Signed edge domination numbers of complete tripartite graphs: Part 2

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by  Abdollah Khodkar

Signed edge domination numbers of complete
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tripartite graphs: Part 2
1
0
2
n Abdollah Khodkar
a Department of Mathematics
J
University of West Georgia
6
1 Carrollton, GA 30118
[email protected]
]
O
C
Abstract
.
h
t The closed neighborhood NG[e] of an edge e in a graph G is the
a set consisting of e and of all edges having an end-vertexin common
m
with e. Let f be a function on E(G), the edge set of G, into the
[ set {−1,1}. If P f(x) ≥ 1 for each edge e ∈ E(G), then f
x∈N[e]
is called a signed edge dominating function of G. The signed edge
1
domination number of G is the minimum weight of a signed edge
v
dominating function of G. In this paper, we find the signed edge
1
7 domination number of the complete tripartite graph Km,n,p, where
4 1≤m≤nandp≥m+n. Thiscompletes thesearch forthesigned
4 edge domination numbersof thecomplete tripartite graphs.
0
. Keywords: domination in graphs, edge domination, signed edge
1
domination
0
MSC 2000: 05C69
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1
:
v 1 Introduction
i
X
Let G be a simple non-empty graph with vertex set V(G) and edge set
r
a E(G). We use [4] for terminology and notation not defined here. Two
edgese ,e ofGarecalledadjacentiftheyaredistinctandhaveacommon
1 2
end-vertex. Theopen neighborhood N (e)ofanedgee∈E(G)isthesetof
G
alledgesadjacenttoe. Itsclosed neighborhood isN [e]=N (e)∪{e}. For
G G
a function f : E(G) → {−1,1} and a subset S of E(G) we define f(S) =
P f(x). If S = N [e] for some e ∈ E, then we denote f(S) by f[e].
x∈S G
The weight of vertex v ∈ V(G) is defined by f(v) = P f(e), where
e∈E(v)
E(v) is the set of all edges at vertex v. A function f : E(G) → {−1,1} is
1
called a signed edge dominating function (SEDF) of G if f[e] ≥ 1 for each
edge e ∈ E(G). The SEDF of a graph was first defined in [5]. The weight
off,denotedw(f), isdefinedtobew(f)=P f(e). The signed edge
e∈E(G)
domination number (SEDN) γ′(G) is defined as γ′(G) = min{w(f) | f is
s s
anSEDFofG}. AnSEDFf iscalledaγ′(G)-functionifω(f)=γ′(G). In
s s
[5] it was conjectured that γ′(G) ≤ |V(G)|−1 for every graph G of order
s
at least 2.
ThesignededgedominationnumbersofthecompletegraphK andthe
n
completebipartitegraphK weredeterminedin[6]and[1],respectively.
m,n
In [3], the signed edge domination number of K was calculated when
m,n,p
1≤m≤ n≤p ≤m+n. For completeness, we state the main theorem of
[3].
Theorem 1. Let m,n andp be positive integersand m≤n≤p≤m+n.
Let (m,n,p)6∈{(1,1,1), (2,3,5)}.
A. Let m, n and p be even.
1. If m+n+p≡0 (mod 4), then γ′(K )=(m+n+p)/2.
s m,n,p
2. If m+n+p≡2 (mod 4), then γ′(K )=(m+n+p+2)/2.
s m,n,p
B. Let m,n and p be odd.
1. If m+n+p≡1 (mod 4), then γ′(K )=(m+n+p+1)/2.
s m,n,p
2. If m+n+p≡3 (mod 4), then γ′(K )=(m+n+p+3)/2.
s m,n,p
C. Let m,n be odd and p be even or m,n be even and p be odd.
1. If m+n≡0 (mod 4), then γ′(K )=(m+n)/2+p+1.
s m,n,p
2. If m+n≡2 (mod 4), then γ′(K )=(m+n)/2+p.
s m,n,p
D. Let m,p be odd and n be even or m,p be even and n be odd.
1. If m+p≡0 (mod 4), then γ′(K )=(m+p)/2+n+1.
s m,n,p
2. If m+p≡2 (mod 4), then γ′(K )=(m+p)/2+n.
s m,n,p
E. Let n,p be odd and m be even or n,p be even and m be odd.
1. If n+p≡0 (mod 4), then γ′(K )=(n+p)/2+m+1.
s m,n,p
2. If n+p≡2 (mod 4), then γ′(K )=(n+p)/2+m.
s m,n,p
In addition, γ′(K )=1 and γ′(K )=5.
