Volume 2009, Article ID 542040,10pages doi:10.1155/2009/542040
Research Article
Dominating Sets and Domination Polynomials
of Paths
Saeid Alikhani
1, 2and Yee-Hock Peng
2, 31Department of Mathematics, Faculty of Science, Yazd University, 89195-741, Yazd, Iran 2Institute for Mathematical Research, University Putra Malaysia, 43400 UPM Serdang, Malaysia 3Department of Mathematics, University Putra Malaysia, 43400 UPM Serdang, Malaysia
Correspondence should be addressed to Saeid Alikhani,alikhani@yazduni.ac.ir
Received 13 November 2008; Accepted 12 March 2009
Recommended by Alexander Rosa
LetG V, Ebe a simple graph. A setS⊆V is a dominating set ofG, if every vertex inV\Sis adjacent to at least one vertex inS. LetPi
nbe the family of all dominating sets of a pathPnwith
cardinalityi, and letdPn, j |Pjn|. In this paper, we constructPin, and obtain a recursive formula
fordPn, i. Using this recursive formula, we consider the polynomialDPn, x inn/3dPn, ixi,
which we call domination polynomial of paths and obtain some properties of this polynomial.
Copyrightq2009 S. Alikhani and Y.-H. Peng. This 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.
1. Introduction
Let G V, E be a simple graph of order |V| n. For any vertex v ∈ V, the open neighborhood ofvis the setNv {u∈V |u v ∈E}and the closed neighborhood ofvis the setNv Nv∪ {v}. For a setS⊆V, the open neighborhood ofSisNS ∪v∈SNv
and the closed neighborhood ofS isNS NS∪S. A setS ⊆ V is a dominating set of G, if NS V, or equivalently, every vertex in V\S is adjacent to at least one vertex inS. The domination numberγGis the minimum cardinality of a dominating set inG. A dominating set with cardinalityγGis called aγ-set, and the family ofγ-sets is denoted by
ΓG. For a detailed treatment of this parameter, the reader is referred to1. It is well known and generally accepted that the problem of determining the dominating sets of an arbitrary graph is a difficult one see2. A path is a connected graph in which two vertices have degree 1 and the remaining vertices have degree 2. LetPn be the path withnvertices. Let Pi
n be the family of dominating sets of a pathPn with cardinalityiand letdPn, i |Pni|.
We call the polynomialDPn, x
n
in/3dPn, ixi, the domination polynomial of the path Pn. For a detailed treatment of the domination polynomial of a graph, the reader is referred
In the next section we construct the families of the dominating sets of paths by a recursive method. In Section 3, we use the results obtained in Section 2 to study the domination polynomial of paths.
As usual we usex, for the smallest integer greater than or equal tox. In this article we denote the set{1,2, . . . , n}simply byn.
2. Dominating sets of paths
Let Pi
n be the family of dominating sets of Pn with cardinality i. We will investigate
dominating sets of path. We need the following lemmas to prove our main results in this article.
Lemma 2.1see4, page 371. γPn n/3.
ByLemma 2.1and the definition of domination number, one has the following lemma.
Lemma 2.2. Pi
j∅, if and only ifi > jori <j/3.
A simple path is a path in which all its internal vertices have degree two. The following lemma follows from observation.
Lemma 2.3. If a graphGcontains a simple path of length 3k−1, then every dominating set ofG must contain at leastkvertices of the path.
To find a dominating set of Pn with cardinality i, we do not need to consider
dominating sets ofPn−4with cardinalityi−1. We show this inLemma 2.4. Therefore, we only
need to considerPin−−11,Pin−−12, and Pin−−13. The families of these dominating sets can be empty or otherwise. Thus, we have eight combinations of whether these three families are empty or not. Two of these combinations are not possible seeLemma 2.5i and ii. Also, the combination thatPni−−11 Pni−−12 Pni−−13 ∅does not need to be considered because it implies Pi
n ∅seeLemma 2.5iii. Thus we only need to consider five combinations or cases. We
consider these cases inTheorem 2.7.
Lemma 2.4. IfY ∈ Pni−−14, and there existsx∈nsuch thatY ∪ {x} ∈ Pi
n, thenY ∈ Pin−−13.
