CC BY
6
ИРКУТСКИЙ ГОСУДАРСТВЕННЫЙ УНИВЕРСИТЕТ ПУТЕЙ СООБЩЕНИЯ
5. Chunhua LI, Baogen XU. Fuzzy good congruences on abundant semigroups // Fuzzy Systems and Mathematics. 2008. V. 22(4). P. 74-78.
6. Fountain J. B. Abundant semigroups // Proc. London Math. Soc. 1982. V. 44 P. 103-129.
7. Mordeson J. N., Malik D. S., Kuroki N. Fuzzy semigroups. Springer-verlag Berlin Heidelberg : New York, 2003.
8. Samhan M. Fuzzy Congruences on semigroups // Inform. Sci. V. 74(1993). P. 165-175.
9. Petrich M. Completely regular semigroups. New York : Jhon Wiley and Sons Inc, 1999.
10. Murali V. Fuzzy equivalence relations // Fuzzy Sets and Systems. V. 30(1989). P. 155-163.
11. Nemitz W. C. Fuzzy relations and fuzzy functions // Fuzzy Sets and Systems. V. 19(1986). P. 177191.
12. Xiaojiang GUO, Xiangyun XIE. A note on left type-A semigroups // Semigroup Forum, 1999. V.58P. 313-316.
13. Xiaojiang GUO. Idempotent-separating good congruence on an IC quasi-adequate semigroup. // Pure Math. V. 16(1999). P. 57-65.
14. Yanfeng LUO, Xiaojiang GUO. Good congruences on a typeosemigroup // Communications in Algebra, 1999. V. 27(12). P. 6199-6211.
15. Yijia TAN. Fuzzy congruences on a regular semigroup // Fuzzy Sets and Systems, 2001 V. 117. P. 447-453.
Baogen Xu
Y^K 519.1
ON REVERSE SIGNED DOMINATION NUMBERS OF GRAPHS
1. Introduction
We use Bondy and Murty [1] for terminology and notation not defined here and consider simple graphs only.
Let G=(V,E) be a graph, ifv e V(G), then NG
(v) denotes the open neighbourhood of v in G, and NG [v] = Ng (v) U{v}for the close neighbourhood, dG (v) = |Ng (v)| is the degree of v in G. For simplicity, sometimes, NG (v), NG [v] and dG (v) are denoted by N(v), N[v]and d(v), respectively. IfS c V(G), then G[S] denotes the subgraph of G induced by S. S(G) and A(G) denote the minimum and maximum degree
of G, respectively. G will denote the complement of G.
In recent years, some kinds of domination in graphs have been investigated [2~5]. T.W. Haynes, etc.[3] survey the major research accomplishments on domination theory.
Definition 1.1.[5] Let G = (V, E) be a graph, a signed dominating function (SDF) of G is a function
f: V(G) ^{-1,+1} satisfying £ f (v) > 1for all verveN [u ]
tices u e V(G), and the signed domination number of
G is defined as ys(G)= min{ £ f(v) If is an SDF
veV (G)
of G}.
In this paper, we initiate the study of a new graph parameter.
Definition 1.2. Let G = (V, E) be a graph, a function f : V ^ {-1,+1} is called a reverse signed dominating function (RSDF) of G if £ f (v) < 0
veN [u ]
holds for every vertex u e V . The reverse signed domination number of G is defined as
yrs (G ) = max { £ f (v) If is a RSDF of G}.
veV (G)
By the above definition, we have the following lemma.
Lemma 1.3. Let G be a graph of order n, then
(1) Yrs (G) = -n if and only if G = Kn ;
(2) For any two disjoint graphs GjandG2,
Yrs (G1 U G2) = Yrs (G1) + Yrs (G2);
(3) Yrs (G) - n(mod 2) .
Let G be a graph, if f be a RSDF of a graph G, and S c V(G), for convenience, we write
f (S) = £ f (v).
veS
In [5] we determined the smallest signed domination number for a graph G of order n.
