On Nonnegative Signed Domination in Graphs ... - Semantic Scholar
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On Nonnegative Signed Domination in Graphs and its Algorithmic Complexity *Zhongsheng Huang Dept. of Mathematics and Information Science, Langfang Teachers College, 065000 Langfang, China *Corresponding Author Email: [email protected]
Wensheng Li, Zhifang Feng and Huaming Xing Dept. of Mathematics and Information Science, Langfang Teachers College, 065000 Langfang, China Email: [email protected], [email protected], [email protected] Abstract—Let G = (V, E) be a simple graph with vertex set V and edge set E. A function f from V to a set {-1, 1} is said to be a nonnegative signed dominating function (NNSDF) if the sum of its function values over any closed neighborhood is at least zero. The weight of f is the sum of function values of vertices in V. The nonnegative signed domination number for a graph G equals the minimum weight of a nonnegative signed dominating function of G. In this paper, exact values are found for cycles, stars, wheels, spiders and complete equally bipartite graphs and we present some lower bounds for nonnegative signed domination number in terms of the order and the maximum and minimum degree. Fothermore, we show that the decision problem corresponding to the problem of computing the nonnegative signed domination number is NP-complete. Index Terms—Nonnegative signed domination number, Lower bounds, Algorithmic complexity, NP-complete
I. INTRODUCTION Let G = (V , E ) be a simple graph with vertex set V (G ) and edge set E (G ) . The open neighborhood of v , denoted by N (v) , is a set {u|uv ∈ E} and the closed neighborhood of v , denoted by N [v] , is a set N (v ) {v} . We write d (v) for the degree of a vertex v and the maximum and minimum degree of G are denoted by Δ and δ respectively. If A, B ⊆ V (G ) , A B = ∅ , let E[ A, B] = {uv | uv ∈ E (G ), u ∈ A, v ∈ B} . In this paper we shall use the terminology of [1]. In many application areas, e.g., computational biology [2, 3, 13], social network analysis [4], and web mining [5], computational linguistics [6], this data is naturally structural and can often be interpreted as relational objects, i.e., as graphs. In applied mathematics, there is an important concept on graphs: signed dominating. A function f : V → {−1,1} is a signed dominating function if for every vertex v ∈ V , f ( N [v]) = u∈N [ v ] f (u ) ≥ 1 .
For S ⊂ V , let f ( S ) = v∈S f (v) . The weight of f ,
w( f ) , is defined as f (V ) .The signed domination
number γ s (G ) of G is the minimum weight of a signed
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dominating function on G . In [7], Dunbar introduced this concept and it has been studied by several authors [8, 9, 16-18]. In the following we give the definition of the nonnegative signed domination number of a graph. Definition 1. Let G = (V , E ) be a graph. A nonnegative signed dominating function of G (shortly NNSDF) is a function f : V → {−1,1} such that f ( N [v]) ≥ 0 for all v ∈ V ． The nonnegative signed domination number of G , denoted by γ sNN (G ) , is the minimum weight of a nonnegative signed dominating function of G . A NNSDF f is called a γ sNN − function
if w( f ) = γ sNN (G ) . We say f is a minimal nonnegative signed dominating function if there does not exist a NNSDF g , g ≠ f , for which g (v) ≤ f (v) for every v ∈V . We introduce the following notation which we shall frequently use in the proofs that follow. For a given NNSDF f on a graph G , we will let V + and V − be the set of vertices in G that are assigned the value +1 and −1 , respectively, under f . For example, the Hajós graph H , shown in Fig. 1, has a nonnegative signed dominating function f of weight 0 as illustrated in Fig. 1, and thus γ sNN ( H ) ≥ 0 = w( f ) . In fact, γ sNN ( H ) = 0 .
Figure 1. The Hajós graph H with γ sNN ( H ) = 0 .
II. PROPERTIES OF NONNEGATIVE SIGNED DOMINATION First we state two lemmas.
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Lemma 1. Let f be a NNSDF on G and let S ⊂ V . Then f ( S ) ≡ | S | (mod 2) .
Proof. Let S + = {v | f (v) = 1, v ∈ S} and S − = {v | f (v) =
Proof. We mark the center of Wn +1 by v0 and mark the vertices of Cn by v1 , v2 ,..., vn in turn. Let f be a NNSDF on Wn +1 . Then w( f ) = f ( N [v0 ]) ≥ 0 , which implies that
−1, v ∈ S } . Then | S + | + | S − |=| S | and | S + | − | S − |=
γ sNN ( K1, n ) ≥ 0 . By Lemma 2, γ sNN ( K1, n ) ≥ 1 when n is
| f ( S ) . Therefore f ( S )+ | S |= 2 | S + | . Then the assertion follows. □ Lemma 2. Always γ sNN (G ) ≡ n(mod 2) . Theorem 1. For any path Pn , we have
even. Now consider the mapping g : V → {−1,1} such that g (vi ) = 1 when i is even and g (vi ) = −1 when i is odd. Then g is a NNSDF of G and w( g ) = 0 when n is odd and w( g ) = 1 when n is even. In conclusion,
γ sNN ( Pn ) = n − 2 n 3 .
γ sNN (Wn +1 ) = 0 when n is odd and γ sNN (Wn +1 ) = 1 when
Proof. Let f be a NNSDF on G . Then there aren’t two
n is even.□ Lemma 3. Let v be a supporting vertex of degree 2. Let u be the leaf supported by v and w be the other vertex adjacent to v . Let f be a γ sNN − function of G , Then f ( N [v]) = 1 .
−
vertices of V adjacent. If not so, we assume that u, v ∈ V − and u is adjacent to v . Then u is adjacent to at most one vertex of V + . Therefore f ( N [u ]) ≤ −1 , which contradicts to the assumption that f is a NNSDF. Then we have | V − |≤ n 3 . So w( f ) = n − 2 | V − | ≥ n − 2 n 3 , which implies γ sNN ( Pn ) ≥ n − 2 n 3 .
On the other hand, we mark the vertices of V by v0 , v1 ,..., vn −1 in turn. Considering the mapping g : V → {−1,1} such that g (vi ) = −1 when i ≡ 1(mod 3)
and otherwise g (vi ) = 1 . Then g is a NNSDF of G and
γ sNN ( Pn ) ≤ w( g ) = n − 2 n 3 . Therefore γ sNN ( Pn ) = n − 2 n 3 .□
Proof. Let f be a γ sNN − function of G . Then f ( N [v]) ≥ 0 . By Lemma 1, we have f ( N [v]) ≥ 1 . Therefore f ( N [v]) = 1 or f ( N [v]) = 3 . If f ( N [v ]) = 3 , consider the mapping g : V → {−1,1} by g (u ) = −1 and g ( x) = f ( x) for every x ∈ V \ {u} . Then g is a NNSDF and w( g ) = w( f ) − 2 < γ sNN (G ) , a contradiction. Then the assertion follows. □ A subdivision of the edge uv is obtained by removing edge uv , adding a new vertex w , and then adding edges uw and vw . A spider, S n , of order 2n + 1 , is the graph formed by subdividing n edges of a star K1, n . The vertex of degree n in S n is called the central vertex. Theorem 4. Let S n be a spider with n(n ≥ 3) . Then
γ sNN ( S n ) = 1 . Figure 2. The path P4 with γ sNN ( P4 ) = 0
Theorem 2. For any star K1, n , we have γ sNN ( K1, n ) = 0
Proof. We mark the central vertex of S n by v0 and mark the other vertices of S n by v1 , v2 , , vn , vn +1 , , v2 n such that vi is only adjacent to v0 and vn + i for 1 ≤ i ≤ n . Let
when n is odd and γ sNN ( K1, n ) = 1 when n is even.
f be a γ sNN − function on S n . Then γ sNN ( Sn ) = w( f ) =
Proof. We mark the vertices of K1,n by v0 , v1 , v2 ,..., vn and
f ( N [vn +1 ]) + + f ( N [v2 n ]) + f (v0 ) ≥ f (v0 ) .
v0 is the center of K1,n . Let f be a γ sNN − function
of K1, n . Then γ sNN ( K1, n ) = w( f ) = f ( N [v0 ]) ≥ 0 and by Lemma 2, γ sNN ( K1, n ) ≥ 1 when n is even. Consider the mapping g : V → {−1,1} such that g (vi ) = 1 when i is even and g (vi ) = −1 when i is odd. Then g[vi ] ≥ 0 for i = 0 , g[vi ] = 2 ≥ 0 when i ≥ 2 is even and g[vi ] = 0 when i ≥ 1 is odd. Therefore g is a NNSDF of G and w( g ) = 0 when n is odd and w( g ) = 1 when n is even. In conclusion, γ sNN ( K1, n ) = 0 when n is odd and
γ sNN ( K1, n ) = 1 when n is even.□ Theorem 3. Let Wn +1 = K1 ∨ Cn be a wheel. Then
γ
(Wn +1 ) = 0 when n is odd and γ sNN (Wn +1 ) = 1 when n is even. NN s
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If f (v0 ) = −1 , by Lemma 3, since f ( N [vi ]) = 1 for 1 ≤ i ≤ n , we have f (vi ) = f (vn + i ) = 1 for 1 ≤ i ≤ n . Therefore w( f ) = 2n − 1 . Consider the mapping g : V → {−1,1} by g (v0 ) = 1 , g (vi ) = 1 and g (vn + i ) = −1 for 1 ≤ i ≤ n . Then g is a NNSDF and w( g ) = 1 < 2n − 1 , a contradiction. Therefore f (v0 ) = 1 , which implies
γ sNN ( K1, n ) ≥ 1 . By the mapping g above, we have
γ sNN ( S n ) = 1 .□ A wounded spider S n , m of order n + m + 1 is the graph formed by subdividing m(1 ≤ m ≤ n − 1) edges of a star K1, n (n ≥ 2) . The vertex of degree n in Sn , m is called the central vertex. Theorem 5. For any wounded spider Sn , m with n(n ≥ 4) , we have
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γ
NN s
−m ( Sn , m ) = 1 − m m + 1 − n
m ≤ n 2，n is odd m ≤ n 2，n is even . m>n 2
Proof. We mark the central vertex of S n , m by v0 and mark the other vertices of S n , m by v1 , , vn , vn +1 , , vn + m such that, for 1 ≤ i ≤ m , vi is only adjacent to v0 and vn + i .
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Then g is a NNSDF on K n and w( g ) = 0 when n is even and w( g ) = 1 when n is odd. In conclusion,
γ sNN ( K n ) = 0 when n is even and γ sNN ( K n ) = 1 when n is odd.□ Theorem 7. Let K n, n be a complete equally bipartite graph with n(n ≥ 6) . Then γ sNN ( K n , n ) = 2 when n is odd
Let f be a γ sNN − function on Sn , m . If f (v0 ) = −1 , by
and γ sNN ( K n , n ) = 4 when n is even.
