SaturationNumbersforLinearForests 2P - Hindawi
7
Research Article Saturation Numbers for Linear Forests 2P 4 ∪ tP 2 ∗ Feifei Song , 1 Yan Zou , 2 and Hengdi Su 1 1 Department of Information & Computational Science, He’nan Agricultural University, Zhengzhou 450002, He’nan, China 2 Department of Management Engineering, Zhengzhou University of Aeronautics, Zhengzhou 450046, He’nan, China Correspondence should be addressed to Hengdi Su; [email protected] Received 17 December 2020; Revised 17 February 2021; Accepted 20 April 2021; Published 6 May 2021 Academic Editor: Ant nio M. Lopes Copyright © 2021 Feifei Song et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. For a given graph F, graph G is called F-saturated if G does not contain a copy of F, but G + e has a copy of F, where e ∉ E(G).e saturation number, denoted by sat(n, F),istheminimumsizeofagraphwithorder n in all F-saturated graphs. For n ≥ 6t + 22, this paper considers the saturation number of linear forests 2P 4 ∪ tP 2 and describes the extremal graph. 1. Introduction In this article, we only deal with simple graph. Usually, the path and the complete graph with n vertices are denoted by P n and K n , respectively. For terminology and notations not undefined in this paper, the reader can refer to [1]. For a given graph F, graph G is called F-saturated if G does not contain a copy of F, but G + e has a copy of F, where e ∉ E(G). e famous Tur´ an number [2], denoted by ex(n, F), is the maximum number size of graphs in all F-saturated graphs with size n. As a complement, satu- ration number, denoted by sat(n, F), is the minimum size of graphs in all F-saturated graphs with size n. We use SAT (n, F) which denotes the set of graphs with a mini- mum number size in SAT(n, F). In 1964, Erdős et al. introduced the notion of the sat- uration number and gave the saturation number of K t in [3]. K´ aszonyi et al. gave the general upper bound of sat(n, F) in [4], when F is a kind of forbidden graphs. en, for a wider range of graphs F, saturation number of F has been studied by many scholars, for example, k-edge-connected graph [5], cliques [6, 7], complete bipartite graphs [8, 9], nearly complete graphs [10], books [11], cycles [12–17], trees [18, 19], and forests [20–22]. e reader can see summary of known results in [23]. In [24], Bushaw et al. gave the Tur´ an number for the linear forest. Corresponding to that, Chen et al. concentrated on the saturation numbers in [20]. ey obtained an interesting set of results; some of those are shown as the following results. Theorem 1 (see [20]) (i) Let n be positive integers such that sufficiently large; then, sat (n, P 3 ∪ tP 2 )� 3t and the extremal graph is tK 3 ∪ (n − 3t)K 1 (ii) Let n and t be positive integers such that n is suffi- ciently large and t ≥ 1; then, sat(n, P 4 ∪ tP 2 )� 3t + 7 and the extremal graph is K 5 ∪ (t − 1)K 3 ∪ (n− 3t − 2)K 1 (iii) Let n be positive integers such that it is sufficiently large; then, sat (n, P 5 ∪ P 2 )� 15 and the extremal graph is K 6 ∪ (n − 6)K 1 Let T p, q be obtained from K 2 by attaching p disjoint triangles to one vertex of K 2 and attaching q disjoint triangles to another vertex of K 2 (see Figure 1). In this paper, we prove the following theorem. Theorem 2. Suppose t and n are positive integers such that n ≥ 6t + 12. en, sat (n, 2P 4 ∪ tP 2 )� 3t + 10 and SAT (n, 2P 4 ∪ tP 2 )� T p,q ∪ aK 3 ∪ (n − 2t − a − 8)K 1 : p ≥ 2,q ≥ 2 and p + q + a � t + 3} for n ≥ 6t + 22. In what follows, Section 2 lists several useful results and study property of (2P 4 ∪ tP 2 )-saturated graphs. In Section 3, we give the proof of eorem 2. Hindawi Mathematical Problems in Engineering Volume 2021, Article ID 6613393, 7 pages https://doi.org/10.1155/2021/6613393
Transcript of SaturationNumbersforLinearForests 2P - Hindawi
Research ArticleSaturation Numbers for Linear Forests 2P4 cup tP2
lowast
Feifei Song 1 Yan Zou 2 and Hengdi Su 1
1Department of Information amp Computational Science Hersquonan Agricultural University Zhengzhou 450002 Hersquonan China2Department of Management Engineering Zhengzhou University of Aeronautics Zhengzhou 450046 Hersquonan China
Correspondence should be addressed to Hengdi Su sudi2006435163com
Received 17 December 2020 Revised 17 February 2021 Accepted 20 April 2021 Published 6 May 2021
Academic Editor Ant nio M Lopes
Copyright copy 2021 Feifei Song et al is is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
For a given graph F graph G is called F-saturated if G does not contain a copy of F but G + e has a copy of F where e notin E(G) esaturation number denoted by sat(n F) is theminimum size of a graph with order n in all F-saturated graphs For nge 6t + 22 thispaper considers the saturation number of linear forests 2P4 cup tP2 and describes the extremal graph
1 Introduction
In this article we only deal with simple graph Usually thepath and the complete graph with n vertices are denoted byPn and Kn respectively For terminology and notations notundefined in this paper the reader can refer to [1]
For a given graph F graph G is called F-saturated if G
does not contain a copy of F but G + e has a copy of Fwhere e notin E(G) e famous Turan number [2] denotedby ex(n F) is the maximum number size of graphs in allF-saturated graphs with size n As a complement satu-ration number denoted by sat(n F) is the minimum sizeof graphs in all F-saturated graphs with size n We useSAT (n F) which denotes the set of graphs with a mini-mum number size in SAT(n F)
In 1964 Erdős et al introduced the notion of the sat-uration number and gave the saturation number of Kt in [3]Kaszonyi et al gave the general upper bound of sat(n F) in[4] when F is a kind of forbidden graphs en for a widerrange of graphs F saturation number of F has been studiedby many scholars for example k-edge-connected graph [5]cliques [6 7] complete bipartite graphs [8 9] nearlycomplete graphs [10] books [11] cycles [12ndash17] trees[18 19] and forests [20ndash22] e reader can see summary ofknown results in [23]
In [24] Bushaw et al gave the Turan number for thelinear forest Corresponding to that Chen et al concentratedon the saturation numbers in [20] ey obtained an
interesting set of results some of those are shown as thefollowing results
Theorem 1 (see [20])
(i) Let n be positive integers such that sufficiently largethen sat(n P3 cup tP2) 3t and the extremal graph istK3 cup (n minus 3t)K1
(ii) Let n and t be positive integers such that n is suffi-ciently large and tge 1 then sat(n P4 cup tP2) 3t + 7and the extremal graph is K5 cup (t minus 1)K3 cup (nminus
3t minus 2)K1
(iii) Let n be positive integers such that it is sufficientlylarge then sat(n P5 cupP2) 15 and the extremalgraph is K6 cup (n minus 6)K1
Let Tp q be obtained from K2 by attaching p disjointtriangles to one vertex of K2 and attaching q disjoint trianglesto another vertex of K2 (see Figure 1)
In this paper we prove the following theorem
Theorem 2 Suppose t and n are positive integers such thatnge 6t + 12 5en sat(n 2P4 cup tP2) 3t + 10 and SAT (n
2P4 cup tP2) Tpq cup aK3 cup (n minus 2t minus a minus 8)K11113966 pge 2 qge 2andp + q + a t + 3 for nge 6t + 22
In what follows Section 2 lists several useful results andstudy property of (2P4 cup tP2)-saturated graphs In Section 3we give the proof of 5eorem 2
HindawiMathematical Problems in EngineeringVolume 2021 Article ID 6613393 7 pageshttpsdoiorg10115520216613393
2 Preliminaries
Firstly we introduce some notations For a graph G theedge set and vertex set are denoted by E(G) and V(G) enumber of edges and the order of G are denoted by |E(G)|
and |V(G)| respectively Let NG(x) v isin V(G) xv isinE(G) NG[x] NG(x)cup x and dG(x) be the degree ofx where x isin V(G) Given a subgraph F of G and XsubeV(G)let NG(X) cup xisinXNG(x) minus X NF(X) NG(X)capV(F)NG[X] cup xisinXNG[x] and NF[X] NG[X]capV(F) Wesimple denote dF(x) |NF(x)| and NF(x) NF( x ) ifX x When no confusion occurs we simple identify asubgraph F of G with V(F) For example G minus V(F) issimple and is denoted by G minus F sometimes
For a graph G let o(G) | odd components of G1113864 1113865| anddenote the number of edges in the maximum matching of G
by αprime(G) Suppose ige 0 let Vi(G) x isin V(G) dG(x) i1113864 1113865We list the following useful results
Lemma 1 (see [25]) Let G be a graph then
αprime(G) 12min |G| +|S| minus o(G minus S) SsubeV(G) (1)
Lemma 2 (see [20]) Suppose k1 km ge 2 are integers LetG be a (Pk1
cupPk2cup middot middot middot cupPkm
)-saturated graph If dG(x) 2and NG(x) u v then uv isin E(G)
Next we give some properties of (2P4 cup tP2)-saturatedgraphs
Lemma 3 Suppose G is (2P4 cup tP2)-saturated with V0(G)neempty then V1(G) empty Moreover if w isin V0(G) and x isin V
(G) minus V0(G) let F be any copy of 2P4 cup tP2 in G + xw then
V(F)supeNG(x)cup x w (2)
Proof Suppose for the sake of contradiction that x isin V1(G) Since wy notin E(G) and G is (2P4 cup tP2)-saturated G +
wy has a 2P4 cup tP2 containing wy Let xy isin E(G) with wy
replaced by xy and we can get a new 2P4 cup tP2 in G acontradiction us V1(G) empty
Since xw notin E(G) and G is (2P4 cup tP2)-saturated thenG + xw has a 2P4 cup tP2 containing xw and assumes F If
there exists xprime isin NG(x) minus V(F) with xw replaced by xxprime inF we can obtain a new 2P4 cup tP2 in graph G a contradictionerefore
V(F)supeNG(x)cup x w (3)
Lemma 3 is true
Lemma 4 Suppose G is (2P4 cup tP2)-saturated such thatV0(G)neempty and all the nontrivial components are with atleast 4 vertices Let x isin V(G) such that dG(x) 2 andNG(x) u v 5en dG(u) + dG(v)ge 6 Moreover ifdG(v) 3 then NG(v)subeNG(u)cup u
Additionally there does not exist a degree 2 vertex xprime nex
such that NG(xprime) u v
Proof By Lemma 2 we know uv isin E(G) Since all thenontrivial components are with at least 4 vertices then thetriangle induced by x u v is not a component of G Hencewe can assume vw isin E(G) for some vertex w notin x u v Consequently dG(v)ge 3
If dG(v) 3 it is enough to show uw isin E(G) Ifuw notin E(G) G + uw has a 2P4 cup tP2 containing uw assumeF Assume v notin V(F) replacing uw by uw and we obtain a2P4 cup tP2 in G Hence v isin V(F) If uv isin F or vw isin F orxu isin F then x u v w are selected as a P4 in F and we canreplace it with P4 xuvw which implies there is a 2P4 cup tP2in G So we have uv notin F and vw notin F and xu notin F is to-gether with v isin V(F) and dG(v) 3 implies that xv isin F andxv is selected as a P2 in F Now we can claim that uw isselected as a P4 in F Otherwise both xv and uw are selectedas a P2 in F Replacing xv uw in F with xu vw we obtaina copy of 2P4 cup tP2 inG a contradiction If uw is selected as amiddle edge in P4 then E(F) minus uw + xu contains a2P4 cup tP2 a contradiction If uw is selected as an end edge inP4 then E(F) minus uw minus xv + vw + xu contains a 2P4 cup tP2 acontradiction
If there exists a degree 2 vertex xprime ne x such thatNG(xprime) NG(x) then select an isolated vertex y isin V0(G)
and add the edge uy Let F be a copy of the graph 2P4 cup tP2in G + uy No matter uy is selected as an edge in P2 or P4 inF since dG(x) dG(xprime) 2 we get | x xprime1113864 1113865capV(F)|le 1which contradicts Lemma 3
Lemma 5 Let G be a (2P4 cup tP2)-saturated graph andV0(G)neempty Let Gprime be an induced subgraph of G If Gprime K3 orGprime K1ℓ(ℓ ge 2) then |NG(Gprime)cap (V(G) minus V(Gprime))|ne 1
Proof Suppose for the sake of contradiction thatNG(Gprime)cap (V(G) minus V(Gprime)) x Select an isolated vertexw isin V0(G) and add the edge wx Since G is(2P4 cup tP2)-saturated then G + wx has a 2P4 cup tP2 con-taining wx and assumes F
First consider the case Gprime K3 SupposeV(Gprime) u1 u2 u31113864 1113865 and u1 isin NG(x) For any ui isin NG(x)we have uix notin E(F) where 1le ile 3 Otherwise |V(Gprime) minus
V(F)| 1 and x w cupV(Gprime) are selected as a P4 Howeverthis P4 can be replaced with xu1u2u3 in F and then weobtain a 2P4 cup tP2 inG a contradiction Hence u1 u2 u31113864 1113865 is
p q
Tpq
Figure 1 e graph Tpq
2 Mathematical Problems in Engineering
selected as a P2 in F We can choose u2u3 as P2 and thenreplace wx in F with xu1 we obtain a 2P4 cup tP2 in G acontradiction
Now consider the case Gprime K1ℓ Suppose u is the centerof the star and V(K1ℓ) u u1 uℓ1113864 1113865 Since G is(2P4 cup tP2)-saturated with V0(G)neempty by Lemma 3V1(G) empty en we have uix isin E(G) for 1le ile ℓ HencedG(ui) 2 and NG(ui) u x for all 1le ile ℓ whichcontradicts Lemma 4 Lemma 5 is true
3 Proof of Theorem 2
Before the proof of eorem 2 we discuss the property of(2P4 cup tP2)-saturated graph which has no components oforder 2 3 and 4
Lemma 6 Suppose G is (2P4 cup tP2)-saturated but not(P4 cup (t + 2)P2)-saturated graph Let V0(G)neempty H1
Hk be the nontrivial components of G and H G[cup i 1
i k V(Hi)] If δ(H)ge 2 |H|ge 2t + 8 and |Hi|ge 5(1le ile k) we have |E(G)|ge 3t + 10
Moreover if the equality |E(G)| 3t + 10 holds thenG Tpq cup (n minus 2t minus 8)K1 where p qge 2 and p + q t + 3
Proof Since G is (2P4 cup tP2)-saturated G + e has a P4 cup(t + 2)P2 where e notin E(G) On the contrary G must containa P4 cup (t + 2)P2 since G is not (P4 cup (t + 2)P2)-saturated
Suppose M is a P4 cup (s + 2)P2 in G with E(M)
w1w2 w2w3 w3w41113864 1113865cup u1v1 us+2vs+21113864 1113865 Choose M
which satisfies the following
(i) s is as large as possible(ii) Subject to (i) 1113936
s+2i1(dG(ui) + dG(vi)) is maximality
Observe that sge t By the choice of M V(H) minus V(M) isan independent set or an empty set As G is(2P4 cup tP2)-saturated we have
G V(M) w1 w2 w3 w41113864 11138651113858 1113859 (s + 2)P2 (4)
NGminusV(M) ui vi1113864 1113865( 1113857capNGminusV(M) uj vj1113966 11139671113872 1113873 empty
for 1le ine jle s + 2(5)
Without loss of generality we may assumew1 w2 w3 w41113864 1113865subeH1
Claim 1 k 1
Proof Suppose for the sake of contradiction that kge 2Since |Hi|ge 5 and δ(H)ge 2 we have αprime(Hi)ge 2 for every1le ile k Without loss of generality we can assumeuℓvℓ umvm1113864 1113865subeH2 where 1le ℓ ltmle s + 2
As H2 is connected there exists a path Q from ui1 vi1
1113966 1113967
to ui2 vi2
1113966 1113967 avoiding all vertices of V(M) ui1 vi1
ui2 vi2
1113966 1113967
for some ℓ le i1 lt i2 lem Again G[ ui1 vi1
ui2 vi2
1113966 1113967cupV(Q)]
has a P4 which combining the edges in M minus ui1vi1
ui2vi2
1113966 1113967creates a 2P4 cup sP2 in G a contradiction Claim 1 is true
It follows from Claim 1 that H H1 Hence in the restof the proof of Lemma 6 H1 is denoted by H Clearly|H|ge 2s + 8ge 2t + 8
Claim 2 For each 1le ile s + 2 we have NG( ui vi1113864 1113865)capw1 w2 w3 w41113864 1113865neempty