Proof. Suppose thatY /∈ Pi−1
n−3. SinceY ∈ Pni−−14,Y contains at least one vertex labeledn−5 or n−4. Ifn−4 ∈ Y, thenY ∈ Pi−1
n−3, a contradiction. Hence, n−5 ∈ Y, but then in this case, Y∪ {x}/∈ Pi
n, for anyx∈n, also a contradiction. Lemma 2.5. iIfPi−1
n−1Pni−−13∅, thenPin−−12∅.
iiIfPi−1
n−1/∅andPni−−13/∅, thenPni−−12/∅.
iiiIfPni−−11Pni−−12Pni−−13∅, thenPi
n∅.
Proof. iSincePi−1
n−1 Pni−−13 ∅, byLemma 2.2,i−1> n−1 ori−1 <n−3/3. In either
case we havePi−1 n−2∅.
iiSuppose thatPni−−12∅, so byLemma 2.2, we havei−1> n−2 ori−1<n−2/3. Ifi−1> n−2, theni−1> n−3, and hence,Pi−1
n−3∅, a contradiction. Hencei−1<n−2/3.
Soi−1<n−1/3, and hence,Pi−1
n−1∅, also a contradiction.
iiiSuppose thatPi
n−1∈Y orn−2∈Y, thenY − {n} ∈ Pni−−11, a contradiction. Ifn−3∈Y, thenY − {n} ∈ Pni−−12, a contradiction. Now suppose thatn−1∈Y. Then byLemma 2.3, at least one vertex labeled
n−2, n−3 orn−4 is inY. Ifn−2∈Y orn−3∈Y, thenY− {n−1} ∈ Pi−1
n−2, a contradiction. If n−4∈Y, thenY− {n−1} ∈ Pin−−13, a contradiction. ThereforePi
n∅. Lemma 2.6. IfPi
n/∅, then
iPi−1
n−1Pni−−12∅, andPni−−13/∅if and only ifn3kandikfor somek∈N;
iiPi−1
n−2Pni−−13∅andPni−−11/∅if and only ifin;
iiiPni−−11 ∅,Pni−−12 /∅andPni−−13/∅if and only ifn 3k 2 andi3k 2/3for somek∈N;
ivPni−−11/∅,Pni−−12/∅andPni−−13∅if and only ifin−1;
vPi−1 n−1/∅,P
i−1
n−2/∅andPni−−13/∅if and only ifn−1/3 1≤i≤n−2.
Proof. i ⇒SincePni−−11 Pni−−12 ∅, byLemma 2.2,i−1 > n−1 ori−1 < n−2/3. If
i−1> n−1, theni > n, and byLemma 2.2,Pin∅, a contradiction. Soi <n−2/3 1, and sincePi
n/∅, togethern/3 ≤i < n−2/3 1, which give usn3kandikfor some k∈N.
⇐Ifn 3k andi k for somek ∈ N, then byLemma 2.2,Pni−−11 Pni−−12 ∅, and Pi−1
n−3/∅.
ii ⇒SincePin−−12 Pni−−13 ∅, byLemma 2.2,i−1 > n−2 or i−1 < n−3/3. If
i−1<n−3/3, theni−1<n−1/3, and hencePni−−11∅, a contradiction. So
i > n−1. Also sincePi−1
n−1/∅,i−1≤n−1. Thereforein.
⇐Ifin, then byLemma 2.2,Pni−−12Pni−−13∅andPni−−11/∅.
iii ⇒SincePni−−11 ∅, byLemma 2.2,i−1> n−1 ori−1<n−1/3. Ifi−1> n−1, then
i−1> n−2 and byLemma 2.2,Pni−−12Pni−−13∅, a contradiction. Soi <n−1/3 1, buti−1≥ n−2/3becausePi−1
n−2/∅. Hence,n−2/3 1≤i <n−1/3 1.
Thereforen3k 2 andik 13k 2/3for somek∈N.
⇐Ifn3k 2 andi3k 2/3for somek ∈N, then byLemma 2.2,Pi−1 n−1 Pk
3k 1∅,P i−1
n−2/∅, andPni−−13/∅.
iv ⇒SincePin−−13 ∅, byLemma 2.2,i−1> n−3 ori−1<n−3/3. SincePin−−12/∅, byLemma 2.2,n−2/3 1≤i≤n−1. Thereforei−1<n−3/3is not possible. Hencei−1> n−3. Thusin−1 orn, buti /nbecausePi−1
n−3∅. Soin−1.
⇐Ifin−1, then byLemma 2.2,Pi−1
n−1/∅,Pni−−12/∅, andPin−−13 ∅.
v ⇒ Since Pni−−11/∅,Pni−−12/∅, and Pin−−13/∅, then by applying Lemma 2.2, n −
1/3 ≤i−1≤n−1,n−2/3 ≤i−1≤n−2, andn−3/3 ≤i−1 ≤n−3.So n−1/3 ≤i−1≤n−3 and hencen−1/3 1≤i≤n−2.
