No finite $5$-regular matchstick graph exists

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by  Sascha Kurz

No finite 5-regular matchstick graph exists
SaschaKurza
aFakulta¨tfu¨rMathematik,PhysikundInformatik,Universita¨tBayreuth,Germany
Abstract
A graph G = (V,E) is called a unit-distance graph in the plane if there is an injective embedding
of V in the plane such that every pair of adjacent vertices are at unit distance apart. If additionally
the corresponding edges are non-crossing and all vertices have the same degree r we talk of a regular
4 matchstick graph. Due to Euler’s polyhedron formula we have r ≤ 5. The smallest known 4-regular
1
matchstickgraphisthesocalledHarborthgraphconsistingof52vertices. Inthisarticleweprovethatno
0
finite5-regularmatchstickgraphexists.
2

n Keywords: unit-distancegraphs
a 2000MSC: 52C99(cid:63),05C62
J

8

1. Prologue
]
O
One of the possibly best known problems in combinatorial geometry asks how often the same distance
C
can occur among n points in the plane. Via scaling we can assume that the most frequent distance has
.
h length 1. Given any set P of points in the plane, we can define the so called unit-distance graph in the
t
a plane, connecting two elements of P by an edge if their distance is one. The known bounds for the
m maximumnumberu(n)ofedgesofaunit-distancegraphintheplane,seee.g. [3],aregivenby
[
Ω(cid:16)nelocglologgnn(cid:17)≤u(n)≤O(cid:16)n43(cid:17).
1
v
3 Forn≤14theexactnumbersofu(n)weredeterminedin[10],seealso[3].
9 Ifweadditionallyrequirethattheedgesarenon-crossing,thenweobtainanotherclassofgeometrical
7
andcombinatorialobjects:
1
.
1 Definition1.1. A matchstick graph M consists of a graph G = (V,E) and an injective embedding
0 g :V →R2intheplanewhichfulfillthefollowingconditions:
4
1 (1) Gisaplanar(simple)graph,
:
v
(2) foralledges{i,j}∈E wehave(cid:107)g(i),g(j)(cid:107) =1,where(cid:107)x,y(cid:107) denotestheEuclideandistance
i 2 2
X betweenthepointsxandy,
r
a (3) g(i)(cid:54)=g(j)fori(cid:54)=j,and
(4) if {i ,j },{i ,j } ∈ E for pairwise different vertices i ,i ,j ,j ∈ V then the line segments
1 1 2 2 1 2 1 2
g(i )g(j )andg(i )g(j )donothaveacommonpoint.
1 1 2 2
Emailaddress:[email protected](SaschaKurz)
PreprintsubmittedtoElsevier January9,2014
Formatchstickgraphstheknownboundsforthemaximumnumberu˜(n)ofedges,seee.g. [3],aregiven
by
√ √
(cid:4) (cid:5) (cid:0) (cid:1)
3n− 12n−3 ≤u˜(n)≤3n−O n ,
wherethelowerboundisconjecturedtobeexact.
Wecallamatchstickgraphr-regularifeveryvertexhasdegreer.In[7]theauthorsconsiderr-regular
matchstickgraphswiththeminimumnumberm(r)ofvertices.Obviouslywehavem(0)=1,m(1)=2,
andm(2)=3,correspondingtoasinglevertex,asingleedge,andatriangle,respectively.
Thedeterminationofm(3)isanentertainingamusement. Atfirstweobservethatm(3)mustbeeven
since every graph contains an even number of vertices with odd degree. For n ≤ 8 vertices the set of
3-regularplanargraphsisfairlyassessable,consistingofonegraphoforder,i.e. thenumberofvertices,
4,onegraphoforder6,andthreegraphsoforder8. Utilizingareaargumentsrulesoutallbutonegraph
oforder8,sothatwehavem(3) = 8. Fordegreer = 4theexactdeterminationofm(4)isunsettledso
far. ThesmallestknownexampleisthesocalledHarborthgraph,seee.g. [6],yieldingm(4)≤52.