Lemma 3, since f ( N [vi ]) = 1 for 1 ≤ i ≤ m , we have f (vi ) = f (vn + i ) = 1 for 1 ≤ i ≤ m . Furthermore, by Definition 1, we have f ( N [vi ]) ≥ 0 for m + 1 ≤ i ≤ n , so f (vi ) = 1 for m + 1 ≤ i ≤ n . Therefore
We mark the vertices of V (G1 ) by v1 , v2 ,..., vn and mark the vertices of V (G2 ) by u1 , u2 ,..., un . Let f be a
γ sNN ( Sn , m ) = n + m − 1 .
VG+1 = {v | f (v) = 1, v ∈ V (G1 )} ,
Consider the mapping g : V → {−1,1} by g (vi ) = 1 for 0 ≤ i ≤ n − 1 and g (vi ) = −1 for n ≤ i ≤ n + m . Then g is a NNSDF and w( g ) = n − m − 1 < n + m − 1 , a contradiction. Therefore f (v0 ) = 1 . In case when 1 ≤ m ≤ n 2 , we have γ sNN ( S n ) = w( f ) = f ( N [v0 ]) + f ( N [vn +1 ]) + f ( N [vn + 2 ]) + + f ( N [vn + m ]) − f (v1 ) − − f (vm ) . If n is even, by Lemma 1, we have f ( N [v0 ]) ≥ 1 . Therefore γ sNN ( Sn ) ≥ 1 − m . Consider the mapping g : V → {−1,1} by g (vi ) = 1 for 0 ≤ i ≤ n 2 and g (vi ) = −1 for n 2 + 1 ≤ i ≤ n + m . Then w( g ) = 1 − m , which implies γ sNN ( K1, n ) = 1 − m . If n is odd, we have
γ sNN ( Sn ) ≥ − m . Consider the mapping g : V → {−1,1} by g (vi ) = 1
for 0 ≤ i ≤ (n − 1) 2 and g (vi ) = −1 for (n + 1) 2 ≤ i ≤ n + m . Then g is a NNSDF and w( g ) = −m , which implies γ sNN ( K1, n ) = −m .
In case when m > n 2 , we have γ sNN ( S n ) = w( f ) = f (v0 ) + f ( N [vn +1 ]) + + f ( N [vn + m ]) + f (vm +1 ) + + f (vn ) ≥ 1 − (n − m) . Consider the mapping g : V → {−1,1} by g (vi ) = 1 for 0 ≤ i ≤ m and g (vi ) = −1 for m + 1 ≤ i ≤ n + m . Then g is a NNSDF and w( g ) = m + 1 − n , which implies that
γ sNN ( K1, n ) = m + 1 − n . In conclusion, the assertion follows. Theorem 6. Let K n be a complete graph. Then
γ
( K n ) = 0 when n is even and γ ( K n ) = 1 when n is odd. Proof. We mark the vertices of K n by v1 , v2 ,..., vn . Let NN s
NN s
f be a γ sNN − function on K n . Then w( f ) = f ( N [v1 ]) ≥ 0 , which implies that γ sNN ( K n ) ≥ 0 . By Lemma 2,
γ sNN ( K n ) ≥ 1 when n is odd. Now consider the mapping g : V → {−1,1} such that g (vi ) = 1 when i is odd and g (vi ) = −1 when i is even.
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Proof. Let K n , n = G1 ∨ G2 . Then | V (G1 ) |=| V (G2 ) |= n .
γ sNN − function on K n, n . Let VG−1 = {v | f (v) = −1, v ∈ V (G1 )} , VG+2 = {v | f (v) = 1, v ∈ V (G2 )} , VG−2 = {v | f (v) = −1, v ∈ V (G2 )} .
If any of VG−1 and VG−2 is empty, we assume that VG−1 = ∅ without loss of generality, i.e. V (G1 ) = VG+1 .
Then f ( N [v1 ]) = f (v1 ) + f (V (G2 )) ≥ 0 . Therefore f (V (G2 )) ≥ − f (v1 ) = −1 . Then we have
γ sNN ( K n, n ) = f (V (G1 )) + f (V (G2 )) ≥ n − 1 . Consider the mapping g : V → {−1,1} such that g (vi ) = g (ui ) = −1 for 1 ≤ i ≤ n 2 − 1 and g (vi ) = g (ui ) = 1 for n 2 ≤ i ≤ n . Then g is a NNSDF on K n, n and w( g ) = 2 ( n − n 2 + 1) − ( n 2 − 1)
= 4 + 2(n − 2 n 2 ) ≤ 4 < n − 1 . Therefore non of VG−1 and VG−2 is empty . Without loss of
generality, we assume that v1 ∈ VG−1 . Then f (V (G2 )) ≥ − f (v1 ) = 1 . In the similar way, we have f (V (G1 )) ≥ 1 .
Therefore γ sNN ( K n , n ) ≥ 2 . If n is odd, we consider the mapping g : V → {−1,1} such that g (vi ) = g (ui ) = −1 for 1 ≤ i ≤ (n − 1) 2 and g (vi ) = g (ui ) = 1 for (n + 1) 2 ≤ i ≤ n . Then g is a NNSDF on K n, n and w( g ) = 2 , which implies that
γ sNN ( K n , n ) = 2 . If n is even, by Lemma 1, we have f (V (G2 )) ≥ 2 and f (V (G1 )) ≥ 2 . Therefore γ sNN ( K n , n ) ≥ 4 . Consider the
mapping g : V → {−1,1} such that g (vi ) = g (ui ) = −1 for 1 ≤ i ≤ n 2 − 1 and g (vi ) = g (ui ) = 1 for n 2 ≤ i ≤ n . Then g is a NNSDF on K n, n and w( g ) = 4 , which implies that γ sNN ( K n , n ) = 4 . From the above, the assertion follows.
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Theorem 8. A NNSDF f on a graph G is minimal if and only if for every vertex v ∈ V with f (v) = 1 , there exists a vertex u ∈ N [v ] with f ( N [u ]) < 2 . Proof. Let f be a minimal NNSDF and assume that there is a vertex v with f (v) = 1 and f ( N [u ]) ≥ 2 for every u ∈ N [v ] . Define a new function g : V → {−1,1} by g (v) = −1 < f (v) = 1 and g (u ) = f (u ) for all u ≠ v . Then for all u ∈ N [v ] , g (u ) = f (u ) − 2 ≥ 0 . For all w ∉ N [v] , g ( N [ w]) = f ( N [ w]) ≥ 0 . Thus g is a NNSDF. Since g < f , the minimality of f is contradicted. Conversely, let f be a NNSDF such that for all v ∈ V with f (v) = 1 , there exists a vertex u ∈ N [v] with f ( N [u ]) < 2 . Assume f is not minimal, i.e. there is a NNSDF g such that g < f . Then g ( w) ≤ f ( w) for all w ∈ V , and there exists at least one vertex v ∈ V with g (v) < f (v) . Therefore f (v) = 1 , and by assumption, there exists a vertex u ∈ N [v] with f ( N [u ]) < 2 . Since g ( w) ≤ f ( w) for all w ∈ V and g (v) < f (v) , we know that g ( N [u ]) = f ( N [u ]) − 2 < 0 . This contradicts the fact that g is a NNSDF. Therefore f is a minimal NNSDF.
Δ | V + | −δ | V − | +γ sNN (G ) ≥ 0 , i.e. (n + γ sNN (G ))Δ 2 − (n − γ sNN (G ))δ 2 + γ sNN (G ) ≥ 0 . Then we have δ −Δ n .□ γ sNN (G ) ≥ Δ +δ + 2 Theorem 11. Let G be a graph with n vertices and
m edges. Then γ sNN (G ) ≥ 4n + 1 − n − 1 . Proof. Let
f be a NNSDF on G . Let E (G[V − ]) = s ,
where G[V − ] is the induced graph of V − . E[V + , V − ] is defined as Theorem 9. Then | E (G[V + ]) |= m − s − | E[V + , V − ] | , where G[V + ] is the induced graph of V + . Let dV − (u ) =| {v | u ∈ V − , uv ∈ E (G[V − ])} | and dV + (u ) =| {v | u ∈ V + , uv ∈ E (G[V + ])} | .
Then
u∈V −
and
u∈V −
| N (u ) V + | =| E[V + , V − ] |= u∈V + | N (u ) V − |
dV − (u ) = u∈V − | N (u ) V − | = 2 | E (G[V − ]) |= 2s .
Since f is a NNSDF on G , for each u ∈ V − , we have
III. LOWER BOUNDS OF NONNEGATIVE SIGNED DOMINATION NUMBER Theorem 9. For any graph G with maximum degree Δ(G ) ≤ 2 , we have γ sNN (G ) ≥ 0 .
Proof. Let f be a γ sNN − function of G . If V − = ∅ ,
| N (u ) V − |≤ | N (u ) V + | −1 , and for u ∈ V + , we have | N (u ) V − |≤ | N (u ) V + | +1 . Therefore 0 ≤ 2s = u∈V − | N (u ) V − |
≤ u∈V − (| N (u ) V + | −1)
then the result follows. So assume V − ≠ ∅ . Let E[V + , V − ] = {uv | u ∈ V + , v ∈ V − } and k = | E[V + , V − ] | .
= u∈V − | N (u ) V + |− | V − | = u∈V + | N (u ) V − |− | V − |
For each v ∈ V − , since f ( N [v]) ≥ 0 , v is at least adjacent
≤ u∈V + (| N (u ) V + | +1)− | V − |
to one vertex of V + . Then we have k ≥| V − | . For
= u∈V + | N (u ) V + |+ | V + | − | V − |
each v ∈ V + , if v is adjacent to q vertices of V − , then v
= u∈V + dV + (u )+ | V + | − | V − |
+
is adjacent to at least q − 1 vertices of V . Since Δ(G ) ≤ 2 , we have k ≤ | V + | . Therefore w( f ) = | V + | − | V | ≥ 0 , which implies γ (G ) ≥ 0 .□ Theorem 10. For any graph G with maximum degree Δ and minimum degree δ , we have δ −Δ γ sNN (G ) ≥ n. Δ +δ + 2 Proof. Let f be a γ sNN − function of G . Then −
NN s
| V + | + | V − |= n and | V + | − | V − |= γ sNN (G ) . We have V + = (n + γ sNN (G )) 2 and V − = (n − γ sNN (G )) 2 . By Definition 1, for every v ∈ V we have f ( N [v]) ≥ 0 , Then v∈V (d (v) + 1) f (v) ≥ 0 . Therefore
v∈V +
d (v ) f (v) + v∈V − d (v) f (v) + γ sNN (G ) ≥ 0 .