⇐Ifn−1/3 1≤i≤n−2,then the result follows fromLemma 2.2.
By Lemma 2.4, for the construction of Pi
n, it’s sufficient to consider Pni−−11,Pin−−12, and Pi−1
Theorem 2.7. For everyn≥4 andi≥ n/3.
iIfPi−1
n−1Pin−−12 ∅andPni−−13/∅, thenPin{{2,5, . . . , n−4, n−1}}.
iiIfPin−−12Pni−−13 ∅, andPni−−11/∅, thenPi
n{n}.
iiiIfPi−1
n−1∅,Pni−−12/∅andPni−−13/∅, then
Pi
n
{2,5, . . . , n−3, n}∪X∪ {n−1} | X ∈ Pi−1 n−3
. 2.1
ivIfPin−−13∅,Pni−−12/∅andPni−−11/∅, thenPi
n{n− {x} |x∈n}.
vIfPi−1
n−1/∅,Pin−−12/∅andPin−−13/∅, then
Pi
n
{n} ∪X1,{n−1} ∪X2 | X1∈ Pni−−11, X2∈ Pin−−12
∪{n−1} ∪X | X∈ Pni−−13\ Pin−−12
∪{n} ∪X | X∈ Pni−−13∩ Pni−−12.
2.2
Proof. iPi−1
n−1 Pni−−12 ∅, andPin−−13/∅. ByLemma 2.6i,n 3kandikfor somek∈N.
ThereforePni Pn/3n {{2,5, . . . , n−4, n−1}}.
iiPni−−12Pni−−13∅, andPni−−11/∅. ByLemma 2.6ii, we havein. SoPi
nPnn{n}.
iiiPi−1
n−1∅,Pni−−12/∅andPni−−13/∅. ByLemma 2.6iii,n3k 2 andi3k 2/3 k 1 for somek∈N. SinceX{2,5, . . . ,3k−1} ∈ Pk3k,X∪ {3k 2} ∈ Pk3k12. Also, if
X∈ Pk
3k−1, thenX∪ {3k 1} ∈ P3kk 12. Therefore we have
{2,5, . . . ,3k−1,3k 2}∪X∪ {3k 1} |X∈ P3kk−1⊆ Pk3k12. 2.3
Now letY ∈ Pk 1
3k 2. Then 3k 2 or 3k 1 is inY. If 3k 2∈Y, then by Lemma 3, at least
one vertex labeled 3k 1,3kor 3k−1 is inY. If 3k 1 or 3kis inY, thenY−{3k 2} ∈ Pk
3k 1, a contradiction becauseP3kk 1∅. Hence, 3k−1∈Y, 3k /∈Y, and 3k 1/∈Y.
ThereforeY X∪{3k 2}for someX∈ Pk
3k, that isY {2,5, . . . ,3k−1,3k 2}. Now
suppose that 3k 1 ∈Y and 3k 2/∈Y. ByLemma 2.3, at least one vertex labeled 3k,3k−1, or 3k−2 is in Y. If 3k∈Y, thenY− {3k 1} ∈ Pk3k{{2,5, . . . ,3k−1}}, a contradiction because 3k /∈Xfor allX ∈ Pk
3k. Therefore 3k−1 or 3k−2 is inY, but
3k /∈Y. ThusY X∪ {3k 1}for someX∈ Pk 3k−1. So
Pk 1 3k 2 ⊆
{2,5, . . . ,3k−1,3k 2}∪{3k 1} ∪X |X∈ Pk 3k−1
. 2.4
ivPni−−13∅,Pin−−11/∅, andPni−−12/∅. ByLemma 2.6iv,in−1. ThereforePi
nPnn−1 {n− {x} |x∈n}.
vPni−−11/∅,Pni−−12/∅, andPin−−13/∅. LetX1 ∈ Pin−−11, so at least one vertex labeledn−1
LetX2∈ Pni−−12, thenn−2 orn−3 is inX2. Ifn−2 orn−3∈X2, thenX2∪ {n−1} ∈ Pin.
Now letX3 ∈ Pni−−13, thenn−3 orn−4 is inX3. Ifn−3 ∈X3, thenX3∪ {x} ∈ Pin, for x∈ {n, n−1}. Ifn−4∈X3, thenX3∪ {n−1} ∈ Pni. Therefore we have
{n} ∪X1,{n−1} ∪X2|X1∈ Pni−−11, X2 ∈ Pni−−12
∪{n−1} ∪X|X ∈ Pni−−13\ Pin−−12
∪{n} ∪X|X∈ Pni−−13∩ Pni−−12⊆ Pin.