DuetotheEulerianpolyhedronformulaeveryfiniteplanargraphcontainsavertexofdegreeatmost
fivesothatwehavem(r) = ∞forr ≥ 6. Wewouldliketoremarkthattheregulartriangularlatticeis
theuniqueexampleofaninfinitematchstickgraphwithdegreeatleast6.
For degree 5 it is announced at several places that no finite 5-regular matchstick graph does exist,
seee.g. IvarsPeterson’sMathTrek“MatchSticksintheSummer”[9]mostlikelyreferringtopersonal
communication with Heiko Harborth, the discoverer of the 4-regular matchstick graph. Indeed Heiko
Harborthposedthequestionwhetherthereexistsa5-regularmatchstickgraph,ornotatanOberwolfach
meetingandwasawareofapreprintclaimingthenon-existenceproof(personalcommunication). Aftera
whilehefoundoutthatthereweresomemistakesintheproof,sothattohisknowledgethereisnocorrect
proofofthenon-existence. Askingtheauthorofthispreprintaboutthestateofthisproblemhereplied
too,thattheproblemisstillopen(atthispointintimetheconcerningwasuntraceable).
WolframResearch’sMathWorld,seehttp://mathworld.wolfram.com/MatchstickGraph.html,refersto
ErichFriedman,whohimselfmaintainsawebpage[5]statingthenon-existenceofa5-regularmatchstick
graph. Asking Erich Friedman for a proof of this claim he responded that unfortunately he lost his
referencebutalsothinksthatsuchagraphcannotexist.
In[2]theauthorslistfivepublicationswheretheyhavementionedanunpublishedproofforthenon-
existenceofa5-regularmatchstickgraphandstatethatuptotheirknowledgetheproblemisopen.
After the submission of this paper one of the referees came up with a very short proof for the non-
existence[8]. Inthispaperwewouldliketogiveadifferent,admittedlymorecomplicated,proofusing
atopologicalequationthatneedssomeexplanation. Soinsomepartsthis(unpublished)papercoincides
with[8].
In the meantime Heiko Harborth could recover the lost preprint and send it to the original author
andmyself. Curiouslyenough,afterrectifyingsomeannoyingminormistakes,thepreprintturnsoutto
bebasicallycorrect. Sotheauthorofthisarticletooktheopportunitytoretypeandslightlymodifythe
originalmanuscript,see[1]. Iteventurnedoutthattheunderlyingmethodcanbeeasilyadoptedtoprove
the following stronger result: No finite matchstick graph with minimum degree 5 does exist. So up to
nowthereareatleastthreeproofsofTheorem2.1whosepairwiseintersectionsarenon-empty.
We would like to mention the recent proof of the Higuchi conjecture [4] – a result of similar flavor
whererelatedtechniquesareapplied.
2. MainTheorem
Theorem2.1. Nofinite5-regularmatchstickgraphdoesexist.
InordertoproveTheorem2.1wedenotethenumberofi-gons,i.e. afaceconsistingofivertices,ofa
givenmatchstickgraphMbyF ,wherewealsocounttheouterface. IntheexampleinFigure1wehave
i
F =3,F =2,F =1,andF =0forallotheri.
3 4 7 i
2
Figure1:Amatchstickgraph.
Lemma2.2. Forafinite5-regularmatchstickgraphwehave
∞
(cid:88)
(10−3i)F =F −2F −5F −8F −11F −···=20.
i 3 4 5 6 7
i=3
PROOF. DuetotheEulerianpolyhedralformulawehaveV +F −E =2,whereV denotesthenumber
ofvertices,F thenumberoffaces,andE thenumberofedges. Thenumberoffacesisgivenby
∞
(cid:88)
F = F .
i
i=3
Sinceeveryedgeispartoftwofacesandeveryvertexispartof5faceswehave
∞ ∞
(cid:88) (cid:88)
2E = iF and 5V = iF .
i i
i=3 i=3
Insertingyieldstheproposedformula.