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Since u∈V + dV + (u ) ≤| V + | (| V + | −1) , we have 0 ≤| V + | (| V + | −1)+ | V + | − | V − |=| V + |2 + | V + | −n .
Solving the inequality, we have | V + |≥ Therefore w( f ) =| V + | − | V − |= 2 | V + | −n ≥ implies that γ
NN s
1 ( 4n + 1 − 1) . 2
4n + 1 − 1 − n , which
(G ) ≥ 4n + 1 − n − 1 .□
IV. COMPLEXITY ISSUES FOR NONNEGATIVE SIGNED DOMINATION The following decision problem for the domination number of a graph is known to be NP - complete, even when restricted to bipartite graphs or chordal graphs. Problem: DOMINATION(DM)
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INSTANCE: A graph G = (V , E ) and a positive integer k ≤ | V | . QUESTION: Is γ (G ) ≤ k ? We will demonstrate a polynomial time reduction of this problem to our nonnegative signed domination problem. Problem: NONNEGATIVE SIGNED DOMINATION(NNSD) INSTANCE: A graph H = (V , E ) and a positive integer j ≤ | V | . QUESTION: Is γ sNN ( H ) ≤ j ? Theorem 12. Problem NNSD is NP -complete, even when restricted to bipartite or chordal graphs. Proof. It is obvious that NNSD is a member of NP since we can, in polynomial time, guess at a function f : V → {−1,1} and verify that f has a weight at most j and is a NNSDF. We next show how a polynomial time algorithm for NNSD could be used to solve DM in polynomial time. Given a graph G = (V , E ) and a positive integer k construct the graph H by adding to each vertex v of G a set of deg G (v) paths of length three. Let m = | E | and n = | V | . We have | V ( H ) |= n + 3 v∈V (G ) deg G (v) = n + 6m
and
| E ( H ) |= m + 3 v∈V (G ) deg G (v) = 7m ,
and H can be constructed in polynomial time. Note that if G is a bipartite or chordal graph, then so too is H . We note that, for any nonnegative signed dominating function g , for each v ∈ V (G ) , if g (v) = −1 and g (u ) = −1 for every u ∈ N (v) V (G ) , then, because | N G [v] |=| N H [v] − N G [v] | +1 = deg G (v) + 1 , we would have g ( N H [v]) ≤ −1 . That is, if g : V ( H ) → {−1,1} is a NNSDF for H , then {v ∈ V (G ) | g (v) = 1} is a dominating set for G . Lemma 4. Let f be a NNSDF on H and w( f ) = γ sNN ( H ) . Then, for each path added to G , there exactly two vertices assigned value 1. Proof. Let P(v) be the path added to v ∈ V (G ) . We mark the vertices of P(v) by v1 , v2 and v3 according to the distance to v such that d (v, v1 ) = 1 , d (v, v2 ) = 2 and d (v, v3 ) = 3 . If there is no vertex of P (v) assigned value 1, then f ( N [v3 ]) = f (v2 ) + f (v3 ) = −2 , which is contradiction. If there is exactly one vertex of P (v) assigned value 1, then f ( N [v2 ]) = f (v1 ) + f (v2 ) + f (v3 ) = −1 , which is contradiction. If all the vertices of P(v) are assigned value 1, let g : V ( H ) → {−1,1} such that g (v3 ) = −1 and g ( w) = f ( w) , w ∈ V ( H ) − {v3 } . Then g
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is a NNSDF on H and w( g ) = w( f ) − 1 = γ sNN ( H ) − 1 , which is conflict to the assumption. From the above, we can get the assertion. By Lemma 4, if we let j = (2n − n) + k − (n − k ) = 2k , then γ (G ) ≤ k if and only if γ sNN ( H ) ≤ j . This completes the proof of Theorem 12.□ Problem DM is polynomial for fixed k . To see this, let G = (V , E ) be a graph with | V |= n . If k ≥ n , then V is a dominating set of G of cardinality at most k . On the other hand, if k < n , then consider all the r -subsets of k n V , where r = 1, 2, , k . There are r =1 of these r subsets, which is bounded above by the polynomial
k r =1
p r . It takes a polynomial amount of time to verify
that a set is or is not a dominating set. These remarks show that it takes a polynomial amount of time to verify whether G has a dominating set of cardinality at most k when k is fixed. Hence for fixed k , problem DM ∈ P . In contrast, we now show that for a fixed k , the problem NNSD can be NP -complete. To see this, we will demonstrate a polynomial time reduction of the nonnegative signed domination problem to the following zero nonnegative signed domination problem. Problem: ZERO NONNEGATIVE SIGNED DOMINATION(ZRSD) INSTANCE: A graph G = (V , E ) . QUESTION: Does G have a NNSDF of weight at most 0? Theorem 13. Problem ZNNSD is NP -complete, even when restricted to bipartite or chordal graphs. Proof. It is obvious that ZNNSD is a member of NP , since we can, in polynomial time guess at a function f : V (G ) → {−1,1} and verify that f has weight at most 0 and is a NNSDF. We next show how a polynomial time algorithm for ZNNSD could be used to solve NNSD in polynomial time. Let L be a graph of Figure 3. Then L has a NNSDF of weight -1 as illustrated. In fact, γ sNN ( L) = −1 . Note that L is chordal.
Figure 3. Given a graph H = (V , E ) and a positive integer j , let G = H ij=1 Li , where Li ≅ L for i = 1, 2, , j . It is clear that G can be constructed in polynomial time. Note that if H is chordal, then so too is G . We now show that γ sNN ( H ) ≤ j if and only if
γ sNN (G ) ≤ 0 . Suppose first that γ sNN ( H ) ≤ j and that f
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is a NNSDF of H of weight γ sNN ( H ) . Let fi be any NNSDF of weight −1 for Li for i = 1, 2, , j . Define g : V (G ) → {−1,1} by g ( w) = f i ( w) if w ∈ V ( Li ) for i = 1, , j , while g (v) = f (v) for v ∈ V ( H ) . Then g is a NNSDF of G of weight γ sNN ( H ) + j (−1) ≤ j − j = 0 ,
Proof. Let f be a inverse signed total dominating function of Pn . Then for any four vertices of Pn in turn, there are at most two of the values of f on these four vertices equal to +1. Therefore γ st0 ( Pn ) ≤ 0 .□ Theorem 14. For any path Pn , then
hence γ sNN (G ) ≤ 0 . Conversely, suppose that γ sNN (G ) ≤ 0
γ st0 ( Pn ) = 0 when n ≡ 0(mod 4) ;
and that g is a NNSDF of weight γ sNN (G ) . Let f be the restriction of g on V ( H ) and let fi be the restriction of
γ st0 ( Pn ) = −1 when n ≡ 1 or 3(mod 4) ;
g on V ( Li ) for i = 1, 2, , j . Then γ sNN ( H ) + j (−1) =
+ i =1 f i (V ( Li )) = g (V (G )) = γ sNN (G ) ≤ 0 + i =1 f i (V ( Li )) j
j
= g (V (G )) = γ sNN (G ) ≤ 0 , so that γ sNN ( H ) ≤ j .
Let F be the graph of Figure 4. Then F has a NNSDF of weight -1 as illustrated. In fact, γ sNN ( F ) = −1 . Note that F is bipartite. Given a graph H = (V , E ) and a positive integer j , let G = H ij=1 Fi , where Fi ≅ F for i = 1, 2, , j . It is clear that G can be constructed in polynomial time. Note that if H is bipartite, then so too is G . Proceeding now as in the preceeding paragraph, we may show that γ sNN ( H ) ≤ j if and only if γ sNN (G ) ≤ 0 . This completes the proof of the theorem.
γ st0 ( Pn ) = −2 when n ≡ 2(mod 4) . Proof. We mark the vertices of Pn by v1 , v2 , v3 , , vn in turn. It is easy to verify the conclusion when n ≤ 4 . Case 1 n ≡ 0(mod 4)(n > 4) . By Lemma 5, γ st0 ( Pn ) ≤ 0 . Consider the function f : V → {−1,1} such that +1, i ≡ 0(mod 4) or i ≡ 1(mod 4)； f (vi ) = −1, i ≡ 2(mod 4) or i ≡ 3(mod 4). Then f is an inverse signed total dominating function
of Pn and γ st0 ( Pn ) ≥ w( f ) = 0 . Therefore γ st0 ( Pn ) = 0 when n ≡ 0(mod 4)(n > 4) . Case 2 n ≡ 1(mod 4) (n > 4) . Consider the function f : V → {−1,1} such that +1, i ≤ n − 2 and i ≡ 0 or 1(mod 4); −1, i ≤ n − 2 and i ≡ 2 or 3(mod 4); f (vi ) = i = n − 1; −1, +1, i = n. Then f is an inverse signed total dominating function
and γ st0 ( Pn ) ≥ w( f ) = −1 .On the other hand, Let S = {v3 ,
Figure 4. V.
v4 , , vn − 3 } ,then S ≡ 0(mod 4) .Let g be a γ st0 − function
PROPERTIES OF INVERSE SIGNED TOTAL DOMINATION
In this part, we do some work on the extension of nonnegative signed domination, i.e , the inverse signed total domination number of a graph. We obtained the exact values of the inverse signed total domination number of special graph classes. Definition 2. For a simple graph G with δ ≥ 1 a function f : V → {−1,1} is said to be a inverse signed total dominating function if for every v ∈ V (G ) , f ( N (v)) ≤ 0 . The inverse signed total domination number γ st0 (G ) = max { w( f ) | f is a inverse signed total dominating function of G }．We say f is a maximal inverse signed total dominating function if there does not exist a inverse signed total domination function g , g ≠ f , for which g (v) ≥ f (v) for every v ∈ V . A inverse signed total dominating function f is called a γ - function of G if 0 st
w( f ) = γ st0 (G ) ． Lemma 5. For any path Pn , if n ≡ 0(mod 4) , then
γ st0 ( Pn ) ≤ 0 . © 2013 ACADEMY PUBLISHER
of Pn . By Lemma 5, we have g ( S ) ≤ 0 .By Definition 2, g (v2 ) = g ( N (v1 )) ≤ 0 and g (vn −1 ) = g ( N (vn )) ≤ 0 .Thus g (v2 ) = −1 and g (vn −1 ) = −1 . Therefore w( g ) = g (v1 ) + g (v2 ) + g ( S ) + g ( N (vn −1 )) + g (vn −1 ) ≤ 1 − 1 +0 + 0 −1 = −1 . Then γ st0 ( Pn ) ≤ −1 .