2.5
Now, letY ∈ Pi
n, thenn∈ Y orn−1 ∈ Y. Ifn ∈ Y, then byLemma 2.3, at least one vertex
labeledn−1, n−2, orn−3 is inY. Ifn−1∈Yorn−2∈Y, thenY X∪{n}for someX ∈ Pi−1 n−1.
Ifn−3∈Y, n−2/∈Y, andn−1/∈Y, thenY X∪ {n}for someX∈ Pn−2
i−1 ∩ Pni−−13. Now suppose
thatn−1∈Yandn /∈Y, then by Lemma 3, at least one vertex labeledn−2, n−3 orn−4 is in
Y. Ifn−2∈Y orn−3∈Y, thenY X∪ {n−1}for someX ∈ Pi−1
n−2. Ifn−4∈Y, n−3/∈Yand n−2/∈Y, thenY X∪ {n−1}for someX∈ Pi−1
n−3\ Pin−−12. So
Pi n⊆
{n} ∪X1,{n−1} ∪X2 |X1∈ Pin−−11, X2∈ Pni−−12
∪{n−1} ∪X|X∈ Pi−1 n−3\ Pni−−12
∪{n} ∪X |X∈ Pni−−13∩ Pin−−12.
2.6
Example 2.8. ConsiderP6withVP6 6. We useTheorem 2.7to constructPi6for 2≤i≤6.
SinceP1
5 P41∅andP31{{2}}, byTheorem 2.7,P62{{2,5}}.
SinceP5
5 {5},P54 ∅, andP53 ∅, we getP66{6}.
Since P45 {{1,2,3,4},{1,2,3,5},{1,3,4,5},{2,3,4,5},{1,2,4,5}}, P44 {4}, and P4
3 ∅, then byTheorem 2.7,
P5 6
6− {x} |x∈6
{1,2,3,4,6},{1,2,3,5,6},{1,3,4,5,6},{2,3,4,5,6},{1,2,4,5,6},{1,2,3,4,5}, 2.7
and, for the construction ofP3
6, byTheorem 2.7,
P3 6
X1∪ {6}, X2∪ {5} |X1∈ P25, X2∈ P42
∪{1,2} ∪ {5},{1,3} ∪ {6},{2,3} ∪ {6}
{1,3,5},{1,3,6},{2,3,6},{2,3,5},{1,4,6},{1,4,5},{2,5,6},{2,4,6},{2,4,5},{1,2,5}.
2.8
Finally, since P3
5 {{1,3,5},{1,2,4},{2,4,5},{2,3,4},{2,3,5},{1,4,5},{1,3,4},{1,2,5}}, P3
4 {{1,2,3},{1,2,4},{2,3,4},{1,3,4}}, andP33{3}, then
P4 6
X1∪ {6}, X2∪ {5} |X1∈ P35, X2∈ P34
∪X∪ {6} |X∈ P33
{1,2,3,5},{1,2,3,6},{1,2,4,6},{1,3,5,6},{1,3,4,6},{1,3,4,5},{1,2,5,6},
{1,2,4,5},{2,3,4,6},{1,4,5,6},{2,3,4,5},{2,4,5,6},{2,3,5,6}.
Table 1:dPn, j, the number of dominating set ofPnwith cardinalityj.
j 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
n
1 1
2 2 1
3 1 3 1
4 0 4 4 1
5 0 3 8 5 1
6 0 1 10 13 6 1
7 0 0 8 22 19 7 1
8 0 0 4 26 40 26 8 1
9 0 0 1 22 61 65 34 9 1
10 0 0 0 13 70 120 98 43 10 1
11 0 0 0 5 61 171 211 140 53 11 1
12 0 0 0 1 40 192 356 343 192 64 12 1
13 0 0 0 0 19 171 483 665 526 255 76 13 1
14 0 0 0 0 6 120 534 1050 1148 771 330 89 14 1
15 0 0 0 0 1 65 483 1373 2058 1866 1090 418 103 15 1
3. Domination Polynomial of a path
LetDPn, x
n
in/3dPn, ixibe the domination polynomial of a pathPn. In this section
we study this polynomial.
