Journal of Applied Mathematics and Physics, 2020, 8, 1168-1179 https://www.scirp.org/journal/jamp ISSN Online: 2327-4379 ISSN Print: 2327-4352
On the Uphill Domination Polynomial of Graphs
Thekra Alsalomy, Anwar Saleh, Najat Muthana, Wafa Al Shammakh
Department of Mathematics, Faculty of Science, University of Jeddah, Jeddah, Saudi Arabia
How to cite this paper: Alsalomy, T., Abstract Saleh, A., Muthana, N. and Al Shammakh,
W. (2020) On the Uphill Domination Po- A path π = [vv12,,, vk ] in a graph G= ( VE, ) is an uphill path if lynomial of Graphs. Journal of Applied deg v≤ deg v 1 ≤≤ik S⊆ VG Mathematics and Physics, 8, 1168-1179. ( ii) ( +1 ) for every . A subset ( ) is an uphill https://doi.org/10.4236/jamp.2020.86088 dominating set if every vertex vi ∈ VG( ) lies on an uphill path originating
Received: April 20, 2019 from some vertex in S. The uphill domination number of G is denoted by
Accepted: June 20, 2020 γ up (G) and is the minimum cardinality of the uphill dominating set of G. In Published: June 23, 2020 this paper, we introduce the uphill domination polynomial of a graph G. The
Copyright © 2020 by author(s) and uphill domination polynomial of a graph G of n vertices is the polynomial Scientific Research Publishing Inc. = n i UP( G,, x) ∑iG=γ ( ) up( G i) x , where up( G, i) is the number of uphill do- This work is licensed under the Creative up Commons Attribution International minating sets of size i in G, and γ up (G) is the uphill domination number of G, License (CC BY 4.0). http://creativecommons.org/licenses/by/4.0/ we compute the uphill domination polynomial and its roots for some families of Open Access standard graphs. Also, UP( G, x) for some graph operations is obtained.
Keywords Domination, Uphill Domination, Uphill Domination Polynomial
1. Introduction
In this paper, we are concerned with simple graphs which are finite, undirected with no loops nor multiple edges. Throughout this paper, we let VG( ) = n and EG( ) = m. In a graph G= ( VE, ) , the degree of v∈ VG( ) denoted by deg( v) is the number of edges that incident with v. A path in G is an alternat- ing sequence of distinct vertices. A path is an uphill path if for every 1 ≤≤ik
we have deg( vii) ≤ deg( v +1 ) [1]. S nk=22 + The bistar graph kk11, with 1 vertices is obtained by joining the S non-pendant vertices of two copies of star graph k1 by new edge. The corona
of two graphs G1 and G2 with n1 and n2 vertices, respectively, denoted by
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GGG= 12 is obtained by taking one copy of G1 and n1 copies of G2 and
joining the ith vertex of G1 with an edge to every vertex in the ith copy of G2 .
The corona GK 1 (in particular) is the graph constructed by a copy of G, where for each vertex v∈ VG( ) a new vertex v′ and a pendant edge vv′ are
added. The tadpole graph Tsk, is a graph consisting of a cycle graph Cs on at
least three vertices and a path graph Pk on k vertices connected with bridge.
The wheel graph Ws is a graph formed by connecting a single vertex to all ver-
tices of a cycle graph Cs . The book graph is a Cartesian product Bmm= SP × 2 ,
where Sm is the star graph with m +1 vertices and P2 is the path graph on two vertices. Also, the windmill graph Wd( s, k ) is a graph constructed for
s ≥ 2 and k ≥ 2 by joining k copies of the complete graph Ks at a shared universal vertex. The dutch windmill graph D( sk, ) is the graph obtained by
taking k copies of the cycle graph Cs with a vertex in common. Also, the
friendship Fk is a graph that constructed by joining k copies of the cycle graph
C3 and observes that Fk is a special case of D( sk, ) . Finlay, the firefly graph
Fstk,, with stk,,≥ 0 and n=22 s + tk ++ 1 vertices is defined by consisting of s triangles, t pendent paths of length 2 and k pendent edges, sharing a com- mon vertex. Any terminology not mentioed here we refer the reader to [2]. A set SV⊆ of vertices in a graph G is called a dominating set if every ver- tex vV∈ is either vS∈ or v is adjacent to an element of S, The uphill do- minating set “UDS” is a set SV⊆ having the property that every vertex vV∈ lies on an uphill path originating from some vertex in S. The uphill do-
mination number of a graph G is denoted by γ up (G) and is defined to be the minimum cardinality of the UDS of G. Moreover, it’s customary to denote the
UDS having the minimum cardinality by γ up (G) -set, for more details in domi- nation see [3] and [4]. Representing a graph by using a polynomial is one of the algebraic representa- tions of a graph to study some of algebraic properties and graph’s structure. In general graph polynomials are a well-developed area which is very useful for analyzing properties of the graphs. The domination polynomial [5] and the uphill domination of a graph [6], mo- tivated us to introduce and study the uphill domination polynomial and the uphill domination roots of a graph.
