stratification and domination in graphs with minimum degree two

Discrete Mathematics 301 (2005) 175–194www.elsevier.com/locate/disc

Stratification and domination in graphs withminimum degree two

Michael A. Henning1, J.E. Maritz2

School of Mathematical Sciences, University of KwaZulu-Natal, Private Bag X01, Scottsville,Pietermaritzburg, 3209 South Africa

Received 30 October 2003; accepted 30 June 2005Available online 9 September 2005

Abstract

A graphG is 2-stratified if its vertex set is partitioned into two classes (each of which is a stratumor a color class). We color the vertices in one color class red and the other color class blue. LetFbe a 2-stratified graph with one fixed blue vertexv specified. We say thatF is rooted atv. TheF-domination number of a graphG is the minimum number of red vertices ofG in a red-blue coloring ofthe vertices ofG such that every blue vertexv ofG belongs to a copy ofF rooted atv. We investigatetheF-domination number whenF is a 2-stratified pathP3 on three vertices rooted at a blue vertexwhich is an end-vertex of theP3 and is adjacent to a blue vertex with the remaining vertex coloredred. We show that for a connected graph of ordernwith minimum degree at least two this parameteris bounded above by(n − 1)/2 with the exception of five graphs (one each of orders four, five andsix and two of order eight). Forn�9, we characterize those graphs that achieve the upper bound of(n− 1)/2.© 2005 Elsevier B.V. All rights reserved.

Keywords:2-Stratified graphs; Domination; Bounds

1. Introduction

In this paper we continue the study of stratification and domination in graphs startedby Chartrand et al.[3] and studied further in[11]. A graphG whose vertex set has been

E-mail address:[email protected](M. Henning).1Research supported in part by the South African National Research Foundation and the University of

KwaZulu-Natal.2Research supported in part by the South African National Research Foundation.

0012-365X/$ - see front matter © 2005 Elsevier B.V. All rights reserved.doi:10.1016/j.disc.2005.06.021

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mailto:[email protected]

176 M.A. Henning, J.E. Maritz / Discrete Mathematics 301 (2005) 175–194

partitioned into two setsV1 andV2 is called a2-stratified graph. The setsV1 andV2 arecalled thestrataor sometimes thecolor classesofG. We ordinarily color the vertices ofV1red and the vertices ofV2 blue.In [13], Rashidi studied a number of problems involving stratified graphs; while distance

in stratified graphs was investigated in[1,2,4].A setS ⊆ V (G) of a graphG is adominating setif every vertex not inS is adjacent to a

vertex inS. The domination number ofG, denoted by�(G), is the minimum cardinality of adominating set.A dominating set ofGof cardinality�(G) is called a�-setofG. The conceptof domination in graphs, with its many variations, is now well studied in graph theory. Thebook by Chartrand and Lesniak[5] includes a chapter on domination. For a more thoroughstudy of domination in graphs, see Haynes et al.[8,9].In [3] a new mathematical framework for studying domination is presented. It is shown

that the domination number and many domination related parameters can be interpretedas restricted 2-stratifications or 2-colorings, with the red vertices forming the dominat-ing set. This framework places the domination number in a new perspective and sug-gests many other parameters of a graph which are related in some way to the dominationnumber.More precisely, letF be a 2-stratified graph with one fixed blue vertexv specified. We

say thatF is rootedat the blue vertexv. An F-coloring of a graphG is defined in[3] tobe a red–blue coloring of the vertices ofG such that every blue vertexv of G belongs to acopy ofF (not necessarily induced inG) rooted atv. TheF-domination number�F (G) ofG is the minimum number of red vertices ofG in anF-coloring ofG. In [3], anF-coloringof G that colors�F (G) vertices red is called a�F -coloringof G. The set of red vertices ina �F -coloring is called a�F -set. IfG has ordern andG has no copy ofF, then certainly�F (G)= n.For notation and graph theory terminology we, in general, follow[8]. Specifically, let

G = (V ,E) be a graph with vertex setV of ordern and edge setE. For a setS ⊆ V , thesubgraph induced bySis denoted byG[S]. The minimum degree (resp., maximum degree)among the vertices ofG is denoted by�(G) (resp.,�(G)).A star is the treeK1,n−1 of ordern�2. A subdivided staris a star where each edge

is subdivided exactly once. Acycle on n vertices is denoted byCn and apath on nvertices byPn. A daisywith k�2 petals is a connected graph that can be constructedfrom k�2 disjoint cycles by identifying a set ofk vertices, one from each cycle, intoone vertex. In particular, if thek cycles have lengthsn1, n2, . . . , nk, we denote the daisyby D(n1, n2, . . . , nk). Further, ifn1 = n2 = · · · = nk, then we writeD(n1, n2, . . . , nk)simply asDk(n1). The daisiesD(3,5), D(4,4) and D3(5) = D(5,5,5) are shownin Fig. 1.

Fig. 1. The daisiesD(3,5),D(4,4) andD3(5).

M.A. Henning, J.E. Maritz / Discrete Mathematics 301 (2005) 175–194 177

Fig. 2. The dumb-bellsDb(5,4,0) andDb(5,5,1).

For integersn1�n2�3 andk�0, we define adumb-bellDb(n1, n2, k) to be the graphof ordern= n1 + n2 + k obtained from the cyclesCn1 andCn2 by joining a vertex ofCn1to a vertex ofCn2 and subdividing the resulting edgek times. The dumb-bellsDb(5,4,0)andDb(5,5,1) are shown inFig. 2.

2. Known results

If F is aK2 rooted at a blue vertexv that is adjacent to a red vertex, then it is shown in[3] that�F (G)= �(G). Thus domination can be interpreted as restricted 2-stratifications or2-colorings, with the red vertices forming the dominating set. Clearly, thisF-coloring is theonly well-defined one for connected graphsF with order 2.LetF be a 2-stratifiedP3 rooted at a blue vertexv. The five possible choices for the graph

F are shown inFig. 3. (The red vertices inFig. 3are darkened.)The following result is established in[3].

Theorem 1 (Chartrand and Haynes[3] ). If G is a connected graph of order at least3,then fori ∈ {1,2,4,5}, the parameter�Fi (G) is given by the following table:

i 1 2 4 5

�Fi (G) = �t(G) �(G) �r(G) �2(G)

where�t(G) denotes the total domination number(see[8]), �r(G) denotes the restraineddomination number(see[6,8]), and�2(G) denotes the2-domination number(see[7,8]).

The parameter�F3(G) (seeFig. 3) appears to be new and is investigated further in[11].If T is a starK1,n−1 of ordern�3, then�F3(T ) = n since the central vertex ofTmust becolored red in anyF3-coloring ofT. If T is a tree of diameter at least 3, then it is shown in[11] that�F3(T )�2n/3 and the trees achieving equality are characterized.As pointed out in[3],F3-domination is not the sameas the distance domination parameter

calledk-step dominationintroduced in[12]. A setS ⊆ V is ak-step dominating setif forevery vertexu ∈ V − S, there exists a path of lengthk from u to some vertex inS.

Fig. 3. The five possible choices for 2-stratifiedP3.


Thek-step domination numberis the minimum cardinality of anyk-step dominating set ofG. The difference in 2-step domination andF3-domination is that inF3-domination everyblue vertex must have a blue–blue–red path (of length two) to some red vertex. Thus, everyF3-dominating set is a 2-step dominating set, but not every 2-step dominating set is anF3-dominating set. IfT is a starK1,n−1 of ordern�3, then�F3(T ) = n while the 2-stepdomination number ofT equals 2 (the set consisting of the central vertex and any leaf ofTis a 2-step dominating set ofT). A survey of results on distance domination in graphs can befound in[8, Section 7.4]. For a more comprehensive survey, the reader is referred to[10].

