Partitioning a triangle-free planar graph into a forest and a forest of bounded degree

0.21 MB·

by  François Dross

Partitioning a triangle-free planar graph into a forest and
a forest of bounded degree
François Drossa, Mickael Montassiera, and Alexandre Pinloua,b
a
Université de Montpellier, CNRS, LIRMM
b
6 Université Paul-Valéry Montpellier 3, Département MIAp
1
0 161 rue Ada, 34095 Montpellier Cedex 5, France
2
{francois.dross,mickael.montassier,alexandre.pinlou}@lirmm.fr

n
a
J January 8, 2016

7

]
Abstract
M
D An (F,Fd)-partition of a graph is a vertex-partition into two sets F and Fd such
that the graph induced by F is a forest and the one induced by Fd is a forest with
.
s maximum degree at most d. Weprovethat everytriangle-free planar graph admits an
c (F,F5)-partition. Moreover we show that if for some integer d there exists a triangle-
[ free planar graph that does not admit an (F,Fd)-partition, then it is an NP-complete

1 problem to decide whethera triangle-free planar graph admits such a partition.
v
3
1 Introduction
2
5
1 We onlyconsiderfinite simplegraphs,withneither loopsnormulti-edges. Planargraphswe
0
consider are supposed to be embedded in the plane. Consider i classes of graphs G ,...,G .
. 1 i
1 A(G ,...,G )-partitionofagraphGisavertex-partitionintoisetsV ,...,V suchthat,for
1 i 1 i
0
all1≤j ≤i, the graphG[V ]induced byV belongs to G . Inthe followingwe willconsider
j j j
6
the following classes of graphs:
1
:
v • F the class of forests,
i
X
• F the class of forests with maximum degree at most d,
d
r
a • D the class of d-degenerate graphs (recall that a d-degenerate graph is a graph such
d
that all subgraphs have a vertex of degree at most d),
• ∆ the class of graphs with maximum degree at most d,
d
• I the class of empty graphs (i.e. graphs with no edges).
Forexample,an(I,F,D )-partitionofGis a vertex-partitioninto three sets V ,V ,V such
2 1 2 3
that G[V ] is an empty graph, G[V ] is a forest, and G[V ] is a 2-degenerate graph.
1 2 3
The Four Colour Theorem [1, 2] states that every planar graph G admits a proper 4-
colouring,thatisthatGcanbepartitionedintofouremptygraphs,i.e. Ghasan(I,I,I,I)-
partition. Borodin [3] proved that every planar graph admits an acyclic colouring with at
mostfivecolours(anacycliccolouringisapropercolouringinwhicheverytwocolourclasses
induceaforest). Thisimpliesthateveryplanargraphadmitsan(I,F,F)-partition. Poh[8]
proved that every planar graph admits an (F ,F ,F )-partition. Thomassen proved that
2 2 2
everyplanargraphadmitsan(F,D )-partition[10],andan(I,D )-partition[11]. However,
2 3
1
Classes Vertex-partitions References
(I,I,I,I) The Four Color Theorem [1, 2]
(I,F,F) Borodin [3]
Planar graphs (F ,F ,F ) Poh [8]
2 2 2
(F,D ) Thomassen [10]
2
(I,D ) Thomassen [11]
3
(I,I,I) Grötzsch [6]
(F,F) Folklore
Planar graphs with girth 4
(F ,F) Present paper (Theorem 3)
5
(I,F) Open question (Question 1)
Planar graphs with girth 5 (I,F) Borodin and Glebov [4]
Table 1: Known results.
there are planar graphs that do not admit any (F,F)-partition [5]. Borodinand Glebov [4]
provedthateveryplanargraphofgirthatleast5(thatiseveryplanargraphwithnotriangles
nor cycles of length 4) admits an (I,F)-partition.
Wefocusontriangle-freeplanargraphs. RaspaudandWang[9]provedthateveryplanar
graph with no triangles at distance at most 2 (and thus in particular every triangle-free
planargraph)admits an(F,F)-partition. However,it is not knownwhether every triangle-
free planar graph admits an (I,F)-partition. We pose the following questions:
Question 1. Does every triangle-free planar graph admit an (I,F)-partition?