2. Uphill Domination Polynomial
Definition 2.1. For any graph G of n vertices, the uphill domination poly- nomial of G is defined by n UP(,) G x= ∑ up( G,, i) xi iG=γup ( )
where up( G, i) is the number of uphill dominating sets of size i in G. The set of roots of UP( G, x) is called uphill domination roots of graph G and denoted by
ZGup ( ) . Example 2.2. The uphill domination polynomial of House graph H (as shown
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in Figure 1) with 6 vertices and γ up (H ) = 2 is given by 2 3 4 56 UP( H, x) = 2795 x ++++ x x x x . Furthermore, ZHup ( ) ={0, −− 1, 2} . The following theorem gives the sufficient condition for the uphill domina- tion polynomial of r-regular graph. Theorem 2.3. Let G be connected graph with n ≥ 2 vertices. Then, n UP( G,1 x) =+−( x) 1 if and only if G is r-regular graph. Proof. Let G be a connected graph of n ≥ 2 vertices. Suppose that the uphill domination polynomial of G is given by
n n 2 n UP( G,1 x) =( + x) −= 1 nx + x + + x . 2 Since the first coefficient of the polynomial is n, then it is easily verified that for every v∈ VG( ) , the singleton vertex set {v} is an UDS in G. Assume that G is not r-regular graph. Hence there exists a vertex u∈ VG( ) such that deg( u) = s ≠ r . Now, we have two cases: Case 1: If sr> , then the set {u} is not UDS which contradict that every singleton vertex set is an UDS in G. Case 2: If sr< , then for all uv≠ with deg( v) = r , we get the set {v} is not UDS which is also contradict that every singleton vertex set is an UDS in G. Thus, G must be r-regular graph. On the other hand, suppose that G is r-regular graph with n ≥ 2 vertices. We
have γ up (G) = 1 , then there exist n UDS of size one, while for i = 2 there are n UDS and so on. Thus, we can write the uphill domination polynomial as 2
nn23 nn n UP( G, x) = nx + x + x ++ x =+(11 x) −. 22 n
Corollary 2.4. Let G ba a graph with s vertices. If G is a cycle Cs or com- s plete graph Ks , then UP( G,1 x) =+−( x) 1. Corollary 2.5. The uphill domination polynomial for the regular graph sk GCC=sk × with sk vertices is given by UP( G,1 x) =+−( x) 1. Corollary 2.6. [6] Let G be a graph with m components. Then, m γγup (GG) = ∑ up( i ). j=1 Proposition 2.7. If a graph G with n vertices consists of m components
GG12, ,, Gm , then
Figure 1. The House graph.
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m UP( G, x) = ∏ UP( Gi ,. x) i=1 Proof. By using mathematical induction we found that for m = 1 the state- ment is true and the proof is trivial. Suppose that the statement is true when mk= such that k UP( G, x) = ∏ UP( Gi ,. x) i=1 Now, we prove that the statement is true when mk= +1. Let G consists of
k +1 components that mean GG= 12 G Gk + 1. If the set {rr12,, , rk + 1} represent the uphill domination number for the components of G respectively,
such that γ up(Gr i) = i ∀≤≤11ik +. Then, by Corollary (2.6) it easily to see that γγ= = γ =++ = up (G) up Gi ∑ up( Gr i ) 11 rk + r. 11≤≤ik + 11≤≤ik + Thus, up( G, r) is exactly equal the number of way for choosing an UDS of
size r1 in G1 and an UDS of size r2 in G2 and so on. Hence, up( G, r) is r the coefficient of x in UP( G12,, x) UP( G x) UP( Gk + 1 , x) and in UP( G, x) . In the same argument we can proof for all up( G, j) , where r≤≤ jn that k +1 up( G, j) = up( G11 , j) up( Gki+ , j) = ∏ up( G ,. j) i=1 Thus, for mk= +1 the statement is true and the proof is done.