3. Main results

In this paper we continue the study of theF3-domination number of a graph. We havetwo immediate aims: firstly to establish an upper bound on theF3-domination number ofa connected graph with minimum degree at least two in terms of the order of the graphand to characterize those graphs achieving equality in this bound. Secondly, to characterizeconnectedgraphsof sufficiently largeorderwithmaximumpossibleF3-dominationnumber.We will refer to a graphG as anF3-minimal graphif G is edge-minimal with respect

to satisfying the following three conditions: (i)�(G)�2, (ii) G is connected, and (iii)�F3(G)�(n − 1)/2, wheren is the order ofG. To achieve our aims, we characterizeF3-minimal graphs. To do this, we define four families of graphs.Let A1(4) = Db(5,4,0) andA1(5) = Db(5,5,1) be the two dumb-bells shown in

Fig. 2. For k�2, let Ak(4) be the graph obtained from a daisyDk(5) by adding a 4-cycle and joining the central vertex of the daisy to a vertex of the added 4-cycle. The graphA2(4) is shown inFig. 4(a). Fork�2, letAk(5) be the graph obtained from a daisyDk(5)by adding a 5-cycle and then adding a new vertex and joining it to the central vertex ofthe daisy and to a vertex of the added 5-cycle. The graphA2(5) is shown inFig. 4(b). LetA = {Ak(4) | k�1} ∪ {Ak(5) | k�1}.LetB= {B1, B2, B3, B4, B5} whereB1,B2,B3,B4 andB5 are the five graphs shown in

Fig. 5. We call each graph in the familyB abad graph.Next we define a subfamilyC of cycles and a subfamilyD of daisies by

C = {C3, C4, C5, C7, C8, C11}and

D = {Dk(5) | k�2} ∪ {D(3,5),D(4,4)}.

(a) (b)

Fig. 4. The graphsA2(4) andA2(5) in the familyA.


Fig. 5. The five bad graphsB1, B2, B3, B4 andB5.

Fig. 6. The graphsH1 andH2.

The following result, a proof of which is given in Section 5, characterizesF3-minimalgraphs.

Theorem 2. A graph G is anF3-minimal graph if and only ifG ∈ A ∪ B ∪ C ∪ D.

LetH1 (resp.,H2) be the graph obtained fromC8 (resp.,C11) by adding an edge joiningtwo vertices at distance four apart on the cycle. The graphsH1 andH2 are shown inFig. 6.As a consequence of Theorem 2 we have our first main result, a proof of which is given

in Section 6.

Theorem 3. If G is a connected graph of order n with�(G)�2, then�F3(G)�(n− 1)/2unlessG ∈ {B2, C4, C8, H1}, in which case�F3(G) = n/2, or G = C5, in which case�F3(G)= (n+ 1)/2.

Our second main result provides a characterization of connected graphs with minimumdegree at least two and order at least nine that have maximum possibleF3-dominationnumber. A proof of Theorem 4 is given in Section 7.

Theorem 4. IfG is a connectedgraphof ordern�9with�(G)�2,then�F3(G)�(n−1)/2with equality if and only ifG ∈ A ∪ (D− {D(3,5),D(4,4)}) orG ∈ {B4, B5, C11, H2}.

4. Preliminary results

Our aim in this section is to establish some preliminary results that we will need laterwhen proving our main results. We begin with the following observation, a proof of whichis presented in Section 4.1.

Observation 5. Let G be a connected graph with�(G)�2 and let F be obtained from Gby subdividing any edge four times. Then, �F3(F )��F3(G)+ 2.

Nextweestablish the valueof�F3(Cn) for a cycleCn.Aproof ofProposition6 is presentedin Section 4.2.


Proposition 6. For n�3, �F3(Cn)= (n− 1)/3� + n/3� − �n/3�.Equivalently,Proposition6states that�F3(Cn)= n/3�+1 if n ≡ 2(mod 3)and�F3(Cn)= n/3� otherwise. For an example of a�F3-coloring of ann-cycleCn: v1, v2, . . ., vn, v1, let

R = {vi | i ≡ 1(mod 3)}, and so|R| = n/3�. If n ≡ 2(mod 3), then coloring the verticesof R ∪ {vn} red and coloring all other vertices blue produces anF3-coloring of Cn. Ifn /≡ 2(mod 3), then coloring the vertices ofR red and coloring all other vertices blueproduces anF3-coloring ofCn. As an immediate consequence of Proposition 6 we cancharacterize theF3-minimal graphs that are cycles.

Corollary 7. A cycle G is anF3-minimal graph if and only ifG ∈ C.

Next we characterize theF3-minimal graphs that are daisies. Proofs of Propositions 8and 9 are presented in Sections 4.3 and 4.4, respectively.

Proposition 8. If G is a daisy of order n with two petals, then�F3(G)�(n − 1)/2. Fur-thermore, �F3(G)= (n− 1)/2 if and only ifG=D(3,5) or G=D(4,4) or G=D(5,5).Proposition 9. If G is a daisy of order n, then�F3(G)�(n−1)/2.Furthermore, �F3(G)=(n− 1)/2 if and only ifG=D(3,5),G=D(4,4) or G=Dk(5) for somek�2.

As an immediate consequence of Proposition 9 we can characterize theF3-minimalgraphs that are daisies.

Corollary 10. A daisy G is anF3-minimal graph if and only ifG ∈ D.

Next we characterize theF3-minimal graphs that are dumb-bells. We begin with thefollowing result, a proof of which is presented in Section 4.5.

Proposition 11. If G is a dumb-bell of order n, then�F3(G)�(n − 1)/2 with equality ifand only ifG ∈ {A1(4), A1(5)}.Corollary 12. A dumb-bell G is anF3-minimal graph if and only ifG ∈ {A1(4), A1(5)}.The following two observations about graphs in the familiesA∪B∪ C∪D will prove

to be useful.

Observation 13. LetG ∈ A ∪ B ∪ C ∪ D have order n. Then, G is a connected graphwith �(G)= 2,and

�F3(G)=

n+ 1

2if G= C5,

n

2if G ∈ {B2, C4, C8},

n− 1

2otherwise.

Corollary 14. Each graph inA ∪ B ∪ C ∪ D is anF3-minimal graph.


Observation 15. LetG ∈ A∪B∪C∪D.Then for any vertexv of G, there is a minimumF3-coloring in whichv is colored blue and in which every blue vertex is adjacent to a redvertex. Further for any vertexv of G, except for the central vertex of a daisyDk(5), thereis a minimumF3-coloring of G in whichv is colored red.

We close our preliminary results with a characterization ofF3-minimal graphs of smallorder. A proof of Lemma 16 is presented in Section 4.6.

Lemma 16. If G is an F3-minimal graph of order n, 3�n�6, thenG ∈ {B1, B2, C3,C4, C5}.

4.1. Proof of Observation 5

Let uv be the edge ofG that is subdivided four times to obtain the graphF, and letu, u1, u2, u3, u4, v be the resulting path inF. Any minimum F3-coloring of G can beextended to anF3-coloring ofF as follows. If bothu andv are colored red, then coloru1andu4 red andu2 andu3 blue. If exactly one ofu andv, sayu, is colored red, then coloru3 andu4 red andu1 andu2 blue. Suppose bothu andv are colored blue. If each ofuandv has a neighbor colored red, then coloru2 andu3 red andu1 andu4 blue. If exactlyone ofu andv, sayu, has a neighbor colored red, then coloru2 red, coloru1, u3 andu4blue, and recolorv red (note that in anyF3-coloring of a graphH that colors a vertexwand all its neighbors blue, we can recolorw red and leave all other vertices unchangedto produce a newF3-coloring ofH). If neitheru nor v has a neighbor colored red, thencoloru1 andu4 red andu2 andu3 blue. In all cases, we produce anF3-coloring ofF thatcolors exactly two more vertices red than does the originalF3-coloring ofG. It follows that�F3(F )��F3(G)+ 2.