Proof Suppose for the sake of contradiction that thereexists 1le ile s + 2 such that NG( ui vi1113864 1113865)cap w11113864 w2 w3
w4 empty is together with (4) implies that
NG ui( 1113857capV(M) vi1113864 1113865 andNG vi( 1113857capV(M) ui1113864 1113865 (6)
We are now ready to show that
dG ui( 1113857 + dG vi( 1113857ge 5 (7)
Otherwise dG(ui) dG(vi) 2 since δ(H)ge 2 It fol-lows from (6) that we can assume w isin NHminusV(M)(ui) ByLemma 2 wvi isin E(G) en M minus uivi1113864 1113865 + uiw1113864 1113865 is a copy ofP4 cup (s + 2)P2 in G However by Lemma 4 we havedG(w)ge 4 and hence dG(ui) + dG(w)ge 6gt dG(ui)+
dG(vi) 4 which contradicts the choice of M en wehave (7) which holds
Combining (6) and (7) there exist z1 ne z2 such thatz1 isin NHminusV(M)(ui) and z2 isin NHminusV(M)(vi) en the sub-graph G[ z1 ui vi z21113864 1113865] will have a P4 which combining theedges in M minus uivi1113864 1113865 creates a 2P4 cup (s + 1)P2 in G Sincesge t we obtain a contradiction Claim 2 is true
Claim 3 There does not exist 1le i jle s + 2 such that
NG ui vi1113864 1113865( 1113857cap w1 w21113864 1113865neempty andNG uj vj1113966 11139671113872 1113873cap w3 w41113864 1113865neempty
(8)
Proof Suppose for the sake of contradiction that there exist1le i jle s + 2 such that NG( ui vi1113864 1113865)cap w1 w21113864 1113865neempty andNG( uj vj1113966 1113967)cap w3 w41113864 1113865neempty By Claim 2 we may assumeine j However the subgraphs G[ ui vi w1 w21113864 1113865] andG[ uj vj w3 w41113966 1113967] both contain P4 which combining theedges in M minus uivi ujvj w1w2 w2w3 w3w41113966 1113967 creates a2P4 cup sP2 in G a contradiction Hence Claim 3 is true
Claim 4 For each 1le ile s + 2 we have |NG( ui vi1113864 1113865)capw1 w2 w3 w41113864 1113865| 1 or |E(G)|gt 3t + 10
Proof Suppose for the sake of contradiction that Claim 4 isfalse en by Claim 3 and the symmetry we may assume|E(G)|le 3t + 10 and
NG u1 v11113864 1113865( 1113857cap w1 w2 w3 w41113864 1113865 w1 w21113864 1113865 (9)
en by Claim 3 we have
NG uj vj1113966 11139671113872 1113873cap w1 w2 w3 w41113864 1113865sube w1 w21113864 1113865
for each 1le jle s + 2(10)
Case 1 NG(u1)cap w1 w21113864 1113865neempty and NG(v1)cap w1 w21113864 1113865neempty
Mathematical Problems in Engineering 3
Combining (9) without loss of generality in this case wemay assume u1w1 isin E(G) and v1w2 isin E(G) Now weconsider ujvj where 2le jle s + 2
If w1 isin NG( uj vj1113966 1113967) then subgraphs G[ uj vj w1 u11113966 1113967]
and G[ v1 w2 w3 w41113864 1113865] both contain P4 which combiningthe edges in M minus u1v1 ujvj w1w2 w2w3 w3w41113966 1113967 creates a2P4 cup sP2 in G a contradiction Hence we havew1 notin NG( uj vj1113966 1113967) Together with Claim 2 and (10) weobtain
NG uj vj1113966 11139671113872 1113873cap w1 w2 w3 w41113864 1113865 w21113864 1113865 (11)
Without loss of generality we may assume ujw2 isin E(G)If NHminusV(M)(vj)neempty assume Gprime G[NHminusV(M) [ uj1113966
vj]] In order to avoid a 2P4 cup tP2 in G we get that Gprime K3or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G) minus V(Gprime)) w21113864 1113865which contradicts Lemma 5 Hence NHminusV(M)(vj) emptyis together with δ(H)ge 2 and (11) implies that dG(vj) 2and vjw2 isin E(G) As the argument of vj we haveNHminusV(M)(uj) empty Hence dG(uj) 2 en we can con-clude that
dG uj1113872 1113873 dG vj1113872 1113873 2 andNG uj1113872 1113873capNG vj1113872 1113873 w21113864 1113865
for any 1le jle s + 2
(12)
And by the choice of M dG(w3) dG(w4) 2 Henceby Lemma 2 w2w4 isin E(G) Combining (12) implies thatdG(w2)ge 2(s + 3) If dG(w1) 2 by Lemma 2 u1w2 isinE(G) Hence dG(u1) + dG(v1) + dG(w1)ge 7 If dG(w1)ge 3we also have dG(u1) + dG(v1) + dG(w1)ge 7 is togetherwith dG(w2)ge 2(s + 3) |H|ge 2t + 8 sge t and δ(H)ge 2implies that 2|E(H)|ge 2(s + 3) + 7+ 2(2t + 4)gt 6t + 20 acontradiction
Case 2 NG(u1)cap w1 w21113864 1113865 empty or NG(v1)cap w1 w21113864 1113865 emptyCombining (9) without loss of generality in this case we
may assume NG(v1)cap w1 w2 w3 w41113864 1113865 empty is combiningδ(H)ge 2 and (4) we get NHminusV(M)(v1)neempty and |H|ge 2t + 9Assume v1x isin E(G) where x isin H minus V(M) By our as-sumption NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w1 w21113864 1113865 we getu1w1 isin E(G) and u1w2 isin E(G) Now we consider ujvjwhere 2le jle s + 2
If w2 isin NG( uj vj1113966 1113967) then the subgraphs G[ uj1113966 vj
w2 w3] and G[ x v1 u1 w11113864 1113865] both contain P4 whichcombining the edges in M minus u1v1 ujvj w1w21113966 w2w3 w3w4creates a 2P4 cup sP2 in G a contradiction Hence we havew2 notin NG( uj vj1113966 1113967) is together with Claim 2 and (9) impliesthat NG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w11113864 1113865 for any 2le jles + 2Without loss of generality wemay assume ujw1 isin E(G)en we have NG(x)cap w1 w2 w3 w41113864 1113865 empty So by thechoice of M (5) and δ(H)ge 2 we get xu1 isin E(G) Hence
dG u1( 1113857ge 4 (13)
In order to avoid a 2P4 cup tP2 in G we get thatNHminusV(M)(vj) empty is together with δ(H)ge 2 andNG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w11113864 1113865 implies that
vjw1 isin E(G) and dG(vj) 2 Similarly we have dG(uj) 2erefore ujw1 isin E(G) and vjw1 isin E(G) for any2le jle s + 2 Hence combining sge t implies that
dG w1( 1113857ge 2(s + 2)ge 2t + 4 (14)
And by the choice of M we have dG(w3) dG(w4) 2en by Lemma 2 w2w4 isin E(G) which implies that
dG w2( 1113857ge 4 (15)
Recall that |H|ge 2t + 9 and δ(H)ge 2 Together with(13)ndash(15) we can get that 2|E(H)| ge 2t + 4 + 4 + 4+
2(2t + 6)gt 6t + 20 a contradiction Claim 4 is true
Claim 5 If there exists 1le ile s + 2 such thatdG(ui) + dG(vi)le 4 then we have |E(G)|ge 3t + 10 More-over the equality |E(G)| 3t + 10 holds if and only ifG Tpq cup (n minus 2t minus 8)K1 where p qge 2 and p + q t + 3
Proof Without loss of generality we may assumedG(u1) + dG(v1)le 4 Since δ(H)ge 2 we havedG(u1) dG(v1) 2 By Claim 4 and the symmetry wecomplete the Claim 5 by following two cases
Case 1 NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w11113864 1113865Since dG(u1) dG(v1) 2 by Lemma 2 u1w1 isin E(G)
and v1w1 isin E(G) By the choice of M dG(w3)
dG(w4) 2 Also by Lemma 2 w2w4 isin E(G) Now weconsider ujvj where 2le jle s + 2 It follows from Claim3 and our assumption that NG( uj vj1113966 1113967)cap w1 w2 w31113864
w4sube w1 w21113864 1113865 where 2le jle s + 2 is together withClaim 4 implies that NG( uj vj1113966 1113967)cap w1 w21113864 w3
w4 wi1113864 1113865 where i 1 or 2If NHminusV(M)(vj)neempty assume Gprime G[NHminusV(M)[ uj1113966
vj]] In order to avoid a 2P4 cup tP2 in G we get thatGprime K3 or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G)minus
V(Gprime)) wi1113864 1113865 which contradicts Lemma 5 HenceNHminusV(M)(vj) empty Since δ(H)ge 2 we get vjwi isin E(G)
and dG(vj) 2 By the similar argument we haveNHminusV(M)(uj) empty and hence dG(uj) 2 ereforefor each 1le jle s + 2 we can conclude that NG(uj)
capNG(vj) w11113864 1113865 or NG(uj)capNG(vj) w21113864 1113865 To-gether with w2w4 isin E(G) we have dG(w1)+ dG(w2)ge2(s + 3) + 2 2s + 8 Recall that sge t |H|ge 2t + 8 andδ(H)ge 2 we can get that 2|E(H)| ge 2s+ 8+ 2(2t+
6)ge 6t + 20 And if |E(G)| 3t + 10 then G
Tpq cup (n minus 2t minus 8)K1 where p + q t + 3 However ifmin p q1113864 1113865le 1 Tpq is not a (2P4 cup tP2)-saturated graphsince the addition of an edge between an isolated vertexof G and the vertex of degree Δ(G) does not result in a2P4 cup tP2 Hence p qge 2 Claim 5 holdsCase 2 NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w21113864 1113865
Since dG(u1) dG(v1) 2 by Lemma 2 u1w2 isin E(G)
and v1w2 isin E(G) By the choice of M we havedG(w3) dG(w4) 2 Also by Lemma 2 w2w4 isin E(G)Now we consider ujvj where 2le jle s + 2 It follows fromClaim 3 and our assumption that NG( uj vj1113966 1113967)cap w11113864 w2
w3 w4sube w1 w21113864 1113865
4 Mathematical Problems in Engineering
If there exists 1le jle s + 2 such that NG( uj1113966 vj) capw1 w2 w3 w41113864 1113865 w11113864 1113865 we can claim that dG(uj)
dG(vj) 2 Otherwise assume Gprime G[NHminusV(M)[ uj vj1113966 1113967]]In order to avoid a 2P4 cup tP2 in G we get that Gprime K3 orGprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G) minus V(Gprime)) w11113864 1113865which contradicts Lemma 5 en as the similar argumentin the Case 1 we can obtain |E(H)|ge 3t + 10 and if|E(G)| 3t + 10 then G Tpq where p qge 2 and p + q
t + 3 en NG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w21113864 1113865 for all1le jle s + 2 Without loss of generality we may assumeujw2 isin E(G) If NHminusV(M)(vj)neempty assume Gprime G[[NHminusV(M)[ uj vj1113966 1113967]]] In order to avoid a 2P4 cup tP2 in Gwe get that Gprime K3 or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap(V(G) minus V(Gprime)) w21113864 1113865 which contradicts Lemma 5Hence NHminusV(M)(vj) empty Since δ(H)ge 2 we get vjw2 isinE(G) and dG(vj) 2 By the similar argument we haveNHminusV(M)(uj) empty and hence dG(uj) 2 ereforeujw2 isin E(G) and vjw2 isin E(G) for any 2le jle s + 2 Hence
dG w2( 1113857ge 2(s + 3) + 1ge 2t + 7 (16)
is together with δ(H)ge 2 |H|ge 2t + 8 and (16) im-plies that 2|E(H)| ge 2t + 7 + 2(2t + 7)gt 6t + 20 is com-pletes the proof of Claim 5
If there exists 1le ile s + 2 such that dG(ui) + dG(vi)le 4by Claim 5 the Lemma is proven by Claim 5 Hence we canassume dG(ui) + dG(vi)ge 5 for each 1le ile s + 2 On thecontrary by (4) and Claim 4 we can assumedV(M)(ui) + dV(M)(vi)le 4 Hence there existsxi isin NG( uj vj1113966 1113967)cap (V(H) minus V(M)) for any 1le ile s + 2And by (5) xi nexj for 1le ine jle s + 2 Hence|H|ge 3(s + 2) + 4ge 3t + 10 Since δ(H)ge 2 we get|E(H)|ge 3t + 10 And if the equality |E(G)| 3t + 10 holdsthen δ(H) Δ(H) 2 and |H| 3t + 10 which contradictsLemma 4 Lemma 6 is true
Lemma 7 Suppose G is (2P4 cup tP2)-saturated and (P4 cup(t + 2)P2)-saturated with V0(G)neempty Let H1 Hk be thenontrivial components of G and H G[cup i 1 i
kV(Hi)] If δ(H)ge 2 and |H|ge 2t + 8 with |Hi|ge5(1le ile k) we have |E(G)|gt 3t + 10
Proof Suppose for the sake of contradiction that|E(G)|le 3t + 10 Since G is (2P4 cup tP2)-saturated thereexists a 2P4 cup tP2 in G + e for any e notin E(G) Henceαprime(H)ge t + 3 If αprime(H)ge t + 4 G must contain a copy of(t + 4)P2 Since δ(H)ge 2 and |Hi|ge 5(1le ile k) it is clearlythat H has a P4 cup (t + 2)P2 which contradicts G is(P4 cup (t + 2)P2)-saturated So we have αprime(H) t + 3 ByLemma 1 we obtain
t + 3 12min |H| +|X| minus o(G minus X) XsubeV(H) (17)
en we can choose a maximal subset SsubeV(H) whichsatisfies
t + 3 12
(|H| +|S| minus o(H minus S)) (18)
Suppose H1prime Hpprime are the components of H minus S We
have the following claims
Claim 6 H[ScupV(Hiprime)] is the complete graph where
1le ilep
Proof Suppose for the sake of contradiction thatH[ScupV(Hi
prime)] is not a complete graph en there isx y isin ScupV(Hi
prime) with xney and xy notin E(H) LetHprime H + xy We have |Hprime| |H| ando(Hprime minus S) o(H minus S) Combining (18) we obtain
t + 3 12
Hprime1113868111386811138681113868
1113868111386811138681113868 +|S| minus o(Hprime minus S)1113874 1113875 (19)
By Lemma 1 we obtain αprime(Hprime)le t + 3 Hence G + xy
has no (t + 4)P2 contrary to G which is (2P4 cup tP2)-satu-rated Claim 6 is true
Claim 7 Sneempty
Proof Suppose for the sake of contradiction that S emptyHence H1prime Hp
prime are all components of H Using Claim 6we have Hi
prime is a complete graph with |Hiprime|ge 5(1le ilep)
Hence we have δ(H)ge 4 and
2|E(H)| 1113944xisinV(H)
dH(x)ge 4|H| (20)
Combining |H|ge 2t + 8 we get |E(H)|gt 3t + 10 HenceClaim 7 is true
Let w isin V0(G) and x isin S By Claim 6NH(x) V(H) minus x Let F be a 2P4 cup tP2 in G + xw usingLemma 3 we get V(F)supe x w cupNG(x) Hence we have|H| + 1 |NH(x)cup x w |le |V(F)| 2t + 8 contrary to G
is (2P4 cup tP2)-saturated Hence Lemma 7 is trueNow we prove eorem 2
Proof of eorem 2 Clearly Tpq cup aK3 cup (nminus 2tminus aminus
8)K1 isin SAT(n 2P4 cup tP2) where pge 2 qge 2 and p + q+
a t + 3 Let G isin SAT (n 2P4 cup tP2) then |E(G)|le 3t + 10Since nge 6t + 12 |V0(G)|ge 2 Using Lemma 3 we haveV1(G)neempty Clearly in graph G all components of order 3 or4 are complete If there exists one of the components sayH1 K4 then the addition of an edge e between an isolatedvertex of G and a vertex of H1 results in a 2P4 cup tP2 in GHowever the edge e can be replaced by the edge in H1resulting in a 2P4 cup tP2 isin G a contradiction Hence com-ponent of order 4 in G does not exist
Now we consider the graph Gprime which is obtained bydeleting all components of order 3 from G Let a be thenumber of components of order 3 in G and we obtain
G Gprime cup aK3 (21)
Note |V0(Gprime)| |V0(G)|ge 2 Let e be an edge joiningtwo isolated vertices in V0(G) then there exists a 2P4 cup tP2in G + e So Gprime has 2P4 Combining G isin SAT (n 2P4 cup tP2)we get ale t minus 1 Let tprime t minus a en tprime ge 1
As G isin SAT (n 2P4 cup tP2) we have Gprime isin SAT(nprime2P4 cup tP2) where nprime n minus 3a Suppose Hprime is a graph
Mathematical Problems in Engineering 5
spanned by all nontrivial components of Gprime Using Lemma 3we have δ(Hprime)ge 2 Clearly any component in Hprime has orderat least 5 Moreover graph Gprime is (2P4 cup tprimeP2)-saturated suchthat |E(Gprime)| |E(G)| minus 3ale 3tprime + 10 and V0(Gprime)neempty ByLemma 6 and Lemma 7 we have |Hprime|le 2tprime + 7 or Hprime Tpq cup (nprime minus 2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3If |Hprime|le 2tprime + 7 as V0(Gprime)neempty and Gprime isin SAT(nprime
2P4 cup tprimeP2) Hprime K2tprime+7 Now E(Hprime) 2tprime + 7
21113888 1113889gt
3tprime + 21 and we have |E(G)|gt 3tprime + 21 + 3a 3t+
21gt 3t + 10 a contradiction Hence Hprime Tpq cup (nprimeminus2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3 Combining(21) we get G Tpq cup aK3 cup (n minus 2t minus a minus 8)K1 wherepge 2 qge 2 and p + q + a t + 3 Hence eorem 2 istrue
4 Conclusion
Forests and trees have attracted wide attention for severalreasons Firstly their simplicity has made somemore preciseresults possible Secondly they can provide inspiration forsome larger results At present we only know the saturationnumber of a few species of trees but the saturation numberof most trees is unknown
In [20] Chen et al focused on the saturation numbersfor the linear forests ey obtained an interesting set ofresults and proposed some conjectures We pave the way forthe following conjectures proposed by Chen et al in [20]
Conjecture 1 (see [20]) Let tge 2 be an integer For n suf-ficiently large sat(n tP3) lfloor(n + 6t minus 6)2rfloor