Definition2.3. Thefacesetf(v)ofavertexv containedinaplanargraphisamultisetcontainingthe
numberofcornersoftheadjacentfaces.
IntheexampleinFigure1wehavef(v) = {3,3,3,4,4}. Fora5-regularmatchstickgraphtheface
sets f(v) of the vertices have cardinality 5, i. e. there exist integers a ,...,a ≥ 3 with f(v) =
v,1 v,5
{a ,...,a }. Due to the angle sum of 2π at a vertex we have a ≥ 4 for at least one index i for
v,1 v,5 v,i
everyvertexv ∈V.
Definition2.4. Foranintegera≥3wesetc(a)= 10−3a andusethistodefinethecontributionc(v)of
a
avertexwithfacesetf(v)={a ,...,a }as
v,1 v,5
5 5
c(v)=c({a ,...,a })=(cid:88)c(a )=(cid:88)10−3av,i.
v,1 v,5 v,i a
v,i
i=1 i=1
(cid:80)
ForU ⊆V wesetc(U)= c(u).
u∈U
NowwecanrelatethisdefinitiontoLemma2.2andstatetwoeasylemmas.
Lemma2.5. IfMisa(finite)5-regularmatchstickgraphwehavec(V)=20.
PROOF. Forj ≥ 3wehave|{av,i =j |v ∈V, 1≤i≤5}| = jFj. Applyingthistothedefinitionof
c(V)yieldsthelefthandsideoftheformulaofLemma2.2.
3
Wewouldliketoremarkthatforaninfinite5-regularmatchstickgraphwewouldhavec(V)=0.
Lemma2.6. Thefacesetswithnon-negativecontributionaregivenby
5
c({3,3,3,3,4})= , c({3,3,3,4,4})=0,
6
1
c({3,3,3,3,5})= , c({3,3,3,3,6})=0.
3
OurstrategyforprovingTheorem2.1willbetopartitionthevertexsetV intosubsetseachhavingnon-
positivecontribution,whichyieldsacontradictiontoLemma2.5.
TodetermineasuitablepartitionofV intosubsetsforagivenmatchstickgraphMweconsiderthe
setTQoftrianglesandquadrangleswithinternalanglesin(cid:8)1π,2π(cid:9)anddefineanequivalencerelation
3 3
∼onTQ. Wheneverthereexistx,y ∈TQsharinganedgewerequirex∼y. Wecompletethisrelation
toanequivalencerelationbytakingthetransitiveclosure.
Sinceweprovethatnofinite5-regularmatchstickgraphsexistsitisimpossibletogiveexamplesto
illustrateourdefinitions. Thereforewedefinethem(mostly)formatchstickgraphsMwherethedegrees
ofitsverticesareatmost5.
Definition2.7. Foragiven(possibleinfinite)matchstickgraphMwithvertexdegreesatmost5wecall
anequivalenceclassx∼={y ∈TQ|y ∼x}oftheabovedefinedequivalencerelationaTQ-class.
So a TQ-class is an edge-connected union of triangles and quadrangles whose vertices are situated
onasuitablecommonregulartriangularlattice. InFigure2wehavedepictedanexample,wherewehave
markedtheTQ-classesbydifferentfacecolors.
Figure2:TQ-classesofamatchstickgraph.
InthenextlemmawesummarizesomeeasyfactsonTQ-classes.
Lemma2.8. LetBbeaTQ-class,thenthefollowingholds:
(1) ThefacesofBareedge-to-edgeconnected,meaningthatthedualgraphisconnected.
(2) TheverticesofBaresituatedonasuitableregulartriangularlattice.
