontradiction.
5 Structure of support vertices
Let H(G) be the subgraph of G induced by the edges incident to at least a support vertex. We
proveseveralpropertiesofsupportverticesandofthegraphH(G).
Lemma 3. Each positive vertex is of degree k and each support vertex is adjacent to exactly one positive
vertex.
Proof. By Lemma 2, G does not contain Configurations (C ), (C ) and (C ). So a support vertex
2 3 5
is adjacent to a vertex of degree k (Configurations (C ), (C ) and (C ) correspond respectively
2 3 5
to support vertices of Type (S ), (S ) and (S )). By definition, a support vertex has at most one
1 2 3
neighbor ofdegreeat least 4, thus it is adjacent to exactly onevertex ofdegreeat least 4 and this
vertex has in fact degree k. So all the positive vertices are of degree k and a support vertex is
adjacenttoexactlyonepositivevertex.
Lemma 4. Each cycle of H(G) with an odd number of support vertices contains a subpath s v s v s
1 1 2 2 3
wheres ,s ,s aresupportverticesoftype(S )andv ,v arevertices ofdegree2.
1 2 3 3 1 2
Proof. LetC be cycle ofH(G) withan oddnumberofsupportvertices. Cycle C doesnotcontain
just one support vertex, as all its edges have to be adjacent to a support vertex (there is no loop
normultipleedgeinH(G)). SoC containsatleastthreesupportvertices.
SupposethatC containsnopositivevertices. Thenitcontainsnosupportverticesoftype(S )
1
or(S )assuchverticesareofdegree2,soalltheirneighborswouldbeonC,andtheyareadjacent
2
toapositivevertexbyLemma3. SoC containsonlysupportverticesoftype(S ). Lets ,s ,s be
3 1 2 3
three supportvertices of C appearing consecutively along C. A supportvertex of Type(S ) is of
3
degree 3, adjacent to two vertices of degree 2 and to a positive vertex. So the neighbors of s on
i
C are vertices of degree2 that are not supportvertices. As H(G) contains only edgesincident to
supportvertices,thereexistv ,v ofdegree2suchthats v s v s isasubpathofC.
1 2 1 1 2 2 3
SupposenowthatCcontainssomepositivevertices. Letp ,...,p bethesetofpositivevertices
1 ℓ
ofC appearinginthisorderalongC whilewalkinginachosendirection(subscriptareunderstood
modulo ℓ). Let Q , 1 ≤ i ≤ ℓ, be the subpath of C between p and p (in the same choosen
i i i+1
direction along C). (Note that if ℓ = 1, then Q = C is not really a subpath.) As C contains an
1
odd number of support vertices, there exists i such that Q contains an odd number of support
i
9
vertices. If Q contains just one supportvertex v, then Q has length 2, since H(G) contains only
i i
edges incident to support vertices. So v is adjacent to two different positive vertices (or has a
multipleedgeifℓ = 1),acontradictiontoLemma3. SoQ containsatleast3supportvertices. Let
i
s ,s ,s bethreesupportverticesofQ appearingconsecutivelyalongQ .
1 2 3 i i
Ifoneofthes isofType(S ), letxbesuchavertex. WiththenotationofFigure2,vertexxis
i 1
ofdegree2, soits twoneighborsu,a are on C, with ua positivevertex and a a supportvertexof
Type (S ). Then vertex a is of degree 2 so its neighbor b distinct from x is also on C. Vertex b is
1
positivesoQ isthepathu,x,a,bandcontainsjusttwosupportvertices,acontradiction.
i
Ifoneofthes isofType(S ), letxbesuchavertex. WiththenotationofFigure2,vertexxis
i 2
of degree 2, so its two neighbors u,a are on C, with u a positive vertex and a a vertex of degree
3. Vertexa is notadjacent toverticesofdegreek soby Lemma3, it isnotasupportvertex. Letc′
betheneighborofaonC thatisdistinctfromx. AsalltheedgesofH(G)areincidenttosupport
vertices, c′ is a supportvertex. Since c′ is adjacent to a vertexofdegree3 it is a supportvertexof
Type(S )andcanplaytheroleofcofFigure2. Thencisofdegree2anditsneighboronC distinct
2
fromaisapositivevertexd. SoQ isthepathu,x,a,c,dandcontainsjusttwosupportvertices,a
i
contradiction.
So s ,s ,s are all of Type (S ). A support vertex of Type (S ) is of degree 3, adjacent to two
1 2 3 3 3
vertices of degree 2 and to a positive vertex. So the neighbors of s on C are vertices v ,v of
2 1 2
degree2that are notsupportvertices. AsH(G) contains onlyedgesincidenttosupportvertices,
wecanassumew.l.o.g. thats v s v s isasubpathofC.
1 1 2 2 3
Lemma 5. H(G) does not contain a 2-connected subgraph of size at least three with exactly two support
vertices.
Proof. SupposebycontradictionthatH(G)containsa2-connectedsubgraphC ofsize≥ 3thathas
exactlytwosupportverticesS = {s ,s }. WecolorbyminimalityG\(S∪{v ∈ N (S)|d (v) ≤ 3}).
1 2 G G
(Notethat by Lemma 3, the set{v ∈ N (S)|d (v) ≤ 3} correspondsto vertexa of Figure2 if the
G G
supportvertexisofType(S )or(S )andtoverticesa,cifthesupportvertexisofType(S ).)
1 2 3
We first show how to color S. For that purposewe consider three cases correspondingto the
typeofs .
1
• s isofType(S ). Thens is ofdegree2, has apositiveneighboruandasupportneighbora
1 1 1
ofType(S ). Ass is ofdegree2, both itsneighborsare in C. Soa is asupportvertexofC,
1 1
thusa = s . Thenuisofdegreek,hastwoneighborss ,s thatarenotcolored,sos ands
2 1 2 1 2
haveatmostk constraints,andwecancolorthem.
• s isofType(S ). Thens isofdegree2,hasapositiveneighboruandanotherneighboraof
1 2 1
degree3. Vertexa is not asupportvertexbyLemma3 since it has noneighborofdegreek.
Ass isofdegree2,allitsneighborsareinC. VerticesuandaareinC thatis2-connectedso
1
theyhaveatleasttwoneighborsinC. Sincetheyarenotsupportvertices,alltheirneighbors
inC aresupportvertices. Sobothuandaareadjacenttos . Vertexs issupport,itisadjacent
2 2
to a that is ofdegree3, sos is ofType(S ). Thenu is ofdegreek, has twoneighborss ,s
2 2 1 2
thatarenotcolored,sos ands haveatmostk constraints,andwecancolorthem.
1 2
• s is of Type (S ). Then s is of degree3, has a positive neighbor u and two otherneighbors
1 3 1
w,w′ of degree 2. Vertices w,w′ are not support vertices by Lemma 3 since they have no
neighborofdegreek. Ass isofdegree3,twoofu,w,w′ areinC. LetY betheneighborsof
1
10