Conjecture 2 (see [20]) Let tge 2 be an integer For n suf-ficiently large
sat n tP4( 1113857
n + 12t minus 122
if n is even
n + 12t minus 112
if n is odd
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
(22)
Conjecture 3 (see [20]) For n sufficiently large kge 4and kle ℓ le lceil(3k minus 2)2rceil sat(n Pk cupPℓ) n minus 1113860nak1113861 + 3where
ak 3 middot 2mminus 1
minus 2 if k 2m
4 middot 2mminus 1minus 2 if k 2m + 1
⎧⎨
⎩ (23)
Data Availability
No data used to support the findings of the study
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was partially supported by NSFC (nos 12001172and 11771172) and supported by Natural Science Founda-tion of Henan Province (no 202300410212) and Science andtechnology innovation fund of Henan Agricultural Uni-versity (nos KJCX2019A15 and KJCX2017A12)
References
[1] J A Bondy and U S R Murty Graph 5eory with Appli-caitons American Elsevier New York NY USA 1976
[2] P Turan ldquoEine Extremalaufgabe aus der GraphentheorierdquoMat Fiz Lapok vol 48 pp 436ndash452 1941
[3] P Erdős A Hajnal and J W Moon ldquoA problem in graphtheoryrdquo 5e American Mathematical Monthly vol 71pp 1107ndash1110 1964
[4] L Kaszonyi and Z Tuza ldquoSaturated graphs with minimalnumber of edgesrdquo Journal of Graph 5eory vol 10 no 2pp 203ndash210 1986
[5] H Lei O Suil D BWest and X Zhu ldquoExtremal problems onsaturation for the family ofk-edge-connected graphsrdquoDiscreteApplied Mathematics vol 260 pp 278ndash283 2019
[6] T Bohman M Fonoberova and O Pikhurko ldquoe saturationfunction of complete partite graphsrdquo Journal of Combina-torics vol 1 no 2 pp 149ndash170 2010
[7] J R Faudree M Ferrara R J Gould andM S Jacobson ldquotKp-saturated graphs of minimum sizerdquo Discrete Mathematicsvol 309 no 19 pp 5870ndash5876 2009
[8] B Bollobas ldquoOn a conjecture of erdos hajnal and moonrdquo5eAmerican Mathematical Monthly vol 74 no 2 pp 178-1791967
[9] S Huang H Lei Y Shi and J Zhang ldquoe saturation numberof K33rdquo 2019 httparxivorgabs191004967
[10] R J Faudree and R J Gould ldquoSaturation numbers for nearlycomplete graphsrdquo Graphs and Combinatorics vol 29 no 3pp 429ndash448 2013
[11] G Chen R J Faudree and R J Gould ldquoSaturation numbersof booksrdquo 5e Electronic Journal of Combinatorics vol 15no 1 pp 118ndash129 2008
[12] Y-C Chen ldquoMinimumC5-saturated graphsrdquo Journal ofGraph 5eory vol 61 no 2 pp 111ndash126 2009
[13] Y-C Chen ldquoAll minimum C5-saturated graphsrdquo Journal ofGraph 5eory vol 67 no 1 pp 9ndash26 2011
[14] L T Ollmann ldquo-saturated graphs with a minimal number ofedgesrdquo in Proceedings of the 5ird Southeastern Conference onCombinatorics Graph 5eory and Computing Florida At-lantic University Boca Raton FL USA 1972
[15] C Timmons ldquo$$C_2k$$-Saturated graphs with No shortodd cyclesrdquo Graphs and Combinatorics vol 35 no 5pp 1023ndash1034 2019
[16] Z Tuza ldquo-saturated graphs of minimum sizerdquo Acta Uni-versitatis Carolinae Mathematica et Physica vol 30 no 2pp 161ndash167 1989
[17] M Zhang S Luo and M Shigeno ldquoOn the number of edgesin a minimum $$C_6$$ C 6 -saturated graphrdquo Graphs andCombinatorics vol 31 no 4 pp 1085ndash1106 2015
[18] J R Faudree R J Faudree R J Gould and M S JacobsonldquoSaturation numbers for treesrdquo 5e Electronic Journal ofCombinatorics vol 16 no 1 pp 91ndash109 2009
[19] F F Song and J J Zhou ldquoe partite saturation number ofspiderrdquo Applied Mathematics and Computation vol 394Article ID 125793 2021
6 Mathematical Problems in Engineering
[20] G Chen J R Faudree R J Faudree R J GouldM S Jacobson and C Magnant ldquoResults and problemson saturation numbers for linear forestsrdquo Bulletin of theInstitute of Combinatorics and its Applications vol 75pp 29ndash46 2015
[21] Q Fan and C Wang ldquoSaturation numbers for linear forests$$P_5 cup tP_2$$ P 5 cup t P 2rdquo Graphs and Combinatoricsvol 31 no 6 pp 2193ndash2200 2015
[22] F F Song ldquoSaturation numbers for linear forestsrdquo UtilitasMathematica vol 104 pp 175ndash186 2017
[23] J R Faudree R J Faudree and J R Schmitt ldquoA survey ofminimum saturation graphsrdquo 5e Electronic Journal ofCombinatorics vol 18 p 36pp 2011
[24] N Bushaw and N Kettle ldquoTuran numbers of multiple pathsand equibipartite forestsrdquo Combinatorics Probability andComputing vol 20 no 6 pp 837ndash853 2011
[25] C Berge ldquoSur le couplage maximum dprimeun grapherdquo Comptesrendus de lrsquoAcademie des Sciences vol 247 pp 258-259 1958
Mathematical Problems in Engineering 7
2 Preliminaries
Firstly we introduce some notations For a graph G theedge set and vertex set are denoted by E(G) and V(G) enumber of edges and the order of G are denoted by |E(G)|
and |V(G)| respectively Let NG(x) v isin V(G) xv isinE(G) NG[x] NG(x)cup x and dG(x) be the degree ofx where x isin V(G) Given a subgraph F of G and XsubeV(G)let NG(X) cup xisinXNG(x) minus X NF(X) NG(X)capV(F)NG[X] cup xisinXNG[x] and NF[X] NG[X]capV(F) Wesimple denote dF(x) |NF(x)| and NF(x) NF( x ) ifX x When no confusion occurs we simple identify asubgraph F of G with V(F) For example G minus V(F) issimple and is denoted by G minus F sometimes
For a graph G let o(G) | odd components of G1113864 1113865| anddenote the number of edges in the maximum matching of G
by αprime(G) Suppose ige 0 let Vi(G) x isin V(G) dG(x) i1113864 1113865We list the following useful results
Lemma 1 (see [25]) Let G be a graph then
αprime(G) 12min |G| +|S| minus o(G minus S) SsubeV(G) (1)
Lemma 2 (see [20]) Suppose k1 km ge 2 are integers LetG be a (Pk1
cupPk2cup middot middot middot cupPkm
)-saturated graph If dG(x) 2and NG(x) u v then uv isin E(G)
Next we give some properties of (2P4 cup tP2)-saturatedgraphs
Lemma 3 Suppose G is (2P4 cup tP2)-saturated with V0(G)neempty then V1(G) empty Moreover if w isin V0(G) and x isin V
(G) minus V0(G) let F be any copy of 2P4 cup tP2 in G + xw then
V(F)supeNG(x)cup x w (2)
Proof Suppose for the sake of contradiction that x isin V1(G) Since wy notin E(G) and G is (2P4 cup tP2)-saturated G +
wy has a 2P4 cup tP2 containing wy Let xy isin E(G) with wy
replaced by xy and we can get a new 2P4 cup tP2 in G acontradiction us V1(G) empty
Since xw notin E(G) and G is (2P4 cup tP2)-saturated thenG + xw has a 2P4 cup tP2 containing xw and assumes F If
there exists xprime isin NG(x) minus V(F) with xw replaced by xxprime inF we can obtain a new 2P4 cup tP2 in graph G a contradictionerefore
V(F)supeNG(x)cup x w (3)
Lemma 3 is true
Lemma 4 Suppose G is (2P4 cup tP2)-saturated such thatV0(G)neempty and all the nontrivial components are with atleast 4 vertices Let x isin V(G) such that dG(x) 2 andNG(x) u v 5en dG(u) + dG(v)ge 6 Moreover ifdG(v) 3 then NG(v)subeNG(u)cup u
Additionally there does not exist a degree 2 vertex xprime nex
such that NG(xprime) u v
Proof By Lemma 2 we know uv isin E(G) Since all thenontrivial components are with at least 4 vertices then thetriangle induced by x u v is not a component of G Hencewe can assume vw isin E(G) for some vertex w notin x u v Consequently dG(v)ge 3
If dG(v) 3 it is enough to show uw isin E(G) Ifuw notin E(G) G + uw has a 2P4 cup tP2 containing uw assumeF Assume v notin V(F) replacing uw by uw and we obtain a2P4 cup tP2 in G Hence v isin V(F) If uv isin F or vw isin F orxu isin F then x u v w are selected as a P4 in F and we canreplace it with P4 xuvw which implies there is a 2P4 cup tP2in G So we have uv notin F and vw notin F and xu notin F is to-gether with v isin V(F) and dG(v) 3 implies that xv isin F andxv is selected as a P2 in F Now we can claim that uw isselected as a P4 in F Otherwise both xv and uw are selectedas a P2 in F Replacing xv uw in F with xu vw we obtaina copy of 2P4 cup tP2 inG a contradiction If uw is selected as amiddle edge in P4 then E(F) minus uw + xu contains a2P4 cup tP2 a contradiction If uw is selected as an end edge inP4 then E(F) minus uw minus xv + vw + xu contains a 2P4 cup tP2 acontradiction
If there exists a degree 2 vertex xprime ne x such thatNG(xprime) NG(x) then select an isolated vertex y isin V0(G)
and add the edge uy Let F be a copy of the graph 2P4 cup tP2in G + uy No matter uy is selected as an edge in P2 or P4 inF since dG(x) dG(xprime) 2 we get | x xprime1113864 1113865capV(F)|le 1which contradicts Lemma 3
Lemma 5 Let G be a (2P4 cup tP2)-saturated graph andV0(G)neempty Let Gprime be an induced subgraph of G If Gprime K3 orGprime K1ℓ(ℓ ge 2) then |NG(Gprime)cap (V(G) minus V(Gprime))|ne 1
Proof Suppose for the sake of contradiction thatNG(Gprime)cap (V(G) minus V(Gprime)) x Select an isolated vertexw isin V0(G) and add the edge wx Since G is(2P4 cup tP2)-saturated then G + wx has a 2P4 cup tP2 con-taining wx and assumes F
First consider the case Gprime K3 SupposeV(Gprime) u1 u2 u31113864 1113865 and u1 isin NG(x) For any ui isin NG(x)we have uix notin E(F) where 1le ile 3 Otherwise |V(Gprime) minus
V(F)| 1 and x w cupV(Gprime) are selected as a P4 Howeverthis P4 can be replaced with xu1u2u3 in F and then weobtain a 2P4 cup tP2 inG a contradiction Hence u1 u2 u31113864 1113865 is
p q
Tpq
Figure 1 e graph Tpq
2 Mathematical Problems in Engineering
selected as a P2 in F We can choose u2u3 as P2 and thenreplace wx in F with xu1 we obtain a 2P4 cup tP2 in G acontradiction
Now consider the case Gprime K1ℓ Suppose u is the centerof the star and V(K1ℓ) u u1 uℓ1113864 1113865 Since G is(2P4 cup tP2)-saturated with V0(G)neempty by Lemma 3V1(G) empty en we have uix isin E(G) for 1le ile ℓ HencedG(ui) 2 and NG(ui) u x for all 1le ile ℓ whichcontradicts Lemma 4 Lemma 5 is true
3 Proof of Theorem 2
Before the proof of eorem 2 we discuss the property of(2P4 cup tP2)-saturated graph which has no components oforder 2 3 and 4
Lemma 6 Suppose G is (2P4 cup tP2)-saturated but not(P4 cup (t + 2)P2)-saturated graph Let V0(G)neempty H1
Hk be the nontrivial components of G and H G[cup i 1
i k V(Hi)] If δ(H)ge 2 |H|ge 2t + 8 and |Hi|ge 5(1le ile k) we have |E(G)|ge 3t + 10
Moreover if the equality |E(G)| 3t + 10 holds thenG Tpq cup (n minus 2t minus 8)K1 where p qge 2 and p + q t + 3
Proof Since G is (2P4 cup tP2)-saturated G + e has a P4 cup(t + 2)P2 where e notin E(G) On the contrary G must containa P4 cup (t + 2)P2 since G is not (P4 cup (t + 2)P2)-saturated
Suppose M is a P4 cup (s + 2)P2 in G with E(M)
w1w2 w2w3 w3w41113864 1113865cup u1v1 us+2vs+21113864 1113865 Choose M
which satisfies the following
(i) s is as large as possible(ii) Subject to (i) 1113936
s+2i1(dG(ui) + dG(vi)) is maximality
Observe that sge t By the choice of M V(H) minus V(M) isan independent set or an empty set As G is(2P4 cup tP2)-saturated we have
G V(M) w1 w2 w3 w41113864 11138651113858 1113859 (s + 2)P2 (4)
NGminusV(M) ui vi1113864 1113865( 1113857capNGminusV(M) uj vj1113966 11139671113872 1113873 empty
for 1le ine jle s + 2(5)
Without loss of generality we may assumew1 w2 w3 w41113864 1113865subeH1
Claim 1 k 1
Proof Suppose for the sake of contradiction that kge 2Since |Hi|ge 5 and δ(H)ge 2 we have αprime(Hi)ge 2 for every1le ile k Without loss of generality we can assumeuℓvℓ umvm1113864 1113865subeH2 where 1le ℓ ltmle s + 2
As H2 is connected there exists a path Q from ui1 vi1
1113966 1113967
to ui2 vi2
1113966 1113967 avoiding all vertices of V(M) ui1 vi1
ui2 vi2
1113966 1113967
for some ℓ le i1 lt i2 lem Again G[ ui1 vi1
ui2 vi2
1113966 1113967cupV(Q)]
has a P4 which combining the edges in M minus ui1vi1
ui2vi2
1113966 1113967creates a 2P4 cup sP2 in G a contradiction Claim 1 is true
It follows from Claim 1 that H H1 Hence in the restof the proof of Lemma 6 H1 is denoted by H Clearly|H|ge 2s + 8ge 2t + 8
Claim 2 For each 1le ile s + 2 we have NG( ui vi1113864 1113865)capw1 w2 w3 w41113864 1113865neempty
Proof Suppose for the sake of contradiction that thereexists 1le ile s + 2 such that NG( ui vi1113864 1113865)cap w11113864 w2 w3
w4 empty is together with (4) implies that
NG ui( 1113857capV(M) vi1113864 1113865 andNG vi( 1113857capV(M) ui1113864 1113865 (6)
We are now ready to show that
dG ui( 1113857 + dG vi( 1113857ge 5 (7)
Otherwise dG(ui) dG(vi) 2 since δ(H)ge 2 It fol-lows from (6) that we can assume w isin NHminusV(M)(ui) ByLemma 2 wvi isin E(G) en M minus uivi1113864 1113865 + uiw1113864 1113865 is a copy ofP4 cup (s + 2)P2 in G However by Lemma 4 we havedG(w)ge 4 and hence dG(ui) + dG(w)ge 6gt dG(ui)+
dG(vi) 4 which contradicts the choice of M en wehave (7) which holds
Combining (6) and (7) there exist z1 ne z2 such thatz1 isin NHminusV(M)(ui) and z2 isin NHminusV(M)(vi) en the sub-graph G[ z1 ui vi z21113864 1113865] will have a P4 which combining theedges in M minus uivi1113864 1113865 creates a 2P4 cup (s + 1)P2 in G Sincesge t we obtain a contradiction Claim 2 is true
Claim 3 There does not exist 1le i jle s + 2 such that
NG ui vi1113864 1113865( 1113857cap w1 w21113864 1113865neempty andNG uj vj1113966 11139671113872 1113873cap w3 w41113864 1113865neempty
(8)
Proof Suppose for the sake of contradiction that there exist1le i jle s + 2 such that NG( ui vi1113864 1113865)cap w1 w21113864 1113865neempty andNG( uj vj1113966 1113967)cap w3 w41113864 1113865neempty By Claim 2 we may assumeine j However the subgraphs G[ ui vi w1 w21113864 1113865] andG[ uj vj w3 w41113966 1113967] both contain P4 which combining theedges in M minus uivi ujvj w1w2 w2w3 w3w41113966 1113967 creates a2P4 cup sP2 in G a contradiction Hence Claim 3 is true
Claim 4 For each 1le ile s + 2 we have |NG( ui vi1113864 1113865)capw1 w2 w3 w41113864 1113865| 1 or |E(G)|gt 3t + 10
Proof Suppose for the sake of contradiction that Claim 4 isfalse en by Claim 3 and the symmetry we may assume|E(G)|le 3t + 10 and
NG u1 v11113864 1113865( 1113857cap w1 w2 w3 w41113864 1113865 w1 w21113864 1113865 (9)
en by Claim 3 we have
NG uj vj1113966 11139671113872 1113873cap w1 w2 w3 w41113864 1113865sube w1 w21113864 1113865
for each 1le jle s + 2(10)
Case 1 NG(u1)cap w1 w21113864 1113865neempty and NG(v1)cap w1 w21113864 1113865neempty
Mathematical Problems in Engineering 3
Combining (9) without loss of generality in this case wemay assume u1w1 isin E(G) and v1w2 isin E(G) Now weconsider ujvj where 2le jle s + 2
If w1 isin NG( uj vj1113966 1113967) then subgraphs G[ uj vj w1 u11113966 1113967]
and G[ v1 w2 w3 w41113864 1113865] both contain P4 which combiningthe edges in M minus u1v1 ujvj w1w2 w2w3 w3w41113966 1113967 creates a2P4 cup sP2 in G a contradiction Hence we havew1 notin NG( uj vj1113966 1113967) Together with Claim 2 and (10) weobtain
NG uj vj1113966 11139671113872 1113873cap w1 w2 w3 w41113864 1113865 w21113864 1113865 (11)