4.2. Proof of Proposition 6

We proceed by induction on the ordern of a cycleCn. The result is straightforwardto verify for n ∈ {3,4,5}. Assume then thatn�6 and that the result of the propositionis true for all cycles on fewer thann vertices. Consider a cycleC: v1, v2, . . . , vn, v1. LetC be a�F3-coloring ofC. Since every blue vertex is rooted at a copy ofF3, every bluevertex on the cycle is adjacent to a red vertex and a blue vertex. Renaming if necessary,we may assume thatC colorsv2 andv3 blue, and therefore colorsv1 andv4 red. LetC′be the cycle obtained fromC by deleting the verticesv1, v2 andv3 and adding the edgev4vn, i.e.,C′ = (C − {v1, v2, v3}) ∪ {v4vn}. ThenC′ is a cycle of ordern − 3�3 and therestriction ofC to C′ is anF3-coloring ofC′ that colors�F3(C) − 1 vertices red. Hence,�F3(C

′)��F3(C) − 1. On the other hand, any�F3-coloringC′ of C′ can be extended to a

�F3-coloring ofC by coloring exactly one ofv1, v2 andv3 red: IfC′ colorsvn andv4 blue,

then colorv2 red andv1 andv3 blue. IfC′ colorsvn red andv4 blue or ifC′ colorsvn and

v4 red, then colorv1 andv2 blue andv3 red. IfC′ colorsvn blue andv4 red, then colorv1

red andv2 andv3 blue. Thus,�F3(C)��F3(C′)+ 1. Consequently,�F3(C)= �F3(C

′)+ 1.SinceC�Cn andC′�Cn−3, the result now follows by applying the inductive hypothesisto the cycleC′.



LetG=D(n1+1, n2+1), and son=n1+n2+1. Letv denote the vertex of degree 4 inGand letF1 andF2 denote the two cycles passing throughv, whereFi�Cni+1 for i=1,2.Let F1 be the cyclev, v1, v2, . . . , vn1, v and letF2 be the cyclev, u1, u2, . . . , un2, v. Weconsider four possibilities.Case1. ni ≡ 2(mod 3) for i = 1 or i = 2.Wemay assumen1 ≡ 2(mod 3). LetR1={vi | i ≡ 0(mod 3)}, and so|R1|= (n1−2)/3.

Let C2 be a�F3-coloring ofF2 that colorsv red. By Proposition 6, ifn2 /≡ 1(mod 3),thenC2 colors at most(n2 + 3)/3 vertices red, while ifn2 ≡ 1(mod 3), thenC2 colors(n2+5)/3 vertices red.We can extendC2 to anF3-coloringC ofGby coloring the verticesin R1 red and all remaining uncolored vertices ofF1 blue. Ifn2 /≡ 1(mod 3), thenC colorsat most(n1 − 2)/3+ (n2 + 3)/3= n/3<(n− 1)/2 vertices red. Ifn2 ≡ 1(mod 3), thenn�7 andC colors(n1 − 2)/3+ (n2 + 5)/3= (n + 2)/3�(n − 1)/2 vertices red withstrict inequality ifn>7. Hence,�F3(G)< (n − 1)/2 unlessG = D(3,5), in which case�F3(G)= 3= (n− 1)/2.Case2. ni ≡ 0(mod 3) for i = 1,2.Then,n�7. LetR1= {vi | i ≡ 1(mod 3)}, and so|R1| = n1/3. LetC2 be a�F3-coloring

of F2 that colorsv red. By Proposition 6,C2 colors(n2 + 3)/3 vertices red.We can extendC2 to anF3-coloringC ofGby coloring the vertices inR1 red and all remaining uncoloredvertices ofF1 blue. Then,C colors(n1 + n2 + 3)/3= (n+ 2)/3�(n− 1)/2 vertices redwith strict inequality ifn>7. Hence,�F3(G)< (n − 1)/2 unlessG = D(4,4), in whichcase�F3(G)= 3= (n− 1)/2.Case3. n1 ≡ 0(mod 3) andn2 ≡ 1(mod 3).Then,n�8. LetR1 = {vi | i ≡ 2(mod 3)}, and so|R1| = n1/3. LetR2 = {ui | i ≡

1(mod 3)}, and so|R2| = (n2 + 2)/3. Then coloring the vertices inR1 ∪ R2 red and allremaining uncolored vertices blue produces anF3-coloring ofG that colors(n1 + n2 +2)/3= (n+ 1)/3<(n− 1)/2 vertices red.Case4. ni ≡ 1(mod 3) for i = 1,2.Then,n�9. LetR1 = {vi | i ≡ 1(mod 3)}, and so|R1| = (n1 + 2)/3. LetR2 = {ui |

i ≡ 2(mod 3)} ∪ {un2−1}, and so|R2| = (n2 + 2)/3. Then coloring the vertices inR1 ∪R2red and all remaining uncolored vertices blue produces anF3-coloring ofG that colors(n1+ n2+4)/3= (n+3)/3�(n−1)/2 vertices red with strict inequality ifn>9. Hence,�F3(G)< (n− 1)/2 unlessG=D(5,5), in which case�F3(G)= 4= (n− 1)/2.


We proceed by induction on the ordern�5 of a daisyG to show that�F3(G)�(n−1)/2.If n=5, thenG=D(3,3) and�F3(G)=1<(n−1)/2, while if n=6, thenG=D(3,4) and�F3(G)= 2<(n− 1)/2. This establishes the base cases. Assume, then, thatn�7 and thatif G′ is a daisy of ordern′<n, then�F3(G

′)�(n′ − 1)/2. LetG be a daisy of ordernwithk�2 petals. Ifk = 2, then the result follows from Proposition 8. Hence we may assumek�3. Letv denote the vertex of degree 2k in G, and letF1, F2, . . . , Fk denote thek cyclespassing throughv, whereFi�Cni+1 for i = 1,2, . . . , k. Thus,n= 1+ ∑k

i=1ni . LetF1 bethe cyclev, v1, v2, . . . , vn1, v.


LetG′ =D(n2, . . . , nk). Then,G′ is a daisy of ordern′ =n−n1. Applying the inductivehypothesis toG′, �F3(G

′)�(n′ − 1)/2= (n− n1 − 1)/2. LetC′ be a�F3-coloring ofG′.