ForbrevityweassociatewithaTQ-classBthesetofitsverticesV(B)andthesetofitsedgesE(B)so
thatwecanutilizethenotationc(B)forthecontributionofthevertexsetofB. Nowouraimistoshow
that the contribution c(B) for every TQ-class B of a 5-regular matchstick graph is at most zero. Since
someverticesmaybecontainedinmorethanoneTQ-classthecorrespondingvertexsetscannotbeused
directly to partition the vertex set V. Instead we choose the union C = ∪ B of all TQ-classes. Here
i i
4
everyvertexv ∈V occursatmostonceinC,neverthelessitmaybecontainedinseveralTQ-classesB .
i
Inordertoprovec(C) ≤ 0wehavetoperformsomebookkeepingduringtheproofofc(B) ≤ 0,which
willbethetopicofthenextsection. Oncewehavenproventhiswecanstate:
Proofofthemaintheorem. LetMbeafinite5-regularmatchstickgraphandCbetheunionofallTQ-
classesofM. UsingCorollary3.7andthefactthatallverticeswithpositivecontributionarecontained
inaTQ-class,seeLemma2.6,weconclude
(cid:88) (cid:88)
c(V) = c(C)+ c(v) ≤ 0+ 0 = 0.
v∈V\C v∈V\C
ThiscontradictsLemma2.5. (cid:3)
3. ContributionofTQ-classesof5-regularmatchstickgraphs
In this section we want to study the contribution c(B) of a finite TQ-class B in a 5-regular matchstick
graph M. Since we show that no finite 5-regular matchstick graph exists M has to be infinite in this
context. NeverthelesstheremaybesomefiniteTQ-classesinM.
Butsinceweareonlyinterestedinfinitematchstickgraphsweintroduceanotherconceptandconsider
incompletefinitepartsof5-regularmatchstickgraphs.
Definition3.1. LetMbeafinitematchstickgraphwithmaximumvertexdegreeatmost5andBaTQ-
class of M which induces a planar graph G containing the vertices, edges, and faces of B. We call a
vertex v in G (or B) an inner vertex if all faces being adjacent to v in G are contained in B. All other
verticesv inG (orB)arecalledoutervertices. NowwecancalltheTQ-classBprospective5-regular
exactlyifallinnerverticesofG havedegree5andallouterverticesofG havedegreeatmost5.
Figure3:Aprospective5-regularTQ-classBofamatchstickgraphManditsinducedplanargraphG.
InFigure3wehavedepictedamatchstickgraphwithmaximumvertexdegreeatmost5onthelefthand
side. Thefacesofthe,inthiscaseuniquely,containedTQ-classBarefilledwithgreencolor. Ifwebuild
upaplanargraphoutofthevertices,edges,andfacesofBweobtaintherighthandsideofFigure3. In
G vertexv istheonlyvertexwhichisnotadjacenttotheouterface. Thusv isaninnervertexinB and
theremaining7verticesofBareoutervertices.
SinceeveryTQ-classB ofagivenfinite5-regularmatchstickgraphMisprospective5-regular,we
nowstudyprospective5-regularTQ-classes. Thereforewewanttospecifythembysomeparametersσ,
k,τ,b andb .(Forbrevityweforegotousenotationsasσ(B),...wheneverthecorrespondingTQ-class
1 2
isevidentfromthecontext.)
Theparameterσ standsforthecontributionofthefacesofB. Tobecomemorepreciselywehaveto
introduceanewtechnicalnotation. Agivenvertexvisadjacenttofivefacesdenotedas(v,1),...,(v,5),
5
where face (v,i) is an a -gon. Since there exist pairs of vertices u (cid:54)= v and indices 1 ≤ i,j ≤ 5
v,i
with (v,i) = (u,j) we introduce the notation [v,i] addressing the arc of face (v,i) at vertex v. The
contributionofsuchafacearc[v,i]isdefinedtobec(a ). IfAisthesetofallfacearcs[v,i]wherethe
v,i
(cid:80)
face(v,i)iscontainedinB,thenσisgivenby c(α).