In conclusion, γ st0 ( Pn ) = −1 when n ≡ 1(mod 4) (n > 4) . Case 3 n ≡ 2(mod 4) (n > 4) . Consider the function f : V → {−1,1} such that +1, i ≤ n − 2 and i ≡ 0 or 1(mod 4); f (vi ) = −1 i ≤ n − 2 and i ≡ 2 or 3(mod 4); −1 i > n − 2. Then f is an inverse signed total dominating function
of Pn . Thus γ st0 ( Pn ) ≥ w( f ) = −2 . On the other hand, let S = {v4 , v5 , , vn − 3 } , then S ≡ 0(mod 4) . Let g be a
γ st0 − function of Pn , then, by Lemma 5, g ( S ) ≤ 0 . By Definition 2, g (v2 ) = g ( N (v1 )) ≤ 0 and g (vn −1 ) =
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g ( N (vn )) ≤ 0 .Thus g (v2 ) = −1 and g (vn −1 ) = −1 . Since g is an inverse signed total dominating function of Pn ,
we have g ( N (v2 )) ≤ 0 and g ( N (vn −1 )) ≤ 0 i.e. g (v1 ) + g (v3 ) ≤ 0 and g (vn − 2 ) + g (vn ) ≤ 0 . Therefore, w( g ) = g (v1 ) + g (v2 ) + g (v3 ) + g ( S ) + g (vn − 2 ) + g (vn −1 ) + g (vn ) ≤ 0 − 1 + 0 + 0 − 1 = −2 . Thus γ ( Pn ) ≤ −2 . 0 st
In conclusion, γ ( Pn ) = −2 when n ≡ 2(mod 4)(n > 4) . Case 4 n ≡ 3(mod 4) (n > 4) . Consider the function f : V → {−1,1} such that 0 st
+1, i = 0 or 1(mod 4); f (vi ) = −1, i = 2 or 3(mod 4). Then f is an inverse signed total dominating function
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−1, i = 3 or 0(mod 4)； f (vi ) = +1, i = 1or 2(mod 4); −1, i = 0. Then f is a inverse signed total dominating function
of Wn +1 , and w( f ) = −1 .Thus γ st0 (Wn +1 ) ≥ w( f ) = −1 . In conclusion, γ st0 (Wn +1 ) = −1 when n ≡ 0(mod 4) . Case 2 n ≡ 1(mod 4) . Consider the function f : V → {−1,1} such that f (vi ) = +1 when i ≡ 1 or 2(mod 4) , otherwise f (vi ) = −1 . Then f is a inverse signed total dominating function of Wn +1 , and w( f ) = −2 . Thus γ st0 (Wn +1 ) ≥ w( f ) = −2 .
a γ st0 − function of Pn . By Lemma 5, g ( S ) ≤ 0 . By Definition 2, g (vn −1 ) = g ( N (vn )) ≤ 0 . Thus g (vn −1 ) = −1 . Since g is an inverse signed total dominating function of
In conclusion, γ st0 (Wn +1 ) = −2 when n ≡ 1(mod 4) . Case 3 n ≡ 2(mod 4) . Consider the function f : V → {−1,1} such that f (vi ) = −1 when i = n − 1 or n ; f (vi ) = +1 when i ≡ 1 or 2(mod 4) ; otherwise f (vi ) = −1 . Then f is a inverse signed total dominating function of Wn +1 , and w( f ) = −3 .
Pn , we have
Thus γ st0 (Wn +1 ) ≥ w( f ) = −3 . On the other hand, let g be
of Pn . Thus γ ( Pn ) ≥ w( f ) = −1 . On the other hand, 0 st
let S = {v1 , v2 , , vn − 3 } , then S ≡ 0(mod 4) . Let g be
g ( N (vn −1 )) ≤ 0 ,i.e. g (vn − 2 ) + g (vn ) ≤ 0 .
Thus w( g ) = g ( S ) + g (vn − 2 ) + g (vn −1 ) + g (vn ) ≤ 0 − 1 + 0 = −1 . Therefore γ st0 ( Pn ) ≤ −1 . In conclusion, γ st0 ( Pn ) = −1 when n ≡ 3(mod 4)(n > 4) .□ The method to prove the theorem is similar to that of theorem 14. Theorem 15. For any cycle Cn ( n ≥ 3 ), then
γ st0 (Cn ) = 0 when n ≡ 0(mod 4) ; γ st0 (Cn ) = −1 when n ≡ 1 or 3 (mod 4) ; γ (Cn ) = −2 when n ≡ 2(mod 4) . 0 st
Theorem 16. Let Wn +1 = K1 ∨ Cn be a wheel, Then
γ st0 (Wn +1 ) = −1 when n ≡ 0(mod 4) ; γ st0 (Wn +1 ) = −2 when n ≡ 1 or 3(mod 4) ; γ st0 (Wn +1 ) = −3 when n ≡ 2(mod 4) . Proof. We mark the center of Wn +1 by v0 and mark the vertices of Cn by v1 , v2 , , vn in turn. Let g be a
γ st0 − function on Wn +1 . Then γ st0 (Wn +1 ) = g (v0 ) + g (Cn ) . By Definition 2, g ( N (v0 )) = g (Cn ) ≤ 0 , and for every v ∈ V (Cn ) , g ( N (v)) ≤ 0 .Since N (v) = 3 is odd, g ( N (v)) ≤ −1 . Thus
v∈V ( Cn )
g ( N (v)) ≤ −n , i.e. 2 g (Cn ) + ng (v0 )
≤ − n .Then γ st0 (Wn +1 ) = g (Cn ) + g (v0 ) ≤ [(n − 2) g (Cn ) − n] n ≤ −1 ,
and γ (Wn +1 ) ≤ −1 when n is even and γ (Wn +1 ) ≤ −2 when n is odd. Case 1 n ≡ 0(mod 4) . Consider the function f : V → {−1,1} such that 0 st
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0 st
a γ st0 − function of Wn +1 . Then there exists at least two adjacent vertices of V (Cn ) assigned value −1 and the number of the other vertices, which construct a path Pn − 2 , of V (Cn ) is multiple of four. Thus w( g ( Pn − 2 )) ≤ 0 , and
γ st0 (Wn +1 ) = g ( Pn − 2 ) − 3 ≤ −3 . In conclusion, γ st0 (Wn +1 ) = −3 when n ≡ 2(mod 4) . Case 4 n ≡ 3(mod 4) . Consider the function f : V → {−1,1} such that −1, i = 3 or 0(mod 4); f (vi ) = +1, i = 1 or 2(mod 4); −1, i = 0. Then f is a inverse signed total dominating function
of Wn +1 , and w( f ) = −2 . Thus γ st0 (Wn +1 ) ≥ w( f ) = −2 . In conclusion, γ st0 (Wn +1 ) = −2 when n ≡ 3(mod 4) .□ ACKNOWLEDGMENT This work was supported by the Educational Commission of Hebei Province (No.ZH2011122, Z2011157), the Natural Science Foundation of Hebei Province (No. A2012408002) and the Youth Foundation of Langfang Teachers College (No. LSZQ201001). REFERENCES [1] G. Chartrand, L. Lesniak, Graphs and Digraphs, second ed. Wadsworth and Brooks Cole, Monterey, 1986. [2] A.L. Barabàsi and Z.N. Oltvai, “Network biology: understanding the cell’s functional organization,” Nature Reviews Genetics, vol.5, pp. 101–113, 2004. [3] F. Emmert-Streib, M. Dehmer, J. Kilian, “Classification of large graphs by a local tree decomposition in: H.R. Arabnia,” A. Scime (Eds.), Proceedings of the 2005
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International Conference on Data Mining (DMIN’05), pp. 200–207, 2005. A.L. Baraba ´si, H. Jeong, Z. Neda, E. Ravasz, A. Schubert, T. Vicsek, “Evolution of the social network of scientific collaborations,” Physica, vol. A 311, pp. 590–614, 2002. R. Gleim, A.Mehler,M. Dehmer, “Web corpus mining by instance of wikipedia,” in: Proceedings of the EACL 2006Workshop onWeb as Corpus, Trento/Italy , pp. 67–74, 2006. A. Mehler, “in: Search of a Bridge between Network Analysis in Computational Linguistics and Computational Biology-A Conceptual Note,” In: Proceedings of the 2006 International Conference on Bioinformatics & Computational Biology (BIO-COMP’06), Las Vegas/USA, pp. 496–500, 2006. J.E. Dunbar, S.T. Hedetniemi, M.A. Henning and P.J. Slater, “Signed Domination in Graphs,” Combinatorics, Graph Theory, Applications, vol.1, pp. 311–322, 1995. J.H. Hattingh, M.A. Henning and P.J. Slater, “On the Algorithmic Complexity of Signed Domination in Graphs,” Australas. J. Combin. , vol.12, pp.101–112, 1995. O. Favaron, “Signed Domination in Regular Graphs,” Discrete Math. , vol.158, 287–293, 1996. D.J. Cook, L.B. Holder, Mining Graph Data, Wiley Publishing, 2006. M. Dehmer and F. Emmert-Streib, “Structural Similarity of Directed Universal Hierarchical Graphs: A Low Computa-tional Complexity Approach,” Apllied Mathematics and Computation, vol.194, pp. 67–74 , 2006. A. Mehler, R. Gleim, M. Dehmer, “Towards structuresensitive hypertext categorization,”in: Proceedings of the 29th Annual Conference of the German Classification Society, Magde-burg, Germany, 2005. A.L. Baraba ´si, Z.N. Oltvai, “Network biology: understanding the cell’s functional organization,” Nature Reviews Genetics, vol.5, pp.101–113, 2004. F. Harary, Structural Models. An Introduction to the Theory of Directed Graphs, Wiley, New York, 1965. C.M. Lee, “On the complexity of signed and minus total domination in graphs,” Information Processing Letters, vol.109, pp.1177–1181, 2009. M.J. Gao and E.F. Shan, “The Signed Total Domination Number of Graphs,” ARS Combinatoria, vol.98, pp. 15–24 , 2011. H.C. Wang, J. Tong and L. Volkmann, “A Note on Signed Total 2-independence in Graphs,” Utilitas Mathematica, vol.85, pp.213–223, 2011. A.H. Berliner, R.A. Brualdi, L. Deaett, K.P. Kiernan, S.A. Meyer and M.W. Schroeder, “Signed Domination of Graphs and (0,1)-Matrices,” Combinatorics and Graphs. , vol.531, pp.19–42 , 2010. M. Atapour, S.M.S. Tabriz and A. Khodkar, “Global signed total domination in graphs,” Publicationes Mathematicae-Debrecen, vol.79, pp.7–22, 2011. A. Poghosyan, V. Zverovich “Global signed total domination in graphs,” Discrete Mathematics, vol. 310, pp. 20912099, 2010. S.M. Sheikholeslami, L. Volkmann, “Signed total(k,k)domatic number of digraphs,” Aequationes Mathematicae. , vol. 83, pp. 87–96, 2012. C.M. Lee, “Signed and minus total domination on subclasses of bipartite graphs,” Ars Combinatora, vol.100, pp. 129–149, 2011. A.K. Dewdney, “Fast Turing Reductions Between Problems in NP 4,” Report 71, University of Western Ontario., 1981.