s 1,1,1 s 2,3,5
2
In this paper, we find the signed edge domination number of the com-
plete tripartite graph K , when m,n ≥ 1 and p ≥ m+n. In Section
m,n,p
2, we present some crucial results which will be employed in the rest of
this paper. In Section 3, we prove that if f is a γ′(K )-function, then
s m,n,p
f(w)≥−1foreveryvertexwinthelargestpartitesetofK . InSection
m,n,p
4,wepresentgeneralconstructionsfortheSEDFsofK withminimum
m,n,p
weight. In Section 5, we calculate the signed edge domination numbers of
K andK2,2,p,wherepiseven. Thesecasesdonotfollowtheconstruc-
1,n,p
tions givenin Section 4. In addition, we notice that γ′(K )=2n+3
s 1,n,n+3
when n is odd. So there is an infinite family of graphs which achieve the
upper bound given in Xu’s conjecture (see [5]). The main theorem of this
paper is presented in Section 6.
2 Preliminary results
Letf beaSEDFofG. Anedgee∈E(G)iscalledanegative edge(positive
edge)iff(e)=−1(f(e)=1). LetuvbeanedgeofGandsupposethatx,y
are the number of negative edges at vertices u and v, respectively. Then
f[uv]=deg(u)+deg(v)−2x−2y−f(uv)≥1. (1)
Hence,ife=uv,thenf(u)+f(v)≥0foreveryedgee∈E(G). Inaddition,
if f(u)+f(v)=0 or 1, then f(e)=−1.
The following result can be found in [2]. Since there are typos in the
proof given in [2] we modify the proof and present it here.
Lemma2. LetGbeagraph,u,v ∈V(G)andN(u)\{v}=N(v)\{u}. Let
f beaSEDFofG. ThenthereexistsaSEDFofG,sayg,withw(g)=w(f)
such that the difference between the number of negative edges at u and at
v is at most 1.
Proof. Let V(G) = {v = u,v = v,v ,...,v } and let x , 1 ≤ i ≤ n, be
1 2 3 n i
the number of edges e at v with f(e)=−1.
i
If x ≤x −2, then there exists a vertex v such that f(v v )= 1 and
1 2 ℓ 1 ℓ
f(v v ) = −1 for some ℓ ∈ {3,4,...,n}. Define g : E(G) → {−1,1} by
2 ℓ
g(v v )=−1, g(v v )=1 and g(e)=f(e) for e∈E(G)\{v v ,v v }. Let
1 ℓ 2 ℓ 1 ℓ 2 ℓ
y , 1 ≤ i ≤ n, be the number of edges e at v with g(e) = −1. Obviously,
i i
y =x +1,y =x −1andy =x for3≤i≤n. Inaddition,w(g)=w(f).
1 1 2 2 i i
We prove that g is a SEDF of G.
3
If v v ∈E(G) and j 6∈{1,2,ℓ}
1 j
g[v v ] = deg(v )+deg(v )−2y −2y −g(v v )
1 j 1 j 1 j 1 j
= deg(v )+deg(v )−2x −2−2x −f(v v )
1 j 1 j 1 j
≥ deg(v )+deg(v )−2x +2−2x −f(v v )
2 j 2 j 1 j
≥ deg(v )+deg(v )−2x −2x −f(v v )
2 j 2 j 2 j
≥ k by (1),
g[v v ] = deg(v )+deg(v )−2y −2y −g(v v )
2 j 2 j 2 j 2 j
= deg(v )+deg(v )−2x +2−2x −f(v v )
2 j 2 j 2 j
≥ k+2 by (1).
Similarly,for e∈E(G)\{v v ,v v |j 6∈{1,2,ℓ}},we obtain g[e]=f[e]≥
1 j 2 j
k. In addition, if v v ∈E(G), then
1 2
g[v v ] = deg(v )+deg(v )−2y −2y −g(v v )
1 2 1 2 1 2 1 2
= deg(v )+deg(v )−2x −2x −f(v v )
1 2 1 2 1 2
≥ k by (1).