α∈A
Let us look at the four TQ-classes of Figure 2 (it is easy to check that they are all prospective 5-
regular). TheblueTQ-classconsistsofasingletriangle. Soithasthreetriangle-arcsandacontribution
ofσ = 1 + 1 + 1 =1. Itiseasytofigureoutthatingeneralatrianglecontributes1andaquadrangle
blue 3 3 3
contributes−2toσsothatwehaveσ =3,σ =0,andσ =16.
red green brown
Wehavealreadyremarkedthateverye∈E edgeofaplanargraphiscontainedintwofaces. When-
everbothfacesarecontainedinagivenTQ-classB wesaythateisaninneredge. Ifonlyonefaceis
containedinBwecalleanouteredge. BykwecounttheouteredgesofaTQ-classB. Inourexample
wehavek =3,k =5,k =10,andk =20.
blue red green brown
Ifweconsidertheverticesandedgesofaprospective5-regularTQ-classasasubgraph,thenevery
vertexv hasadegreeδ(v)atmost5. Byτ(v)wedenotethegap5−δ(v)(freevalencies)andbyτ(B)
thesumoverallτ(v)wherevisinB. Soinourexamplewehaveτ =9,τ =11,τ =20,and
blue red green
τ =22.
brown
Ina5-regularplanargrapheveryvertexviscontainedinexactlyfivedifferentfaces(v,i). Whenever
allfivefacesarecontainedinagivenTQ-classBwesaythatvinBisaninnervertex. Ifthenumberof
faces(v,i)whicharecontainedinB isbetweenoneandfourwecallv anoutervertex. Thiscoincides
withourpreviousdefinitionofinnerandouterverticesofaprospective5-regularTQ-class. Clearlywe
haveτ(v)=0forallinnerverticesv. Ife={v,u}isanouteredgeinBthenvanduareoutervertices
ofB. Fortheotherdirectionwehavethatforanoutervertexv inB thereexistatleasttwoouteredges
e ,e inBbeingadjacenttov. Morepreciselythenumberofouteredgesbeingadjacenttov iseither2
1 2
or4.
Figure4:Verticesofspecialtype.
Foragivenvertexv inB weconsiderthefaces(v,i)whicharecontainedinagivenTQ-classB as
verticesofagraphH. TwoverticesofH areconnectedviaanedgewheneverthecorrespondingfaces
have a common edge in M. If the graph H is connected we call v of normal type. Otherwise we say
thatvertexvisofspecialtype. InFigure4wedepictallcasesofverticesofspecialtypeuptosymmetry.
Herevertexv ismarkedblueandthefacesofB aremarkedgreen. Alternativelywecouldalsodefinea
vertexofspecialtypeasanoutervertexofaprospective5-regularTQ-classwhichisadjacenttoexactly
4outeredgesofB. Wewouldliketoremarkthatfromalocalpointofviewatavertexofspecialtypeit
seemsthatthefacesoftheTQ-classarenotedgeconnected.
Byb wecountthenumberofverticesv inB ofspecialtype. InourexampleofFigure2novertex
1
6
ofspecialtypeexistsandwehaveb =0. InFigure5wehavedepictedanotherexampleofaTQ-class,
1
wherewehaveb =1–thebluevertex.
1
Figure5:AnotherTQ-classofamatchstickgraph.
If a face f ∈ B of a TQ-class is a quadrangle then there are two inner angles of 2π and two inner
3
angles of π. By b we count the number of inner angles of 2π where the corresponding vertex v is an
3 2 3
outervertex. Itmayhappenthatanoutervertexv ispartoftwoquadranglesinB eachhavinganinner
angle of 2π at v. Thus for a prospective 5-regular TQ-class B we may write b (v) ∈ {0,1,2} for the
3 2
numberofinneranglesof 2π atanoutervertexv. Forinnerverticeswesetb (v)=0sothatwecanset
(cid:80) 3 2
b = b (v). IntheexampleofFigure2wehaveb =4inthegreenclass,b =2inthebrown
2 v∈V(B) 2 2 2
class,andb =0intheredandblueclass. Weremarkthatduetoananglesumof2πthereisexactlyone
2
facewithaninnerangleof 2π ateveryinnervertexofB.