Note that ifC′ colorsv blue, thenv must be adjacent to at least one vertex colored redunderC′. We extendC′ to anF3-coloring ofG as follows. Ifn1 ≡ 2(mod 3), letR = {vi |i ≡ 0(mod 3)}, and so|R| = (n1 − 2)/3. If n1 ≡ 0(mod 3) andv is colored blue inC′, letR={vi | i ≡ 2(mod 3)}, and so|R|=n1/3. In all other cases, letR={vi | i ≡ 1(mod 3)},and so|R| = n1/3�. Then,C′ can be extended to anF3-coloringC of G by coloring thevertices inR red and all remaining uncolored vertices ofF1 blue. Ifn1 ≡ 2(mod 3), thenCcolors at most|R| + (n′ − 1)/2= (n1− 2)/3+ (n− n1− 1)/2<(n− 1)/2 vertices red. Ifn1 ≡ 0(mod 3), thenC colors atmostn1/3+(n−n1−1)/2<(n−1)/2 vertices red. Ifn1 ≡1(mod 3), thenC colors at most(n1+2)/3+ (n−n1−1)/2= (3n−n1+1)/6�(n−1)/2vertices red with strict inequality ifn1>4. Hence in all cases,C colors strictly less than(n−1)/2 vertices red unlessn1=4 (and so,F1=C5) and�F3(G′)=(n′ −1)/2.An identicalargument shows that ifni �= 4 for somei, 1� i�k, then there is anF3-coloring ofG thatcolors strictly less than(n−1)/2 vertices red. Thus we have shown that�F3(G)< (n−1)/2unlessG=Dk(5), in which case�F3(G)�(n− 1)/2.Weshownext that�F3(G)=(n−1)/2 if andonly ifG=D(3,5),G=D(4,4)orG=Dk(5)

for somek�2. By Proposition 8, the result is proven ifG is a daisy with two petals. Hencewe may assumeG has at least three petals. If�F3(G)= (n− 1)/2, then we have shown thatG=Dk(5). Conversely, supposeG=Dk(5). ThenGhasordern=4k+1andanyF3-coloringofGcolors at least two vertices from eachFi , 1� i�k, red, and so�F3(G)�2k=(n−1)/2.On the other hand, ifv denotes the vertex of degree 2k inGand ifv, v1, v2, v3, v4, v denotesa 5-cycle inG, then coloring all vertices in(N [v]−{v1, v4})∪{v2, v3}blue andall remainingvertices red, produces anF3-coloring ofG that colors exactly 2k = (n− 1)/2 vertices red,and so�F3(G)�(n− 1)/2. Consequently,�F3(G)= (n− 1)/2.


We proceed by induction on the ordern�6. If n=6, thenG=Db(3,3,0) and�F3(G)=2= (n − 2)/2. Letn�7, and assume that the result is true for all dumb-bells of ordern′,wheren′<n. LetG=Db(n1, n2, k) be a dumb-bell of ordern= n1 + n2 + k. SupposeGcontains a path on six vertices each internal vertex of which has degree 2 inG and whoseend-vertices, sayuandv, are not adjacent. LetG′ be the graph obtained fromGby removingthe four internal vertices of this path and adding the edgeuv. Then,G′ is a dumb-bell ofordern′=n−4. ByObservation 5,�F3(G)��F3(G

′)+2.Applying the inductive hypothesistoG′, �F3(G

′)�(n′ −1)/2. If �F3(G′)< (n′ −1)/2, then�F3(G)< (n−1)/2. On the other

hand, if �F3(G′) = (n′ − 1)/2, then by the inductive hypothesis,G′ ∈ {A1(4), A1(5)}.

Now G is obtained fromG′ by subdividing the edgeuv of G′ four times. Irrespectiveof whether the edgeuv is a cycle edge or a bridge ofG′, it is straightforward to checkthat �F3(G)�(n − 3)/2. Hence we may assume thatG contains no path on six verticeseach internal vertex of which has degree 2 inG and whose end-vertices are not adjacent,for otherwise�F3(G)< (n − 1)/2. With this assumption, 3�n1, n2�6 and 0�k�3. Itis now a simple exercise to check that�F3(G)�(n − 1)/2 with equality if and only if(n1, n2, k) ∈ {(5,4,0), (5,5,1)}.


4.6. Proof of Lemma 16

LetG= (V ,E). Letu be a vertex of maximum degree inG. If n ∈ {3,4}, thenG= Cn.Supposen=5. If �(G)=4, then coloringu red and coloring all other vertices blue producesanF3-coloring ofG, and so�F3(G)= 1<(n− 1)/2, a contradiction. If�(G)= 3, then itfollows from Observation 13 thatG= B1. If �(G)= 2, thenG= C5.Supposen= 6. If �(G)= 2, then�F3(G)�2<(n− 1)/2, a contradiction. If�(G)= 4,

letV −N [u]= {v}. Then,uandv have at least two common neighbors. Coloringv and anyneighbor ofu red and coloring all other vertices blue produces anF3-coloring ofG, andso�F3(G)�2<(n− 1)/2, a contradiction. If�(G)= 5, then coloringu red and coloringall other vertices blue produces anF3-coloring ofG, and so�F3(G) = 1<(n − 1)/2, acontradiction. Hence,�(G) = 3. Let V − N [u] = {v,w}. If v andw have a commonneighborx, let y ∈ N(u) − {x}. Coloringv andy red and coloring all other vertices blueproduces anF3-coloring ofG, and so�F3(G)�2<(n − 1)/2, a contradiction. Hence,vandw have no common neighbor, whenceG= B2.

5. Proof of Theorem 2

The sufficiency follows from Corollary 14. To prove the necessary, we proceed by in-duction on the ordern�3 of anF3-minimal graph. By Lemma 16, the result is true forn�6. Letn�7, and assume that the result is true for allF3-minimal graphs of order lessthann. Let G = (V ,E) be anF3-minimal graph of ordern. If e is an edge ofG, then�F3(G− e)��F3(G). Hence, by the minimality ofG, we have the following observation.

Observation 17. If e ∈ E, then either e is a bridge of G or�(G− e)= 1.

Since theF3-domination number of a graph cannot decrease if edges are removed, thenext result is a consequence of the inductive hypothesis.

Observation 18. If G′ is a connected subgraph of G of ordern′<n with �(G′)�2, theneitherG′ ∈ A ∪ B ∪ C ∪ D or �F3(G

′)< (n′ − 1)/2.

The following observation will prove useful.

Observation 19. LetG′ be a graph and letv be a vertex ofG′ all of whose neighbors havedegree at most2 inG′.Then in anyF3-coloring ofG′, at least one vertex inN [v] is coloredred.

Proof. If every vertex inN [v] is colored blue in someF3-coloring ofG′, then there mustbe a red vertex at distance 2 fromv. But then the neighbor ofv that is adjacent to such ared vertex is not rooted at a copy ofF3, a contradiction. �

We now return to the proof of Theorem 2. IfG = Cn, then, by Corollary 7,G ∈ C. IfG is a daisy, then by Corollary 10,G ∈ D. If G is a dumb-bell, then, by Corollary 12,G ∈ {A1(4), A1(5)}. So we may assume thatG is neither a cycle, nor a daisy, nor a dumb-


bell. Hence,Gcontains at least two vertices of degree at least 3. LetS={v ∈ V |degv�3}.Then,|S|�2 and each vertex ofV − S has degree 2 inG.For eachv ∈ S, we define the 2-graph ofv to be the component ofG − (S − {v}) that

containsv. So each vertex of the 2-graph ofv has degree 2 inG, except forv. Furthermore,the 2-graph ofv consists of edge-disjoint cycles throughv, which we call 2-graph cycles,and paths emanating fromv, which we call 2-graph paths.Using the inductive hypothesis, we shall prove the following lemma, a proof of which is

given in Section 5.1.

Lemma 20. If S is not an independent set, thenG= Ak(4) for somek�2.

By Lemma 20, we may assume thatS is an independent set, for otherwiseG ∈ A. Letu andv be two vertices ofSthat are joined by a pathu, u1, . . . , um, v every internal vertexof which has degree 2 inG. By assumption,d(u, v)�2, whencem�1. If m is large, thenthe following result, a proof of which is presented in Section 5.2, shows thatG= B4.

Lemma 21. If m�4, thenG= B4.

By Lemma21, wemay assume thatevery2-graph path has length atmost3. In particular,m�3. LetPdenote thepathu1, . . . , um. LetH=G−V (P ). Then,Hhasordern′=n−mand�(H)�2. Possibly,H is disconnected in which caseH has two components, one containingu and the otherv. SinceS is an independent set, we observe that each neighbor of a vertexof Shas degree 2 inG. In particular each neighbor ofu andv in H has degree 2. Thus anyF3-coloring ofH that colorsu (resp.,v) blue must color at least one neighbor ofu (resp.,v) red. A proof of the following lemma is given in Section 5.3.