УПРАВЛЕНИЕ В ТЕХНИЧЕСКИХ СИСТЕМАХ. МОДЕЛИРОВАНИЕ
©
Lemma 1.4. [5] For any graph G of ordern > 1,
ZrAG) < n - 2
then
-1 + v 1 + 4n
(2.1)
Ys (G) > 2
-1 w 1 + 8n
- n .
And this bound is sharp.
Case 2.5(G) = 0 . Since when G = Kn , by Lemma 1.4(1), we have Yrs (G) = -n, it implies that (2.1) holds for all integers n > 1.
Thus, we may suppose that G = K U H,
In this paper we obtain some upper bounds of
Yrs (G) for general graphs, and determine the exact where 5(H) > 1and \V(H^ = n -1 > 2We know
that
. By Lemma
values of yrs (G )for some special classes of graphs G, such as the complete graphs Kn, paths Pn, cycles Cn and complete bipartite graphs Kmn etc.. In addition, we pose some open problems and conjectures.
2. Some bounds of reverse signed domination number
In this section, we give some upper bounds of Yrs (G) for general graphs G .
Theorem 2.1. For any graph G of order n > 1,
then
-1W1 + 4n
Yrs (G) < n - 2
2
And this bound is sharp.
Proof: Let f be such a RSDF of G that f (V (G)) = Yrs (G). Define
A = {veV(G)|f (v) = 1}, B = {v e V(G)| f (v) = -1}, A = s and |B| = t. Obviously, s +1 = n and yrs (G) = s -1 = n - 2t. Case 1.5(G) > 1. For each vertex u e A, since f (N[u]) < 0, then u is joined to at least one vertex of B, which implies |{uv e E(G)|u e A,v e B}| > s . Thus, there exists a vertex v e B so that
N(v) n A\ >s
least
by Definition 1.2 Dv is joined to at
n -1
-1 vertices of B, t = |5| >
s n -1
t t
t
that is, t2 +1 - n > 0, it implies t >
-1+ V1 + 4n
note that t is an integer, so t > that Yrs (G) = n - 2t, we have
-1 + V1 + 4n
note
from
Yrs(H) < (n - q) - 2 1.3, we < -q + (n - q) - 2
Case 1
-1 + д/1 + 4(n - q)
have Yrs(G ) = Y„ ( Kq ) + Y„(H ) < -1 + yl 1 + 4(n - q)
2
. It implies that
(2.1) holds.
Combining Case 1 and 2, we have shown that (2.1) holds for all graphs G of n > 1.
Next we construct a graph G such that the following equality holds:
-1+ V1 + 4n
Yrs(G) = n - 2
When n = 1,
let G = K
note
(2.2 ) that
Yrs (Kj) = -1, clearly, (2.2) holds.
Next we may suppose that n > 2. -1 + V1 + 4n
Let t =
2
and s = n -1. Note that
n > 2, and hence, t > 1 .Since t2 +1 - n > 0, that is, t2 > n -1 = s, thus, we construct a graph G as follows: Let G be the graph obtained from Kt by adding exactly s pendant-edges such that each vertex of Kt adds at most t pendant-edges. Clearly,
|V(G)| = s +1 = n .We define a function
f: V(G) ^{-1,+1} as follows:
[-1 when v eV (Kt);
[+1 when v gV (Kt).
It is easy to see that f is a RSDF of G. Thus, Yrs(G) > f ((V(G)) = n - 2t, combining with (2.1) , we
have Yrs(G) = n - 2t. This proof is complete.