3
Nowwewanttoprovetheequation
τ −k 5 5
σ−k+ + b + b =0. (1)
3 3 1 3 2
relatingtheparametersσ,k,τ,b andb .Thisequationevenholdsformoregeneralobjectsthanprospec-
1 2
tive 5-regular TQ-classes. A TR-class B is a union of triangles and quadrangles with inner angles in
(cid:8)π,2π(cid:9)onaregulartriangulargridsuchthateachinnervertexhasdegree5andeachoutervertexhas
3 3
degree at most 5. Clearly we can transfer the definitions of the parameters σ, k, τ, b and b from
1 2
(prospective5-regular)TQ-classestoTR-classes. ThecrucialdifferencebetweenTQ-classesandTR-
classesisthatthedualgraphofaTR-class(consideredasaplanargraph)isnotconnectedingeneral.
Figure6:ExamplesofTR-classes.
InFigure6wedepictsomeextraordinaryexamplesofTR-classestodemonstratethewholevariety
beingcoveredbythedefinitionofaTR-class. HerethefacesofaTR-classBarefilledingreen. Those
faces which are not contained in B are left in white. In the left example of Figure 6 we have σ = 3,
k = 9, τ = 12, b = 3, andb = 0. Inthesecondexamplewehaveσ = 4, k = 12, τ = 16, b = 4,
1 2 1
and b = 0. In the third example of Figure 6 we have σ = 5, k = 11, τ = 14, b = 3, and b = 0.
2 1 2
ForthecentralvertexvoftherightexampleofFigure6wehaveb (v)= 2. Additionallyvisofspecial
2
type. Theparametersaregivenbyσ = −4,k = 8,τ = 19,b = 1,andb = 4. Thusinallfourcases
1 2
7
Equation(1)isvalid. Ifweconsiderthosefourexamplesasasingleexampleonthesametriangulargrid
againEquation(1)isvalid.
Lemma3.2. ForaTR-classBEquation(1)isvalid.
PROOF. WeprovebyinductiononthenumberoffacesofB.Forasingletrianglewehaveσ =1,k =3,
τ =9,andb =b =0. Thusthelefthandsideof(1)sumstozero. Foraquadranglewehaveσ =−2,
1 2
k =4,τ =12,b =0,andb =2,againsumminguptozeroin(1). WeremarkthatEquation(1)isalso
1 2
validforanemptyTR-class.
For the induction step we now assume that Equality (1) is valid and we show that it remains valid
afteraddinganothersingletriangleorquadrangletoB. Hereweonlyconsiderthoseadditionswhichdo
notalterfacesofB(onemightthinkofinsertinganedgeintoaquadrangleofBresultingintwotriangles
anddestroyingtheinitialquadrangle). Ingeneralitmighthappenthataddingtheedgesofonenewface
producesasecondnewfaceinonestep. Anexamplearisesbyaddinganedgeintoaquadranglewhichis
notcontainedinB,seethesecondandthethirdexampleofFigure6. Here,inonestepweonlyaddone
oftheresultingtrianglestoB. IftheothertriangleshouldalsobeaddedtoBwedothisinasecondstep
whereweonlyhaveanewfaceandnonewverticesoredges.
Figure7:Addingatriangle.
Adding a triangle results in eight (main) cases, see Figure 7, and adding a quadrangle results in
nineteen(main)cases,seeFigure8. Inthedifferent(main)caseswedepictthenewedgesbydottedlines,
thenewverticesbyemptycircles,andtheoldedgesandverticesinblack. Wehaveacloserlookonthe
changeoftheparameters. IfpistheparameterofB andp(cid:48) thecorrespondingparameterafteraddinga
triangle(oraquadrangle),thenwedenotethechangeby∆(p):=p(cid:48)−p. Inordertoprovetheinduction
stepitsufficestoverify
∆(τ)−∆(k) 5 5
∆(σ)−∆(k)+ + ∆(b )+ ∆(b )=0. (2)
3 3 1 3 2
We remark that all black edges in the (main) cases of Figure 7 and Figure 8 are outer edges before
adding the new triangle or quadrangle. After adding the new face the black edges become inner edges
andthedottededgesbecomeouteredges.