Without loss of generality we may assume ujw2 isin E(G)If NHminusV(M)(vj)neempty assume Gprime G[NHminusV(M) [ uj1113966
vj]] In order to avoid a 2P4 cup tP2 in G we get that Gprime K3or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G) minus V(Gprime)) w21113864 1113865which contradicts Lemma 5 Hence NHminusV(M)(vj) emptyis together with δ(H)ge 2 and (11) implies that dG(vj) 2and vjw2 isin E(G) As the argument of vj we haveNHminusV(M)(uj) empty Hence dG(uj) 2 en we can con-clude that
dG uj1113872 1113873 dG vj1113872 1113873 2 andNG uj1113872 1113873capNG vj1113872 1113873 w21113864 1113865
for any 1le jle s + 2
(12)
And by the choice of M dG(w3) dG(w4) 2 Henceby Lemma 2 w2w4 isin E(G) Combining (12) implies thatdG(w2)ge 2(s + 3) If dG(w1) 2 by Lemma 2 u1w2 isinE(G) Hence dG(u1) + dG(v1) + dG(w1)ge 7 If dG(w1)ge 3we also have dG(u1) + dG(v1) + dG(w1)ge 7 is togetherwith dG(w2)ge 2(s + 3) |H|ge 2t + 8 sge t and δ(H)ge 2implies that 2|E(H)|ge 2(s + 3) + 7+ 2(2t + 4)gt 6t + 20 acontradiction
Case 2 NG(u1)cap w1 w21113864 1113865 empty or NG(v1)cap w1 w21113864 1113865 emptyCombining (9) without loss of generality in this case we
may assume NG(v1)cap w1 w2 w3 w41113864 1113865 empty is combiningδ(H)ge 2 and (4) we get NHminusV(M)(v1)neempty and |H|ge 2t + 9Assume v1x isin E(G) where x isin H minus V(M) By our as-sumption NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w1 w21113864 1113865 we getu1w1 isin E(G) and u1w2 isin E(G) Now we consider ujvjwhere 2le jle s + 2
If w2 isin NG( uj vj1113966 1113967) then the subgraphs G[ uj1113966 vj
w2 w3] and G[ x v1 u1 w11113864 1113865] both contain P4 whichcombining the edges in M minus u1v1 ujvj w1w21113966 w2w3 w3w4creates a 2P4 cup sP2 in G a contradiction Hence we havew2 notin NG( uj vj1113966 1113967) is together with Claim 2 and (9) impliesthat NG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w11113864 1113865 for any 2le jles + 2Without loss of generality wemay assume ujw1 isin E(G)en we have NG(x)cap w1 w2 w3 w41113864 1113865 empty So by thechoice of M (5) and δ(H)ge 2 we get xu1 isin E(G) Hence
dG u1( 1113857ge 4 (13)
In order to avoid a 2P4 cup tP2 in G we get thatNHminusV(M)(vj) empty is together with δ(H)ge 2 andNG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w11113864 1113865 implies that
vjw1 isin E(G) and dG(vj) 2 Similarly we have dG(uj) 2erefore ujw1 isin E(G) and vjw1 isin E(G) for any2le jle s + 2 Hence combining sge t implies that
dG w1( 1113857ge 2(s + 2)ge 2t + 4 (14)
And by the choice of M we have dG(w3) dG(w4) 2en by Lemma 2 w2w4 isin E(G) which implies that
dG w2( 1113857ge 4 (15)
Recall that |H|ge 2t + 9 and δ(H)ge 2 Together with(13)ndash(15) we can get that 2|E(H)| ge 2t + 4 + 4 + 4+
2(2t + 6)gt 6t + 20 a contradiction Claim 4 is true
Claim 5 If there exists 1le ile s + 2 such thatdG(ui) + dG(vi)le 4 then we have |E(G)|ge 3t + 10 More-over the equality |E(G)| 3t + 10 holds if and only ifG Tpq cup (n minus 2t minus 8)K1 where p qge 2 and p + q t + 3
Proof Without loss of generality we may assumedG(u1) + dG(v1)le 4 Since δ(H)ge 2 we havedG(u1) dG(v1) 2 By Claim 4 and the symmetry wecomplete the Claim 5 by following two cases
Case 1 NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w11113864 1113865Since dG(u1) dG(v1) 2 by Lemma 2 u1w1 isin E(G)
and v1w1 isin E(G) By the choice of M dG(w3)
dG(w4) 2 Also by Lemma 2 w2w4 isin E(G) Now weconsider ujvj where 2le jle s + 2 It follows from Claim3 and our assumption that NG( uj vj1113966 1113967)cap w1 w2 w31113864
w4sube w1 w21113864 1113865 where 2le jle s + 2 is together withClaim 4 implies that NG( uj vj1113966 1113967)cap w1 w21113864 w3
w4 wi1113864 1113865 where i 1 or 2If NHminusV(M)(vj)neempty assume Gprime G[NHminusV(M)[ uj1113966
vj]] In order to avoid a 2P4 cup tP2 in G we get thatGprime K3 or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G)minus
V(Gprime)) wi1113864 1113865 which contradicts Lemma 5 HenceNHminusV(M)(vj) empty Since δ(H)ge 2 we get vjwi isin E(G)
and dG(vj) 2 By the similar argument we haveNHminusV(M)(uj) empty and hence dG(uj) 2 ereforefor each 1le jle s + 2 we can conclude that NG(uj)
capNG(vj) w11113864 1113865 or NG(uj)capNG(vj) w21113864 1113865 To-gether with w2w4 isin E(G) we have dG(w1)+ dG(w2)ge2(s + 3) + 2 2s + 8 Recall that sge t |H|ge 2t + 8 andδ(H)ge 2 we can get that 2|E(H)| ge 2s+ 8+ 2(2t+
6)ge 6t + 20 And if |E(G)| 3t + 10 then G
Tpq cup (n minus 2t minus 8)K1 where p + q t + 3 However ifmin p q1113864 1113865le 1 Tpq is not a (2P4 cup tP2)-saturated graphsince the addition of an edge between an isolated vertexof G and the vertex of degree Δ(G) does not result in a2P4 cup tP2 Hence p qge 2 Claim 5 holdsCase 2 NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w21113864 1113865
Since dG(u1) dG(v1) 2 by Lemma 2 u1w2 isin E(G)
and v1w2 isin E(G) By the choice of M we havedG(w3) dG(w4) 2 Also by Lemma 2 w2w4 isin E(G)Now we consider ujvj where 2le jle s + 2 It follows fromClaim 3 and our assumption that NG( uj vj1113966 1113967)cap w11113864 w2
w3 w4sube w1 w21113864 1113865
4 Mathematical Problems in Engineering
If there exists 1le jle s + 2 such that NG( uj1113966 vj) capw1 w2 w3 w41113864 1113865 w11113864 1113865 we can claim that dG(uj)
dG(vj) 2 Otherwise assume Gprime G[NHminusV(M)[ uj vj1113966 1113967]]In order to avoid a 2P4 cup tP2 in G we get that Gprime K3 orGprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G) minus V(Gprime)) w11113864 1113865which contradicts Lemma 5 en as the similar argumentin the Case 1 we can obtain |E(H)|ge 3t + 10 and if|E(G)| 3t + 10 then G Tpq where p qge 2 and p + q
t + 3 en NG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w21113864 1113865 for all1le jle s + 2 Without loss of generality we may assumeujw2 isin E(G) If NHminusV(M)(vj)neempty assume Gprime G[[NHminusV(M)[ uj vj1113966 1113967]]] In order to avoid a 2P4 cup tP2 in Gwe get that Gprime K3 or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap(V(G) minus V(Gprime)) w21113864 1113865 which contradicts Lemma 5Hence NHminusV(M)(vj) empty Since δ(H)ge 2 we get vjw2 isinE(G) and dG(vj) 2 By the similar argument we haveNHminusV(M)(uj) empty and hence dG(uj) 2 ereforeujw2 isin E(G) and vjw2 isin E(G) for any 2le jle s + 2 Hence
dG w2( 1113857ge 2(s + 3) + 1ge 2t + 7 (16)
is together with δ(H)ge 2 |H|ge 2t + 8 and (16) im-plies that 2|E(H)| ge 2t + 7 + 2(2t + 7)gt 6t + 20 is com-pletes the proof of Claim 5
If there exists 1le ile s + 2 such that dG(ui) + dG(vi)le 4by Claim 5 the Lemma is proven by Claim 5 Hence we canassume dG(ui) + dG(vi)ge 5 for each 1le ile s + 2 On thecontrary by (4) and Claim 4 we can assumedV(M)(ui) + dV(M)(vi)le 4 Hence there existsxi isin NG( uj vj1113966 1113967)cap (V(H) minus V(M)) for any 1le ile s + 2And by (5) xi nexj for 1le ine jle s + 2 Hence|H|ge 3(s + 2) + 4ge 3t + 10 Since δ(H)ge 2 we get|E(H)|ge 3t + 10 And if the equality |E(G)| 3t + 10 holdsthen δ(H) Δ(H) 2 and |H| 3t + 10 which contradictsLemma 4 Lemma 6 is true
Lemma 7 Suppose G is (2P4 cup tP2)-saturated and (P4 cup(t + 2)P2)-saturated with V0(G)neempty Let H1 Hk be thenontrivial components of G and H G[cup i 1 i
kV(Hi)] If δ(H)ge 2 and |H|ge 2t + 8 with |Hi|ge5(1le ile k) we have |E(G)|gt 3t + 10
Proof Suppose for the sake of contradiction that|E(G)|le 3t + 10 Since G is (2P4 cup tP2)-saturated thereexists a 2P4 cup tP2 in G + e for any e notin E(G) Henceαprime(H)ge t + 3 If αprime(H)ge t + 4 G must contain a copy of(t + 4)P2 Since δ(H)ge 2 and |Hi|ge 5(1le ile k) it is clearlythat H has a P4 cup (t + 2)P2 which contradicts G is(P4 cup (t + 2)P2)-saturated So we have αprime(H) t + 3 ByLemma 1 we obtain
t + 3 12min |H| +|X| minus o(G minus X) XsubeV(H) (17)
en we can choose a maximal subset SsubeV(H) whichsatisfies
t + 3 12
(|H| +|S| minus o(H minus S)) (18)
Suppose H1prime Hpprime are the components of H minus S We
have the following claims
Claim 6 H[ScupV(Hiprime)] is the complete graph where
1le ilep
Proof Suppose for the sake of contradiction thatH[ScupV(Hi
prime)] is not a complete graph en there isx y isin ScupV(Hi
prime) with xney and xy notin E(H) LetHprime H + xy We have |Hprime| |H| ando(Hprime minus S) o(H minus S) Combining (18) we obtain
t + 3 12
Hprime1113868111386811138681113868
1113868111386811138681113868 +|S| minus o(Hprime minus S)1113874 1113875 (19)
By Lemma 1 we obtain αprime(Hprime)le t + 3 Hence G + xy
has no (t + 4)P2 contrary to G which is (2P4 cup tP2)-satu-rated Claim 6 is true
Claim 7 Sneempty
Proof Suppose for the sake of contradiction that S emptyHence H1prime Hp
prime are all components of H Using Claim 6we have Hi
prime is a complete graph with |Hiprime|ge 5(1le ilep)
Hence we have δ(H)ge 4 and
2|E(H)| 1113944xisinV(H)
dH(x)ge 4|H| (20)
Combining |H|ge 2t + 8 we get |E(H)|gt 3t + 10 HenceClaim 7 is true
Let w isin V0(G) and x isin S By Claim 6NH(x) V(H) minus x Let F be a 2P4 cup tP2 in G + xw usingLemma 3 we get V(F)supe x w cupNG(x) Hence we have|H| + 1 |NH(x)cup x w |le |V(F)| 2t + 8 contrary to G
is (2P4 cup tP2)-saturated Hence Lemma 7 is trueNow we prove eorem 2
Proof of eorem 2 Clearly Tpq cup aK3 cup (nminus 2tminus aminus
8)K1 isin SAT(n 2P4 cup tP2) where pge 2 qge 2 and p + q+
a t + 3 Let G isin SAT (n 2P4 cup tP2) then |E(G)|le 3t + 10Since nge 6t + 12 |V0(G)|ge 2 Using Lemma 3 we haveV1(G)neempty Clearly in graph G all components of order 3 or4 are complete If there exists one of the components sayH1 K4 then the addition of an edge e between an isolatedvertex of G and a vertex of H1 results in a 2P4 cup tP2 in GHowever the edge e can be replaced by the edge in H1resulting in a 2P4 cup tP2 isin G a contradiction Hence com-ponent of order 4 in G does not exist
Now we consider the graph Gprime which is obtained bydeleting all components of order 3 from G Let a be thenumber of components of order 3 in G and we obtain
G Gprime cup aK3 (21)
Note |V0(Gprime)| |V0(G)|ge 2 Let e be an edge joiningtwo isolated vertices in V0(G) then there exists a 2P4 cup tP2in G + e So Gprime has 2P4 Combining G isin SAT (n 2P4 cup tP2)we get ale t minus 1 Let tprime t minus a en tprime ge 1
As G isin SAT (n 2P4 cup tP2) we have Gprime isin SAT(nprime2P4 cup tP2) where nprime n minus 3a Suppose Hprime is a graph
Mathematical Problems in Engineering 5
spanned by all nontrivial components of Gprime Using Lemma 3we have δ(Hprime)ge 2 Clearly any component in Hprime has orderat least 5 Moreover graph Gprime is (2P4 cup tprimeP2)-saturated suchthat |E(Gprime)| |E(G)| minus 3ale 3tprime + 10 and V0(Gprime)neempty ByLemma 6 and Lemma 7 we have |Hprime|le 2tprime + 7 or Hprime Tpq cup (nprime minus 2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3If |Hprime|le 2tprime + 7 as V0(Gprime)neempty and Gprime isin SAT(nprime
2P4 cup tprimeP2) Hprime K2tprime+7 Now E(Hprime) 2tprime + 7
21113888 1113889gt
3tprime + 21 and we have |E(G)|gt 3tprime + 21 + 3a 3t+
21gt 3t + 10 a contradiction Hence Hprime Tpq cup (nprimeminus2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3 Combining(21) we get G Tpq cup aK3 cup (n minus 2t minus a minus 8)K1 wherepge 2 qge 2 and p + q + a t + 3 Hence eorem 2 istrue
4 Conclusion
Forests and trees have attracted wide attention for severalreasons Firstly their simplicity has made somemore preciseresults possible Secondly they can provide inspiration forsome larger results At present we only know the saturationnumber of a few species of trees but the saturation numberof most trees is unknown
In [20] Chen et al focused on the saturation numbersfor the linear forests ey obtained an interesting set ofresults and proposed some conjectures We pave the way forthe following conjectures proposed by Chen et al in [20]
Conjecture 1 (see [20]) Let tge 2 be an integer For n suf-ficiently large sat(n tP3) lfloor(n + 6t minus 6)2rfloor
Conjecture 2 (see [20]) Let tge 2 be an integer For n suf-ficiently large
sat n tP4( 1113857
n + 12t minus 122
if n is even
n + 12t minus 112
if n is odd
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
(22)
Conjecture 3 (see [20]) For n sufficiently large kge 4and kle ℓ le lceil(3k minus 2)2rceil sat(n Pk cupPℓ) n minus 1113860nak1113861 + 3where
ak 3 middot 2mminus 1
minus 2 if k 2m
4 middot 2mminus 1minus 2 if k 2m + 1
⎧⎨
⎩ (23)
Data Availability
No data used to support the findings of the study
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was partially supported by NSFC (nos 12001172and 11771172) and supported by Natural Science Founda-tion of Henan Province (no 202300410212) and Science andtechnology innovation fund of Henan Agricultural Uni-versity (nos KJCX2019A15 and KJCX2017A12)
References
[1] J A Bondy and U S R Murty Graph 5eory with Appli-caitons American Elsevier New York NY USA 1976
[2] P Turan ldquoEine Extremalaufgabe aus der GraphentheorierdquoMat Fiz Lapok vol 48 pp 436ndash452 1941
[3] P Erdős A Hajnal and J W Moon ldquoA problem in graphtheoryrdquo 5e American Mathematical Monthly vol 71pp 1107ndash1110 1964
[4] L Kaszonyi and Z Tuza ldquoSaturated graphs with minimalnumber of edgesrdquo Journal of Graph 5eory vol 10 no 2pp 203ndash210 1986
[5] H Lei O Suil D BWest and X Zhu ldquoExtremal problems onsaturation for the family ofk-edge-connected graphsrdquoDiscreteApplied Mathematics vol 260 pp 278ndash283 2019
[6] T Bohman M Fonoberova and O Pikhurko ldquoe saturationfunction of complete partite graphsrdquo Journal of Combina-torics vol 1 no 2 pp 149ndash170 2010
[7] J R Faudree M Ferrara R J Gould andM S Jacobson ldquotKp-saturated graphs of minimum sizerdquo Discrete Mathematicsvol 309 no 19 pp 5870ndash5876 2009
[8] B Bollobas ldquoOn a conjecture of erdos hajnal and moonrdquo5eAmerican Mathematical Monthly vol 74 no 2 pp 178-1791967
[9] S Huang H Lei Y Shi and J Zhang ldquoe saturation numberof K33rdquo 2019 httparxivorgabs191004967
[10] R J Faudree and R J Gould ldquoSaturation numbers for nearlycomplete graphsrdquo Graphs and Combinatorics vol 29 no 3pp 429ndash448 2013
[11] G Chen R J Faudree and R J Gould ldquoSaturation numbersof booksrdquo 5e Electronic Journal of Combinatorics vol 15no 1 pp 118ndash129 2008
[12] Y-C Chen ldquoMinimumC5-saturated graphsrdquo Journal ofGraph 5eory vol 61 no 2 pp 111ndash126 2009
[13] Y-C Chen ldquoAll minimum C5-saturated graphsrdquo Journal ofGraph 5eory vol 67 no 1 pp 9ndash26 2011
[14] L T Ollmann ldquo-saturated graphs with a minimal number ofedgesrdquo in Proceedings of the 5ird Southeastern Conference onCombinatorics Graph 5eory and Computing Florida At-lantic University Boca Raton FL USA 1972
[15] C Timmons ldquo$$C_2k$$-Saturated graphs with No shortodd cyclesrdquo Graphs and Combinatorics vol 35 no 5pp 1023ndash1034 2019