Question 2. More generally, what is the lowest d such that every triangle-free planar graph
admits an (F,F )-partition?
d
Note that proving d=0 in Question 2 would prove Question 1. The main result of this
paper is the following:
Theorem 3. Every triangle-free planar graph admits an (F,F )-partition.
5
This implies that d ≤ 5 in Question 2. Our proof uses the discharging method. It
is constructive and immediately yields an algorithm for finding an (F,F )-partition of a
5
triangle-free planar graph in quadratic time.
Note that Montassier and Ochem [7] proved that not every triangle-free planar graph
can be partitioned into two graphs of bounded degree (which shows that our result is tight
in some sense).
Finally, we show that if for some d, there exists a triangle-free planar graph that does
not admit an (F,F )-partition, then deciding whether a triangle-free planar graph admits
d
such a partition is NP-complete. That is, if the answer to Question 2 is some k > 0, then
for all 0≤d<k, deciding whether a triangle-free planar graph admits an (F,F )-partition
d
is NP-complete. We prove this by reduction to Planar 3-Sat.
All presented results on vertex-partitionof planar graphs are summarized in Table 1.
Theorem 3 will be proved in Section 2. Section 3 is devoted to complexity results.
Notation
Let G=(V,E) be a plane graph (i.e. planar graph together with its embedding).
For a set S ⊂ V, let G−S be the graph constructed from G by removing the vertices
of S and all the edges incident to some vertex of S. If x ∈ V, then we denote G−{x} by
G−x. For a set S of vertices suchthat S∩V =∅, let G+S be the graphconstructed from
G by adding the vertices of S. If x∈/ V, then we denote G+{x} by G+x. For a set E′ of
pairs of vertices of G such that E′∩E =∅, let G+E′ be the graph constructed from G by
2
adding the edges of E′. If e is a pair of vertices of G and e∈/ E, then we denote G+{e} by
G+e. For a set W ⊂V, we denote by G[W] the subgraph of G induced by W.
We call a vertex of degree k, at least k and at most k, a k-vertex, a k+-vertex and a
k−-vertex respectively, and by extension, for any fixed vertex v, we call a neighbour of v
of degree k, at least k and at most k, a k-neighbour, a k+-neighbour, and a k−-neighbour
of v respectively. When there is some ambiguity on the graph, we call a neighbour of v
in G a G-neighbour of v. We call a cycle of length ℓ, at least ℓ and at most ℓ a ℓ-cycle,
a ℓ+-cycle, and a ℓ−-cycle respectively, and by extension a face of length ℓ, at least ℓ and
at most ℓ a ℓ-face, a ℓ+-face, and a ℓ−-face respectively. We say that a vertex of G is big
if it is a 8+-vertex, and small otherwise. By extension, a big neighbour of a vertex v is a
8+-neighbour of v, and a small neighbour of v is a 7−-neighbour of v.
Twoneighboursuandwofavertexvareconsecutiveifuvwformsapathontheboundary
of a face.
2 Proof of Theorem 3
We prove Theorem 3 by contradiction. Let G=(V,E) be a counter-example to Theorem 3
of minimum order.
Graph G is connected, otherwise at least one of its connected components would be a
counter-example to Theorem 3, contradicting the minimality of G.
Let us consider any plane embedding of G. Let us prove a series of lemmas on the
structure of G, that correspond to forbidden configurations in G.
Lemma 4. There are no 2−-vertices in G.
Proof. Suppose there is a 2−-vertex v in G. By minimality of G, G−v admits an (F,F )-
5
partition (F,D). If v is a 1−-vertex, then G[F ∪{v}]∈F. Suppose v is a 2-vertex. If both
of its neighbours are in F, then G[D∪{v}]∈F . Otherwise, G[F ∪{v}]∈F. In all cases,
5
one can obtain an (F,F )-partition of G, a contradiction.
5
Lemma 5. Every 3-vertex in G has at least one big neighbour.