Inordertoshortenourcalculationandcasedistinctionwehavealookatthesituationfromavertex
pointofview.Forafixedvertexa(inFigure7)uptosymmetrytherearefourdifferent(vertex)cases:Ifa
isanew(empty)vertexthenalsothetwoadjacentedgesofthenewtrianglehavetobedotted. Otherwise
ifaisanold(black)vertexthenumberofdottedadjacentedgesofthenewtrianglecanbetwo,one,or
zero. These four possibilities show up as (main) cases (1), (2), (5), and (7) in Figure 7 and we denote
themas(vertex)cases(i),(ii),(iii),and(iv). Clearly,alsoforvertexbandvertexvwehavethesamefour
possibilities. Sowerestrictourconsiderationsonvertexa.
Bya wedenotethevertexabeforetheadditionofthenewfaceandbya wedenotethevertexa
1 2
after the addition of the new face. In (vertex) case (i), occurring in (main) case (1), vertex a does not
1
exist and a is an outer vertex. Obviously a is of normal type and b (a ) = 0. In (vertex) case (ii),
2 2 2 2
occurring in (main) case (2) both a and a are outer vertices. We have b (a ) = b (a ). Since a is
1 2 2 1 2 2
adjacenttotworededgesofthenewfacevertexa isofspecialtypeanda isofnormaltype,sothatthe
2 1
b -count increases by one. In (vertex) case (iii) both a and a are outer vertices. So b (a ) = b (a )
1 1 2 2 1 2 2
8
(main)case ∆(σ) ∆(k) ∆(τ) ∆(b )+∆(b )
1 2
(1) 1 3 9 0
(2) 1 3 4 1
(3) 1 3 -1 2
(4) 1 3 -6 3
(5) 1 1 1 0
(6) 1 1 -4 1
(7) 1 -1 -2 -1
(8) 1 -3 0 -3
Table1:∆(·)-valuesfortheeight(main)casesofFigure7.
Figure8:Addingaquadrangle.
anda isofspecialtypeifandonlyifa isofspecialtype. Thustheb -andtheb -countsdonotchange.
1 2 1 2
In(vertex)case(iv)twodifferentthingscouldhappenatvertexa. Atfirstwementionthata isanouter
1
vertex. Ifalsoa isanoutervertexthena isofspecialtypeanda isofnormaltype. Inthissituationwe
2 1 2
haveb (a ) = b (a ). Theotherpossibilityisthata isaninnervertex. Sincethedegreeofa is5this
2 1 2 2 2 2
isonlypossibleifb (a )=1andifa isofnormaltype. Andsincea isaninnervertexitisofnormal
2 1 1 2
typeandwehaveb (a )=0. Thusthesumoftheb -countandtheb -countdecreasesbyone.
2 2 1 2
Withtheaboveweareabletogivethe∆(·)-valuesforthe(main)casesofFigure7inTable1. Itis
easytocheckthatinalleight(main)casesEquation(2)isvalid.
Next we can deal with the nineteen (main) cases of Figure 8. In every (main) case the situation of
vertexaandvertexwhasanequivalentinFigure7. Souptosymmetrywehavetoconsiderthesituation
atvertexv. Againtherearefourpossibilitiestoconsidershowingupin(main)cases(9),(14),(21),and
(27) of Figure 8. We denote the corresponding (vertex) cases by (v), (vi), (vii), and (viii), respectively.
Similarlyasbeforewedenotethevertexvbeforetheadditionofthequadranglebyv andafterwardsby
1
v .