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Zhongsheng Huang received his B.S. degree in Mathematics Education and M.S. degree in Applied Mathematics from Hebei Normal University in 1998 and 2007, respectively. He worked as an associate professor in Dept. of Mathematics and Information Science, Langfang Teachers College. His interesting research areas are graph theory and algebraic combinatorics.
Wensheng Li received the B.S. degree in Information and Computing Science from the Hebei University of Technology, Hebei, China, in 2003. He obtained the M.S. degree in Computational Mathematics from the Dalian University of Technology, Dalian, China, in 2006. He worked as a lecturer in Department of Mathematics and Infor-mation Science, Langfang Teachers College. His research interest is in the area of graph theory.
Zhifang Feng was born in Shijiazhuang, China. She has completed her B.S. and M.S. degree in Mathematics and Applied Mathematics from Hebei University in 2004 and 2007, respectively. She is now an lecturer in the Dept. of Mathematics and Information Science, Langfang Teachers College, Langfang, China. Her current research includes the theory of probability and graph theory.
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On Nonnegative Signed Domination in Graphs and its Algorithmic Complexity *Zhongsheng Huang Dept. of Mathematics and Information Science, Langfang Teachers College, 065000 Langfang, China *Corresponding Author Email: [email protected]
Wensheng Li, Zhifang Feng and Huaming Xing Dept. of Mathematics and Information Science, Langfang Teachers College, 065000 Langfang, China Email: [email protected], [email protected], [email protected] Abstract—Let G = (V, E) be a simple graph with vertex set V and edge set E. A function f from V to a set {-1, 1} is said to be a nonnegative signed dominating function (NNSDF) if the sum of its function values over any closed neighborhood is at least zero. The weight of f is the sum of function values of vertices in V. The nonnegative signed domination number for a graph G equals the minimum weight of a nonnegative signed dominating function of G. In this paper, exact values are found for cycles, stars, wheels, spiders and complete equally bipartite graphs and we present some lower bounds for nonnegative signed domination number in terms of the order and the maximum and minimum degree. Fothermore, we show that the decision problem corresponding to the problem of computing the nonnegative signed domination number is NP-complete. Index Terms—Nonnegative signed domination number, Lower bounds, Algorithmic complexity, NP-complete
I. INTRODUCTION Let G = (V , E ) be a simple graph with vertex set V (G ) and edge set E (G ) . The open neighborhood of v , denoted by N (v) , is a set {u|uv ∈ E} and the closed neighborhood of v , denoted by N [v] , is a set N (v ) {v} . We write d (v) for the degree of a vertex v and the maximum and minimum degree of G are denoted by Δ and δ respectively. If A, B ⊆ V (G ) , A B = ∅ , let E[ A, B] = {uv | uv ∈ E (G ), u ∈ A, v ∈ B} . In this paper we shall use the terminology of [1]. In many application areas, e.g., computational biology [2, 3, 13], social network analysis [4], and web mining [5], computational linguistics [6], this data is naturally structural and can often be interpreted as relational objects, i.e., as graphs. In applied mathematics, there is an important concept on graphs: signed dominating. A function f : V → {−1,1} is a signed dominating function if for every vertex v ∈ V , f ( N [v]) = u∈N [ v ] f (u ) ≥ 1 .
For S ⊂ V , let f ( S ) = v∈S f (v) . The weight of f ,
w( f ) , is defined as f (V ) .The signed domination
number γ s (G ) of G is the minimum weight of a signed
© 2013 ACADEMY PUBLISHER doi:10.4304/jnw.8.2.365-372
dominating function on G . In [7], Dunbar introduced this concept and it has been studied by several authors [8, 9, 16-18]. In the following we give the definition of the nonnegative signed domination number of a graph. Definition 1. Let G = (V , E ) be a graph. A nonnegative signed dominating function of G (shortly NNSDF) is a function f : V → {−1,1} such that f ( N [v]) ≥ 0 for all v ∈ V ． The nonnegative signed domination number of G , denoted by γ sNN (G ) , is the minimum weight of a nonnegative signed dominating function of G . A NNSDF f is called a γ sNN − function
if w( f ) = γ sNN (G ) . We say f is a minimal nonnegative signed dominating function if there does not exist a NNSDF g , g ≠ f , for which g (v) ≤ f (v) for every v ∈V . We introduce the following notation which we shall frequently use in the proofs that follow. For a given NNSDF f on a graph G , we will let V + and V − be the set of vertices in G that are assigned the value +1 and −1 , respectively, under f . For example, the Hajós graph H , shown in Fig. 1, has a nonnegative signed dominating function f of weight 0 as illustrated in Fig. 1, and thus γ sNN ( H ) ≥ 0 = w( f ) . In fact, γ sNN ( H ) = 0 .
Figure 1. The Hajós graph H with γ sNN ( H ) = 0 .
II. PROPERTIES OF NONNEGATIVE SIGNED DOMINATION First we state two lemmas.
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Lemma 1. Let f be a NNSDF on G and let S ⊂ V . Then f ( S ) ≡ | S | (mod 2) .
Proof. Let S + = {v | f (v) = 1, v ∈ S} and S − = {v | f (v) =
Proof. We mark the center of Wn +1 by v0 and mark the vertices of Cn by v1 , v2 ,..., vn in turn. Let f be a NNSDF on Wn +1 . Then w( f ) = f ( N [v0 ]) ≥ 0 , which implies that
−1, v ∈ S } . Then | S + | + | S − |=| S | and | S + | − | S − |=
γ sNN ( K1, n ) ≥ 0 . By Lemma 2, γ sNN ( K1, n ) ≥ 1 when n is
| f ( S ) . Therefore f ( S )+ | S |= 2 | S + | . Then the assertion follows. □ Lemma 2. Always γ sNN (G ) ≡ n(mod 2) . Theorem 1. For any path Pn , we have
even. Now consider the mapping g : V → {−1,1} such that g (vi ) = 1 when i is even and g (vi ) = −1 when i is odd. Then g is a NNSDF of G and w( g ) = 0 when n is odd and w( g ) = 1 when n is even. In conclusion,
γ sNN ( Pn ) = n − 2 n 3 .
γ sNN (Wn +1 ) = 0 when n is odd and γ sNN (Wn +1 ) = 1 when
Proof. Let f be a NNSDF on G . Then there aren’t two
n is even.□ Lemma 3. Let v be a supporting vertex of degree 2. Let u be the leaf supported by v and w be the other vertex adjacent to v . Let f be a γ sNN − function of G , Then f ( N [v]) = 1 .
−
vertices of V adjacent. If not so, we assume that u, v ∈ V − and u is adjacent to v . Then u is adjacent to at most one vertex of V + . Therefore f ( N [u ]) ≤ −1 , which contradicts to the assumption that f is a NNSDF. Then we have | V − |≤ n 3 . So w( f ) = n − 2 | V − | ≥ n − 2 n 3 , which implies γ sNN ( Pn ) ≥ n − 2 n 3 .
On the other hand, we mark the vertices of V by v0 , v1 ,..., vn −1 in turn. Considering the mapping g : V → {−1,1} such that g (vi ) = −1 when i ≡ 1(mod 3)
and otherwise g (vi ) = 1 . Then g is a NNSDF of G and
γ sNN ( Pn ) ≤ w( g ) = n − 2 n 3 . Therefore γ sNN ( Pn ) = n − 2 n 3 .□
Proof. Let f be a γ sNN − function of G . Then f ( N [v]) ≥ 0 . By Lemma 1, we have f ( N [v]) ≥ 1 . Therefore f ( N [v]) = 1 or f ( N [v]) = 3 . If f ( N [v ]) = 3 , consider the mapping g : V → {−1,1} by g (u ) = −1 and g ( x) = f ( x) for every x ∈ V \ {u} . Then g is a NNSDF and w( g ) = w( f ) − 2 < γ sNN (G ) , a contradiction. Then the assertion follows. □ A subdivision of the edge uv is obtained by removing edge uv , adding a new vertex w , and then adding edges uw and vw . A spider, S n , of order 2n + 1 , is the graph formed by subdividing n edges of a star K1, n . The vertex of degree n in S n is called the central vertex. Theorem 4. Let S n be a spider with n(n ≥ 3) . Then
γ sNN ( S n ) = 1 . Figure 2. The path P4 with γ sNN ( P4 ) = 0
Theorem 2. For any star K1, n , we have γ sNN ( K1, n ) = 0
Proof. We mark the central vertex of S n by v0 and mark the other vertices of S n by v1 , v2 , , vn , vn +1 , , v2 n such that vi is only adjacent to v0 and vn + i for 1 ≤ i ≤ n . Let
when n is odd and γ sNN ( K1, n ) = 1 when n is even.
f be a γ sNN − function on S n . Then γ sNN ( Sn ) = w( f ) =
Proof. We mark the vertices of K1,n by v0 , v1 , v2 ,..., vn and
f ( N [vn +1 ]) + + f ( N [v2 n ]) + f (v0 ) ≥ f (v0 ) .
v0 is the center of K1,n . Let f be a γ sNN − function
of K1, n . Then γ sNN ( K1, n ) = w( f ) = f ( N [v0 ]) ≥ 0 and by Lemma 2, γ sNN ( K1, n ) ≥ 1 when n is even. Consider the mapping g : V → {−1,1} such that g (vi ) = 1 when i is even and g (vi ) = −1 when i is odd. Then g[vi ] ≥ 0 for i = 0 , g[vi ] = 2 ≥ 0 when i ≥ 2 is even and g[vi ] = 0 when i ≥ 1 is odd. Therefore g is a NNSDF of G and w( g ) = 0 when n is odd and w( g ) = 1 when n is even. In conclusion, γ sNN ( K1, n ) = 0 when n is odd and
γ sNN ( K1, n ) = 1 when n is even.□ Theorem 3. Let Wn +1 = K1 ∨ Cn be a wheel. Then
γ
(Wn +1 ) = 0 when n is odd and γ sNN (Wn +1 ) = 1 when n is even. NN s
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If f (v0 ) = −1 , by Lemma 3, since f ( N [vi ]) = 1 for 1 ≤ i ≤ n , we have f (vi ) = f (vn + i ) = 1 for 1 ≤ i ≤ n . Therefore w( f ) = 2n − 1 . Consider the mapping g : V → {−1,1} by g (v0 ) = 1 , g (vi ) = 1 and g (vn + i ) = −1 for 1 ≤ i ≤ n . Then g is a NNSDF and w( g ) = 1 < 2n − 1 , a contradiction. Therefore f (v0 ) = 1 , which implies
γ sNN ( K1, n ) ≥ 1 . By the mapping g above, we have
γ sNN ( S n ) = 1 .□ A wounded spider S n , m of order n + m + 1 is the graph formed by subdividing m(1 ≤ m ≤ n − 1) edges of a star K1, n (n ≥ 2) . The vertex of degree n in Sn , m is called the central vertex. Theorem 5. For any wounded spider Sn , m with n(n ≥ 4) , we have
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γ
NN s
−m ( Sn , m ) = 1 − m m + 1 − n
m ≤ n 2，n is odd m ≤ n 2，n is even . m>n 2
Proof. We mark the central vertex of S n , m by v0 and mark the other vertices of S n , m by v1 , , vn , vn +1 , , vn + m such that, for 1 ≤ i ≤ m , vi is only adjacent to v0 and vn + i .