Proof. Supposethereis a3-vertexv inGthathasthreesmallneighbours. Byminimalityof
G, G−v admits an (F,F )-partition (F,D). If at least two neighbours of v are in D, then
5
G[F ∪{v}] ∈ F. If no neighbour of v is in D, then G[D∪{v}]∈ F . Suppose exactly one
5
neighbouruofv isinD. Ifatmostoneofthe neighboursofuis inF,thenG[F∪{u}]∈F,
and G[D\{u}∪{v}]∈F . Otherwise, since u is small, at most four of the neighbours of u
5
are in D, thus G[D∪{v}] ∈ F . In all cases, one can obtain an (F,F )-partition of G, a
5 5
contradiction.
Lemma 6. Every 4-vertex or 5-vertex in G has at least one 4+-neighbour.
Proof. Suppose there is a 4-vertex or 5-vertex v in G that has no 4+-neighbour. Let the u
i
be the neighbours of v, for i ∈ {0,...,3} or i ∈ {0,...,4}. Let G′ = G−v− {u }. By
Si i
minimality of G, G′ admits an (F,F )-partition (F,D). Add v to D, and for all u , add u
5 i i
toD if its twoneighboursdistinct fromv areinF,andaddu toF otherwise. Vertex v has
i
at most five neighbours in D, and each of the u that is in D has one neighbour in D. Each
i
of the u that is in F has at most one neighbour in F. We have an (F,F )-partition of G,
i 5
a contradiction.
We will need the following observation in the next two lemmas.
Observation 7. Let v v v v be a face of G, u a neighbour of v and u a neighbour of
0 1 2 3 0 0 1
v . Either u and v are at distance at least 3, or u and v are at distance at least 3.
1 0 2 1 3
3
b
1
v
1
s
0
s
1
v
0
b
0
Figure 1: The forbidden configuration of Lemma 8. The big vertices are represented with
big circles, and the small vertices with small circles. The filled circles represent vertices
whose incident edges are all represented.
By contradiction, suppose that u and v are at distance at most two, and that u and
0 2 1
v are at distance at most two. Since G is triangle-free,a shortestpath fromu to v (resp.
3 0 2
from u to v ) does not contain any of the u and v except for its extremities. Then by
1 3 i i
planarity there exists a vertex w adjacent to u , v , u and v . In particular v v w is a
0 2 1 3 2 3
triangle, a contradiction.
Lemma 8. The following configuration does not occur in G: two adjacent 3-vertices v and
0
v such that for i∈{0,1}, v has a big neighbour b and a small neighbour s , and such that
1 i i i
v v s b bounds a face of G.
0 1 1 0
Proof. Suppose such a configuration exists in G. See Figure 1 for an illustration of this
configuration. Observe that all the vertices defined in the statement are distinct (since G
is triangle-free). By Observation 7, either b and b are at distance at least 3, or s and s
0 1 0 1
are at distance at least 3. For the remaining of the proof, we no longer need the fact that
b s ∈ E(G). We forget this assumption, and only remember that either b and b are at
0 1 0 1
distance at least 3, or s and s are at distance at least 3. This provides some symmetry in
0 1
the graph.
Let G =G−{v ,v }+b b andG =G−{v ,v }+s s . By whatprecedes,either G
0 0 1 0 1 1 0 1 0 1 0
or G is triangle-free, thus there exists a j such that G is a triangle-free planar graph. By
1 j
minimality of G, G admits an (F,F )-partition (F,D).
j 5
Let us first prove that if we do not have b and b in D, and s and s in F, then the
0 1 0 1
conditionsG[F]∈F andG[D]∈F leadtoa contradiction. We willsee thatwecanalways
5
extend the (F,F )-partition of G to G.
5 j
• If at least three of the b and s are in D, then G[F ∪{v ,v }]∈F.
i i 0 1
• If all of the b and s are in F, then G[D∪{v ,v }]∈F .
i i 0 1 5
• Suppose now that exactly three of the b and s are in F. W.l.o.g., b ∈D or s ∈D.
i i 0 0
We have G[F ∪{v }]∈F and G[D∪{v }]∈F .
0 1 5
• Suppose now thatexactlytwo oftheb ands areinF. Ifb ands arein F (resp. b
i i 0 0 1
and s are in F), then G[D∪{v }]∈F and G[F ∪{v }]∈F (resp. G[F ∪{v }]∈F
1 0 5 1 0
and G[D∪{v }]∈F ).