#
Theorem 2.2. Let G be a graph with |V (G)| = n and E (G )| = m, then
f (v) = ■
2
ИРКУТСКИЙ ГОСУДАРСТВЕННЫЙ УНИВЕРСИТЕТ ПУТЕЙ СООБЩЕНИЯ
Yrs (G) <
2m - n 2
\E (G[B])| > Combining
m > |E(A, + |E(G[B])| > 3n - 2m
2 2 with (2.3), 3n - 4t
2
we that
t >■
4
■. Thus, we have
Yra (G) = n - 2t <
2m - n 2
m
= E (G)| =
3r - r
+ r =-
2
f as follows:
f (v) =
-1 when v gK (Kr);
[+1 when v iV (Kr); It is easy to see that f is a RSDF ofGr. Thus,
yJGr) > f ((V(Gr)) = r2 - r = 2m - n
Combining with
2
(2.4),
2m — n
Yrs (Gr) =-. This proof is complete.
we have.
#
Corollary 2.3. For any tree T of order n, then
And this bound is the best possible.
Proof. Let f be such a RSDF of G that
f (V(G)) = yrs (G). Define A = {v e V(G)| f (v) = 1} ,
B = {v eV(G)| f (v) = -1} , |A| = s and |b| = t ,obviously, s +1 = n and Yrs (G) = n - 2t .And let E(A, B) = {uv e E(G) |u e A,v e B}.
As each vertex in A must be joined to at least one vertex in B, it implies
|E(A,B)| > s = n -1. (2.3)
For each vertex v e B, |N(v) n B| +1 > |N(v) H A|, so we have
s < |E(A,B)| = £|N(v) n A| <
veB
<£ (| N (v) n B| +1) = 2| E (G[ B])| +1,
veB
thus we have
s -1 n - 2t
n-2
Yrs (T) < ——, and this bound is the best possible.
Proof. By Theorem 2.2, we have yrs(T) <
n - 2 2
Next we construct a class of trees T so that this equality holds. Let H = K1t be a star of order t +1. Let T be
the tree obtained from H by adding exactly dH (v) +1 pendant-edges at each vertex v of H. Thus we have n = |V(T)| = (t +1) + £ (dH (v) +1) =
veV (H )
= t +1 + 21E (H )| +1V (H )| = 4t + 2 Define a function f as follows:
f (v) =
-1 when v eV (H); + 1 when v i¥ (H);
It is easy to check that f is a RSDF of T. Thus,
n-2
Yrs(T) > f ((V(T)) = n - 2(t +1) = . This proof is
complete.
#
have is,
(2.4)
Next we construct a class of graphs G so that the equality in Theorem 2.2 holds.
Let Gr be the graph obtained from K r by adding exactly r pendant-edges at each vertex of Kr (r = 1,2,3,......) .Obviously, n = |V(Gr )| = r2 + r ,
.We define a function
Theorem 2.4. For any graph G of order n (n > 2), then
y (G) <-n .
/rA ' A + S + 2
Where A and 5 denote the maximum and minimum degree of G, respectively.
Proof. Let f be such a RSDF of G that f (V (G)) = Yrs(G). Define
A = {v e V(G)|f (v) = 1} , B = {v e V(G)| f (v) = -1} ,
|A| = s and B| = t ,obviously, s +1 = n and Yrs(G) = s -1 .So, we have
s = nirJV and t = nzln^ (2.5) 22 We know from Definition 1.2 that f (N[u]) < 0
holds for every vertex u e V(G), thus
£ f (N[u]) < 0, that is, £ (d(u) +1) f (u) < 0 .Note
ueV (G) ueV(G)
that £ f (u) = yrs (G), thus
ueV (G )
£ d (u )f (u) + £ d (u )f(u) <-YrS (G), it implies
ueA ueB
5\A -A|B\ <-yrs(G) .By (2.5) we have 5(n + yrs(G))-A(n-yrs(G)) <-2yrs(G), and hence A-5
Y„ (G) <-n .We have completed the proof of
" A+5+2 f f
Theorem 2.4. #
The following statements are immediate from Theorem 2.4.
УПРАВЛЕНИЕ В ТЕХНИЧЕСКИХ СИСТЕМАХ. МОДЕЛИРОВАНИЕ
©
Corollary 2.5. For any regular graph G, then
yrs (G) < 0.