[16] Z Tuza ldquo-saturated graphs of minimum sizerdquo Acta Uni-versitatis Carolinae Mathematica et Physica vol 30 no 2pp 161ndash167 1989
[17] M Zhang S Luo and M Shigeno ldquoOn the number of edgesin a minimum $$C_6$$ C 6 -saturated graphrdquo Graphs andCombinatorics vol 31 no 4 pp 1085ndash1106 2015
[18] J R Faudree R J Faudree R J Gould and M S JacobsonldquoSaturation numbers for treesrdquo 5e Electronic Journal ofCombinatorics vol 16 no 1 pp 91ndash109 2009
[19] F F Song and J J Zhou ldquoe partite saturation number ofspiderrdquo Applied Mathematics and Computation vol 394Article ID 125793 2021
6 Mathematical Problems in Engineering
[20] G Chen J R Faudree R J Faudree R J GouldM S Jacobson and C Magnant ldquoResults and problemson saturation numbers for linear forestsrdquo Bulletin of theInstitute of Combinatorics and its Applications vol 75pp 29ndash46 2015
[21] Q Fan and C Wang ldquoSaturation numbers for linear forests$$P_5 cup tP_2$$ P 5 cup t P 2rdquo Graphs and Combinatoricsvol 31 no 6 pp 2193ndash2200 2015
[22] F F Song ldquoSaturation numbers for linear forestsrdquo UtilitasMathematica vol 104 pp 175ndash186 2017
[23] J R Faudree R J Faudree and J R Schmitt ldquoA survey ofminimum saturation graphsrdquo 5e Electronic Journal ofCombinatorics vol 18 p 36pp 2011
[24] N Bushaw and N Kettle ldquoTuran numbers of multiple pathsand equibipartite forestsrdquo Combinatorics Probability andComputing vol 20 no 6 pp 837ndash853 2011
[25] C Berge ldquoSur le couplage maximum dprimeun grapherdquo Comptesrendus de lrsquoAcademie des Sciences vol 247 pp 258-259 1958
Mathematical Problems in Engineering 7
selected as a P2 in F We can choose u2u3 as P2 and thenreplace wx in F with xu1 we obtain a 2P4 cup tP2 in G acontradiction
Now consider the case Gprime K1ℓ Suppose u is the centerof the star and V(K1ℓ) u u1 uℓ1113864 1113865 Since G is(2P4 cup tP2)-saturated with V0(G)neempty by Lemma 3V1(G) empty en we have uix isin E(G) for 1le ile ℓ HencedG(ui) 2 and NG(ui) u x for all 1le ile ℓ whichcontradicts Lemma 4 Lemma 5 is true
3 Proof of Theorem 2
Before the proof of eorem 2 we discuss the property of(2P4 cup tP2)-saturated graph which has no components oforder 2 3 and 4
Lemma 6 Suppose G is (2P4 cup tP2)-saturated but not(P4 cup (t + 2)P2)-saturated graph Let V0(G)neempty H1
Hk be the nontrivial components of G and H G[cup i 1
i k V(Hi)] If δ(H)ge 2 |H|ge 2t + 8 and |Hi|ge 5(1le ile k) we have |E(G)|ge 3t + 10
Moreover if the equality |E(G)| 3t + 10 holds thenG Tpq cup (n minus 2t minus 8)K1 where p qge 2 and p + q t + 3
Proof Since G is (2P4 cup tP2)-saturated G + e has a P4 cup(t + 2)P2 where e notin E(G) On the contrary G must containa P4 cup (t + 2)P2 since G is not (P4 cup (t + 2)P2)-saturated
Suppose M is a P4 cup (s + 2)P2 in G with E(M)
w1w2 w2w3 w3w41113864 1113865cup u1v1 us+2vs+21113864 1113865 Choose M
which satisfies the following
(i) s is as large as possible(ii) Subject to (i) 1113936
s+2i1(dG(ui) + dG(vi)) is maximality
Observe that sge t By the choice of M V(H) minus V(M) isan independent set or an empty set As G is(2P4 cup tP2)-saturated we have
G V(M) w1 w2 w3 w41113864 11138651113858 1113859 (s + 2)P2 (4)
NGminusV(M) ui vi1113864 1113865( 1113857capNGminusV(M) uj vj1113966 11139671113872 1113873 empty
for 1le ine jle s + 2(5)
Without loss of generality we may assumew1 w2 w3 w41113864 1113865subeH1
Claim 1 k 1
Proof Suppose for the sake of contradiction that kge 2Since |Hi|ge 5 and δ(H)ge 2 we have αprime(Hi)ge 2 for every1le ile k Without loss of generality we can assumeuℓvℓ umvm1113864 1113865subeH2 where 1le ℓ ltmle s + 2
As H2 is connected there exists a path Q from ui1 vi1
1113966 1113967
to ui2 vi2
1113966 1113967 avoiding all vertices of V(M) ui1 vi1
ui2 vi2
1113966 1113967
for some ℓ le i1 lt i2 lem Again G[ ui1 vi1
ui2 vi2
1113966 1113967cupV(Q)]
has a P4 which combining the edges in M minus ui1vi1
ui2vi2
1113966 1113967creates a 2P4 cup sP2 in G a contradiction Claim 1 is true
It follows from Claim 1 that H H1 Hence in the restof the proof of Lemma 6 H1 is denoted by H Clearly|H|ge 2s + 8ge 2t + 8
Claim 2 For each 1le ile s + 2 we have NG( ui vi1113864 1113865)capw1 w2 w3 w41113864 1113865neempty
Proof Suppose for the sake of contradiction that thereexists 1le ile s + 2 such that NG( ui vi1113864 1113865)cap w11113864 w2 w3
w4 empty is together with (4) implies that
NG ui( 1113857capV(M) vi1113864 1113865 andNG vi( 1113857capV(M) ui1113864 1113865 (6)
We are now ready to show that
dG ui( 1113857 + dG vi( 1113857ge 5 (7)
Otherwise dG(ui) dG(vi) 2 since δ(H)ge 2 It fol-lows from (6) that we can assume w isin NHminusV(M)(ui) ByLemma 2 wvi isin E(G) en M minus uivi1113864 1113865 + uiw1113864 1113865 is a copy ofP4 cup (s + 2)P2 in G However by Lemma 4 we havedG(w)ge 4 and hence dG(ui) + dG(w)ge 6gt dG(ui)+
dG(vi) 4 which contradicts the choice of M en wehave (7) which holds
Combining (6) and (7) there exist z1 ne z2 such thatz1 isin NHminusV(M)(ui) and z2 isin NHminusV(M)(vi) en the sub-graph G[ z1 ui vi z21113864 1113865] will have a P4 which combining theedges in M minus uivi1113864 1113865 creates a 2P4 cup (s + 1)P2 in G Sincesge t we obtain a contradiction Claim 2 is true
Claim 3 There does not exist 1le i jle s + 2 such that
NG ui vi1113864 1113865( 1113857cap w1 w21113864 1113865neempty andNG uj vj1113966 11139671113872 1113873cap w3 w41113864 1113865neempty
(8)
Proof Suppose for the sake of contradiction that there exist1le i jle s + 2 such that NG( ui vi1113864 1113865)cap w1 w21113864 1113865neempty andNG( uj vj1113966 1113967)cap w3 w41113864 1113865neempty By Claim 2 we may assumeine j However the subgraphs G[ ui vi w1 w21113864 1113865] andG[ uj vj w3 w41113966 1113967] both contain P4 which combining theedges in M minus uivi ujvj w1w2 w2w3 w3w41113966 1113967 creates a2P4 cup sP2 in G a contradiction Hence Claim 3 is true
Claim 4 For each 1le ile s + 2 we have |NG( ui vi1113864 1113865)capw1 w2 w3 w41113864 1113865| 1 or |E(G)|gt 3t + 10
Proof Suppose for the sake of contradiction that Claim 4 isfalse en by Claim 3 and the symmetry we may assume|E(G)|le 3t + 10 and
NG u1 v11113864 1113865( 1113857cap w1 w2 w3 w41113864 1113865 w1 w21113864 1113865 (9)
en by Claim 3 we have
NG uj vj1113966 11139671113872 1113873cap w1 w2 w3 w41113864 1113865sube w1 w21113864 1113865
for each 1le jle s + 2(10)
Case 1 NG(u1)cap w1 w21113864 1113865neempty and NG(v1)cap w1 w21113864 1113865neempty
Mathematical Problems in Engineering 3
Combining (9) without loss of generality in this case wemay assume u1w1 isin E(G) and v1w2 isin E(G) Now weconsider ujvj where 2le jle s + 2
If w1 isin NG( uj vj1113966 1113967) then subgraphs G[ uj vj w1 u11113966 1113967]
and G[ v1 w2 w3 w41113864 1113865] both contain P4 which combiningthe edges in M minus u1v1 ujvj w1w2 w2w3 w3w41113966 1113967 creates a2P4 cup sP2 in G a contradiction Hence we havew1 notin NG( uj vj1113966 1113967) Together with Claim 2 and (10) weobtain
NG uj vj1113966 11139671113872 1113873cap w1 w2 w3 w41113864 1113865 w21113864 1113865 (11)
Without loss of generality we may assume ujw2 isin E(G)If NHminusV(M)(vj)neempty assume Gprime G[NHminusV(M) [ uj1113966
vj]] In order to avoid a 2P4 cup tP2 in G we get that Gprime K3or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G) minus V(Gprime)) w21113864 1113865which contradicts Lemma 5 Hence NHminusV(M)(vj) emptyis together with δ(H)ge 2 and (11) implies that dG(vj) 2and vjw2 isin E(G) As the argument of vj we haveNHminusV(M)(uj) empty Hence dG(uj) 2 en we can con-clude that
dG uj1113872 1113873 dG vj1113872 1113873 2 andNG uj1113872 1113873capNG vj1113872 1113873 w21113864 1113865
for any 1le jle s + 2
(12)
And by the choice of M dG(w3) dG(w4) 2 Henceby Lemma 2 w2w4 isin E(G) Combining (12) implies thatdG(w2)ge 2(s + 3) If dG(w1) 2 by Lemma 2 u1w2 isinE(G) Hence dG(u1) + dG(v1) + dG(w1)ge 7 If dG(w1)ge 3we also have dG(u1) + dG(v1) + dG(w1)ge 7 is togetherwith dG(w2)ge 2(s + 3) |H|ge 2t + 8 sge t and δ(H)ge 2implies that 2|E(H)|ge 2(s + 3) + 7+ 2(2t + 4)gt 6t + 20 acontradiction
Case 2 NG(u1)cap w1 w21113864 1113865 empty or NG(v1)cap w1 w21113864 1113865 emptyCombining (9) without loss of generality in this case we
may assume NG(v1)cap w1 w2 w3 w41113864 1113865 empty is combiningδ(H)ge 2 and (4) we get NHminusV(M)(v1)neempty and |H|ge 2t + 9Assume v1x isin E(G) where x isin H minus V(M) By our as-sumption NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w1 w21113864 1113865 we getu1w1 isin E(G) and u1w2 isin E(G) Now we consider ujvjwhere 2le jle s + 2
If w2 isin NG( uj vj1113966 1113967) then the subgraphs G[ uj1113966 vj
w2 w3] and G[ x v1 u1 w11113864 1113865] both contain P4 whichcombining the edges in M minus u1v1 ujvj w1w21113966 w2w3 w3w4creates a 2P4 cup sP2 in G a contradiction Hence we havew2 notin NG( uj vj1113966 1113967) is together with Claim 2 and (9) impliesthat NG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w11113864 1113865 for any 2le jles + 2Without loss of generality wemay assume ujw1 isin E(G)en we have NG(x)cap w1 w2 w3 w41113864 1113865 empty So by thechoice of M (5) and δ(H)ge 2 we get xu1 isin E(G) Hence
dG u1( 1113857ge 4 (13)
In order to avoid a 2P4 cup tP2 in G we get thatNHminusV(M)(vj) empty is together with δ(H)ge 2 andNG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w11113864 1113865 implies that
vjw1 isin E(G) and dG(vj) 2 Similarly we have dG(uj) 2erefore ujw1 isin E(G) and vjw1 isin E(G) for any2le jle s + 2 Hence combining sge t implies that
dG w1( 1113857ge 2(s + 2)ge 2t + 4 (14)
And by the choice of M we have dG(w3) dG(w4) 2en by Lemma 2 w2w4 isin E(G) which implies that
dG w2( 1113857ge 4 (15)
Recall that |H|ge 2t + 9 and δ(H)ge 2 Together with(13)ndash(15) we can get that 2|E(H)| ge 2t + 4 + 4 + 4+
2(2t + 6)gt 6t + 20 a contradiction Claim 4 is true
Claim 5 If there exists 1le ile s + 2 such thatdG(ui) + dG(vi)le 4 then we have |E(G)|ge 3t + 10 More-over the equality |E(G)| 3t + 10 holds if and only ifG Tpq cup (n minus 2t minus 8)K1 where p qge 2 and p + q t + 3
Proof Without loss of generality we may assumedG(u1) + dG(v1)le 4 Since δ(H)ge 2 we havedG(u1) dG(v1) 2 By Claim 4 and the symmetry wecomplete the Claim 5 by following two cases
Case 1 NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w11113864 1113865Since dG(u1) dG(v1) 2 by Lemma 2 u1w1 isin E(G)
and v1w1 isin E(G) By the choice of M dG(w3)
dG(w4) 2 Also by Lemma 2 w2w4 isin E(G) Now weconsider ujvj where 2le jle s + 2 It follows from Claim3 and our assumption that NG( uj vj1113966 1113967)cap w1 w2 w31113864
w4sube w1 w21113864 1113865 where 2le jle s + 2 is together withClaim 4 implies that NG( uj vj1113966 1113967)cap w1 w21113864 w3
w4 wi1113864 1113865 where i 1 or 2If NHminusV(M)(vj)neempty assume Gprime G[NHminusV(M)[ uj1113966
vj]] In order to avoid a 2P4 cup tP2 in G we get thatGprime K3 or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G)minus
V(Gprime)) wi1113864 1113865 which contradicts Lemma 5 HenceNHminusV(M)(vj) empty Since δ(H)ge 2 we get vjwi isin E(G)
and dG(vj) 2 By the similar argument we haveNHminusV(M)(uj) empty and hence dG(uj) 2 ereforefor each 1le jle s + 2 we can conclude that NG(uj)
capNG(vj) w11113864 1113865 or NG(uj)capNG(vj) w21113864 1113865 To-gether with w2w4 isin E(G) we have dG(w1)+ dG(w2)ge2(s + 3) + 2 2s + 8 Recall that sge t |H|ge 2t + 8 andδ(H)ge 2 we can get that 2|E(H)| ge 2s+ 8+ 2(2t+
6)ge 6t + 20 And if |E(G)| 3t + 10 then G
Tpq cup (n minus 2t minus 8)K1 where p + q t + 3 However ifmin p q1113864 1113865le 1 Tpq is not a (2P4 cup tP2)-saturated graphsince the addition of an edge between an isolated vertexof G and the vertex of degree Δ(G) does not result in a2P4 cup tP2 Hence p qge 2 Claim 5 holdsCase 2 NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w21113864 1113865
Since dG(u1) dG(v1) 2 by Lemma 2 u1w2 isin E(G)
and v1w2 isin E(G) By the choice of M we havedG(w3) dG(w4) 2 Also by Lemma 2 w2w4 isin E(G)Now we consider ujvj where 2le jle s + 2 It follows fromClaim 3 and our assumption that NG( uj vj1113966 1113967)cap w11113864 w2
w3 w4sube w1 w21113864 1113865
4 Mathematical Problems in Engineering
If there exists 1le jle s + 2 such that NG( uj1113966 vj) capw1 w2 w3 w41113864 1113865 w11113864 1113865 we can claim that dG(uj)
dG(vj) 2 Otherwise assume Gprime G[NHminusV(M)[ uj vj1113966 1113967]]In order to avoid a 2P4 cup tP2 in G we get that Gprime K3 orGprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G) minus V(Gprime)) w11113864 1113865which contradicts Lemma 5 en as the similar argumentin the Case 1 we can obtain |E(H)|ge 3t + 10 and if|E(G)| 3t + 10 then G Tpq where p qge 2 and p + q
t + 3 en NG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w21113864 1113865 for all1le jle s + 2 Without loss of generality we may assumeujw2 isin E(G) If NHminusV(M)(vj)neempty assume Gprime G[[NHminusV(M)[ uj vj1113966 1113967]]] In order to avoid a 2P4 cup tP2 in Gwe get that Gprime K3 or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap(V(G) minus V(Gprime)) w21113864 1113865 which contradicts Lemma 5Hence NHminusV(M)(vj) empty Since δ(H)ge 2 we get vjw2 isinE(G) and dG(vj) 2 By the similar argument we haveNHminusV(M)(uj) empty and hence dG(uj) 2 ereforeujw2 isin E(G) and vjw2 isin E(G) for any 2le jle s + 2 Hence
dG w2( 1113857ge 2(s + 3) + 1ge 2t + 7 (16)
is together with δ(H)ge 2 |H|ge 2t + 8 and (16) im-plies that 2|E(H)| ge 2t + 7 + 2(2t + 7)gt 6t + 20 is com-pletes the proof of Claim 5
If there exists 1le ile s + 2 such that dG(ui) + dG(vi)le 4by Claim 5 the Lemma is proven by Claim 5 Hence we canassume dG(ui) + dG(vi)ge 5 for each 1le ile s + 2 On thecontrary by (4) and Claim 4 we can assumedV(M)(ui) + dV(M)(vi)le 4 Hence there existsxi isin NG( uj vj1113966 1113967)cap (V(H) minus V(M)) for any 1le ile s + 2And by (5) xi nexj for 1le ine jle s + 2 Hence|H|ge 3(s + 2) + 4ge 3t + 10 Since δ(H)ge 2 we get|E(H)|ge 3t + 10 And if the equality |E(G)| 3t + 10 holdsthen δ(H) Δ(H) 2 and |H| 3t + 10 which contradictsLemma 4 Lemma 6 is true
Lemma 7 Suppose G is (2P4 cup tP2)-saturated and (P4 cup(t + 2)P2)-saturated with V0(G)neempty Let H1 Hk be thenontrivial components of G and H G[cup i 1 i
kV(Hi)] If δ(H)ge 2 and |H|ge 2t + 8 with |Hi|ge5(1le ile k) we have |E(G)|gt 3t + 10
Proof Suppose for the sake of contradiction that|E(G)|le 3t + 10 Since G is (2P4 cup tP2)-saturated thereexists a 2P4 cup tP2 in G + e for any e notin E(G) Henceαprime(H)ge t + 3 If αprime(H)ge t + 4 G must contain a copy of(t + 4)P2 Since δ(H)ge 2 and |Hi|ge 5(1le ile k) it is clearlythat H has a P4 cup (t + 2)P2 which contradicts G is(P4 cup (t + 2)P2)-saturated So we have αprime(H) t + 3 ByLemma 1 we obtain