1 5
Now w.l.o.g. b ∈ F and s ∈ D. If s has at most one G-neighbour in F, then
0 0 0
G[F ∪{s }]∈F, we can replace F by F ∪{s } and D by D\{s }, and we fall into a
0 0 0
previous case. We can thus assume that s has at least two of its G-neighbours in F,
0
and thus it has at most four of its G-neighbours in D. Therefore G[D∪{v }] ∈ F ,
0 5
and G[F ∪{v }]∈F.
1
4
In all cases, G has an (F,F )-partition, a contradiction.
5
Remainsthecasewhereb andb areinD,ands ands areinF. Inthecasewherewe
0 1 0 1
addedthe edgeb b (i.e. thecasej =0),wehaveG[D∪{v ,v }]∈F ,sinceG[D∪{v ,v }]
0 1 0 1 5 0 1
is equal to G [D] where an edge is subdivided twice. Similarily, in the case where we added
0
the edge s s (i.e. the case j = 1), we have G[F ∪{v ,v }] ∈ F, since G[F ∪{v ,v }] is
0 1 0 1 0 1
equal to G [F] where an edge is subdivided twice. Again, G has an (F,F )-partition, a
0 5
contradiction.
w w
0 1
s v
0 1
v s
0 1
b
Figure 2: The forbidden configuration of Lemma 9.
Lemma 9. The following configuration does not occur in G: a 3-vertex v adjacent to a
0
4-vertex v such that v has a big neighbour b and a small neighbour s , and v has three
1 0 0 1
other small neighbours s , w , and w such that v v s b bounds a face of G and s has
1 0 1 0 1 1 1
degree 3.
Proof. Suppose such a configuration exists in G. See Figure 2 for an illustration of this
configuration. Observethat all the vertices defined in the statement are distinct (since G is
triangle-free). By Observation7, either b and w are at distance at least 3, or s and s are
0 0 1
at distance at least 3. Let G =G−{v ,v }+bw and G =G−{v ,v }+s s . By what
0 0 1 0 1 0 1 0 1
precedes,eitherG orG istriangle-free,thusthereexistsaj suchthatG isatriangle-free
0 1 j
planar graph. By minimality of G, G has an (F,F )-partition (F,D).
j 5
Let us first prove that except in the case where {b,w ,w } ⊂ D and {s ,s } ⊂ F, the
0 1 0 1
conditionsG[F]∈F andG[D]∈F leadtoa contradiction. We willsee thatwecanalways
5
extend the (F,F )-partition of G to G.
5 j
If at least four among the w , the s and b are in D, then G[F ∪{v ,v }]∈F.
i i 0 1
Suppose now that at most three among the w , the s and b are in D. Suppose x ∈
i i
{b,s ,s ,w ,w } is inD. If x has atmostone G-neighbourin F, then G[F ∪{x}]∈F, and
0 1 0 1
wecouldconsiderF∪{x} insteadofF andD\{x}insteadofD. Note that this cannotlead
to the case we excluded ({b,w ,w } ⊂ D and {s ,s } ⊂ F) unless at least four among the
0 1 0 1
w , the s and b are in D. Thus we can assume that for any x among the w and s such
i i i i
that x ∈ D, x has at most four G-neighbours in D, and thus adding one neighbour of x in
D cannot cause x to have at least six neighbours in D. We consider two cases according to
b:
• Suppose b∈F. If at least three of the w and s are in F, then G[D∪{v ,v }]∈F .
i i 0 1 5
Ifatleasttwoamongthe w ands areinD, thenG[F∪{v }]∈F andG[D∪{v }]∈
i 1 1 0
F . Else,at leasttwo among the w and s are in F, andwe may assume that s is in
5 i 1 0
D (otherwise we fall into a previous case), so G[D∪{v }]∈F and G[F ∪{v }]∈F.
1 5 0
• Suppose now that b ∈ D. As s has degree 3, it has at most one G-neighbour in F,
1
and thus as previouslywe couldconsiderF ∪{s } insteadof F and D\{s } insteadof
1 1
5
D. Again,thiscannotleadtothecaseweexcluded({b,w ,w }⊂D and{s ,s }⊂F)
0 1 0 1
unless at least four among the w , the s and b are in D. Therefore we can assume
i i
that s ∈F. The w are not both in D (otherwise we fall into the case we excluded).