A graph G is said to be an Euler graph if dG (v) = 0(mod2) for each vertex v e V(G). Perhaps, one
has found the equality Yrs (Kn) + ys (Kn) = 0.In fact, we have the following
Theorem 2.6. For any graph G, then Yrs (G) + Ys (G) > 0, and the equality holds for all Euler graphs G.
Proof. Let f be an SDF of G so that f (V(G)) = Ys (G). Define g = - f, since f (N[u]) > 1 holds for each vertexu e V(G), then g(N[u]) < -1 < 0, and hence, g is a RSDF of G, yrs (G) > g(V(G)) = - f (V(G)) = -ys (G) .That is
yrs (G) + ys (G) > 0.
When G is an Euler graph, since f (N[u]) ^ 0 for each vertex u e V(G), thus, f is an SDF of G if and only if g = - f is a RSDF of G. We see from Definition 1. 1—1.2 that f (V(G)) = ys (G)if and only if g(V(G)) = ys (G), which implies that Ys (G) + ys (G) = 0 . We have completed the proof of Theorem 2.6.
#
3. The Special graphs
In this section, we give the exact values of Yrs (G) for some special graphs, such as the complete
graphs Kn, paths Pn, cycles Cn and complete bipartite graphs Kmn etc.. It is easy to check that
(-1)" -1
Y s ( K )=-
2
■ holds for all integers n > 1.
yrs ( P ) = 2
Theorem 3.1. If n > 2 is an integer, then
n 3
Proof.
- n
Let V (Pn) = {v„ v2,......, v„} and
E(Pn) = {v,v,.+J1 < i < n -1}. We define a function f: V(Pn) ^ {-1,+1} as follows:
1 when i = 1(mod3) [-1 Otherwise; It is easy to check that f is a RSDF ofPn. So, we have
f (V ) =
yrs (Pn ) > f (V (Pn ) =
- (n -
) = 2
- n
On the other hand, let /be such a RSDF of G that f (V ( Pn )) = rrs ( Pn). Define
M = {u e V(Pn )| f (u ) = +1} , W = {u e V(Pn )| f (u ) = -l} .
We prove that |m| <
. Assume, to the con-
trary, that |m| >
+1, then, there exist two vertices
u, v e M such thatl < dist(u, v) < 2 .Thus, there exists vertex w e V(Pn) such that {u, v} tz N[w], it implies f (N[w]) > 1 .This contradicts that f be such a RSDF of Pn .Thus, we have
Ys, (Pn) = f (V(Pn)) = M -|Щ = 2|M| -n < 2
- n .
Combining with (3.1), we have completed the proof of Theorem 3.1. #
Similar to the proof of Theorem 3.1, we can obtain easily the following corollary.
Corollary 3.2. If n > 3, then
" n - 2"
ys (C ) = 2
- n .
Theorem 3.3. If m > 3 and n > 3 are two in-
(-1)m+1 + (-1)"+1 - 6
teg^ then Yrs (Kmn) =
2
Proof. let f be such a RSDF of G that
f (V(Kmn)) = yrs(Kmn). Write V(Kmn) = V u V2, where IVI = m and IV21 = n . We define
V,+ = {u e V |f (u) = +1} , V,- = {u e V, \f (u) = -l} ,(/ = 1,2) Case 1. If Vj+ ^ (/) and V2+ For any vertex u e Vj+, by Definition 1.2, we have f (W[u]) < 0 ,and hence, |v2+ I -1V2-1 < -1, note that |v2+ I + |v2- I = n , thus we
have IV2+1 -1V2-1 ^ n(mod 2)
(-1)n+1 - 3
which
imply
V2+ - V2- <
So, we have
2
. Analogously,
(-1)m+1 - 3
V+ - V- <
2
Yrs(Kmn) = HI-\К-\+ V2+ - V2- <
(-1)m+1 + (-1)"+1 - 6
Case 2. If V1+ ^ <p and V2+ = (/); For any u e V2-, then f (N[u]) < 0, \v+\- \v;\< 1, note that V1+I + |Vr| = m , and thus Vl-\V1\ = m(mod2), which
(3.1)
imply
V+\-K\<
1+(-1)" 2
Note
that
. . . . (-1)n+1 - 3
V2+ - V2- = 0-n <- (n > 3), so, we have
2
1 + / 1)m+1
Y s Km, я ) = И+ h И1- V- < —--П <
(-1)m+1 + (-1)"+1 - 6
Case 3. If V1+ = p ; and V2+ ^ ф ; it is similar to
Case 2.