Theorem 3.1. iIfPi
nis the family of dominating set with cardinalityiofPn, then
|Pi
n||Pin−−11| |Pni−−12| |Pni−−13|. 3.1
iiFor everyn≥4,
DPn, x x
DPn−1, x DPn−2, x DPn−3, x
, 3.2
with the initial valuesDP1, x x, DP2, x x2 2x, andDP3, x x3 3x2 x.
Proof. iIt follows fromTheorem 2.7.
iiIt follows from Partiand the definition of the domination polynomial.
UsingTheorem 3.1, we obtaindPn, jfor 1 ≤ n ≤ 15 as shown inTable 1. There are
interesting relationships between numbers in this table. In the following theorem, we obtain some properties ofdPn, j.
Theorem 3.2. The following properties hold for the coefficients ofDPn, x:
idP3n, n 1, for everyn∈N.
iidP3n 2, n 1 n 2, for everyn∈N.
iiidP3n 1, n 1 n 2n 3/2−2, for everyn∈N.
vdPn, n 1, for everyn∈N.
vidPn, n−1 n, for everyn≥2.
viidPn, n−2 n2−2 nn−1/2−2, for everyn≥3.
viiidPn, n−3 n3−3n−8 n−4n−3n 4/6, for everyn≥4.
ixdPn, n−4 n−5n3−n2−42n 96/24, for everyn≥5.
x1 dPn, n< dPn 1, n < dPn 2, n<· · · < dP2n−1, n < dP2n, n> dP2n 1, n> · · ·> dP3n−1, n> dP3n, n 1.
xi3jijdPi, j 33jij−−31dPi, j−1.
xiiIfsn njn/3dPn, j, then for everyn≥4,sn sn−1 sn−2 sn−3with initial values s1 1, s23 ands35.
xiiiFor everyn∈N, andk0,1,2, . . . ,2n−1,dP2n−k, n dP2n k, n.
xivFor everyj≥ n/3,dPn 1, j 1−dPn, j 1 dPn, j−dPn−3, j.
Proof. iSincePn
3n {{2,5, . . . ,3n−1}}, we havedP3n, n 1.
iiProof by induction onn. SinceP2
5{{1,4},{2,4},{2,5}}, sodP5,2 3. Therefore
the result is true forn 1. Now suppose that the result is true for all natural numbers less thann, and we prove it forn. By parti,Theorem 3.1, and the induction hypothesis we have
dP3n 2, n 1 dP3n 1, n dP3n, n dP3n−1 2, n n 2.
iiiProof by induction onn. The result is true forn2, becausedP7,3 8 4 4.
Now suppose that the result is true for all natural numbers less thann, and we prove it forn. By parti,ii,Theorem 3.1, and the induction hypothesis we have
dP3n 1, n 1
dP3n, n
dP3n−1, n
dP3n−2, n
1 dP3n−1 2, n
dP3n−1 1, n
1 n 1 n 1n 2 2 −2
n 2n 3
2 −2.
3.3
ivProof by induction on n. Since dP3,2 3, the result is true for n 1. Now
suppose that the result is true for all natural numbers less thann, and we prove it forn. By Theorem 3.1, partsii,iii, and the induction hypothesis we have
dP3n, n 1
dP3n−1, n
dP3n−2, n
dP3n−3, n
n 1 n 1n 2 2 −2
nn−1n 7 6
nn 1n 8
6 .
3.4
vSincePn
n {n}, we have the result.
viSincePn−1
viiBy induction onn. The result is true forn3, becausedP3,1 1. Now suppose
that the result is true for all numbers less that n, and we prove it for n. ByTheorem 3.1, induction hypothesis, partvand partviwe have
dPn, n−2
dPn−1, n−3
dPn−2, n−3
dPn−3, n−3
n−1n−2
2 n−3
nn−1
2 −2.
3.5
viiiBy induction onn. The result is true forn 4, sincedP4,1 0. Now suppose
that the result is true for all natural numbers less than or equalnand we prove it forn 1. By Theorem 3.1, induction hypothesis, partsviiandviwe have
dPn 1, n−2
dPn, n−3
dPn−1, n−3
dPn−2, n−3
n−4n−3n 4
6
n−1n−2
2 −2 n−2
n−3n−2n 5
6 .
3.6
ixBy induction onn. SincedP5,1 0, the result is true forn5. Now suppose that
the result is true for all natural numbers less thann, and we prove it forn. ByTheorem 3.1, induction hypothesis, partsviiiandvii, we have
dPn, n−4
dPn−1, n−5
dPn−2, n−5
dPn−3, n−5
n−6
n−13−n−12−42n 138 24
n−6n−5n 2 6
n−3n−4 2 −2
n−5
n3−n2−42n 96
24 .