Theorem 2.8. For any path Pn with n ≥ 3 vertices, 2 n−2 UP( G,1 x) = x( + x) . Furthermore, ZPup( n ) ={0, − 1} .
Proof. Let G be a path graph Pn with n ≥ 3 . We know that γ up(P n ) = 2 , then there is only one UDS of size two. For i = 3 there are n − 2 UDS of size three and so on. Thus, we get
234nn−−22 n − 2n UP( G, x) = x + x + x ++ x 12 n − 2
n−2 2 n −2 i =xx1 + ∑ i=1 i n−2 2 n −2 i = xx∑ i=0 i n−2 =xx2 (1. + )
n Theorem 2.9. For any graph G. UP( G, x) = x if and only if GK≅ n . n Proof. Let G be a graph with UP( G, x) = x . Since, UP( K1, x) = x , then we can write that UP( G, x) = xn
=xx ⋅ x n times
=UP( K11,, x) ⋅ UP( K x) UP( K 1 , x) n times
= UP( Kn ,. x)
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Thus, GK≅ n . On the other hand if GK≅ n , then by Proposition (2.7) we get UP( G, x) = xn . Corollary 2.10. A graph G has one uphill domination root if and only if
GK≅ n . S nk=22 + Theorem 2.11. Let G be a bistar graph kk11, with 1 vertices. 2 2k1 Then, UP( G,1 x) = x( + x) . Furthermore, ZGup ( ) ={0, − 1} . S nk=22 + Proof. Let G be a bistar graph kk11, with 1 vertices, we have
γ up (Gk) = 2 1 . Then, there is only one UDS of size 2k1 and for ik=211 +
there are two UDS. Finally, for ik=221 += n there is only one UDS. Thus, the result will be as following
UP( G,2 x) =++ x2kk11 x 21++ x 22 k 1
2k1 2 =x12 ++ xx 2 =xx2k1 (1. + )
Theorem 2.12. For any graph GK≅ rs, with rs< and rs+≥3 vertices, s r UP( G,1 x) = x( + x) . Furthermore, ZKup( r, s ) ={0, − 1} .
Proof. Let G is a complete bipartite graph Krs, with rs< , then we have
γ up(Ks r, s ) = . There is only one UDS of size s. Now, for is= +1 there exist r r UDS. For is= + 2 there exist UDS and so on. Thus, we get 2
srr s++12 s rsr + UP( G, x) = x + x + x ++ x 12 r r s r si+ =xx + ∑ i=1 i r sir = xx∑ i=0 i r =xxs (1. + )
Corollary 2.13. For any graph GS≅ r with r +1 vertices, r UP( G,1 x) = x( + x) . Furthermore, ZGup ( ) ={0, − 1} . The generalization of Theorem 0.12 is the following result. GK≅ rr<<< r Theorem 2.14. For any graph rr1,, k where 12 k with k nr− = =rk + k nr∑i=1 i vertices, UP( G,1 x) x( x) . Furthermore, ZK =0, − 1 up( r1,, rk ) { } . K rr<<< r Proof. Let G be a complete k-partite graph rr1,, k with 12 k , we γ Kr= r ir= +1 have up( r1,, rk ) k . There is only one UDS of size k for k there
nr− k are nr− k UDS of size rk +1. Also, for ir=k + 2 there are and so 2 on. Thus,
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rrnr−−kk++12 nr rnr− k n UP( G, x) = xkk + x + x k ++ x 12 nr− k
nr− k r nr− k ri+ =xxkk + ∑ i=1 i
nr− k r nr −k i = xxk ∑ i=0 i nr− =xxrk (1. + ) k
k GK≅ nr= Proposition 2.15. For any graph rr12,,, rk with ∑i=1 i vertices we have the following:
1) If rr12≤≤≤ rkk− 1 < r, such that at least two partite sets of the same size, nr− then UP( G,1 x) = xrk ( + x) k . n 2) If rr12= = = rk , then the graph is regular and UP( G,1 x) =+−( x) 1. k GK≅ nr= Theorem 2.16. For any graph rr12,,, rk with ∑i=1 i vertices, where
rr12≤≤< rkk− 1 = r. Then, n 22rnr − UP G,. x = kkxh ( ) ∑∑ h=1 r1 ≥1 rr12 rr12+= h
K Proof. Let G be a complete k-partite graph rr1,, k with rr≤≤< r = r γ K = 1 12 kk− 1 , then we have up( r1,, rk ) . Let divide the vertices of a
graph into two sets R1 and R2 where R1 contains the vertices of rk and
rk −1 which means R1 is of cardinality 2rk while R21= VG( ) \ R this implies
that R2 is of cardinality nr− 2 k . Thus, we get
22rnrkk − up( G,1) = = 2rk . 10
We have for up( G,2) ,
2222rnrkkkk −− rnr up( G,2) = + . 20 11
Also, for up( G,3) we get
222222rnrkkkkkk−−− rnr rnr up( G,3) =++ . 30 21 12
And so on we get for all up( G, h) , where 1 ≤≤hn
22rnrkk − up( G,. h) = ∑ r1 ≥1 rr12 rr12+= h
Thus, the proof is done.