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Then g is a NNSDF on K n and w( g ) = 0 when n is even and w( g ) = 1 when n is odd. In conclusion,
γ sNN ( K n ) = 0 when n is even and γ sNN ( K n ) = 1 when n is odd.□ Theorem 7. Let K n, n be a complete equally bipartite graph with n(n ≥ 6) . Then γ sNN ( K n , n ) = 2 when n is odd
Let f be a γ sNN − function on Sn , m . If f (v0 ) = −1 , by
and γ sNN ( K n , n ) = 4 when n is even.
Lemma 3, since f ( N [vi ]) = 1 for 1 ≤ i ≤ m , we have f (vi ) = f (vn + i ) = 1 for 1 ≤ i ≤ m . Furthermore, by Definition 1, we have f ( N [vi ]) ≥ 0 for m + 1 ≤ i ≤ n , so f (vi ) = 1 for m + 1 ≤ i ≤ n . Therefore
We mark the vertices of V (G1 ) by v1 , v2 ,..., vn and mark the vertices of V (G2 ) by u1 , u2 ,..., un . Let f be a
γ sNN ( Sn , m ) = n + m − 1 .
VG+1 = {v | f (v) = 1, v ∈ V (G1 )} ,
Consider the mapping g : V → {−1,1} by g (vi ) = 1 for 0 ≤ i ≤ n − 1 and g (vi ) = −1 for n ≤ i ≤ n + m . Then g is a NNSDF and w( g ) = n − m − 1 < n + m − 1 , a contradiction. Therefore f (v0 ) = 1 . In case when 1 ≤ m ≤ n 2 , we have γ sNN ( S n ) = w( f ) = f ( N [v0 ]) + f ( N [vn +1 ]) + f ( N [vn + 2 ]) + + f ( N [vn + m ]) − f (v1 ) − − f (vm ) . If n is even, by Lemma 1, we have f ( N [v0 ]) ≥ 1 . Therefore γ sNN ( Sn ) ≥ 1 − m . Consider the mapping g : V → {−1,1} by g (vi ) = 1 for 0 ≤ i ≤ n 2 and g (vi ) = −1 for n 2 + 1 ≤ i ≤ n + m . Then w( g ) = 1 − m , which implies γ sNN ( K1, n ) = 1 − m . If n is odd, we have
γ sNN ( Sn ) ≥ − m . Consider the mapping g : V → {−1,1} by g (vi ) = 1
for 0 ≤ i ≤ (n − 1) 2 and g (vi ) = −1 for (n + 1) 2 ≤ i ≤ n + m . Then g is a NNSDF and w( g ) = −m , which implies γ sNN ( K1, n ) = −m .
In case when m > n 2 , we have γ sNN ( S n ) = w( f ) = f (v0 ) + f ( N [vn +1 ]) + + f ( N [vn + m ]) + f (vm +1 ) + + f (vn ) ≥ 1 − (n − m) . Consider the mapping g : V → {−1,1} by g (vi ) = 1 for 0 ≤ i ≤ m and g (vi ) = −1 for m + 1 ≤ i ≤ n + m . Then g is a NNSDF and w( g ) = m + 1 − n , which implies that
γ sNN ( K1, n ) = m + 1 − n . In conclusion, the assertion follows. Theorem 6. Let K n be a complete graph. Then
γ
( K n ) = 0 when n is even and γ ( K n ) = 1 when n is odd. Proof. We mark the vertices of K n by v1 , v2 ,..., vn . Let NN s
NN s
f be a γ sNN − function on K n . Then w( f ) = f ( N [v1 ]) ≥ 0 , which implies that γ sNN ( K n ) ≥ 0 . By Lemma 2,
γ sNN ( K n ) ≥ 1 when n is odd. Now consider the mapping g : V → {−1,1} such that g (vi ) = 1 when i is odd and g (vi ) = −1 when i is even.
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Proof. Let K n , n = G1 ∨ G2 . Then | V (G1 ) |=| V (G2 ) |= n .
γ sNN − function on K n, n . Let VG−1 = {v | f (v) = −1, v ∈ V (G1 )} , VG+2 = {v | f (v) = 1, v ∈ V (G2 )} , VG−2 = {v | f (v) = −1, v ∈ V (G2 )} .
If any of VG−1 and VG−2 is empty, we assume that VG−1 = ∅ without loss of generality, i.e. V (G1 ) = VG+1 .
Then f ( N [v1 ]) = f (v1 ) + f (V (G2 )) ≥ 0 . Therefore f (V (G2 )) ≥ − f (v1 ) = −1 . Then we have
γ sNN ( K n, n ) = f (V (G1 )) + f (V (G2 )) ≥ n − 1 . Consider the mapping g : V → {−1,1} such that g (vi ) = g (ui ) = −1 for 1 ≤ i ≤ n 2 − 1 and g (vi ) = g (ui ) = 1 for n 2 ≤ i ≤ n . Then g is a NNSDF on K n, n and w( g ) = 2 ( n − n 2 + 1) − ( n 2 − 1)
= 4 + 2(n − 2 n 2 ) ≤ 4 < n − 1 . Therefore non of VG−1 and VG−2 is empty . Without loss of
generality, we assume that v1 ∈ VG−1 . Then f (V (G2 )) ≥ − f (v1 ) = 1 . In the similar way, we have f (V (G1 )) ≥ 1 .
Therefore γ sNN ( K n , n ) ≥ 2 . If n is odd, we consider the mapping g : V → {−1,1} such that g (vi ) = g (ui ) = −1 for 1 ≤ i ≤ (n − 1) 2 and g (vi ) = g (ui ) = 1 for (n + 1) 2 ≤ i ≤ n . Then g is a NNSDF on K n, n and w( g ) = 2 , which implies that
γ sNN ( K n , n ) = 2 . If n is even, by Lemma 1, we have f (V (G2 )) ≥ 2 and f (V (G1 )) ≥ 2 . Therefore γ sNN ( K n , n ) ≥ 4 . Consider the
mapping g : V → {−1,1} such that g (vi ) = g (ui ) = −1 for 1 ≤ i ≤ n 2 − 1 and g (vi ) = g (ui ) = 1 for n 2 ≤ i ≤ n . Then g is a NNSDF on K n, n and w( g ) = 4 , which implies that γ sNN ( K n , n ) = 4 . From the above, the assertion follows.
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Theorem 8. A NNSDF f on a graph G is minimal if and only if for every vertex v ∈ V with f (v) = 1 , there exists a vertex u ∈ N [v ] with f ( N [u ]) < 2 . Proof. Let f be a minimal NNSDF and assume that there is a vertex v with f (v) = 1 and f ( N [u ]) ≥ 2 for every u ∈ N [v ] . Define a new function g : V → {−1,1} by g (v) = −1 < f (v) = 1 and g (u ) = f (u ) for all u ≠ v . Then for all u ∈ N [v ] , g (u ) = f (u ) − 2 ≥ 0 . For all w ∉ N [v] , g ( N [ w]) = f ( N [ w]) ≥ 0 . Thus g is a NNSDF. Since g < f , the minimality of f is contradicted. Conversely, let f be a NNSDF such that for all v ∈ V with f (v) = 1 , there exists a vertex u ∈ N [v] with f ( N [u ]) < 2 . Assume f is not minimal, i.e. there is a NNSDF g such that g < f . Then g ( w) ≤ f ( w) for all w ∈ V , and there exists at least one vertex v ∈ V with g (v) < f (v) . Therefore f (v) = 1 , and by assumption, there exists a vertex u ∈ N [v] with f ( N [u ]) < 2 . Since g ( w) ≤ f ( w) for all w ∈ V and g (v) < f (v) , we know that g ( N [u ]) = f ( N [u ]) − 2 < 0 . This contradicts the fact that g is a NNSDF. Therefore f is a minimal NNSDF.
Δ | V + | −δ | V − | +γ sNN (G ) ≥ 0 , i.e. (n + γ sNN (G ))Δ 2 − (n − γ sNN (G ))δ 2 + γ sNN (G ) ≥ 0 . Then we have δ −Δ n .□ γ sNN (G ) ≥ Δ +δ + 2 Theorem 11. Let G be a graph with n vertices and
m edges. Then γ sNN (G ) ≥ 4n + 1 − n − 1 . Proof. Let
f be a NNSDF on G . Let E (G[V − ]) = s ,
where G[V − ] is the induced graph of V − . E[V + , V − ] is defined as Theorem 9. Then | E (G[V + ]) |= m − s − | E[V + , V − ] | , where G[V + ] is the induced graph of V + . Let dV − (u ) =| {v | u ∈ V − , uv ∈ E (G[V − ])} | and dV + (u ) =| {v | u ∈ V + , uv ∈ E (G[V + ])} | .
Then
u∈V −
and
u∈V −
| N (u ) V + | =| E[V + , V − ] |= u∈V + | N (u ) V − |
dV − (u ) = u∈V − | N (u ) V − | = 2 | E (G[V − ]) |= 2s .
Since f is a NNSDF on G , for each u ∈ V − , we have
III. LOWER BOUNDS OF NONNEGATIVE SIGNED DOMINATION NUMBER Theorem 9. For any graph G with maximum degree Δ(G ) ≤ 2 , we have γ sNN (G ) ≥ 0 .