Lemma 22. If H is disconnected, thenG= Ak(5) for somek�2.

By Lemma 22, we may assume that removing the vertices inV − S from any 2-graphpath inG produces a connected graph, for otherwiseG ∈ A. In particular,H is connected.In what follows, for each vertexu ∈ S, letGu = G − N [u]. A proof of Lemma 23 is

given in Sections 5.4, 5.5 and 5.6.

Lemma 23. If every2-graph path has length exactly1, then

(a) There is no2-graph cycle in G.(b) If u, v ∈ S, thenN(u)�N(v).(c) �(Gu)= 1 for everyu ∈ S.

A proof of Lemma 24 is given in Section 5.7.

Lemma 24. At least one2-graph path has length2 or 3.

By Lemma 24, wemay assume thatm ∈ {2,3}. By Observation 18,H ∈ A∪B∪C∪Dor �F3(H)< (n

′ − 1)/2. LetCH be a minimumF3-coloring ofH. A proof of the followingtwo lemmas are given in Sections 5.8 and 5.9.


Lemma 25. If m= 3, thenG= B5.

Lemma 26. If m= 2, thenG= B3.

This completes the proof of Theorem 2.


Let e = uv be an edge, whereu, v ∈ S. By Observation 17,emust be a bridge ofG.LetG1 = (V1, E1) andG2 = (V2, E2) be the two components ofG− e whereu ∈ V1. Fori = 1,2, let |Vi | = ni . EachGi satisfies�(Gi)�2 and is connected. Hence by Observation18,Gi ∈ A ∪ B ∪ C ∪ D or �F3(Gi)< (ni − 1)/2 for i = 1,2. If �F3(Gi)< (ni − 1)/2for i = 1,2, then�F3(G)��F3(G1)+ �F3(G2)< (n− 1)/2, a contradiction. Hence at leastone ofG1 andG2, sayG1, must belong toA ∪ B ∪ C ∪ D.

Claim 1. G1 �= C5.

Proof. SupposeG1 = C5. By assumption,G is not a dumb-bell, and soG2 is not a cycle.Thus if�F3(G2)�n2/2, then, by Observation 13,G2=B2. But thenG is not anF3-minimalgraph (either we contradict Observation 17 or�F3(G)< (n− 1)/2), a contradiction. Thus,�F3(G2)�(n2 − 1)/2.Suppose�F3(G2)=(n2−1)/2 (andstillG2 isnot acycle).Then,G2 ∈ (A∪B∪D)−{B2}.

SupposeG2=Dk(5) for somek�2 andv is the central vertex ofG2. Then a minimumF3-coloring ofG2 can be extended to anF3-coloring ofGby coloring the two neighbors ofu inG1 redand the three remainingverticesofG1 blue.Thus,�F3(G)�(n2−1)/2+2=(n−2)/2,a contradiction. Ifv is not the central vertex of a daisyDk(5), then by Observation 15, thereis a minimumF3-coloring ofG2 in whichv is colored red. Such anF3-coloring ofG2 canbe extended to anF3-coloring ofG by coloringu and its two neighbors inG1 blue andcoloring the remaining two vertices ofG1 red. Thus,�F3(G)�(n2−1)/2+2= (n−2)/2,a contradiction. Hence,�F3(G2)�(n2 − 2)/2.If �F3(G2)�(n2−3)/2, then�F3(G)��F3(G1)+�F3(G2)�3+(n2−3)/2=(n−2)/2,

a contradiction. Hence,�F3(G2) = (n2 − 2)/2. If there exists a minimumF3-coloring ofG2 in which v or a neighbor ofv is colored red, then such anF3-coloring ofG2 canbe extended to anF3-coloring ofG by coloring exactly two vertices ofG1 red, and so�F3(G)�(n2 − 2)/2 + 2 = (n − 3)/2, a contradiction. On the other hand, suppose thatevery minimumF3-coloring ofG2 colorsv and all its neighbors blue. Then at least oneneighborw of v in G2 must have degree at least 3, and sow ∈ S. By Observation 17, theedgevw must be a bridge ofG. LetH1 andH2 be the two components ofG − vw wherev ∈ V (H1). Fori=1,2, letHi have ordern′

i . EachHi satisfies�(Hi)�2 and is connected.Hence by Observation 18,Hi ∈ A ∪ B ∪ C ∪ D or �F3(Hi)< (n

′i − 1)/2 for i = 1,2.

SinceH1 /∈ {B2, C4, C5, C8}, �F3(H1)�(n′1 − 1)/2. If H2 ∈ {B2, C4, C5, C8}, then there

would exists aminimumF3-coloring ofG2 in whichw is colored red, contrary to our earlierassumption that there is no suchminimumF3-coloring ofG2. Hence,�F3(H2)�(n′

2−1)/2.Thus,�F3(G)��F3(H1)+ �F3(H2)< (n− 1)/2, a contradiction. �


By Claim 1,G1 �= C5. Similarly,G2 �= C5.

Claim 2. G1 /∈ {B2, C8}.

Proof. SupposeG1 ∈ {B2, C8}. If G2 ∈ {B2, C4, C8}, thenG is not anF3-minimalgraph (either we contradict Observation 17 or�F3(G)< (n − 1)/2), a contradiction. If�F3(G2)�(n2 − 2)/2, then�F3(G)�n1/2 + (n2 − 2)/2<(n − 1)/2, a contradiction.Hence,�F3(G2) = (n2 − 1)/2, and soG2 ∈ (A ∪ B ∪ C ∪ D) − {B2, C4, C5, C8}. ByObservation 15, there is a minimumF3-coloringC1 of G2 in which v is colored blue andis adjacent to a vertex colored red.SupposeG1 = B2. If u is a vertex of degree 2 inG1, then at least one of the two

edges incident withu in G1 joins two vertices ofSbut is not a bridge ofG, contradictingObservation17.Hence the vertexumust bea vertex of degree3 inG1.TheF3-coloringC1 ofG2 can be extended to anF3-coloring ofGas follows: color one neighbor ofuon the 4-cycleinG1 red, color the neighbor ofu inG1 that does not belong to the 4-cycle red, and color theremaining four vertices ofG1 blue.Thus,�F3(G)��F3(G2)+2=(n2−1)/2+2=(n−3)/2,a contradiction.SupposeG1 = C8. TheF3-coloringC1 of G2 can be extended to anF3-coloring ofG

by coloring the two neighbors ofu in G1 red, coloring the vertex at maximum distance 4from u in G1 red, and coloring the remaining five vertices ofG1 (includingu) blue. Thus,�F3(G)�(n2 − 1)/2+ 3= (n− 3)/2, a contradiction. �

By Claim 2,G1 /∈ {B2, C8}. Similarly,G2 /∈ {B2, C8}. If neitherG1 norG2 is a 4-cycle,then fori = 1,2, �F3(Gi)�(ni − 1)/2, and so�F3(G)��F3(G1)+ �F3(G2)< (n− 1)/2,a contradiction. Hence at least one ofG1 andG2, sayG1, is a 4-cycle.If G2=C4, then�F3(G)=(n−2)/2, contradicting the fact thatG is anF3-minimal graph.