Theorem 2.17. For any graph GW≅ s with s +1 vertices and s > 3 , then =+ +−s UP( G,1 x) ( x) ( 1 x) 1.
Proof. Let G be a wheel graph Ws ( s > 3 ), then we have γ up(W s ) = 1. There
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s +1 are s UDS of size one. For i = 2 there are UDS of size two and so on. 2 Thus,
ss++1123 s + 1s+ 1 UP( G, x) =+ sx x + x ++ x 23 s + 1 s+1 s +1 =i −+ ∑xx( 1) i=0 i s+1 =+( xx11) −+( ) s =+( xx1) ( +− 1) 1 .
Corollary 2.18. For any wheel graph Ws and s > 3 we have {0,−− 1, 2} , ifs is even. ZWup( s ) = {0,− 1} , ifs is odd.
3. Uphill Domination Polynomials of Graphs under Some Binary Operations
Theorem 3.1. Let GPP≅×rs be a grid graph with rs vertices and rs,4≥ . rs−4 Then, UP( G,1 x) = x4 ( + x) . Proof. Let G be a grid graph with rs vertices and rs,4≥ , then we have
γ up (G) = 4 . Note that, there is only one UDS of size four. For i = 5 , there are rs − 4 UDS of size five and so on. Thus, we get
45rs −−44 rs rs UP( G, x) = x + x ++ x 14 rs −
rs−4 4 rs −4 i = xx∑ i=0 i rs−4 =xx4 (1. + )
Theorem 3.2. Let GCK≅ rs be a corona graph with rs+ r vertices. Then, r UP( G,1 x) = xrs ( + x) .
Proof. Let GCK≅ rs with rs+ r vertices, we have γ up(C r K s ) = rs . For rs vertices, there is only one UDS of size rs. For rs +1 vertices, there are r UDS and so on. Thus, we get
rs rrrs++1 rs r UP( G, x) = x + x ++ x 1 r r r rs+ i = ∑x i=0 i r rs r i = xx∑ i=0 i r =xxrs (1. + )
Corollary 3.3. Let GCK≅ r 1 be a corona graph with 2r vertices. Then, r UP( G,1 x) = xr ( + x) .
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Theorem 3.2 can generalize in the following result. Theorem 3.4. For any nontrivial connected graph H with r vertices, if rs r G≅ HK s , then, UP( G,1 x) = x( + x) . Proof. The proof similarly to the proof of Theorem 3.2.
Theorem 3.5. Let G be a book graph Bmm= PS2 × with 22m + vertices. Then, mm m−1 m ++ 11 m UP( G,2 x) =++ x m( 2) 2x
21m− mmmi− mi−+22 m mi−+ mi + +∑ 22 ++ 2x i=2 ii −−12 i
2m 2 2m21 mm++ 22 ++12m + 2 x +( 22 mx +) + x . 2
Proof. Suppose we have the book graph Bmm= PS2 × with 22m + vertices,
then we have γ up(PS2 ×= m ) m. Let divide the vertices of Bm into m +1 sets
“as shown in Figure 2” let the set Ri= { uv ii, } i.e., 1 ≤≤im while
Rm+1 = { uv, } . Since γ up(PS2 ×= m ) m, then for up( G, m) we have to take one m vertex from each Ri ( im≠+1 ) so, there exist 2 UDS of size m. For up( G,1 m + ) we have, 22 2 2 2 up( G,1 m +=) +∑ 11 ∑ rmi = +1r1 rm 0 rr1,1m ≥ (m+1) times =2mm+−11 + m( 2.)