Proof. Let f be a γ sNN − function of G . If V − = ∅ ,
| N (u ) V − |≤ | N (u ) V + | −1 , and for u ∈ V + , we have | N (u ) V − |≤ | N (u ) V + | +1 . Therefore 0 ≤ 2s = u∈V − | N (u ) V − |
≤ u∈V − (| N (u ) V + | −1)
then the result follows. So assume V − ≠ ∅ . Let E[V + , V − ] = {uv | u ∈ V + , v ∈ V − } and k = | E[V + , V − ] | .
= u∈V − | N (u ) V + |− | V − | = u∈V + | N (u ) V − |− | V − |
For each v ∈ V − , since f ( N [v]) ≥ 0 , v is at least adjacent
≤ u∈V + (| N (u ) V + | +1)− | V − |
to one vertex of V + . Then we have k ≥| V − | . For
= u∈V + | N (u ) V + |+ | V + | − | V − |
each v ∈ V + , if v is adjacent to q vertices of V − , then v
= u∈V + dV + (u )+ | V + | − | V − |
+
is adjacent to at least q − 1 vertices of V . Since Δ(G ) ≤ 2 , we have k ≤ | V + | . Therefore w( f ) = | V + | − | V | ≥ 0 , which implies γ (G ) ≥ 0 .□ Theorem 10. For any graph G with maximum degree Δ and minimum degree δ , we have δ −Δ γ sNN (G ) ≥ n. Δ +δ + 2 Proof. Let f be a γ sNN − function of G . Then −
NN s
| V + | + | V − |= n and | V + | − | V − |= γ sNN (G ) . We have V + = (n + γ sNN (G )) 2 and V − = (n − γ sNN (G )) 2 . By Definition 1, for every v ∈ V we have f ( N [v]) ≥ 0 , Then v∈V (d (v) + 1) f (v) ≥ 0 . Therefore
v∈V +
d (v ) f (v) + v∈V − d (v) f (v) + γ sNN (G ) ≥ 0 .
So © 2013 ACADEMY PUBLISHER
Since u∈V + dV + (u ) ≤| V + | (| V + | −1) , we have 0 ≤| V + | (| V + | −1)+ | V + | − | V − |=| V + |2 + | V + | −n .
Solving the inequality, we have | V + |≥ Therefore w( f ) =| V + | − | V − |= 2 | V + | −n ≥ implies that γ
NN s
1 ( 4n + 1 − 1) . 2
4n + 1 − 1 − n , which
(G ) ≥ 4n + 1 − n − 1 .□
IV. COMPLEXITY ISSUES FOR NONNEGATIVE SIGNED DOMINATION The following decision problem for the domination number of a graph is known to be NP - complete, even when restricted to bipartite graphs or chordal graphs. Problem: DOMINATION(DM)
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INSTANCE: A graph G = (V , E ) and a positive integer k ≤ | V | . QUESTION: Is γ (G ) ≤ k ? We will demonstrate a polynomial time reduction of this problem to our nonnegative signed domination problem. Problem: NONNEGATIVE SIGNED DOMINATION(NNSD) INSTANCE: A graph H = (V , E ) and a positive integer j ≤ | V | . QUESTION: Is γ sNN ( H ) ≤ j ? Theorem 12. Problem NNSD is NP -complete, even when restricted to bipartite or chordal graphs. Proof. It is obvious that NNSD is a member of NP since we can, in polynomial time, guess at a function f : V → {−1,1} and verify that f has a weight at most j and is a NNSDF. We next show how a polynomial time algorithm for NNSD could be used to solve DM in polynomial time. Given a graph G = (V , E ) and a positive integer k construct the graph H by adding to each vertex v of G a set of deg G (v) paths of length three. Let m = | E | and n = | V | . We have | V ( H ) |= n + 3 v∈V (G ) deg G (v) = n + 6m
and
| E ( H ) |= m + 3 v∈V (G ) deg G (v) = 7m ,
and H can be constructed in polynomial time. Note that if G is a bipartite or chordal graph, then so too is H . We note that, for any nonnegative signed dominating function g , for each v ∈ V (G ) , if g (v) = −1 and g (u ) = −1 for every u ∈ N (v) V (G ) , then, because | N G [v] |=| N H [v] − N G [v] | +1 = deg G (v) + 1 , we would have g ( N H [v]) ≤ −1 . That is, if g : V ( H ) → {−1,1} is a NNSDF for H , then {v ∈ V (G ) | g (v) = 1} is a dominating set for G . Lemma 4. Let f be a NNSDF on H and w( f ) = γ sNN ( H ) . Then, for each path added to G , there exactly two vertices assigned value 1. Proof. Let P(v) be the path added to v ∈ V (G ) . We mark the vertices of P(v) by v1 , v2 and v3 according to the distance to v such that d (v, v1 ) = 1 , d (v, v2 ) = 2 and d (v, v3 ) = 3 . If there is no vertex of P (v) assigned value 1, then f ( N [v3 ]) = f (v2 ) + f (v3 ) = −2 , which is contradiction. If there is exactly one vertex of P (v) assigned value 1, then f ( N [v2 ]) = f (v1 ) + f (v2 ) + f (v3 ) = −1 , which is contradiction. If all the vertices of P(v) are assigned value 1, let g : V ( H ) → {−1,1} such that g (v3 ) = −1 and g ( w) = f ( w) , w ∈ V ( H ) − {v3 } . Then g
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is a NNSDF on H and w( g ) = w( f ) − 1 = γ sNN ( H ) − 1 , which is conflict to the assumption. From the above, we can get the assertion. By Lemma 4, if we let j = (2n − n) + k − (n − k ) = 2k , then γ (G ) ≤ k if and only if γ sNN ( H ) ≤ j . This completes the proof of Theorem 12.□ Problem DM is polynomial for fixed k . To see this, let G = (V , E ) be a graph with | V |= n . If k ≥ n , then V is a dominating set of G of cardinality at most k . On the other hand, if k < n , then consider all the r -subsets of k n V , where r = 1, 2, , k . There are r =1 of these r subsets, which is bounded above by the polynomial
k r =1
p r . It takes a polynomial amount of time to verify
that a set is or is not a dominating set. These remarks show that it takes a polynomial amount of time to verify whether G has a dominating set of cardinality at most k when k is fixed. Hence for fixed k , problem DM ∈ P . In contrast, we now show that for a fixed k , the problem NNSD can be NP -complete. To see this, we will demonstrate a polynomial time reduction of the nonnegative signed domination problem to the following zero nonnegative signed domination problem. Problem: ZERO NONNEGATIVE SIGNED DOMINATION(ZRSD) INSTANCE: A graph G = (V , E ) . QUESTION: Does G have a NNSDF of weight at most 0? Theorem 13. Problem ZNNSD is NP -complete, even when restricted to bipartite or chordal graphs. Proof. It is obvious that ZNNSD is a member of NP , since we can, in polynomial time guess at a function f : V (G ) → {−1,1} and verify that f has weight at most 0 and is a NNSDF. We next show how a polynomial time algorithm for ZNNSD could be used to solve NNSD in polynomial time. Let L be a graph of Figure 3. Then L has a NNSDF of weight -1 as illustrated. In fact, γ sNN ( L) = −1 . Note that L is chordal.
Figure 3. Given a graph H = (V , E ) and a positive integer j , let G = H ij=1 Li , where Li ≅ L for i = 1, 2, , j . It is clear that G can be constructed in polynomial time. Note that if H is chordal, then so too is G . We now show that γ sNN ( H ) ≤ j if and only if
γ sNN (G ) ≤ 0 . Suppose first that γ sNN ( H ) ≤ j and that f
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is a NNSDF of H of weight γ sNN ( H ) . Let fi be any NNSDF of weight −1 for Li for i = 1, 2, , j . Define g : V (G ) → {−1,1} by g ( w) = f i ( w) if w ∈ V ( Li ) for i = 1, , j , while g (v) = f (v) for v ∈ V ( H ) . Then g is a NNSDF of G of weight γ sNN ( H ) + j (−1) ≤ j − j = 0 ,
Proof. Let f be a inverse signed total dominating function of Pn . Then for any four vertices of Pn in turn, there are at most two of the values of f on these four vertices equal to +1. Therefore γ st0 ( Pn ) ≤ 0 .□ Theorem 14. For any path Pn , then
hence γ sNN (G ) ≤ 0 . Conversely, suppose that γ sNN (G ) ≤ 0
γ st0 ( Pn ) = 0 when n ≡ 0(mod 4) ;
and that g is a NNSDF of weight γ sNN (G ) . Let f be the restriction of g on V ( H ) and let fi be the restriction of
γ st0 ( Pn ) = −1 when n ≡ 1 or 3(mod 4) ;
g on V ( Li ) for i = 1, 2, , j . Then γ sNN ( H ) + j (−1) =
+ i =1 f i (V ( Li )) = g (V (G )) = γ sNN (G ) ≤ 0 + i =1 f i (V ( Li )) j
j
= g (V (G )) = γ sNN (G ) ≤ 0 , so that γ sNN ( H ) ≤ j .
Let F be the graph of Figure 4. Then F has a NNSDF of weight -1 as illustrated. In fact, γ sNN ( F ) = −1 . Note that F is bipartite. Given a graph H = (V , E ) and a positive integer j , let G = H ij=1 Fi , where Fi ≅ F for i = 1, 2, , j . It is clear that G can be constructed in polynomial time. Note that if H is bipartite, then so too is G . Proceeding now as in the preceeding paragraph, we may show that γ sNN ( H ) ≤ j if and only if γ sNN (G ) ≤ 0 . This completes the proof of the theorem.