t + 3 12min |H| +|X| minus o(G minus X) XsubeV(H) (17)
en we can choose a maximal subset SsubeV(H) whichsatisfies
t + 3 12
(|H| +|S| minus o(H minus S)) (18)
Suppose H1prime Hpprime are the components of H minus S We
have the following claims
Claim 6 H[ScupV(Hiprime)] is the complete graph where
1le ilep
Proof Suppose for the sake of contradiction thatH[ScupV(Hi
prime)] is not a complete graph en there isx y isin ScupV(Hi
prime) with xney and xy notin E(H) LetHprime H + xy We have |Hprime| |H| ando(Hprime minus S) o(H minus S) Combining (18) we obtain
t + 3 12
Hprime1113868111386811138681113868
1113868111386811138681113868 +|S| minus o(Hprime minus S)1113874 1113875 (19)
By Lemma 1 we obtain αprime(Hprime)le t + 3 Hence G + xy
has no (t + 4)P2 contrary to G which is (2P4 cup tP2)-satu-rated Claim 6 is true
Claim 7 Sneempty
Proof Suppose for the sake of contradiction that S emptyHence H1prime Hp
prime are all components of H Using Claim 6we have Hi
prime is a complete graph with |Hiprime|ge 5(1le ilep)
Hence we have δ(H)ge 4 and
2|E(H)| 1113944xisinV(H)
dH(x)ge 4|H| (20)
Combining |H|ge 2t + 8 we get |E(H)|gt 3t + 10 HenceClaim 7 is true
Let w isin V0(G) and x isin S By Claim 6NH(x) V(H) minus x Let F be a 2P4 cup tP2 in G + xw usingLemma 3 we get V(F)supe x w cupNG(x) Hence we have|H| + 1 |NH(x)cup x w |le |V(F)| 2t + 8 contrary to G
is (2P4 cup tP2)-saturated Hence Lemma 7 is trueNow we prove eorem 2
Proof of eorem 2 Clearly Tpq cup aK3 cup (nminus 2tminus aminus
8)K1 isin SAT(n 2P4 cup tP2) where pge 2 qge 2 and p + q+
a t + 3 Let G isin SAT (n 2P4 cup tP2) then |E(G)|le 3t + 10Since nge 6t + 12 |V0(G)|ge 2 Using Lemma 3 we haveV1(G)neempty Clearly in graph G all components of order 3 or4 are complete If there exists one of the components sayH1 K4 then the addition of an edge e between an isolatedvertex of G and a vertex of H1 results in a 2P4 cup tP2 in GHowever the edge e can be replaced by the edge in H1resulting in a 2P4 cup tP2 isin G a contradiction Hence com-ponent of order 4 in G does not exist
Now we consider the graph Gprime which is obtained bydeleting all components of order 3 from G Let a be thenumber of components of order 3 in G and we obtain
G Gprime cup aK3 (21)
Note |V0(Gprime)| |V0(G)|ge 2 Let e be an edge joiningtwo isolated vertices in V0(G) then there exists a 2P4 cup tP2in G + e So Gprime has 2P4 Combining G isin SAT (n 2P4 cup tP2)we get ale t minus 1 Let tprime t minus a en tprime ge 1
As G isin SAT (n 2P4 cup tP2) we have Gprime isin SAT(nprime2P4 cup tP2) where nprime n minus 3a Suppose Hprime is a graph
Mathematical Problems in Engineering 5
spanned by all nontrivial components of Gprime Using Lemma 3we have δ(Hprime)ge 2 Clearly any component in Hprime has orderat least 5 Moreover graph Gprime is (2P4 cup tprimeP2)-saturated suchthat |E(Gprime)| |E(G)| minus 3ale 3tprime + 10 and V0(Gprime)neempty ByLemma 6 and Lemma 7 we have |Hprime|le 2tprime + 7 or Hprime Tpq cup (nprime minus 2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3If |Hprime|le 2tprime + 7 as V0(Gprime)neempty and Gprime isin SAT(nprime
2P4 cup tprimeP2) Hprime K2tprime+7 Now E(Hprime) 2tprime + 7
21113888 1113889gt
3tprime + 21 and we have |E(G)|gt 3tprime + 21 + 3a 3t+
21gt 3t + 10 a contradiction Hence Hprime Tpq cup (nprimeminus2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3 Combining(21) we get G Tpq cup aK3 cup (n minus 2t minus a minus 8)K1 wherepge 2 qge 2 and p + q + a t + 3 Hence eorem 2 istrue
4 Conclusion
Forests and trees have attracted wide attention for severalreasons Firstly their simplicity has made somemore preciseresults possible Secondly they can provide inspiration forsome larger results At present we only know the saturationnumber of a few species of trees but the saturation numberof most trees is unknown
In [20] Chen et al focused on the saturation numbersfor the linear forests ey obtained an interesting set ofresults and proposed some conjectures We pave the way forthe following conjectures proposed by Chen et al in [20]
Conjecture 1 (see [20]) Let tge 2 be an integer For n suf-ficiently large sat(n tP3) lfloor(n + 6t minus 6)2rfloor
Conjecture 2 (see [20]) Let tge 2 be an integer For n suf-ficiently large
sat n tP4( 1113857
n + 12t minus 122
if n is even
n + 12t minus 112
if n is odd
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
(22)
Conjecture 3 (see [20]) For n sufficiently large kge 4and kle ℓ le lceil(3k minus 2)2rceil sat(n Pk cupPℓ) n minus 1113860nak1113861 + 3where
ak 3 middot 2mminus 1
minus 2 if k 2m
4 middot 2mminus 1minus 2 if k 2m + 1
⎧⎨
⎩ (23)
Data Availability
No data used to support the findings of the study
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was partially supported by NSFC (nos 12001172and 11771172) and supported by Natural Science Founda-tion of Henan Province (no 202300410212) and Science andtechnology innovation fund of Henan Agricultural Uni-versity (nos KJCX2019A15 and KJCX2017A12)
References
[1] J A Bondy and U S R Murty Graph 5eory with Appli-caitons American Elsevier New York NY USA 1976
[2] P Turan ldquoEine Extremalaufgabe aus der GraphentheorierdquoMat Fiz Lapok vol 48 pp 436ndash452 1941
[3] P Erdős A Hajnal and J W Moon ldquoA problem in graphtheoryrdquo 5e American Mathematical Monthly vol 71pp 1107ndash1110 1964
[4] L Kaszonyi and Z Tuza ldquoSaturated graphs with minimalnumber of edgesrdquo Journal of Graph 5eory vol 10 no 2pp 203ndash210 1986
[5] H Lei O Suil D BWest and X Zhu ldquoExtremal problems onsaturation for the family ofk-edge-connected graphsrdquoDiscreteApplied Mathematics vol 260 pp 278ndash283 2019
[6] T Bohman M Fonoberova and O Pikhurko ldquoe saturationfunction of complete partite graphsrdquo Journal of Combina-torics vol 1 no 2 pp 149ndash170 2010
[7] J R Faudree M Ferrara R J Gould andM S Jacobson ldquotKp-saturated graphs of minimum sizerdquo Discrete Mathematicsvol 309 no 19 pp 5870ndash5876 2009
[8] B Bollobas ldquoOn a conjecture of erdos hajnal and moonrdquo5eAmerican Mathematical Monthly vol 74 no 2 pp 178-1791967
[9] S Huang H Lei Y Shi and J Zhang ldquoe saturation numberof K33rdquo 2019 httparxivorgabs191004967
[10] R J Faudree and R J Gould ldquoSaturation numbers for nearlycomplete graphsrdquo Graphs and Combinatorics vol 29 no 3pp 429ndash448 2013
[11] G Chen R J Faudree and R J Gould ldquoSaturation numbersof booksrdquo 5e Electronic Journal of Combinatorics vol 15no 1 pp 118ndash129 2008
[12] Y-C Chen ldquoMinimumC5-saturated graphsrdquo Journal ofGraph 5eory vol 61 no 2 pp 111ndash126 2009
[13] Y-C Chen ldquoAll minimum C5-saturated graphsrdquo Journal ofGraph 5eory vol 67 no 1 pp 9ndash26 2011
[14] L T Ollmann ldquo-saturated graphs with a minimal number ofedgesrdquo in Proceedings of the 5ird Southeastern Conference onCombinatorics Graph 5eory and Computing Florida At-lantic University Boca Raton FL USA 1972
[15] C Timmons ldquo$$C_2k$$-Saturated graphs with No shortodd cyclesrdquo Graphs and Combinatorics vol 35 no 5pp 1023ndash1034 2019
[16] Z Tuza ldquo-saturated graphs of minimum sizerdquo Acta Uni-versitatis Carolinae Mathematica et Physica vol 30 no 2pp 161ndash167 1989
[17] M Zhang S Luo and M Shigeno ldquoOn the number of edgesin a minimum $$C_6$$ C 6 -saturated graphrdquo Graphs andCombinatorics vol 31 no 4 pp 1085ndash1106 2015
[18] J R Faudree R J Faudree R J Gould and M S JacobsonldquoSaturation numbers for treesrdquo 5e Electronic Journal ofCombinatorics vol 16 no 1 pp 91ndash109 2009
[19] F F Song and J J Zhou ldquoe partite saturation number ofspiderrdquo Applied Mathematics and Computation vol 394Article ID 125793 2021
6 Mathematical Problems in Engineering
[20] G Chen J R Faudree R J Faudree R J GouldM S Jacobson and C Magnant ldquoResults and problemson saturation numbers for linear forestsrdquo Bulletin of theInstitute of Combinatorics and its Applications vol 75pp 29ndash46 2015
[21] Q Fan and C Wang ldquoSaturation numbers for linear forests$$P_5 cup tP_2$$ P 5 cup t P 2rdquo Graphs and Combinatoricsvol 31 no 6 pp 2193ndash2200 2015
[22] F F Song ldquoSaturation numbers for linear forestsrdquo UtilitasMathematica vol 104 pp 175ndash186 2017
[23] J R Faudree R J Faudree and J R Schmitt ldquoA survey ofminimum saturation graphsrdquo 5e Electronic Journal ofCombinatorics vol 18 p 36pp 2011
[24] N Bushaw and N Kettle ldquoTuran numbers of multiple pathsand equibipartite forestsrdquo Combinatorics Probability andComputing vol 20 no 6 pp 837ndash853 2011
[25] C Berge ldquoSur le couplage maximum dprimeun grapherdquo Comptesrendus de lrsquoAcademie des Sciences vol 247 pp 258-259 1958
Mathematical Problems in Engineering 7
Combining (9) without loss of generality in this case wemay assume u1w1 isin E(G) and v1w2 isin E(G) Now weconsider ujvj where 2le jle s + 2
If w1 isin NG( uj vj1113966 1113967) then subgraphs G[ uj vj w1 u11113966 1113967]
and G[ v1 w2 w3 w41113864 1113865] both contain P4 which combiningthe edges in M minus u1v1 ujvj w1w2 w2w3 w3w41113966 1113967 creates a2P4 cup sP2 in G a contradiction Hence we havew1 notin NG( uj vj1113966 1113967) Together with Claim 2 and (10) weobtain
NG uj vj1113966 11139671113872 1113873cap w1 w2 w3 w41113864 1113865 w21113864 1113865 (11)
Without loss of generality we may assume ujw2 isin E(G)If NHminusV(M)(vj)neempty assume Gprime G[NHminusV(M) [ uj1113966
vj]] In order to avoid a 2P4 cup tP2 in G we get that Gprime K3or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G) minus V(Gprime)) w21113864 1113865which contradicts Lemma 5 Hence NHminusV(M)(vj) emptyis together with δ(H)ge 2 and (11) implies that dG(vj) 2and vjw2 isin E(G) As the argument of vj we haveNHminusV(M)(uj) empty Hence dG(uj) 2 en we can con-clude that
dG uj1113872 1113873 dG vj1113872 1113873 2 andNG uj1113872 1113873capNG vj1113872 1113873 w21113864 1113865
for any 1le jle s + 2
(12)
And by the choice of M dG(w3) dG(w4) 2 Henceby Lemma 2 w2w4 isin E(G) Combining (12) implies thatdG(w2)ge 2(s + 3) If dG(w1) 2 by Lemma 2 u1w2 isinE(G) Hence dG(u1) + dG(v1) + dG(w1)ge 7 If dG(w1)ge 3we also have dG(u1) + dG(v1) + dG(w1)ge 7 is togetherwith dG(w2)ge 2(s + 3) |H|ge 2t + 8 sge t and δ(H)ge 2implies that 2|E(H)|ge 2(s + 3) + 7+ 2(2t + 4)gt 6t + 20 acontradiction
Case 2 NG(u1)cap w1 w21113864 1113865 empty or NG(v1)cap w1 w21113864 1113865 emptyCombining (9) without loss of generality in this case we
may assume NG(v1)cap w1 w2 w3 w41113864 1113865 empty is combiningδ(H)ge 2 and (4) we get NHminusV(M)(v1)neempty and |H|ge 2t + 9Assume v1x isin E(G) where x isin H minus V(M) By our as-sumption NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w1 w21113864 1113865 we getu1w1 isin E(G) and u1w2 isin E(G) Now we consider ujvjwhere 2le jle s + 2
If w2 isin NG( uj vj1113966 1113967) then the subgraphs G[ uj1113966 vj
w2 w3] and G[ x v1 u1 w11113864 1113865] both contain P4 whichcombining the edges in M minus u1v1 ujvj w1w21113966 w2w3 w3w4creates a 2P4 cup sP2 in G a contradiction Hence we havew2 notin NG( uj vj1113966 1113967) is together with Claim 2 and (9) impliesthat NG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w11113864 1113865 for any 2le jles + 2Without loss of generality wemay assume ujw1 isin E(G)en we have NG(x)cap w1 w2 w3 w41113864 1113865 empty So by thechoice of M (5) and δ(H)ge 2 we get xu1 isin E(G) Hence
dG u1( 1113857ge 4 (13)
In order to avoid a 2P4 cup tP2 in G we get thatNHminusV(M)(vj) empty is together with δ(H)ge 2 andNG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w11113864 1113865 implies that
vjw1 isin E(G) and dG(vj) 2 Similarly we have dG(uj) 2erefore ujw1 isin E(G) and vjw1 isin E(G) for any2le jle s + 2 Hence combining sge t implies that
dG w1( 1113857ge 2(s + 2)ge 2t + 4 (14)
And by the choice of M we have dG(w3) dG(w4) 2en by Lemma 2 w2w4 isin E(G) which implies that
dG w2( 1113857ge 4 (15)
Recall that |H|ge 2t + 9 and δ(H)ge 2 Together with(13)ndash(15) we can get that 2|E(H)| ge 2t + 4 + 4 + 4+
2(2t + 6)gt 6t + 20 a contradiction Claim 4 is true
Claim 5 If there exists 1le ile s + 2 such thatdG(ui) + dG(vi)le 4 then we have |E(G)|ge 3t + 10 More-over the equality |E(G)| 3t + 10 holds if and only ifG Tpq cup (n minus 2t minus 8)K1 where p qge 2 and p + q t + 3
Proof Without loss of generality we may assumedG(u1) + dG(v1)le 4 Since δ(H)ge 2 we havedG(u1) dG(v1) 2 By Claim 4 and the symmetry wecomplete the Claim 5 by following two cases
Case 1 NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w11113864 1113865Since dG(u1) dG(v1) 2 by Lemma 2 u1w1 isin E(G)
and v1w1 isin E(G) By the choice of M dG(w3)
dG(w4) 2 Also by Lemma 2 w2w4 isin E(G) Now weconsider ujvj where 2le jle s + 2 It follows from Claim3 and our assumption that NG( uj vj1113966 1113967)cap w1 w2 w31113864
w4sube w1 w21113864 1113865 where 2le jle s + 2 is together withClaim 4 implies that NG( uj vj1113966 1113967)cap w1 w21113864 w3
w4 wi1113864 1113865 where i 1 or 2If NHminusV(M)(vj)neempty assume Gprime G[NHminusV(M)[ uj1113966
vj]] In order to avoid a 2P4 cup tP2 in G we get thatGprime K3 or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G)minus
V(Gprime)) wi1113864 1113865 which contradicts Lemma 5 HenceNHminusV(M)(vj) empty Since δ(H)ge 2 we get vjwi isin E(G)
and dG(vj) 2 By the similar argument we haveNHminusV(M)(uj) empty and hence dG(uj) 2 ereforefor each 1le jle s + 2 we can conclude that NG(uj)
capNG(vj) w11113864 1113865 or NG(uj)capNG(vj) w21113864 1113865 To-gether with w2w4 isin E(G) we have dG(w1)+ dG(w2)ge2(s + 3) + 2 2s + 8 Recall that sge t |H|ge 2t + 8 andδ(H)ge 2 we can get that 2|E(H)| ge 2s+ 8+ 2(2t+
6)ge 6t + 20 And if |E(G)| 3t + 10 then G
Tpq cup (n minus 2t minus 8)K1 where p + q t + 3 However ifmin p q1113864 1113865le 1 Tpq is not a (2P4 cup tP2)-saturated graphsince the addition of an edge between an isolated vertexof G and the vertex of degree Δ(G) does not result in a2P4 cup tP2 Hence p qge 2 Claim 5 holdsCase 2 NG( u1 v11113864 1113865)cap w1 w2 w3 w41113864 1113865 w21113864 1113865
Since dG(u1) dG(v1) 2 by Lemma 2 u1w2 isin E(G)
and v1w2 isin E(G) By the choice of M we havedG(w3) dG(w4) 2 Also by Lemma 2 w2w4 isin E(G)Now we consider ujvj where 2le jle s + 2 It follows fromClaim 3 and our assumption that NG( uj vj1113966 1113967)cap w11113864 w2
w3 w4sube w1 w21113864 1113865
4 Mathematical Problems in Engineering
If there exists 1le jle s + 2 such that NG( uj1113966 vj) capw1 w2 w3 w41113864 1113865 w11113864 1113865 we can claim that dG(uj)