1 i
We have G[D∪{v }]∈F and G[F ∪{v }]∈F.
1 5 0
In all cases, G has an (F,F )-partition, a contradiction.
5
Remains the case {b,w ,w } ⊂ D and {s ,s } ⊂ F. In the case where we added the
0 1 0 1
edge bw (i.e. the case j = 0), b has at most five G -neighbours in D, and thus at most
0 0
four G-neighbours in D, so G[D∪{v }] ∈ F , and G[F ∪{v }] ∈ F. In the case where we
0 5 1
addedthe edge s s (i.e. the case j =1), we haveG[F ∪{v ,v }]∈F, since G[F ∪{v ,v }]
0 1 0 1 0 1
is equal to G [F] where an edge is subdivided twice. Again, G has an (F,F )-partition, a
0 5
contradiction.
v
3
v v
0 2
b0 v1
w0 b1
Figure 3: Configuration 10.
We define a specific configuration:
Configuration 10. Two 4-faces b v v w and v v v v , suchthat b is a big vertex, v and
0 0 1 0 0 1 2 3 0 0
w are 3-vertices, v is a 4-vertex, v and v are small vertices, and the fourth neighbour of
0 1 2 3
v , say b , is a big vertex. See Figure 3 for an illustration of this configuration.
1 1
v
3
v v
0 2
b0 v1 w1
w0 b1
Figure 4: The forbidden configuration of Lemma 11.
Lemma 11. The following configuration is forbidden: Configuration 10 with the added
condition that there is a 4-face b v v w with w a 3-vertex, v a 4-vertex, and the fourth
1 1 2 1 1 2
neighbour of v , the third neighbour of w , and the third neighbour of w are small vertices.
2 1 0
Proof. Suppose such a configuration exists in G. See Figure 4 for an illustration of this
configuration. Observe that all the vertices named in the statement are distinct since G is
triangle-free and w is a small vertex whereas b is a big one.
1 0
6
Letusprovethateitherb andb areatdistanceatleast3,orw andw ,andw andv
0 1 0 1 0 3
are at distance at least 3. By contradiction, suppose that b and b are at distance at most
0 1
two, and that either w and w are at distance at most two, or w and v are at distance
0 1 0 3
at most 2. Since G is triangle-free, a shortest path from b to b , from w to w or from w
0 1 0 1 0
to v does not go through any of the vertices defined in the statement. Then by planarity
3
there exists a vertex w adjacent to b , b , w and either w or v . In particular b w w is a
0 1 0 1 3 0 0
triangle, a contradiction.
Let G =G−{v ,v }+b b and G =G−{v ,v }+w w +w v . By what precedes,
0 0 1 0 1 1 0 1 0 1 0 3
either G or G is triangle-free, thus there exists a j such that G is a triangle-free planar
0 1 j
graph. By minimality of G, G has an (F,F )-partition (F,D).
j 5
Lets bethethirdneighbourofw ,s bethethirdneighbourofw ands bethefourth
0 0 1 1 2
neighbour of v . They are all small vertices, but there may be some that are equal between
2
themselves, or equal to some vertices we defined previously. However, if one of the s is
i
in {v ,v ,v ,w ,w }, then this s is a 4−-vertex in G (and in particular it has at most 4
0 1 2 0 1 i
neighbours in D).
Suppose first that b and b are both in D.
0 1
1. Supposew isinD. Hereweonlyconsider(F,D)asan(F,F )-partitionofG−{v ,v }.
0 5 0 1
Ifv isalsoinD,thenaddingv andv toF leadstoan(F,F )-partitionofG. Suppose
3 0 1 5
v isinF. Weshownowthatwecanassumethatv isinD. Bycontradiction,suppose
3 2
v is in F. We remove v from F.