Also, for up( G,2 m + ) we get 222222 up( G,2 m +=) ∑∑+ ∑∑rmii=+=21rr11rrmm01rm+ rr11,1mm≥≥rr,1 2 2 2 + ∑ ∑ rmi = r1 rm 2 rr1,1m ≥
mm−2 mm mm =2 ++ 22 2 10
m m−2 mm =2 ++m 2 2. 2
Figure 2. A Book Graph Bm.
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Therefore, for up( G,3 m + ) we have 222222 up( G,3 m +=) ∑∑+ ∑∑rmii=+=32rr11rrmm01rm+ rr11,1mm≥≥rr,1 2 2 2 + ∑ ∑ rmi = +1r1 rm 2 rr1,1m ≥
mm−3 mm mm =2 ++ 22 3 21
mmmm−3 m =22 ++ m2. 32 And so on, we use the same argument until up( G,2 m − 1) . After that, for up( G,2 m) we have 2 22 2 2 2 up( G,2 m) = + ∑ 2 20 ∑ rmi =21 − r1 rm 1 rr1,1m ≥ 2 2 2 + ∑ ∑ rmi =22 − r1 rm 2 rr1,1m ≥
22m =++1m 2 2. 2 Finally, 22m + up( G,2 m += 1) =2m + 2 & up( G ,2 m + 2) = 1. 21m +
Thus, the proof is completed.
Theorem 3.6. Let G be a graph. If GPC≅×ks with sk vertices, then sk ss sk− 2 s UP G,. x = xt ( ) ∑∑ t=2 rr12,1≥ rr12 r3 rr123++= rt
Proof. Let GPC≅×ks with sk vertices, then we have γ up(PC k×= s ) 2 . We
first divide the vertices of G into three sets called them RR12, and R3 , where
R1 (resp. R2 ) is contains the vertices of the outer cycle (resp. inner cycle)
which every vertex is of degree three. The third set R3 contains the vertices of the middle cycles, where every vertex is of degree four. Note that, any UDS
should contain at least one vertex form R1 and one vertex from R2 . Thus, for up( G,2)
sssks − 2 2 up( G,2) = = s . 11 0
For up( G,3) we have ss sk− 2 s up( G,3) = ∑ . rr12,1≥ rr12 r3 rr123++= r3
And so on, we use the same argument for all up( G, t) i.e., 3 ≤≤t sk and
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the proof is done.
Theorem 3.7. Let G ba a tadpole graph Tsk, with sk+ vertices. Then, sk+ ks −1 UP G,1 x=−+ s x2 xt . ( ) ( ) ∑∑ t=3 rrt12+=−1rr12 r2 ≥1
Proof. Let G be a tadpole graph Tsk, with sk+ vertices, we have
γ up(T s, k ) = 2 . We first divide the vertices of Tsk, into three sets called them
RR12, and R3 such that R1 is a singleton set that contains the pendant vertex,
R2 has k vertices each of them is of degree two except one vertex is of degree
three while the last set R3 has s −1 vertices each of them of degree two which
are the vertices that lies in a cycle part of a graph. Notice that, any UDS of Tsk,
should contains the pendant vertex and at least one vertex from R3 . Now, for
up( G,2) we have to take the pendant vertex with one vertex from R3 , so there exist s −1 UDS of size two. For up( G,3) we get k s −1 up( G,3) = ∑ . r3 ≥1 r2 r3 rr23+=2
And so on, we use the same argument for all up( G, t) i.e., 3 ≤≤+t sk and the proof is completed. Theorem 3.8. Let G be a windmill graph Wd( s, k ) with ks( −+11) vertic- es. Then,
ks( −+11) k s −1 s −11 UP G,1 x=−+ s xktx . ( ) ( ) ∑∑ tk= +1 rr1,, k ≥ 1r1 rrkk +1 r11++ rtk+ =
Proof. Let G be a windmill graph with center vertex w, we have γ up (Gk) = . Any minimum uphill domination set must contains one vertex from each copy k of Ks without the center vertex w, that means, we have (s −1) uphill domi-
nating set of size k. Suppose Ri be the set of vertices of the i-th copy of KS
without the center vertex w and Rw be the singleton, with the center vertex w. To get the number of uphill dominating sets of size tkj= + , where