dG(vj) 2 Otherwise assume Gprime G[NHminusV(M)[ uj vj1113966 1113967]]In order to avoid a 2P4 cup tP2 in G we get that Gprime K3 orGprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G) minus V(Gprime)) w11113864 1113865which contradicts Lemma 5 en as the similar argumentin the Case 1 we can obtain |E(H)|ge 3t + 10 and if|E(G)| 3t + 10 then G Tpq where p qge 2 and p + q
t + 3 en NG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w21113864 1113865 for all1le jle s + 2 Without loss of generality we may assumeujw2 isin E(G) If NHminusV(M)(vj)neempty assume Gprime G[[NHminusV(M)[ uj vj1113966 1113967]]] In order to avoid a 2P4 cup tP2 in Gwe get that Gprime K3 or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap(V(G) minus V(Gprime)) w21113864 1113865 which contradicts Lemma 5Hence NHminusV(M)(vj) empty Since δ(H)ge 2 we get vjw2 isinE(G) and dG(vj) 2 By the similar argument we haveNHminusV(M)(uj) empty and hence dG(uj) 2 ereforeujw2 isin E(G) and vjw2 isin E(G) for any 2le jle s + 2 Hence
dG w2( 1113857ge 2(s + 3) + 1ge 2t + 7 (16)
is together with δ(H)ge 2 |H|ge 2t + 8 and (16) im-plies that 2|E(H)| ge 2t + 7 + 2(2t + 7)gt 6t + 20 is com-pletes the proof of Claim 5
If there exists 1le ile s + 2 such that dG(ui) + dG(vi)le 4by Claim 5 the Lemma is proven by Claim 5 Hence we canassume dG(ui) + dG(vi)ge 5 for each 1le ile s + 2 On thecontrary by (4) and Claim 4 we can assumedV(M)(ui) + dV(M)(vi)le 4 Hence there existsxi isin NG( uj vj1113966 1113967)cap (V(H) minus V(M)) for any 1le ile s + 2And by (5) xi nexj for 1le ine jle s + 2 Hence|H|ge 3(s + 2) + 4ge 3t + 10 Since δ(H)ge 2 we get|E(H)|ge 3t + 10 And if the equality |E(G)| 3t + 10 holdsthen δ(H) Δ(H) 2 and |H| 3t + 10 which contradictsLemma 4 Lemma 6 is true
Lemma 7 Suppose G is (2P4 cup tP2)-saturated and (P4 cup(t + 2)P2)-saturated with V0(G)neempty Let H1 Hk be thenontrivial components of G and H G[cup i 1 i
kV(Hi)] If δ(H)ge 2 and |H|ge 2t + 8 with |Hi|ge5(1le ile k) we have |E(G)|gt 3t + 10
Proof Suppose for the sake of contradiction that|E(G)|le 3t + 10 Since G is (2P4 cup tP2)-saturated thereexists a 2P4 cup tP2 in G + e for any e notin E(G) Henceαprime(H)ge t + 3 If αprime(H)ge t + 4 G must contain a copy of(t + 4)P2 Since δ(H)ge 2 and |Hi|ge 5(1le ile k) it is clearlythat H has a P4 cup (t + 2)P2 which contradicts G is(P4 cup (t + 2)P2)-saturated So we have αprime(H) t + 3 ByLemma 1 we obtain
t + 3 12min |H| +|X| minus o(G minus X) XsubeV(H) (17)
en we can choose a maximal subset SsubeV(H) whichsatisfies
t + 3 12
(|H| +|S| minus o(H minus S)) (18)
Suppose H1prime Hpprime are the components of H minus S We
have the following claims
Claim 6 H[ScupV(Hiprime)] is the complete graph where
1le ilep
Proof Suppose for the sake of contradiction thatH[ScupV(Hi
prime)] is not a complete graph en there isx y isin ScupV(Hi
prime) with xney and xy notin E(H) LetHprime H + xy We have |Hprime| |H| ando(Hprime minus S) o(H minus S) Combining (18) we obtain
t + 3 12
Hprime1113868111386811138681113868
1113868111386811138681113868 +|S| minus o(Hprime minus S)1113874 1113875 (19)
By Lemma 1 we obtain αprime(Hprime)le t + 3 Hence G + xy
has no (t + 4)P2 contrary to G which is (2P4 cup tP2)-satu-rated Claim 6 is true
Claim 7 Sneempty
Proof Suppose for the sake of contradiction that S emptyHence H1prime Hp
prime are all components of H Using Claim 6we have Hi
prime is a complete graph with |Hiprime|ge 5(1le ilep)
Hence we have δ(H)ge 4 and
2|E(H)| 1113944xisinV(H)
dH(x)ge 4|H| (20)
Combining |H|ge 2t + 8 we get |E(H)|gt 3t + 10 HenceClaim 7 is true
Let w isin V0(G) and x isin S By Claim 6NH(x) V(H) minus x Let F be a 2P4 cup tP2 in G + xw usingLemma 3 we get V(F)supe x w cupNG(x) Hence we have|H| + 1 |NH(x)cup x w |le |V(F)| 2t + 8 contrary to G
is (2P4 cup tP2)-saturated Hence Lemma 7 is trueNow we prove eorem 2
Proof of eorem 2 Clearly Tpq cup aK3 cup (nminus 2tminus aminus
8)K1 isin SAT(n 2P4 cup tP2) where pge 2 qge 2 and p + q+
a t + 3 Let G isin SAT (n 2P4 cup tP2) then |E(G)|le 3t + 10Since nge 6t + 12 |V0(G)|ge 2 Using Lemma 3 we haveV1(G)neempty Clearly in graph G all components of order 3 or4 are complete If there exists one of the components sayH1 K4 then the addition of an edge e between an isolatedvertex of G and a vertex of H1 results in a 2P4 cup tP2 in GHowever the edge e can be replaced by the edge in H1resulting in a 2P4 cup tP2 isin G a contradiction Hence com-ponent of order 4 in G does not exist
Now we consider the graph Gprime which is obtained bydeleting all components of order 3 from G Let a be thenumber of components of order 3 in G and we obtain
G Gprime cup aK3 (21)
Note |V0(Gprime)| |V0(G)|ge 2 Let e be an edge joiningtwo isolated vertices in V0(G) then there exists a 2P4 cup tP2in G + e So Gprime has 2P4 Combining G isin SAT (n 2P4 cup tP2)we get ale t minus 1 Let tprime t minus a en tprime ge 1
As G isin SAT (n 2P4 cup tP2) we have Gprime isin SAT(nprime2P4 cup tP2) where nprime n minus 3a Suppose Hprime is a graph
Mathematical Problems in Engineering 5
spanned by all nontrivial components of Gprime Using Lemma 3we have δ(Hprime)ge 2 Clearly any component in Hprime has orderat least 5 Moreover graph Gprime is (2P4 cup tprimeP2)-saturated suchthat |E(Gprime)| |E(G)| minus 3ale 3tprime + 10 and V0(Gprime)neempty ByLemma 6 and Lemma 7 we have |Hprime|le 2tprime + 7 or Hprime Tpq cup (nprime minus 2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3If |Hprime|le 2tprime + 7 as V0(Gprime)neempty and Gprime isin SAT(nprime
2P4 cup tprimeP2) Hprime K2tprime+7 Now E(Hprime) 2tprime + 7
21113888 1113889gt
3tprime + 21 and we have |E(G)|gt 3tprime + 21 + 3a 3t+
21gt 3t + 10 a contradiction Hence Hprime Tpq cup (nprimeminus2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3 Combining(21) we get G Tpq cup aK3 cup (n minus 2t minus a minus 8)K1 wherepge 2 qge 2 and p + q + a t + 3 Hence eorem 2 istrue
4 Conclusion
Forests and trees have attracted wide attention for severalreasons Firstly their simplicity has made somemore preciseresults possible Secondly they can provide inspiration forsome larger results At present we only know the saturationnumber of a few species of trees but the saturation numberof most trees is unknown
In [20] Chen et al focused on the saturation numbersfor the linear forests ey obtained an interesting set ofresults and proposed some conjectures We pave the way forthe following conjectures proposed by Chen et al in [20]
Conjecture 1 (see [20]) Let tge 2 be an integer For n suf-ficiently large sat(n tP3) lfloor(n + 6t minus 6)2rfloor
Conjecture 2 (see [20]) Let tge 2 be an integer For n suf-ficiently large
sat n tP4( 1113857
n + 12t minus 122
if n is even
n + 12t minus 112
if n is odd
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
(22)
Conjecture 3 (see [20]) For n sufficiently large kge 4and kle ℓ le lceil(3k minus 2)2rceil sat(n Pk cupPℓ) n minus 1113860nak1113861 + 3where
ak 3 middot 2mminus 1
minus 2 if k 2m
4 middot 2mminus 1minus 2 if k 2m + 1
⎧⎨
⎩ (23)
Data Availability
No data used to support the findings of the study
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was partially supported by NSFC (nos 12001172and 11771172) and supported by Natural Science Founda-tion of Henan Province (no 202300410212) and Science andtechnology innovation fund of Henan Agricultural Uni-versity (nos KJCX2019A15 and KJCX2017A12)
References
[1] J A Bondy and U S R Murty Graph 5eory with Appli-caitons American Elsevier New York NY USA 1976
[2] P Turan ldquoEine Extremalaufgabe aus der GraphentheorierdquoMat Fiz Lapok vol 48 pp 436ndash452 1941
[3] P Erdős A Hajnal and J W Moon ldquoA problem in graphtheoryrdquo 5e American Mathematical Monthly vol 71pp 1107ndash1110 1964
[4] L Kaszonyi and Z Tuza ldquoSaturated graphs with minimalnumber of edgesrdquo Journal of Graph 5eory vol 10 no 2pp 203ndash210 1986
[5] H Lei O Suil D BWest and X Zhu ldquoExtremal problems onsaturation for the family ofk-edge-connected graphsrdquoDiscreteApplied Mathematics vol 260 pp 278ndash283 2019
[6] T Bohman M Fonoberova and O Pikhurko ldquoe saturationfunction of complete partite graphsrdquo Journal of Combina-torics vol 1 no 2 pp 149ndash170 2010
[7] J R Faudree M Ferrara R J Gould andM S Jacobson ldquotKp-saturated graphs of minimum sizerdquo Discrete Mathematicsvol 309 no 19 pp 5870ndash5876 2009
[8] B Bollobas ldquoOn a conjecture of erdos hajnal and moonrdquo5eAmerican Mathematical Monthly vol 74 no 2 pp 178-1791967
[9] S Huang H Lei Y Shi and J Zhang ldquoe saturation numberof K33rdquo 2019 httparxivorgabs191004967
[10] R J Faudree and R J Gould ldquoSaturation numbers for nearlycomplete graphsrdquo Graphs and Combinatorics vol 29 no 3pp 429ndash448 2013
[11] G Chen R J Faudree and R J Gould ldquoSaturation numbersof booksrdquo 5e Electronic Journal of Combinatorics vol 15no 1 pp 118ndash129 2008
[12] Y-C Chen ldquoMinimumC5-saturated graphsrdquo Journal ofGraph 5eory vol 61 no 2 pp 111ndash126 2009
[13] Y-C Chen ldquoAll minimum C5-saturated graphsrdquo Journal ofGraph 5eory vol 67 no 1 pp 9ndash26 2011
[14] L T Ollmann ldquo-saturated graphs with a minimal number ofedgesrdquo in Proceedings of the 5ird Southeastern Conference onCombinatorics Graph 5eory and Computing Florida At-lantic University Boca Raton FL USA 1972
[15] C Timmons ldquo$$C_2k$$-Saturated graphs with No shortodd cyclesrdquo Graphs and Combinatorics vol 35 no 5pp 1023ndash1034 2019
[16] Z Tuza ldquo-saturated graphs of minimum sizerdquo Acta Uni-versitatis Carolinae Mathematica et Physica vol 30 no 2pp 161ndash167 1989
[17] M Zhang S Luo and M Shigeno ldquoOn the number of edgesin a minimum $$C_6$$ C 6 -saturated graphrdquo Graphs andCombinatorics vol 31 no 4 pp 1085ndash1106 2015
[18] J R Faudree R J Faudree R J Gould and M S JacobsonldquoSaturation numbers for treesrdquo 5e Electronic Journal ofCombinatorics vol 16 no 1 pp 91ndash109 2009
[19] F F Song and J J Zhou ldquoe partite saturation number ofspiderrdquo Applied Mathematics and Computation vol 394Article ID 125793 2021
6 Mathematical Problems in Engineering
[20] G Chen J R Faudree R J Faudree R J GouldM S Jacobson and C Magnant ldquoResults and problemson saturation numbers for linear forestsrdquo Bulletin of theInstitute of Combinatorics and its Applications vol 75pp 29ndash46 2015
[21] Q Fan and C Wang ldquoSaturation numbers for linear forests$$P_5 cup tP_2$$ P 5 cup t P 2rdquo Graphs and Combinatoricsvol 31 no 6 pp 2193ndash2200 2015
[22] F F Song ldquoSaturation numbers for linear forestsrdquo UtilitasMathematica vol 104 pp 175ndash186 2017
[23] J R Faudree R J Faudree and J R Schmitt ldquoA survey ofminimum saturation graphsrdquo 5e Electronic Journal ofCombinatorics vol 18 p 36pp 2011
[24] N Bushaw and N Kettle ldquoTuran numbers of multiple pathsand equibipartite forestsrdquo Combinatorics Probability andComputing vol 20 no 6 pp 837ndash853 2011
[25] C Berge ldquoSur le couplage maximum dprimeun grapherdquo Comptesrendus de lrsquoAcademie des Sciences vol 247 pp 258-259 1958
Mathematical Problems in Engineering 7
If there exists 1le jle s + 2 such that NG( uj1113966 vj) capw1 w2 w3 w41113864 1113865 w11113864 1113865 we can claim that dG(uj)
dG(vj) 2 Otherwise assume Gprime G[NHminusV(M)[ uj vj1113966 1113967]]In order to avoid a 2P4 cup tP2 in G we get that Gprime K3 orGprime K1ℓ(ℓ ge 2) and NG(Gprime)cap (V(G) minus V(Gprime)) w11113864 1113865which contradicts Lemma 5 en as the similar argumentin the Case 1 we can obtain |E(H)|ge 3t + 10 and if|E(G)| 3t + 10 then G Tpq where p qge 2 and p + q
t + 3 en NG( uj vj1113966 1113967)cap w1 w2 w3 w41113864 1113865 w21113864 1113865 for all1le jle s + 2 Without loss of generality we may assumeujw2 isin E(G) If NHminusV(M)(vj)neempty assume Gprime G[[NHminusV(M)[ uj vj1113966 1113967]]] In order to avoid a 2P4 cup tP2 in Gwe get that Gprime K3 or Gprime K1ℓ(ℓ ge 2) and NG(Gprime)cap(V(G) minus V(Gprime)) w21113864 1113865 which contradicts Lemma 5Hence NHminusV(M)(vj) empty Since δ(H)ge 2 we get vjw2 isinE(G) and dG(vj) 2 By the similar argument we haveNHminusV(M)(uj) empty and hence dG(uj) 2 ereforeujw2 isin E(G) and vjw2 isin E(G) for any 2le jle s + 2 Hence
dG w2( 1113857ge 2(s + 3) + 1ge 2t + 7 (16)
is together with δ(H)ge 2 |H|ge 2t + 8 and (16) im-plies that 2|E(H)| ge 2t + 7 + 2(2t + 7)gt 6t + 20 is com-pletes the proof of Claim 5
If there exists 1le ile s + 2 such that dG(ui) + dG(vi)le 4by Claim 5 the Lemma is proven by Claim 5 Hence we canassume dG(ui) + dG(vi)ge 5 for each 1le ile s + 2 On thecontrary by (4) and Claim 4 we can assumedV(M)(ui) + dV(M)(vi)le 4 Hence there existsxi isin NG( uj vj1113966 1113967)cap (V(H) minus V(M)) for any 1le ile s + 2And by (5) xi nexj for 1le ine jle s + 2 Hence|H|ge 3(s + 2) + 4ge 3t + 10 Since δ(H)ge 2 we get|E(H)|ge 3t + 10 And if the equality |E(G)| 3t + 10 holdsthen δ(H) Δ(H) 2 and |H| 3t + 10 which contradictsLemma 4 Lemma 6 is true
Lemma 7 Suppose G is (2P4 cup tP2)-saturated and (P4 cup(t + 2)P2)-saturated with V0(G)neempty Let H1 Hk be thenontrivial components of G and H G[cup i 1 i
kV(Hi)] If δ(H)ge 2 and |H|ge 2t + 8 with |Hi|ge5(1le ile k) we have |E(G)|gt 3t + 10
Proof Suppose for the sake of contradiction that|E(G)|le 3t + 10 Since G is (2P4 cup tP2)-saturated thereexists a 2P4 cup tP2 in G + e for any e notin E(G) Henceαprime(H)ge t + 3 If αprime(H)ge t + 4 G must contain a copy of(t + 4)P2 Since δ(H)ge 2 and |Hi|ge 5(1le ile k) it is clearlythat H has a P4 cup (t + 2)P2 which contradicts G is(P4 cup (t + 2)P2)-saturated So we have αprime(H) t + 3 ByLemma 1 we obtain
t + 3 12min |H| +|X| minus o(G minus X) XsubeV(H) (17)
en we can choose a maximal subset SsubeV(H) whichsatisfies
t + 3 12
(|H| +|S| minus o(H minus S)) (18)
Suppose H1prime Hpprime are the components of H minus S We
have the following claims
Claim 6 H[ScupV(Hiprime)] is the complete graph where
1le ilep
Proof Suppose for the sake of contradiction thatH[ScupV(Hi
prime)] is not a complete graph en there isx y isin ScupV(Hi
prime) with xney and xy notin E(H) LetHprime H + xy We have |Hprime| |H| ando(Hprime minus S) o(H minus S) Combining (18) we obtain
t + 3 12
Hprime1113868111386811138681113868
1113868111386811138681113868 +|S| minus o(Hprime minus S)1113874 1113875 (19)
By Lemma 1 we obtain αprime(Hprime)le t + 3 Hence G + xy
has no (t + 4)P2 contrary to G which is (2P4 cup tP2)-satu-rated Claim 6 is true
Claim 7 Sneempty
Proof Suppose for the sake of contradiction that S emptyHence H1prime Hp
prime are all components of H Using Claim 6we have Hi
prime is a complete graph with |Hiprime|ge 5(1le ilep)
Hence we have δ(H)ge 4 and
2|E(H)| 1113944xisinV(H)
dH(x)ge 4|H| (20)
Combining |H|ge 2t + 8 we get |E(H)|gt 3t + 10 HenceClaim 7 is true