If �F3(G2)�(n2 − 2)/2, then�F3(G)�n1/2+ (n2 − 2)/2<(n − 1)/2, a contradiction.Hence,�F3(G2) = (n2 − 1)/2, and soG2 ∈ (A ∪ B ∪ C ∪ D) − {B2, C4, C5, C8}. Ifv is not the central vertex of a daisyDk(5), then by Observation 15, there is a minimumF3-coloring ofG2 in whichv is colored red. Such anF3-coloring ofG2 can be extended toanF3-coloring ofGby coloring the vertex inG1 at distance 2 fromuwith the color red andcoloring the remaining three vertices ofG1 blue. Thus,�F3(G)�(n2−1)/2+1=(n−3)/2,a contradiction. Thus,v must be the central vertex of a daisyDk(5) for somek�2, whenceG= Ak(4).


LetG′ be the graph obtained fromG by removing the verticesu1, u2, u3, u4, and eitheradding the edgeuu5 if m�5 or adding the edgeuv if m = 4. Then,G′ is a connectedgraph of ordern′ = n − 4 with �(G′)�2. By Observation 5,�F3(G)��F3(G

′) + 2. If�F3(G

′)< (n′−1)/2, then�F3(G)< (n−1)/2,acontradiction.Hence,�F3(G′)�(n′−1)/2.

SinceG is anF3-minimal graph, it follows thatG′ is anF3-minimal graph. SinceG is neithera cycle nor a dumb-bell,G′ is not a cycle or a dumb-bell. Further the degree of each vertexofSis unchanged inGandG′, and soG′ has at least two vertices of degree at least 3. Henceapplying the inductive hypothesis toG′, G′ ∈ A ∪ {B1, B2, . . . , B5}. A straightforward


check confirms that ifG′ �= B1, then�F3(G)< (n − 1)/2. Therefore,G′ = B1, whenceG= B4.


Let H1 andH2 be the two components ofH, whereu ∈ V (H1). For i = 1,2, letHihave orderni . EachHi is a connected graph with�(Hi)�2. By Observation 18,Hi ∈A ∪ B ∪ C ∪ D or �F3(Hi)< (ni − 1)/2.

Claim 3. �F3(Hi)�ni/2 for i = 1 or i = 2.

Proof. Suppose�F3(Hi)�(ni − 1)/2 for i = 1,2. LetC12 be a minimumF3-coloring ofH1∪H2. Then the restriction ofC12 toV (Hi) is a minimumF3-coloring ofHi for i=1,2,and soC12 colors at most(n1 + n2)/2− 1 vertices ofH red.Supposem = 3. Then,n1 + n2 = n − 3. If at least one ofu andv, sayu, is colored red

in C12, thenC12 can be extended to anF3-coloring ofG by coloring the vertexu3 red andthe verticesu1 andu2 blue. On the other hand, if bothu andv are colored blue inC12, thenC12 can be extended to anF3-coloring ofG by coloring the vertexu2 red and the verticesu1 andu3 blue. Hence,�F3(G)�(n− 3)/2, a contradiction.Supposem = 2. Then,n1 + n2 = n − 2. If u andv are colored with the same color in

C12, thenC12 can be extended to anF3-coloring ofG by coloring bothu1 andu2 blue,whence�F3(G)�(n− 4)/2, a contradiction. Ifu andv are colored with different colors inC12, sayu is colored red andv blue, thenC12 can be extended to anF3-coloring ofG bycoloringu1 red andu2 blue, and so�F3(G)�(n− 2)/2, a contradiction. Hence,m= 1 andn1 + n2 = n− 1.If uorv is coloredblue inC12, thenC12canbeextended toanF3-coloringofGbycoloring

u1 blue, whence�F3(G)�(n−3)/2, a contradiction. Hence in every minimumF3-coloringof H, the verticesu andv are colored red. There is therefore no minimumF3-coloring ofH1 that colorsu blue, and so it follows from Observation 15 that�F3(H1)�(n1 − 2)/2.Similarly, �F3(H2)�(n2 − 2)/2. Thus,C12 colors at most(n1 + n2 − 4)/2= (n − 5)/2vertices ofH red. The coloringC12 can be extended to anF3-coloring ofG by coloringu1red, and so�F3(G)�(n− 3)/2, a contradiction. �

ByClaim3andObservation 13,wemayassume thatH1 ∈ {B2, C4, C5, C8}.We considereach possibility in turn.

Claim 4. If H1 = C5, thenG= Ak(5) for somek�2.

Proof. SinceG is not a dumb-bell,H2 is not a cycle. IfH2=B2, thenv must be one of thetwo vertices of degree 3 inB2 and it is easy to check that for each value ofm ∈ {1,2,3},�F3(G)< (n−1)/2, a contradiction. Hence, by Observation 18,H2 ∈ A∪ (B−{B2})∪Dor �F3(H2)< (n2 − 1)/2. In particular,�F3(H2)�(n2 − 1)/2. LetC2 be a minimumF3-coloring ofH2.Supposem = 3. If v is colored red in the coloringC2, thenC2 can be extended to an

F3-coloring ofG by coloringu1 and the two vertices inH1 not adjacent touwith the color


red and coloring all remaining uncolored vertices ofG blue. On the other hand, ifv iscolored blue inC2, thenC2 can be extended to anF3-coloring ofG by coloringu2 and thetwo neighbors ofu in H1 with the color red and coloring all remaining uncolored verticesofGblue. In both cases we color at most(n−3)/2 vertices red, and so�F3(G)�(n−3)/2,a contradiction.Supposem= 2. If v is colored red inC2, thenC2 can be extended to anF3-coloring of

G by coloringu and its two neighbors inH1 with the color red and coloring all remaininguncolored vertices ofG blue. On the other hand, ifv is colored blue inC2, thenC2 can beextended to anF3-coloring ofG by coloringu1 and the two vertices inH1 not adjacent touwith the color red and coloring all remaining uncolored vertices ofG blue. In both caseswe color at most(n−2)/2 vertices red, and so�F3(G)�(n−2)/2, a contradiction. Hence,m= 1.If v is colored red inC2, thenC2 can be extended to anF3-coloring ofG by coloring the

two neighbors ofu in H1 with the color red and coloring all remaining uncolored verticesof G blue, whence�F3(G)�(n − 3)/2, a contradiction. Hence, by Observation 15, eitherH2 =Dk(5) for somek�2 with v the central vertex of this daisy, or�F3(H2)�(n2 − 2)/2andv is colored blue inC2. In the latter case,C2 can be extended to anF3-coloring ofG by coloringu and its two neighbors inH1 with the color red and coloring all remain-ing uncolored vertices ofG blue, whence�F3(G)�(n − 2)/2, a contradiction. Hence,H2 = Dk(5) for somek�2 with v the central vertex of this daisy. Thus,G = Ak(5) forsomek�2. �

ByClaim 4, wemay assume that neitherH1 norH2 is a 5-cycle, for otherwise the desiredresult follows. Hence,�F3(H2)�n2/2.

Claim 5. H1 �= B2.