2
In(vertex)case(v), occurringin(main)case(9), v doesnotexistandv isofnormaltypeandwe
1 2
haveb (v ) = 1sothatthesumoftheb -andtheb -countincreasesbyone. In(vertex)case(vi)both
2 2 1 2
v andv areoutervertices. Vertexv isofspecialtypeandv hastobeofnormaltype. Fortheb -value
1 2 2 1 2
we have b (v ) = b (v )+1 so thatthe sumof the b - andthe b -count increasesby two. In(vertex)
2 2 2 1 1 2
case(vii)bothv andv areoutervertices. Vertexv isofspecialtypeifandonlyifv isofspecialtype.
1 2 1 2
Sinceb (v )=b (v )+1thesumoftheb -andtheb -countincreasesbyone. In(vertex)case(viii)If
2 2 2 1 1 2
v isaninnervertexthanv ,v areofnormaltype,b (v ) = 0,andb (v ) = 0. Ifv isanoutervertex
2 1 2 2 1 2 2 2
thenv isofspecialtype,v isofnormaltype,andb (v ) = b (v )+1. Thusinbothcasesthesumof
1 2 2 2 2 1
9
(main)case ∆(σ) ∆(k) ∆(τ) ∆(b )+∆(b )
1 2
(9) -2 4 12 2
(10) -2 4 7 3
(11) -2 4 7 3
(12) -2 4 2 4
(13) -2 4 2 4
(14) -2 4 2 4
(15) -2 4 -3 5
(16) -2 4 -8 6
(17) -2 2 4 2
(18) -2 2 -1 3
(19) -2 2 -1 3
(20) -2 2 -6 4
(21) -2 0 -4 2
(22) -2 0 1 1
(23) -2 0 1 1
(24) -2 0 -4 2
(25) -2 0 -4 2
(26) -2 -2 -2 0
(27) -2 -4 0 -2
Table2:∆(·)-valuesforthenineteen(main)casesofFigure8.
theb -andtheb -countdoesnotchange.
1 2
Withtheaboveweareabletogivethe∆(·)-valuesforthe(main)casesofFigure8inTable2. Itis
easytocheckthatinallnineteen(main)casesEquation(2)isvalid.
Corollary3.3. Fora(finite)prospective5-regularTQ-classofamatchstickgraphwehave
τ −k 5 5
σ−k+ + b + b =0.
3 3 1 3 2
Now we are ready to prove c(B) ≤ 0 for every (finite) prospective 5-regular TQ-class of a matchstick
graphwithmaximumvertexdegreeatmost5. Tobemorepreciselyweneedtoconsiderfinitematchstick
graphsMwithmaximumvertexdegreeatmost5andaprospective5-regularTQ-classBofM. InM
allverticesofBmusthavevertexdegreeexactly5. Inordertobeabletospeakofacontributionc(B)for
everyvertexvinBtheremustbefivefaces(v,i),unequaltotheouterface,completelybeingcontained
inM. Ifafinite5-regularmatchstickgraphwouldexist,itwouldcertainlybesuchagraph. Otherwise
wecanonlylookatlocalpartsofsucha(possibleinfinite)graph.
As mentioned in the previous section we have to perform some bookkeeping in order to prove
c(∪ B ) ≤ 0 for a set of prospective 5-regular TQ-classes. To every face arc [v,i] where v is con-
i i
tained in a given prospective 5-regular TQ-class B we will assign a real weight ω([v,i]) ≥ c([v,i])
whichfulfills
5
(cid:88)(cid:88)
ω([v,i])≤0.
v∈Bi=1
We start with the face arcs [v,i] where the face (v,i) is contained in B. We call those arcs inner
(cid:80)
arcsandsetω([v,i]) = c([v,i]). Withthisthesum ω([v,i])overtheinnerarcsequalsthe
[v,i]innerarcofB
parameterσfromCorollary3.3.
10