3.7
xWe will prove that for everyn,dPi, n< dPi 1, nforn≤i≤2n−1, anddPi, n> dPi 1, nfor 2n≤i≤3n−1. We prove the first inequality by induction onn. The result holds
forn 1. Suppose that result is true for alln ≤ k. Now we prove it forn k 1, that is
dPi, k 1< dPi 1, k 1fork 1≤i≤2k 1. ByTheorem 3.1and the induction hypothesis
we have
dPi, k 1
dPi−1, k
dPi−2, k
dPi−3, k
< dPi, k
dPi−1, k
dPi−2, k
dPi 1, k 1
.
3.8
xi Proof by induction on j. First, suppose that j 2. Then i62dPi,2 12
33i1dPi,1. Now suppose that the result is true for everyj < k, and we prove forj k:
3k
ik dPi, k
3k
ik
dPi−1, k−1
3k
ik
dPi−2, k−1
3k
ik
dPi−3, k−1
3
3k−1
ik−1
dPi−1, k−2
3
3k−1
ik−1
dPi−2, k−2
3
3k−1
ik−1
dPi−3, k−2
3
3k−3
ik−1
dPi, k−1
.
3.9
xiiByTheorem 3.1, we have
sn n
jn/3
dPn, j
n
jn/3
dPn−1, j−1
dPn−2, j−1
dPn−3, j−1
n−1
jn/3−1
dPn−1, j
n−2
jn/3−1
dPn−2, j
n−3
jn/3−1
dPn−3, j−1
sn−1 sn−2 sn−3.
3.10
xiiiProof by induction onn. SincedP1,1 dP3,1, the theorem is true forn1.
Now, suppose that the theorem is true for all numbers less thann, and we will prove it forn. ByTheorem 3.1and the induction hypothesis, we can write
dP2n−k, n
dP2n−k−1, n−1
dP2n−k−2, n−1
dP2n−k−3, n−1
dP2n−1 1−k, n−1
dP2n−1−k, n−1
dP2n−1−1−k, n−1
dP2n−1 k−1, n−1
dP2n−1 k, n−1
dP2n−1 1 k, n−1
dP2n k−3, n−1
dP2n k−2, n−1
dP2n k−1, n−1
dP2n k, n
.
3.11
xivByTheorem 3.1, we have
dPn 1, j 1
−dPn, j 1
dPn, j
dPn−1, j
dPn−2, j
−dPn−1, j
dPn−2, j
dPn−3, j
dPn, j
−dPn−3, j
.
3.12
In the following theorem we use the generating function technique to find|Pi n|.
Theorem 3.3. For everyn∈Nandn/3 ≤i≤n,|Pi
n|is the coefficient ofunviin the expansion of the function
fu, v u
3v1 3v v2 3uv uv2 2u2v u2v2
1−uv−u2v−u3v . 3.13
Proof. Setfu, v ∞n3
∞
i1|Pin|unvi. By recursive formula for|Pin|inTheorem 3.1we can
writefu, vin the following form:
fu, v
∞
n3 ∞
i1 Pi−1
n−1 Pni−−12 Pin−−13unvi
uv
∞
n3 ∞
i1 Pi−1
n−1un−1vi−1 u2v ∞
n3 ∞
i1 Pi−1
n−2un−2vi−1 u3v ∞
n3 ∞
i1 Pi−1
n−3un−3vi−1
uvP20u2 P21u2v P22u2v2 uvfu, v
u2vP10u P11uv P02u2 P21u2v P22u2v2
u2vfu, v u3vP00 P10u P11uv P20u2 P21u2v P22u2v2
u3vfu, v.
3.14
By substituting the values from Table 1 note that |P0
n| 0 for all natural numbers nand |P0
0|1, we have
fu, v1−uv−u2v−u3vu3v1 3v v2 3uv uv2 2u2v u2v2. 3.15
Therefore we have the result.
Acknowledgments
Authors wish to thank the referee for his valuable comments and the research management centerRMCof University Putra Malaysia for their partial financial support.
References
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Monographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, New York, NY, USA, 1998. 2 M. R. Garey and D. S. Johnson, Computers and Intractability: A Guide to the Theory of NP-Completeness,
Series of Books in the Mathematical Science, W. H. Freeman, San Francisco, Calif, USA, 1979.
3 S. Alikhani and Y. H. Peng, “Introduction to domination polynomial of a graph,” to appear in Ars
Combinatoria.