γ st0 ( Pn ) = −2 when n ≡ 2(mod 4) . Proof. We mark the vertices of Pn by v1 , v2 , v3 , , vn in turn. It is easy to verify the conclusion when n ≤ 4 . Case 1 n ≡ 0(mod 4)(n > 4) . By Lemma 5, γ st0 ( Pn ) ≤ 0 . Consider the function f : V → {−1,1} such that +1, i ≡ 0(mod 4) or i ≡ 1(mod 4)； f (vi ) = −1, i ≡ 2(mod 4) or i ≡ 3(mod 4). Then f is an inverse signed total dominating function
of Pn and γ st0 ( Pn ) ≥ w( f ) = 0 . Therefore γ st0 ( Pn ) = 0 when n ≡ 0(mod 4)(n > 4) . Case 2 n ≡ 1(mod 4) (n > 4) . Consider the function f : V → {−1,1} such that +1, i ≤ n − 2 and i ≡ 0 or 1(mod 4); −1, i ≤ n − 2 and i ≡ 2 or 3(mod 4); f (vi ) = i = n − 1; −1, +1, i = n. Then f is an inverse signed total dominating function
and γ st0 ( Pn ) ≥ w( f ) = −1 .On the other hand, Let S = {v3 ,
Figure 4. V.
v4 , , vn − 3 } ,then S ≡ 0(mod 4) .Let g be a γ st0 − function
PROPERTIES OF INVERSE SIGNED TOTAL DOMINATION
In this part, we do some work on the extension of nonnegative signed domination, i.e , the inverse signed total domination number of a graph. We obtained the exact values of the inverse signed total domination number of special graph classes. Definition 2. For a simple graph G with δ ≥ 1 a function f : V → {−1,1} is said to be a inverse signed total dominating function if for every v ∈ V (G ) , f ( N (v)) ≤ 0 . The inverse signed total domination number γ st0 (G ) = max { w( f ) | f is a inverse signed total dominating function of G }．We say f is a maximal inverse signed total dominating function if there does not exist a inverse signed total domination function g , g ≠ f , for which g (v) ≥ f (v) for every v ∈ V . A inverse signed total dominating function f is called a γ - function of G if 0 st
w( f ) = γ st0 (G ) ． Lemma 5. For any path Pn , if n ≡ 0(mod 4) , then
γ st0 ( Pn ) ≤ 0 . © 2013 ACADEMY PUBLISHER
of Pn . By Lemma 5, we have g ( S ) ≤ 0 .By Definition 2, g (v2 ) = g ( N (v1 )) ≤ 0 and g (vn −1 ) = g ( N (vn )) ≤ 0 .Thus g (v2 ) = −1 and g (vn −1 ) = −1 . Therefore w( g ) = g (v1 ) + g (v2 ) + g ( S ) + g ( N (vn −1 )) + g (vn −1 ) ≤ 1 − 1 +0 + 0 −1 = −1 . Then γ st0 ( Pn ) ≤ −1 .
In conclusion, γ st0 ( Pn ) = −1 when n ≡ 1(mod 4) (n > 4) . Case 3 n ≡ 2(mod 4) (n > 4) . Consider the function f : V → {−1,1} such that +1, i ≤ n − 2 and i ≡ 0 or 1(mod 4); f (vi ) = −1 i ≤ n − 2 and i ≡ 2 or 3(mod 4); −1 i > n − 2. Then f is an inverse signed total dominating function
of Pn . Thus γ st0 ( Pn ) ≥ w( f ) = −2 . On the other hand, let S = {v4 , v5 , , vn − 3 } , then S ≡ 0(mod 4) . Let g be a
γ st0 − function of Pn , then, by Lemma 5, g ( S ) ≤ 0 . By Definition 2, g (v2 ) = g ( N (v1 )) ≤ 0 and g (vn −1 ) =
JOURNAL OF NETWORKS, VOL. 8, NO. 2, FEBRUARY 2013
g ( N (vn )) ≤ 0 .Thus g (v2 ) = −1 and g (vn −1 ) = −1 . Since g is an inverse signed total dominating function of Pn ,
we have g ( N (v2 )) ≤ 0 and g ( N (vn −1 )) ≤ 0 i.e. g (v1 ) + g (v3 ) ≤ 0 and g (vn − 2 ) + g (vn ) ≤ 0 . Therefore, w( g ) = g (v1 ) + g (v2 ) + g (v3 ) + g ( S ) + g (vn − 2 ) + g (vn −1 ) + g (vn ) ≤ 0 − 1 + 0 + 0 − 1 = −2 . Thus γ ( Pn ) ≤ −2 . 0 st
In conclusion, γ ( Pn ) = −2 when n ≡ 2(mod 4)(n > 4) . Case 4 n ≡ 3(mod 4) (n > 4) . Consider the function f : V → {−1,1} such that 0 st
+1, i = 0 or 1(mod 4); f (vi ) = −1, i = 2 or 3(mod 4). Then f is an inverse signed total dominating function
371
−1, i = 3 or 0(mod 4)； f (vi ) = +1, i = 1or 2(mod 4); −1, i = 0. Then f is a inverse signed total dominating function
of Wn +1 , and w( f ) = −1 .Thus γ st0 (Wn +1 ) ≥ w( f ) = −1 . In conclusion, γ st0 (Wn +1 ) = −1 when n ≡ 0(mod 4) . Case 2 n ≡ 1(mod 4) . Consider the function f : V → {−1,1} such that f (vi ) = +1 when i ≡ 1 or 2(mod 4) , otherwise f (vi ) = −1 . Then f is a inverse signed total dominating function of Wn +1 , and w( f ) = −2 . Thus γ st0 (Wn +1 ) ≥ w( f ) = −2 .
a γ st0 − function of Pn . By Lemma 5, g ( S ) ≤ 0 . By Definition 2, g (vn −1 ) = g ( N (vn )) ≤ 0 . Thus g (vn −1 ) = −1 . Since g is an inverse signed total dominating function of
In conclusion, γ st0 (Wn +1 ) = −2 when n ≡ 1(mod 4) . Case 3 n ≡ 2(mod 4) . Consider the function f : V → {−1,1} such that f (vi ) = −1 when i = n − 1 or n ; f (vi ) = +1 when i ≡ 1 or 2(mod 4) ; otherwise f (vi ) = −1 . Then f is a inverse signed total dominating function of Wn +1 , and w( f ) = −3 .
Pn , we have
Thus γ st0 (Wn +1 ) ≥ w( f ) = −3 . On the other hand, let g be
of Pn . Thus γ ( Pn ) ≥ w( f ) = −1 . On the other hand, 0 st
let S = {v1 , v2 , , vn − 3 } , then S ≡ 0(mod 4) . Let g be
g ( N (vn −1 )) ≤ 0 ,i.e. g (vn − 2 ) + g (vn ) ≤ 0 .
Thus w( g ) = g ( S ) + g (vn − 2 ) + g (vn −1 ) + g (vn ) ≤ 0 − 1 + 0 = −1 . Therefore γ st0 ( Pn ) ≤ −1 . In conclusion, γ st0 ( Pn ) = −1 when n ≡ 3(mod 4)(n > 4) .□ The method to prove the theorem is similar to that of theorem 14. Theorem 15. For any cycle Cn ( n ≥ 3 ), then
γ st0 (Cn ) = 0 when n ≡ 0(mod 4) ; γ st0 (Cn ) = −1 when n ≡ 1 or 3 (mod 4) ; γ (Cn ) = −2 when n ≡ 2(mod 4) . 0 st
Theorem 16. Let Wn +1 = K1 ∨ Cn be a wheel, Then
γ st0 (Wn +1 ) = −1 when n ≡ 0(mod 4) ; γ st0 (Wn +1 ) = −2 when n ≡ 1 or 3(mod 4) ; γ st0 (Wn +1 ) = −3 when n ≡ 2(mod 4) . Proof. We mark the center of Wn +1 by v0 and mark the vertices of Cn by v1 , v2 , , vn in turn. Let g be a
γ st0 − function on Wn +1 . Then γ st0 (Wn +1 ) = g (v0 ) + g (Cn ) . By Definition 2, g ( N (v0 )) = g (Cn ) ≤ 0 , and for every v ∈ V (Cn ) , g ( N (v)) ≤ 0 .Since N (v) = 3 is odd, g ( N (v)) ≤ −1 . Thus
v∈V ( Cn )
g ( N (v)) ≤ −n , i.e. 2 g (Cn ) + ng (v0 )
≤ − n .Then γ st0 (Wn +1 ) = g (Cn ) + g (v0 ) ≤ [(n − 2) g (Cn ) − n] n ≤ −1 ,
and γ (Wn +1 ) ≤ −1 when n is even and γ (Wn +1 ) ≤ −2 when n is odd. Case 1 n ≡ 0(mod 4) . Consider the function f : V → {−1,1} such that 0 st
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0 st
a γ st0 − function of Wn +1 . Then there exists at least two adjacent vertices of V (Cn ) assigned value −1 and the number of the other vertices, which construct a path Pn − 2 , of V (Cn ) is multiple of four. Thus w( g ( Pn − 2 )) ≤ 0 , and
γ st0 (Wn +1 ) = g ( Pn − 2 ) − 3 ≤ −3 . In conclusion, γ st0 (Wn +1 ) = −3 when n ≡ 2(mod 4) . Case 4 n ≡ 3(mod 4) . Consider the function f : V → {−1,1} such that −1, i = 3 or 0(mod 4); f (vi ) = +1, i = 1 or 2(mod 4); −1, i = 0. Then f is a inverse signed total dominating function
of Wn +1 , and w( f ) = −2 . Thus γ st0 (Wn +1 ) ≥ w( f ) = −2 . In conclusion, γ st0 (Wn +1 ) = −2 when n ≡ 3(mod 4) .□ ACKNOWLEDGMENT This work was supported by the Educational Commission of Hebei Province (No.ZH2011122, Z2011157), the Natural Science Foundation of Hebei Province (No. A2012408002) and the Youth Foundation of Langfang Teachers College (No. LSZQ201001). REFERENCES [1] G. Chartrand, L. Lesniak, Graphs and Digraphs, second ed. Wadsworth and Brooks Cole, Monterey, 1986. [2] A.L. Barabàsi and Z.N. Oltvai, “Network biology: understanding the cell’s functional organization,” Nature Reviews Genetics, vol.5, pp. 101–113, 2004. [3] F. Emmert-Streib, M. Dehmer, J. Kilian, “Classification of large graphs by a local tree decomposition in: H.R. Arabnia,” A. Scime (Eds.), Proceedings of the 2005
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Zhongsheng Huang received his B.S. degree in Mathematics Education and M.S. degree in Applied Mathematics from Hebei Normal University in 1998 and 2007, respectively. He worked as an associate professor in Dept. of Mathematics and Information Science, Langfang Teachers College. His interesting research areas are graph theory and algebraic combinatorics.
Wensheng Li received the B.S. degree in Information and Computing Science from the Hebei University of Technology, Hebei, China, in 2003. He obtained the M.S. degree in Computational Mathematics from the Dalian University of Technology, Dalian, China, in 2006. He worked as a lecturer in Department of Mathematics and Infor-mation Science, Langfang Teachers College. His research interest is in the area of graph theory.
Zhifang Feng was born in Shijiazhuang, China. She has completed her B.S. and M.S. degree in Mathematics and Applied Mathematics from Hebei University in 2004 and 2007, respectively. She is now an lecturer in the Dept. of Mathematics and Information Science, Langfang Teachers College, Langfang, China. Her current research includes the theory of probability and graph theory.