Let w isin V0(G) and x isin S By Claim 6NH(x) V(H) minus x Let F be a 2P4 cup tP2 in G + xw usingLemma 3 we get V(F)supe x w cupNG(x) Hence we have|H| + 1 |NH(x)cup x w |le |V(F)| 2t + 8 contrary to G
is (2P4 cup tP2)-saturated Hence Lemma 7 is trueNow we prove eorem 2
Proof of eorem 2 Clearly Tpq cup aK3 cup (nminus 2tminus aminus
8)K1 isin SAT(n 2P4 cup tP2) where pge 2 qge 2 and p + q+
a t + 3 Let G isin SAT (n 2P4 cup tP2) then |E(G)|le 3t + 10Since nge 6t + 12 |V0(G)|ge 2 Using Lemma 3 we haveV1(G)neempty Clearly in graph G all components of order 3 or4 are complete If there exists one of the components sayH1 K4 then the addition of an edge e between an isolatedvertex of G and a vertex of H1 results in a 2P4 cup tP2 in GHowever the edge e can be replaced by the edge in H1resulting in a 2P4 cup tP2 isin G a contradiction Hence com-ponent of order 4 in G does not exist
Now we consider the graph Gprime which is obtained bydeleting all components of order 3 from G Let a be thenumber of components of order 3 in G and we obtain
G Gprime cup aK3 (21)
Note |V0(Gprime)| |V0(G)|ge 2 Let e be an edge joiningtwo isolated vertices in V0(G) then there exists a 2P4 cup tP2in G + e So Gprime has 2P4 Combining G isin SAT (n 2P4 cup tP2)we get ale t minus 1 Let tprime t minus a en tprime ge 1
As G isin SAT (n 2P4 cup tP2) we have Gprime isin SAT(nprime2P4 cup tP2) where nprime n minus 3a Suppose Hprime is a graph
Mathematical Problems in Engineering 5
spanned by all nontrivial components of Gprime Using Lemma 3we have δ(Hprime)ge 2 Clearly any component in Hprime has orderat least 5 Moreover graph Gprime is (2P4 cup tprimeP2)-saturated suchthat |E(Gprime)| |E(G)| minus 3ale 3tprime + 10 and V0(Gprime)neempty ByLemma 6 and Lemma 7 we have |Hprime|le 2tprime + 7 or Hprime Tpq cup (nprime minus 2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3If |Hprime|le 2tprime + 7 as V0(Gprime)neempty and Gprime isin SAT(nprime
2P4 cup tprimeP2) Hprime K2tprime+7 Now E(Hprime) 2tprime + 7
21113888 1113889gt
3tprime + 21 and we have |E(G)|gt 3tprime + 21 + 3a 3t+
21gt 3t + 10 a contradiction Hence Hprime Tpq cup (nprimeminus2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3 Combining(21) we get G Tpq cup aK3 cup (n minus 2t minus a minus 8)K1 wherepge 2 qge 2 and p + q + a t + 3 Hence eorem 2 istrue
4 Conclusion
Forests and trees have attracted wide attention for severalreasons Firstly their simplicity has made somemore preciseresults possible Secondly they can provide inspiration forsome larger results At present we only know the saturationnumber of a few species of trees but the saturation numberof most trees is unknown
In [20] Chen et al focused on the saturation numbersfor the linear forests ey obtained an interesting set ofresults and proposed some conjectures We pave the way forthe following conjectures proposed by Chen et al in [20]
Conjecture 1 (see [20]) Let tge 2 be an integer For n suf-ficiently large sat(n tP3) lfloor(n + 6t minus 6)2rfloor
Conjecture 2 (see [20]) Let tge 2 be an integer For n suf-ficiently large
sat n tP4( 1113857
n + 12t minus 122
if n is even
n + 12t minus 112
if n is odd
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
(22)
Conjecture 3 (see [20]) For n sufficiently large kge 4and kle ℓ le lceil(3k minus 2)2rceil sat(n Pk cupPℓ) n minus 1113860nak1113861 + 3where
ak 3 middot 2mminus 1
minus 2 if k 2m
4 middot 2mminus 1minus 2 if k 2m + 1
⎧⎨
⎩ (23)
Data Availability
No data used to support the findings of the study
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was partially supported by NSFC (nos 12001172and 11771172) and supported by Natural Science Founda-tion of Henan Province (no 202300410212) and Science andtechnology innovation fund of Henan Agricultural Uni-versity (nos KJCX2019A15 and KJCX2017A12)
References
[1] J A Bondy and U S R Murty Graph 5eory with Appli-caitons American Elsevier New York NY USA 1976
[2] P Turan ldquoEine Extremalaufgabe aus der GraphentheorierdquoMat Fiz Lapok vol 48 pp 436ndash452 1941
[3] P Erdős A Hajnal and J W Moon ldquoA problem in graphtheoryrdquo 5e American Mathematical Monthly vol 71pp 1107ndash1110 1964
[4] L Kaszonyi and Z Tuza ldquoSaturated graphs with minimalnumber of edgesrdquo Journal of Graph 5eory vol 10 no 2pp 203ndash210 1986
[5] H Lei O Suil D BWest and X Zhu ldquoExtremal problems onsaturation for the family ofk-edge-connected graphsrdquoDiscreteApplied Mathematics vol 260 pp 278ndash283 2019
[6] T Bohman M Fonoberova and O Pikhurko ldquoe saturationfunction of complete partite graphsrdquo Journal of Combina-torics vol 1 no 2 pp 149ndash170 2010
[7] J R Faudree M Ferrara R J Gould andM S Jacobson ldquotKp-saturated graphs of minimum sizerdquo Discrete Mathematicsvol 309 no 19 pp 5870ndash5876 2009
[8] B Bollobas ldquoOn a conjecture of erdos hajnal and moonrdquo5eAmerican Mathematical Monthly vol 74 no 2 pp 178-1791967
[9] S Huang H Lei Y Shi and J Zhang ldquoe saturation numberof K33rdquo 2019 httparxivorgabs191004967
[10] R J Faudree and R J Gould ldquoSaturation numbers for nearlycomplete graphsrdquo Graphs and Combinatorics vol 29 no 3pp 429ndash448 2013
[11] G Chen R J Faudree and R J Gould ldquoSaturation numbersof booksrdquo 5e Electronic Journal of Combinatorics vol 15no 1 pp 118ndash129 2008
[12] Y-C Chen ldquoMinimumC5-saturated graphsrdquo Journal ofGraph 5eory vol 61 no 2 pp 111ndash126 2009
[13] Y-C Chen ldquoAll minimum C5-saturated graphsrdquo Journal ofGraph 5eory vol 67 no 1 pp 9ndash26 2011
[14] L T Ollmann ldquo-saturated graphs with a minimal number ofedgesrdquo in Proceedings of the 5ird Southeastern Conference onCombinatorics Graph 5eory and Computing Florida At-lantic University Boca Raton FL USA 1972
[15] C Timmons ldquo$$C_2k$$-Saturated graphs with No shortodd cyclesrdquo Graphs and Combinatorics vol 35 no 5pp 1023ndash1034 2019
[16] Z Tuza ldquo-saturated graphs of minimum sizerdquo Acta Uni-versitatis Carolinae Mathematica et Physica vol 30 no 2pp 161ndash167 1989
[17] M Zhang S Luo and M Shigeno ldquoOn the number of edgesin a minimum $$C_6$$ C 6 -saturated graphrdquo Graphs andCombinatorics vol 31 no 4 pp 1085ndash1106 2015
[18] J R Faudree R J Faudree R J Gould and M S JacobsonldquoSaturation numbers for treesrdquo 5e Electronic Journal ofCombinatorics vol 16 no 1 pp 91ndash109 2009
[19] F F Song and J J Zhou ldquoe partite saturation number ofspiderrdquo Applied Mathematics and Computation vol 394Article ID 125793 2021
6 Mathematical Problems in Engineering
[20] G Chen J R Faudree R J Faudree R J GouldM S Jacobson and C Magnant ldquoResults and problemson saturation numbers for linear forestsrdquo Bulletin of theInstitute of Combinatorics and its Applications vol 75pp 29ndash46 2015
[21] Q Fan and C Wang ldquoSaturation numbers for linear forests$$P_5 cup tP_2$$ P 5 cup t P 2rdquo Graphs and Combinatoricsvol 31 no 6 pp 2193ndash2200 2015
[22] F F Song ldquoSaturation numbers for linear forestsrdquo UtilitasMathematica vol 104 pp 175ndash186 2017
[23] J R Faudree R J Faudree and J R Schmitt ldquoA survey ofminimum saturation graphsrdquo 5e Electronic Journal ofCombinatorics vol 18 p 36pp 2011
[24] N Bushaw and N Kettle ldquoTuran numbers of multiple pathsand equibipartite forestsrdquo Combinatorics Probability andComputing vol 20 no 6 pp 837ndash853 2011
[25] C Berge ldquoSur le couplage maximum dprimeun grapherdquo Comptesrendus de lrsquoAcademie des Sciences vol 247 pp 258-259 1958
Mathematical Problems in Engineering 7
spanned by all nontrivial components of Gprime Using Lemma 3we have δ(Hprime)ge 2 Clearly any component in Hprime has orderat least 5 Moreover graph Gprime is (2P4 cup tprimeP2)-saturated suchthat |E(Gprime)| |E(G)| minus 3ale 3tprime + 10 and V0(Gprime)neempty ByLemma 6 and Lemma 7 we have |Hprime|le 2tprime + 7 or Hprime Tpq cup (nprime minus 2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3If |Hprime|le 2tprime + 7 as V0(Gprime)neempty and Gprime isin SAT(nprime
2P4 cup tprimeP2) Hprime K2tprime+7 Now E(Hprime) 2tprime + 7
21113888 1113889gt
3tprime + 21 and we have |E(G)|gt 3tprime + 21 + 3a 3t+
21gt 3t + 10 a contradiction Hence Hprime Tpq cup (nprimeminus2tprime minus 8)K1 where pge 2 qge 2 and p + q tprime + 3 Combining(21) we get G Tpq cup aK3 cup (n minus 2t minus a minus 8)K1 wherepge 2 qge 2 and p + q + a t + 3 Hence eorem 2 istrue
4 Conclusion
Forests and trees have attracted wide attention for severalreasons Firstly their simplicity has made somemore preciseresults possible Secondly they can provide inspiration forsome larger results At present we only know the saturationnumber of a few species of trees but the saturation numberof most trees is unknown
In [20] Chen et al focused on the saturation numbersfor the linear forests ey obtained an interesting set ofresults and proposed some conjectures We pave the way forthe following conjectures proposed by Chen et al in [20]
Conjecture 1 (see [20]) Let tge 2 be an integer For n suf-ficiently large sat(n tP3) lfloor(n + 6t minus 6)2rfloor
Conjecture 2 (see [20]) Let tge 2 be an integer For n suf-ficiently large
sat n tP4( 1113857
n + 12t minus 122
if n is even
n + 12t minus 112
if n is odd
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
(22)
Conjecture 3 (see [20]) For n sufficiently large kge 4and kle ℓ le lceil(3k minus 2)2rceil sat(n Pk cupPℓ) n minus 1113860nak1113861 + 3where
ak 3 middot 2mminus 1
minus 2 if k 2m
4 middot 2mminus 1minus 2 if k 2m + 1
⎧⎨
⎩ (23)
Data Availability
No data used to support the findings of the study
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was partially supported by NSFC (nos 12001172and 11771172) and supported by Natural Science Founda-tion of Henan Province (no 202300410212) and Science andtechnology innovation fund of Henan Agricultural Uni-versity (nos KJCX2019A15 and KJCX2017A12)
References
[1] J A Bondy and U S R Murty Graph 5eory with Appli-caitons American Elsevier New York NY USA 1976
[2] P Turan ldquoEine Extremalaufgabe aus der GraphentheorierdquoMat Fiz Lapok vol 48 pp 436ndash452 1941
[3] P Erdős A Hajnal and J W Moon ldquoA problem in graphtheoryrdquo 5e American Mathematical Monthly vol 71pp 1107ndash1110 1964
[4] L Kaszonyi and Z Tuza ldquoSaturated graphs with minimalnumber of edgesrdquo Journal of Graph 5eory vol 10 no 2pp 203ndash210 1986
[5] H Lei O Suil D BWest and X Zhu ldquoExtremal problems onsaturation for the family ofk-edge-connected graphsrdquoDiscreteApplied Mathematics vol 260 pp 278ndash283 2019
[6] T Bohman M Fonoberova and O Pikhurko ldquoe saturationfunction of complete partite graphsrdquo Journal of Combina-torics vol 1 no 2 pp 149ndash170 2010
[7] J R Faudree M Ferrara R J Gould andM S Jacobson ldquotKp-saturated graphs of minimum sizerdquo Discrete Mathematicsvol 309 no 19 pp 5870ndash5876 2009
[8] B Bollobas ldquoOn a conjecture of erdos hajnal and moonrdquo5eAmerican Mathematical Monthly vol 74 no 2 pp 178-1791967
[9] S Huang H Lei Y Shi and J Zhang ldquoe saturation numberof K33rdquo 2019 httparxivorgabs191004967
[10] R J Faudree and R J Gould ldquoSaturation numbers for nearlycomplete graphsrdquo Graphs and Combinatorics vol 29 no 3pp 429ndash448 2013
[11] G Chen R J Faudree and R J Gould ldquoSaturation numbersof booksrdquo 5e Electronic Journal of Combinatorics vol 15no 1 pp 118ndash129 2008
[12] Y-C Chen ldquoMinimumC5-saturated graphsrdquo Journal ofGraph 5eory vol 61 no 2 pp 111ndash126 2009
[13] Y-C Chen ldquoAll minimum C5-saturated graphsrdquo Journal ofGraph 5eory vol 67 no 1 pp 9ndash26 2011
[14] L T Ollmann ldquo-saturated graphs with a minimal number ofedgesrdquo in Proceedings of the 5ird Southeastern Conference onCombinatorics Graph 5eory and Computing Florida At-lantic University Boca Raton FL USA 1972
[15] C Timmons ldquo$$C_2k$$-Saturated graphs with No shortodd cyclesrdquo Graphs and Combinatorics vol 35 no 5pp 1023ndash1034 2019
[16] Z Tuza ldquo-saturated graphs of minimum sizerdquo Acta Uni-versitatis Carolinae Mathematica et Physica vol 30 no 2pp 161ndash167 1989
[17] M Zhang S Luo and M Shigeno ldquoOn the number of edgesin a minimum $$C_6$$ C 6 -saturated graphrdquo Graphs andCombinatorics vol 31 no 4 pp 1085ndash1106 2015
[18] J R Faudree R J Faudree R J Gould and M S JacobsonldquoSaturation numbers for treesrdquo 5e Electronic Journal ofCombinatorics vol 16 no 1 pp 91ndash109 2009
[19] F F Song and J J Zhou ldquoe partite saturation number ofspiderrdquo Applied Mathematics and Computation vol 394Article ID 125793 2021
6 Mathematical Problems in Engineering
[20] G Chen J R Faudree R J Faudree R J GouldM S Jacobson and C Magnant ldquoResults and problemson saturation numbers for linear forestsrdquo Bulletin of theInstitute of Combinatorics and its Applications vol 75pp 29ndash46 2015
[21] Q Fan and C Wang ldquoSaturation numbers for linear forests$$P_5 cup tP_2$$ P 5 cup t P 2rdquo Graphs and Combinatoricsvol 31 no 6 pp 2193ndash2200 2015
[22] F F Song ldquoSaturation numbers for linear forestsrdquo UtilitasMathematica vol 104 pp 175ndash186 2017
[23] J R Faudree R J Faudree and J R Schmitt ldquoA survey ofminimum saturation graphsrdquo 5e Electronic Journal ofCombinatorics vol 18 p 36pp 2011
[24] N Bushaw and N Kettle ldquoTuran numbers of multiple pathsand equibipartite forestsrdquo Combinatorics Probability andComputing vol 20 no 6 pp 837ndash853 2011
[25] C Berge ldquoSur le couplage maximum dprimeun grapherdquo Comptesrendus de lrsquoAcademie des Sciences vol 247 pp 258-259 1958
Mathematical Problems in Engineering 7
[20] G Chen J R Faudree R J Faudree R J GouldM S Jacobson and C Magnant ldquoResults and problemson saturation numbers for linear forestsrdquo Bulletin of theInstitute of Combinatorics and its Applications vol 75pp 29ndash46 2015
[21] Q Fan and C Wang ldquoSaturation numbers for linear forests$$P_5 cup tP_2$$ P 5 cup t P 2rdquo Graphs and Combinatoricsvol 31 no 6 pp 2193ndash2200 2015
[22] F F Song ldquoSaturation numbers for linear forestsrdquo UtilitasMathematica vol 104 pp 175ndash186 2017
[23] J R Faudree R J Faudree and J R Schmitt ldquoA survey ofminimum saturation graphsrdquo 5e Electronic Journal ofCombinatorics vol 18 p 36pp 2011
[24] N Bushaw and N Kettle ldquoTuran numbers of multiple pathsand equibipartite forestsrdquo Combinatorics Probability andComputing vol 20 no 6 pp 837ndash853 2011
[25] C Berge ldquoSur le couplage maximum dprimeun grapherdquo Comptesrendus de lrsquoAcademie des Sciences vol 247 pp 258-259 1958
Mathematical Problems in Engineering 7
![Introduction [2p]](https://static.fdocuments.us/doc/165x107/586a17fc1a28abd97c8bbe27/introduction-2p.jpg)