Proof. SupposeH1=B2. LetC2 be a minimumF3-coloring ofH2. Supposem= 3. If v iscolored red in the coloringC2, thenC2 can be extended to anF3-coloring ofG by coloringu1 and the two vertices inH1 not adjacent touwith the color red and coloring all remaininguncolored vertices ofG blue. On the other hand, ifv is colored blue inC2, thenC2 can beextended to anF3-coloring ofG by coloringu2 and two neighbors ofu in H1 that lie ona common 5-cycle with the color red and coloring all remaining uncolored vertices ofGblue. In both cases we color at most(n − 3)/2 vertices red, and so�F3(G)�(n − 3)/2, acontradiction.Supposem = 2. If v is colored red inC2, thenC2 can be extended to anF3-coloring

of G by coloringu and two neighbors ofu in H1 that lie on a common 5-cycle with thecolor red and coloring all remaining uncolored vertices ofG blue. On the other hand, ifv is colored blue inC2, thenC2 can be extended to anF3-coloring ofG by coloringu1and the two vertices inH1 not adjacent tou with the color red and coloring all remaininguncolored vertices ofG blue. In both cases we color at most(n− 2)/2 vertices red, and so�F3(G)�(n− 2)/2, a contradiction.Supposem = 1. If v is colored red inC2, thenC2 can be extended to anF3-coloring

of G by coloring two neighbors ofu in H1 that lie on a common 5-cycle with the colorred and coloring all remaining uncolored vertices ofG blue, and so�F3(G)�(n − 3)/2,


a contradiction. Hencewemayassume that everyminimumF3-coloring ofH2 colorsv blue,for otherwise we reach a contradiction. Thus by Observations 13 and 15,�F3(H2)�(n2 −1)/2. Sincev is colored blue inC2, the coloringC2 can be extended to anF3-coloringof G by coloring u1 and the two vertices inH1 not adjacent tou with the color redand coloring all remaining uncolored vertices ofG blue, whence�F3(G)�(n − 2)/2,a contradiction. �

By Claim 5,H1 �= B2. Similarly,H2 �= B2.

Claim 6. H1 �= C4.

Proof. SupposeH1 = C4. LetH1 be the 4-cycleu,w, x, y, u. SinceG is not a dumb-bell,H2 is not a cycle. Hence,�F3(H2)�(n2 − 1)/2. LetC2 be a minimumF3-coloring ofH2.Supposem= 3. If v is colored red inC2, thenC2 can be extended to anF3-coloring of

G by coloringu1 andx red and coloring all remaining uncolored vertices ofG blue. Onthe other hand, ifv is colored blue inC2, thenC2 can be extended to anF3-coloring ofGby coloring the verticesu2, x andy red, and coloring all remaining uncolored vertices ofGblue. Hence,�F3(G)�(n− 2)/2, a contradiction.Supposem= 2. If v is colored red inC2, thenC2 can be extended to anF3-coloring of

G by coloringu andw red and coloring all remaining uncolored vertices ofG blue. On theother hand, ifv is colored blue inC2, thenC2 can be extended to anF3-coloring ofG bycoloring the verticesu1 andx red, and coloring all remaining uncolored vertices ofG blue.Hence,�F3(G)�(n− 3)/2, a contradiction.Supposem= 1. If v is colored red inC2, thenC2 can be extended to anF3-coloring of

G by coloringx andy red and coloring all remaining uncolored vertices ofG blue. On theother hand, ifv is colored blue inC2, thenC2 can be extended to anF3-coloring ofG bycoloring the verticesu andw red, and coloring all remaining uncolored vertices ofG blue.Hence,�F3(G)�(n− 2)/2, a contradiction. �

We now return to the proof of Lemma 22. By Claim 6,H1 �= C4. Similarly,H2 �= C4.Hence,H1=C8. LetH1 be the 8-cycleu=w1, w2, . . . , w8, u. SinceG is not a dumb-bell,H2 is not a cycle. Hence,�F3(H2)�(n2 − 1)/2. LetC2 be a minimumF3-coloring ofH2.Supposem=3. If v is colored red inC2, thenC2 can be extended to anF3-coloring ofG

by coloring the vertices in the set{u1, w3, w4, w7} red and coloring all remaining uncoloredvertices ofG blue. On the other hand, ifv is colored blue inC2, thenC2 can be extendedto anF3-coloring ofGby coloring the vertices in the set{u2, w2, w5, w8} red, and coloringall remaining uncolored vertices ofG blue. Hence,�F3(G)�(n− 4)/2, a contradiction.Supposem=2. If v is colored red inC2, thenC2 can be extended to anF3-coloring ofG

by coloring the vertices in the set{w1, w2, w5, w8} red and coloring all remaining uncoloredvertices ofG blue. On the other hand, ifv is colored blue inC2, thenC2 can be extendedto anF3-coloring ofGby coloring the vertices in the set{u1, w3, w4, w7} red, and coloringall remaining uncolored vertices ofG blue. Hence,�F3(G)�(n− 3)/2, a contradiction.Supposem= 1. If v is colored red inC2, thenC2 can be extended to anF3-coloring of

G by coloring the vertices in the set{w2, w5, w8} red and coloring all remaining uncoloredvertices ofGblue. On the other hand, ifv is colored blue inC2, thenC2 can be extended to


anF3-coloring ofG by coloring the vertices in the set{w1, w2, w5, w8} red, and coloringall remaining uncolored vertices ofG blue. Hence,�F3(G)�(n − 2)/2, a contradiction.This completes the proof of Lemma 22.

5.4. Proof of Lemma 23(a)

Suppose that there is a 2-graph cycle inG. Since|S|�2, each vertex ofS that has a2-graph cycle also has a 2-graph path. Hence we may assume that the vertexu (definedearlier) has a 2-graph cycleCu of ordern1 + 1. LetHu =G − (V (Cu) − {u}) have ordern2. Then,Hu is a connected graph of ordern2 = n − n1. If degG(u) = 3, then the graphH =G−u1 (defined earlier) is disconnected, contrary to assumption. Hence, degG(u)�4,and so�(Hu)�2. By Observation 18,Hu ∈ A ∪ B ∪ C ∪ D or �F3(Hu)< (n2 − 1)/2.Sincev is a vertex of degree at least 3 inHu, the graphHu is not a cycle. Further by

our earlier assumptions (that every 2-graph path has length exactly 1; that the setS is anindependent set with|S|�2; that removing the vertices not inSof any 2-graph path fromGproduces a connected graph), it follows thatHu /∈A∪ (B−{B1})∪C∪ (D−D(4,4)}.Hence by Observation 13, eitherHu ∈ {B1,D(4,4)} or �F3(Hu)�(n2 − 2)/2.Let C∗ be a minimumF3-coloring ofHu. If n1 �= 4 (i.e., if Cu is not a 5-cycle), then

irrespective of whetheru is colored is red or blue inC∗, the coloringC∗ can be extendedto anF3-coloring ofG by coloring at most(n1 − 1)/2 additional vertices inCu red, andso �F3(G)�(n − 2)/2, a contradiction. Ifn1 = 4, thenC∗ can be extended to anF3-coloring ofG by coloringn1/2 additional vertices red. Thus, if�F3(Hu)�(n2− 2)/2, then�F3(G)�(n−2)/2, a contradiction. Hence,n1=4 andHu ∈ {B1,D(4,4)}. But once again,�F3(G)�(n− 3)/2, a contradiction.

5.5. Proof of Lemma 23(b)

By Lemma 23(a),G is a bipartite graph with partite setsSandV − S. Every vertex inV − S has degree exactly 2, while every vertex inShas degree at least 3.Suppose thatN(u) ⊆ N(v) for some pair of verticesu, v ∈ S. If |S|=2 (and stilln�7),

then coloringuand one neighbor ofu red and coloring all remaining vertices blue producesanF3-coloring ofG, and so�F3(G) = 2�(n − 3)/2, a contradiction. Hence,|S|�3, andso at least one neighbor ofv is not a neighbor ofu.Suppose degG(v)=degG(u)+1. LetG′ =G−N [v]−u have ordern′. Then,n′ �n−6

andG′ is an induced subgraph ofGwith �(G′)�2. SinceG′ is a bipartite graph,G′ has no5-cycles, and so, by the inductive hypothesis,�F3(G

′)�n′/2�(n−6)/2.AnyminimumF3-coloringofG′ canbeextended toanF3-coloringofGby coloringuandoneneighbor ofuredand coloring all remaining uncolored vertices blue. Thus,�F3(G)�2+�F3(G

′)�(n−2)/2,a contradiction.On the other hand, suppose degG(v)�degG(u)+ 2. LetG∗ =G−N [u] have ordern∗.