2 2
Observe that we can assume that v has no G-neighbour in D with five G-neighbours
2
in D. Indeed, suppose v has a G-neighbour in D with five G-neighbours in D. This
2
G-neighbour is a 5+-vertex, so it is s . Moreover, s is not equal to v (because v is
2 2 3 3
in F), and is not equal to any of the other vertices named in the statement (because
of the degree conditions). As s is a small D-vertex, has at least five G-neighbours in
2
D andis adjacentto v that is neither inF norin D, s has atmost one neighbourin
2 2
F. Therefore we can put s in F.
2
Observe that we can assume that v has at most one G-neighbour in D. Suppose v
2 2
has two G-neighbours in D. These G-neighbours are s and w . Vertex w has at
2 1 1
most one neighbour in F, that is s , so we can put w in F.
1 1
Now v has at most one G-neighbour in D, and no G-neighbour of v in D has five
2 2
G-neighbours in D, so we can put v in D. Therefore we can always assume that v
2 2
is in D. Note that we do not need to change where s is in the partition if it is equal
2
to one of the vertices named in the statement. Adding v and v to F leads to an
0 1
(F,F )-partition of G.
5
2. Suppose w is in F, v is in D and w is in D. If s is in D, then putting v , v and
0 3 1 2 0 1
v in F leads to an (F,F )-partition of G. Suppose s is in F. We put v , v and w
2 5 2 0 1 1
in F, and v in D. If this increasesthe number of G-neighboursof v in D above five,
2 3
then since v is small, v has at most one neighbour in F, which is v , and we put v
3 3 0 3
in F. This leads to an (F,F )-partition of G.
5
3. Suppose w is in F, v is in D and w is in F. Suppose s is in F. We put v and
0 3 1 2 0
v in F, and v in D. If this increases the number of G-neighbours of v in D above
1 2 3
five, then since v is small, v has at most one neighbour in F, which is v , and we
3 3 0
put v in F. This leads to an (F,F )-partition of G. Suppose s is in D. If v is
3 5 2 2
not in F, we may put it in F, since it has only one G -neighbour in F, that is w .
j 1
Thereforewecanassumethatv is inF. Ifj =0,thenb hasatmost4G-neighbours
2 1
in D (since it has at most 5 such G -neighbours), so adding v to F and v to D
0 0 1
leads to an (F,F )-partition of G. If j = 1, then adding v and v to F leads to an
5 0 1
(F,F )-partition of G.
5
7
4. Suppose w is in F and v is in F. Suppose j = 0. The vertex b has at most 4
0 3 0
G-neighbours in D (since it has at most 5 such G -neighbours), so we can add v to
0 0
D. If v is in D, then adding v to F leads to an (F,F )-partition of G. If v is in F,
2 1 5 2
then adding v to D makes G[D] equal to G [D] with an edge subdivided twice, and
1 0
this leads to an (F,F )-partition of G. Suppose j = 1. Here we only consider (F,D)
5
asan(F,F )-partitionofG−{v ,v }+w v . Asin1,wecansuppose,uptochanging
5 0 1 0 3
where s and w are in the partition, that v is in D. Note that if s is equal to one
2 1 2 2
of the vertices named in the statement, we do not need to move s in the partition.
2
Adding v and v to F leads to an (F,F )-partition of G.
0 1 5
Nowwemayassumethatatleastoneofb andb isinF. Fromnowonweonlyconsider
0 1
(F,D) as an (F,F )-partition of G−{v ,v }.
5 0 1
• Suppose b is in F and b is in D. In that case we put v and w in D, and v in F.
0 1 0 0 1
Adding v inD (resp. w inD) mayviolatethe degreeconditionofG[D] ; however,if
0 0
ithappens,onecanputv (resp. s )inF. Inanycase,weobtainan(F,F )-partition
3 0 5
of G.
• Suppose b is in D and b is in F. If at least one of w and v is in F, then adding v
0 1 0 2 0
inF andv inD leadstoan(F,F )-partitionofG. Assume w andv arebothinD.
1 5 0 2
If v is in D, then adding v and v in F leads to an (F,F )-partition of G. Assume
3 0 1 5
v is in F. We consider three cases:
3
– Suppose s and w are in F. Adding v in F and v in D leads to an (F,F )-
2 1 0 1 5
partition of G.
– Suppose s is inF andw is inD. Ifs is inD, then we canput w inF andwe
2 1 1 1
fall into the previous case. If s is in F, then adding v in F and v in D leads
1 0 1
to an (F,F )-partition of G.