Case 4. If V1+ = p ; and V2+ = ф ; we have
Y„ ( Km,„) = -VT - V; =-m - n <
(-1)m+1 + (-1)n+1 - 6
Sum up the above Case 1~4, we have
Yrs (Km,n ) <
(-1)m+1 + ( -1)n+1 - 6
(3.2)
On the other hand, we may partition easily
V = A и A2 and V2 = B1 UB2 so that 2 > Щ-|A2| > 1
3 - (-1)m+1
and 2 > B-| B2\> 1. That is, -| A2\ =—— and
3 _ (-1)n+1
B-|B^ = ) . Define f : V(Km,J ^{-1, +1) as
follows:
f (v) =
-1 when v e A1 U B1;
Although some upper bounds of yrs (G) are obtained in this paper, it seems more difficult to give a good lower bound for the reverse signed domination number of G, we have the following
Conjecture 4.3. For any graph G without iso-
3
late vertex, if |V(G)| = n, then yrs (G) > - 5 n .
If true, the lower bound is the best possible. For example, Let G be a graph whose every component is C5 .By Lemma 1.3 and Corollary 3.2, we have
Y (G) = -5\V(G
Conjecture 4.4. For any tree T of ordern > 2,
then Yrs (T) >-
If true, the lower bound is the best possible for all positive integers n - 0(mod 3) . For example, let
1+1 when v e A2 UB2;
It is easy to see that f is a RSDF ofKm n, thus,
Yrs (Kmn ) > f (V (Kmn ) = | A^ +| B,\-| Aj -| B\ =
= (-1)m+1 + (-1)n+1 - 6
= 2 '
Combining with (3.2), we have completed the proof of Theorem 3.3. #
4. Some open problems and conjectures
In this section, we pose some open problems and conjectures relating to our main results. By Theorem 2.6, it is natural to pose the following problem.
Problem 4.1. Characterize all graphs with Yrs (G) + Ys (G) = 0.
It is interesting to consider the relations between Yrs (G) and Yrs (G). We pose the following
Conjecture 4.2. For any graph G of order n, then Ys (G) + Yrs (G) >-(n +1).
If true, the lower bound is the best possible. For example, Let G = Kn .Obviously, the equality holds when n is odd.
n
s = 3 and T be the tree obtained from Ps by adding
exactly one pendant-edge at each vertex of Ps. ( V(Tj)| = 2s). Let T be the tree obtained from Tjby adding exactly one pendant-edge at each vertex of degree one in T . Clearly, |V(T) = 3s = n. It is easy
to see that Yrs (T) = -s.
REFERENCES
1. Bondy J. A., Murty V.S.R. Graph Theory with Applications. Amsterdam : Elsevier, 1976.
2. Cockayne E. J., Mynhart C. M. On a generalization of signed domination functions of graphs // Ars. Combin. V. 43 (1996). P. 235-245.
3. Haynes T. W., Hedetniemi S. T., Slater P. J. Fundamentals of domination in graphs. New York : Marcel Dekker, 1998.
4. Xu B. On minus domination and signed domination in graphs // J. Math. Res. Exposition. V. 4(2003). P. 586-590.
A note on the lower bounds of signed domination number of a graph / Zhang Z., Xu B., Li Y., Liu L. // Discrete Math. V. 195(1999). P. 295-298.
2
2
2