Then,n∗ �n−4 andG∗ is an induced subgraph ofGwith �(G∗)�2. SinceG∗ is a bipartitegraph,G∗ has no 5-cycles, and so, by the inductive hypothesis,�F3(G

∗)�n∗/2�(n−4)/2.Any minimumF3-coloring ofG∗ that colorsv red can be extended to anF3-coloring ofG by coloring one neighbor ofu red (and coloring all remaining uncolored vertices blue),while any minimumF3-coloring ofG∗ that colorsv blue can be extended to anF3-coloring


ofGby coloring the vertexu red. Hence,�F3(G)�1+�F3(G∗)�(n−2)/2, a contradiction.

We deduce, therefore, that for any pair of verticesu, v ∈ S,N(u)�N(v).

5.6. Proof of Lemma 23(c)

Suppose�(Gu)�2 for some vertexu ∈ S. Then,Gu is an induced subgraph (possibledisconnected) ofG. SinceGu is a bipartite graph,Gu is C5-free. Hence by the inductivehypothesis, each component ofGu hasF3-domination number at most one-half its order,and so�F3(Gu)� |V (Gu)|/2�(n− 4)/2.Let Su = {w ∈ S | dG(u,w) = 2}. By Lemma 23(b),|Su|�2. LetCu be a minimum

F3-coloring ofGu. SupposeCu colors a vertexx in Su red. Letx′ be a common neighborof u andx. Then,Cu can be extended to anF3-coloring ofG by coloring one vertex inN(u) − {x′} red (and coloring all other vertices inN [u] blue). On the other hand, ifCucolors no vertex inSu red, then it follows by Observation 19 thatCu can be extended to anF3-coloring ofGby coloring the vertexu red (and coloring all remaining uncolored verticesblue). Hence,�F3(G)�1+ �F3(Gu)�(n− 2)/2, a contradiction.


Suppose that every 2-graph path has length exactly 1. Letu ∈ S. By Lemma23,�(Gu)=1andG is a bipartite graphwith partite setsSandV −S.Wemay assume thatv has degree 1 inGu. Thus,uandv have at least two common neighbors. Letcbe one such common neighborof u andv. Letv′ be the neighbor ofv inGu, and letN(v′)= {v,w}. Then,w ∈ S − {u, v}andw has degree at least 2 inGv. Since�(Gv)=1 by Lemma 23(c), the vertexumust havedegree 1 inGv. Letu′ be the neighbor ofu inGv, and letN(u′)={u, z}. Then,z ∈ S−{u, v}andzhas degree at least 2 inGv (possibly,z=w). LetG′ =G−N [u] −N [v] have ordern′. Then,n′ �n− 6 andG′ is an induced subgraph ofG. In particular,G′ has no 5-cycles.LetC′ be a minimumF3-coloring ofG′. We now consider two possibilities.Supposez �= w. Then,�(G′)�2. It follows from the inductive hypothesis that�F3(G

′)�n′/2�(n− 6)/2. If C′ colors bothw andz red, thenC′ can be extended to anF3-coloringof G by coloring the vertexc red and coloring all remaining uncolored vertices blue. IfC′colors bothw andzblue, then it follows from Observation 19 thatC′ can be extended to anF3-coloring ofG by coloring the verticesc andu red and coloring all remaining uncoloredvertices blue. Suppose, finally, thatC′ colors exactly one ofw andz red. By symmetry,we may assume thatw is colored red. ThenC′ can be extended to anF3-coloring ofG bycoloring the verticesc andu red and coloring all remaining uncolored vertices blue. Thus,�F3(G)�2+ �F3(G

′)�(n− 2)/2.Suppose, on the other hand, thatz = w. If degG(w) = 3, then�(Gw)�2, which con-

tradicts Lemma 23(c). Hence, degG(w)�4. Then,�(G′)�2. It follows from the inductivehypothesis that�F3(G

′)�n′/2�(n−6)/2. IfC′ colorsw red, thenC′ can be extended to anF3-coloring ofGby coloring the vertexc red and coloring all remaining uncolored verticesblue. IfC′ colorsw blue, then it follows from Observation 19 thatC′ can be extended to anF3-coloring ofG by coloring the verticesc andu red and coloring all remaining uncoloredvertices blue. Thus,�F3(G)�2+ �F3(G

′)�(n− 2)/2.



Suppose�F3(H)�(n′ −1)/2= (n−4)/2. If at least one ofuandv, sayu, is colored redin CH , thenCH can be extended to anF3-coloring ofG by coloring the vertexu3 red andthe verticesu1 andu2 blue. On the other hand, if bothu andv are colored blue inCH , thenCH can be extended to anF3-coloring ofG by coloring the vertexu2 red and the verticesu1 andu3 blue. Hence,�F3(G)�(n− 2)/2, a contradiction. Thus,�F3(H)�n′/2, whenceH ∈ {B2, C4, C5, C8}. If H ∈ {B2, C4, C5}, then it is easily checked that�F3(G)< (n −1)/2, a contradiction. IfH = C8 and if u andv are at distance 2 or 3 apart inH, then�F3(G)< (n− 1)/2, a contradiction. Thus,H =C8 and the verticesu andv are at distance4 apart inH, and soG= B5.


Note thatn′ =n−2�5. IfH ∈ {B2, C5, C8}, then it is easily checked that�F3(G)< (n−1)/2, a contradiction. Hence,�F3(H)�(n′ − 1)/2= (n− 3)/2. If u andv are both coloredwith the same color inCH , thenCH can be extended to anF3-coloring ofG by coloringbothu1 andu2 blue, and so�F3(G)< (n−1)/2, a contradiction. Hence everyminimumF3-coloringofH colorsuandvwith different colors.Wemayassume thatu is colored red inCH .If �F3(H)�(n− 4)/2, thenCH can be extended to anF3-coloring ofG by coloringu1 redandu2 blue, and so�F3(G)�(n−2)/2, a contradiction. Hence,�F3(H)= (n′ −1)/2. Sincethe setS is independent and sinceu andv must receive different colors in every minimumF3-coloring ofH, it therefore follows thatH ∈ {B1, B2, B3, A1(5),D(4,4),D2(5)}. IfH �= B1, then it is easily checked that�F3(G)< (n−1)/2, a contradiction. Hence,H =B1,and soG= B3.


Theproof ofTheorem3 follows readily fromTheorem2.Since theF3-dominationnumberof a graph cannot decrease if edges are removed, it follows fromTheorem2 andObservation13 that theF3-domination number ofG is at most(n+1)/2. Further suppose�F3(G)�n/2.Then by removing edges ofG, if necessary, we produce anF3-minimal graphG′. ByTheorem2andObservation 13,G′ ∈ {B2, C4, C5, C8}. In all cases it can be readily checkedthatG=G′ unless possibly ifG′ = C8 in which caseG ∈ {C8, H1}.


Theproof ofTheorem4 follows readily fromTheorem2.Since theF3-dominationnumberof a graph cannot decrease if edges are removed, and sincen�9, it follows from Theorem2 and Observation 13 that theF3-domination number ofG is at most(n − 1)/2. Furthersuppose�F3(G)=(n−1)/2. Then by removing edges ofG, if necessary, we produce anF3-minimal graphG′. By Theorem 2 and Observation 13,G′ ∈ A∪ (D−{D(3,5),D(4,4)})


orG′ ∈ {B4, B5, C11}. In all cases it is straightforward (though tedious) to check thatG=G′unless possibly ifG′ = C11 in which caseG ∈ {C11, H2}.

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stratification and domination in graphs with minimum degree two

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