5
– Suppose s is in D. If s is in D and has five G-neighbours in D distinct from
2 1
w ,thenass issmall,itisdistinctfromalltheverticesnamedinthestatement,
1 1
and we can put it in F. Therefore we can put w in D and v in F. We fall into
1 2
a previous case (at least one of w and v is in F).
0 2
• Suppose b andb areinF. Ifs is inD andhas fiveG-neighboursinD distinct from
0 1 0
w ,thenass issmall,itisdistinctfromalltheverticesnamedinthe statementaside
0 0
from v , and we can put it in F. Therefore we can put w in D. We consider the
3 0
following cases:
– If v and v are in F, then adding v and v to D leads to an (F,F )-partition
2 3 0 1 5
of G.
– If v is in F and v is in D, then adding v to F and v to D leads to an
2 3 0 1
(F,F )-partition of G.
5
– If v is in D and v is in F, then adding v to D and v to F leads to an
2 3 0 1
(F,F )-partition of G.
5
– If v and v are in D, then adding v to D and v to F leads to an (F,F )-
2 3 0 1 5
partition of G. Adding v to D may violate the degree condition of G[D], but in
0
that case we can put v in F.
3
Wenowapplyadischargingprocedure: first,forallj,everyj-vertexvhasachargeequal
to c (v)=j−4,andeveryj-face f hasa chargeequaltoc (f)=j−4. By Euler’sformula,
0 0
the total charge is negative (equal to −8). Observe that, since G is triangle-free, every face
has a non-negative initial charge, and by Lemma 4, the vertices that have negative initial
8
charges are exactly the 3-vertices of G, and they have an initial charge of −1. Here is our
discharging procedure:
Discharging procedure:
• Step 1: Every big vertex gives 1 to each of its small neighbours. Furthermore, for
2
every 4-face uvwx where u and v are big, and w and x are small, v gives 1 to x (and
4
u gives 1 to w).
4
• Step 2: Consider a 4-vertex v that does not correspond to v in Configuration 10.
1
Vertex v gives 1 to eachofits smallneighbours that areconsecutive (as neighboursof
4
v)toexactlyonebigvertex,and 1 toeachofitssmallneighboursthatareconsecutive
2
(as neighbours of v) to two big vertices.
Considerthecasewherev correspondstov inConfiguration10. Weusethenotations
1
of Configuration 10. If w has two big neighbours, then v gives 1 to v and 1 to v .
0 1 4 0 4 2
Otherwise, it gives 1 to w and 1 to v .
4 0 4 0
Everysmall5+-vertexthathasabigneighbourgives 1 toeachofitssmallneighbours,
4
and an additional 1 for each that is consecutive (as neighbours of v) to at least one
4
big vertex. Every small 5+-vertex that has no big neighbour gives 1 to each of its
4
3-neighbours.
• Step 3: For every 4-face uvwx, with u a big vertex, v a 3-vertex, w a 4-vertex, and x
a small vertex such that x gave charge to w in Step 2, w gives 1 to v.
4
• Step 4: Every 5+-face that has a big vertex in its boundary gives 1 to each of the
4
small vertices in its boundary. Every 5+-face that has no big vertex in its boundary
gives 1 to each of the vertices in its boundary.
5
• Step 5: For every 4-face uvwx, with u a big vertex, v a 3-vertex, w a 4-vertex and x
a 3-vertex such that the other face that has vw in its boundary is a 5+-face, w gives
1 to v.
5
For everyvertex or face x ofG, for everyi∈{1,2,3,4,5},let c (x) be the chargeofx at
i
theendofStepi. Observethatduringtheprocedure,nochargesarecreatedandnocharges
disappear; hence the total charge is kept fixed.
We now prove that every vertex and every face has a non-negative charge at the end of
the procedure. That leads to the following contradiction:
0≤ X c5(x)= X c0(x)=−8
x∈V(G)∪F(G) x∈V(G)∪F(G)
Lemma 12. Every face has non-negative charge at the end of the procedure.