j= 1,2, ,( ks( −+ 2) 1) , we need to select ri vertices from each Ri , and rk +1 = k +1 = ≥ = from Rw where ik1, 2, , , ∑i=1 rti and ri 1 for all ik1, 2, , . Hence, s −1 s −11 up( G,. t) = ∑ rr1,, k ≥ 1r1 rrkk +1 r11++ rtk+ =
Thus,
ks( −+11) k s −1 s −11 UP G,1 x=−+ s xktx . ( ) ( ) ∑∑ tk= +1 rr1,, k ≥ 1r1 rrkk +1 r11++ rtk+ =
Proposition 3.9. Let G be a dutch windmill graph D( sk, ) with s > 3 and ks( −+11) vertices. Then,
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UP( D( s,, k) x) = UP( Wd( s ,,. k) x)
Theorem 3.10. Let G be a firefly graph Fstk,, with stk,,≥ 0 ,
n=22 s + tk ++ 1 vertices and γ up (G) = stk ++ = b. Then, sb s s−+11 b Up( G, x) = 2 x + 2( t ++ 12) ( s) x n ss st +1 + h ∑∑ x . hb= +2 rr1,, s ≥ 1 rr12 rrss +1 r11+ + rs+ =−+ h( tk)
Proof. Let G be a firefly graph Fstk,, with n vertices and
γ up (G) = stk ++ = b. First, let us divide the vertices of G into s + 2 sets and
let u be the shared vertex in G. Suppose that R1 ⊂ VG( ) contains the vertices
of the first triangle without u, this implies R1 has two vertices each of them are
of degree two, also we mean by R2 ⊂ VG( ) the set that contains the vertices of
the second triangle without u and so on for all Ri , where 1 ≤≤is. Now, the
subset Rs+1 ⊂ VG( ) contains u in addition the t vertices of the pendant paths
that adjacent to u which means Rs+1 is of cardinality t +1 . Finally,
Rs+2 ⊂ VG( ) contains all the leaves vertices of G which be exactly of cardinality
tk+ . Notice that, any UDS of G should contain all the vertices of Rs+2 with at
least one vertex from each Ri . Thus, for up( G, b) we have 2 21 t++ tk up( G, b) = ∑ s+2 r rr++ r ∑i=1 rbi = 1 ss 12 s 22 t + 1 tk + = 1 10 tk+ s =××2 2 × 2 = 2. s times
For up( G,1 b + ) we get 2 21 t + tk+ up( G,1 b +=) ∑ s+2 r rr+ tk+ ∑i=1 rbi = +1 1 ss 1 2 2 tt++1 22 1 = ∑ + ∑s+1r=( b +−+1) ( tk) r1 rs 0 1 11 i=1 i s times =2ss−1 (st) ++ 2( 1.)
And for up( G,2 b + ) we have 2 21 t + tk+ up( G,2 b +=) ∑ s+2 r rr tk+ ∑ = rbi = +2 1 ss +1 i 1 2 21 t + = ∑ . s+1 r rr+ ∑i=1 ri =( b +2) −+( tk) 1 ss 1
In the same argument we can find all up( G, h) , where b+≤≤2 hn and the proof is completed.
Corollary 3.11. Let G be a friendship graph Fk with 21k + vertices. Then,
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21k + 2 21 UP G, x= 2kk x ++ 22 k k k−+11 x k + xt. ( ) ∑∑ tk= +2 rr1,, k ≥ 1r1 rrkk +1 r11++ rtk+ =
4. Open Problems
Finally, for feature work we state the following definition. Definition 4.1. Two graphs G and H are said to be uphill-equivalent if UP( G,, x) = UP( H x) . The uphill-equivalence classes of G noted by = [G]up { HH: is uphill-equivalent to G} . Example 4.2. = 1) [Kn ]up { HH: is regular graph ofn vertices} . 2) The windmill graph Wd( s, k ) and Dutch windmill graph D( sk, ) are uphill-equivalent. We state the following open problems for feature work: 1) which graphs have two distinct uphill domination roots? 2) which families of graphs have only real uphill domination roots? = 3) which graphs satisfy [GG]up { } ? 4) determine the uphill-equivalence classes for some new families of graphs.
Conflicts of Interest
The authors declare no conflicts of interest regarding the publication of this paper.
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DOI: 10.4236/jamp.2020.86088 1179 Journal of Applied Mathematics and Physics