Proof. At the beginning of the procedure, for every j-face f we have c (f)=j−4 ≥0 (as
0
j ≥ 4). The procedure does not involve 4-faces. Hence if j = 4, then c (f) = c (f)= 0. If
5 0
j = 5, then f gives at most four times 1 if it is incident to a big vertex and at most five
4
times 1 otherwise in Step 4. It follows that c (f) ≥ 0. If j ≥ 6, then f can give 1 to each
5 5 3
of its incident vertices (and so 1 or 1) during Step 4, and c (f)≥j−4− j ≥0.
4 5 5 3
Lemma 13. A 4+-vertex never has negative charge.
Proof. Consider a j-vertex z with j ≥4. At the beginning, c (z)=j−4≥0. We will show
0
that c (z)≥0 for i=1,...,5.
i
• Suppose z is a big vertex. Such a vertex only loses charge in Step 1. Since j ≥ 8, we
have c (z) ≥ j. In Step 1, vertex z loses 1 for each of its small neighbours, and at
0 2 2
most 1 foreachofits bigneighbours. Thereforeithasmorechargethanwhatitgives,
2
and thus it keeps a non-negative charge.
9
• Suppose z is a small 5+-vertex. It does not lose charge in Steps 1, 3, 4 and 5.
Supposez hasabigneighbour. Ithasatmostj−1smallneighbours,andithascharge
at least 1(j−1) at the beginning of the procedure, since j ≥5. Moreover,it receives
4
1 from each of its big neighbours in Step 1. Therefore it does not give more charge
2
that it has in Step 2.
Suppose now that z has no big neighbour. If z is a 5-vertex, then by Lemma 6,
it has at most four 3-vertices, and c (z) ≥ 1 − 41 ≥ 0. If z is a 6+-vertex, then
2 4
c (z)≥j−4−j1 ≥0.
2 4
• Suppose z is a 4-vertex. It does not lose charge in Steps 1 and 4. Suppose z gives
charge in Step 2. Consider first that z does not correspond to v in Configuration 10.
1
If z is adjacent to a small vertex that is consecutive (as a neighbour of z) to two big
neighbours, then z gives at most twice 1 in Step 2 and received twice 1 in Step 1;
2 2
hence c (z) ≥ 0. Otherwise, z gives at most twice 1 in Step 2, and received at least
2 4
once 1 in Step 1; hence c (z)≥ 0. Let us now consider the case where z corresponds
2 2
to v in Configuration 10. The vertex z has a big neighbour that gave 1 to z in Step
1 2
1, and z gives 1 to two of its neighbours in Step 2. Therefore z received in Step 1 at
4
least as much as what it gives in Step 2.
u x u x u′
v z v z v′
Figure 5: Some configurations that appear in Lemma 13.
Suppose z gives charge in Step 3. There is a 4-face uvzx with u a big vertex, v a
3-vertex, and x a small vertex such that x gave charge to z in Step 2. Suppose z is
consecutive to exactly one big vertex (as neighbours of x). The vertex x gave at least
1 toz inStep2,andthereisexactlyonesuchfacewiththe samez andx(i.e. thereis
4
no pair (u′,v′) distinct from (u,v) that verifies the properties we stated for (u,v))(see
Figure 5, left). Therefore z can give 1 to v in Step 3. Suppose z is consecutive to
4
exactly two big vertices (as neighbours of x). The vertex x gave 1 to z in Step 2,
2
and there are at most two such faces with the same z and x (i.e. there is at most one
pair (u′,v′) distinct from (u,v) that verifies the properties we stated for (u,v)) (see
Figure 5, right). Therefore z can give 1 to each of the corresponding v’s in Step 3.
4
Therefore z received in Step 2 at least as much as what it gives in Step 3.
Suppose z gives charge in Step 5. There is a 4-face uvzx, with u a big vertex, v a
3-vertex,and x a 3-vertex suchthat the other face, say f, that has vz in its boundary
is a 5+-face. Vertex z received at least 1 from f in Step 4, and it gives 1 to v. There
5 5
is a problem only if there is another 4-face u′v′zx′, such that vzv′ is on the boundary
of f, u′ is a big vertex, and x′ and v′ are 3-vertices. But then z would have four
3-neighbours,contradictingLemma 6. Thereforez receivedin Step 4 at leastas much
as what it gives in Step 5.
In all cases, z never has negative charge.
Lemma 14. At the end of the procedure, every 3-vertex has non-negative charge.
10