Hence, g is a SEDF of G. If g satisfies the required condition, the proof is
complete. Otherwise, by repeating this process we can obtain the required
function.
Corollary 3. Let G be a complete multipartite graph. There exists a
γ′ (G)-function f suchthatthe difference betweenthe number ofnegative
sk
edges at every two vertices in the same partite set is at most 1.
3 SEDFs of complete tripartite graphs with
vertices of negative weight
Consider the complete tripartite graph K with partite sets U,V and
m,n,p
W. Throughout this section we assume |U| = m, |V| = n and |W| = p,
where 1 ≤ m ≤ n ≤ p. In this section, we study SEDFs f of K such
m,n,p
that f(w)<0 for some w ∈W.
Lemma 4. Let m,n,p be all even or all odd and 1≤m≤n≤p. If f is a
γ′(K )-function, then f(w)≥0 for every vertex of w ∈W.
s m,n,p
Proof. The proofis by contradiction. By Corollary3, we may assumethat
the difference between the number of negative edges at every two vertices
inthe same partite set ofK is atmost 1. Assume f(w)=−2k,where
m,n,p
1 ≤ k ≤ (m+n)/2, for some w ∈ W, and f(w′) ≥ −2k for all w′ ∈ W.
Then there are (m+n+2k)/2 negative edges and(m+n−2k)/2positive
edges atw. Thereforethe weightof (m+n+2k)/2verticesin U∪V must
4
be at least 2k and the weight of (m+n−2k)/2 vertices in U ∪V must
be at least 2k+2. Since m ≤ n and f is a γ′(K )-function, we can
s m,n,p
assume f(u) = 2k for every u ∈ U. So there are (n+p−2k)/2 negative
edges at every vertex u∈U. Let U ∪V , where V ⊆V, consist of vertices
1 1
of weight 2k and let V =V \V consist of vertices of weight 2k+2. Since
2 1
(m+n+2k)/2>mandf isaγ′(K )-function,itfollowsthatthereisa
s m,n,p
vertexv ∈V ofweight2k. Indeed, |V |=(n−m+2k)/2. Thereforethere
1 1
are (m+p−2k)/2 negative edges at v. Let W ⊆W consist of vertices of
1
weight−2k. Since everyvertexin V must be joined to everyvertexin W
1 1
with a negative edge by (1), it follows that |W |≤(m+p−2k)/2. Hence,
1
|W\W |≥(p−m+2k)/2andeveryvertexinthis sethasweight−2k+2.
1
Notethat(n+p−2k)/2−(m+p−2k)/2=(n−m)/2. LetW beasubset
2
of W \W with (n−m)/2 vertices and the edges between W and U are
1 2
all negative edges. Let W =W \(W ∪W ). Then
3 1 2
|W |≥p−[(m+p−2k)/2+(n−m)/2]=(p−n+2k)/2≥1.
3
Now let w′ ∈ W . Then the edges between w′ and U ∪V are all positive
3 1
edges. Thereforthenumberofnegativeedgesatw′isatmost(n+m−2k)/2.
On the other hand, for every vertex w ∈ W there are (n+m +2k)/2
1
negativeedges. Since k ≥1,the differencebetweenthe numberofnegative
edges at w and at w′ is 2k ≥2, which is a contradiction.
The proof of the following result is similar to the proof of Lemma 4.
Lemma 5. Let m,n be even and p be odd or m,n be odd and p be even,
where 1 ≤ m ≤ n < p. If f is a γ′(K )-function, then f(w) ≥ 0 for
s m,n,p
every vertex of w ∈W.
Proof. Theproofisbycontradiction. Letm,nbeevenandpbe odd. (The
case m,n odd and p even is similar.) By Corollary 3, we may assume that
the difference between the number of negative edges at every two vertices
in the same partite set of K is at most 1. Assume f(w) = −2k,
m,n,p
where 1 ≤ k ≤ (m + n)/2, for some w ∈ W, and f(w′) ≥ −2k for all
w′ ∈W. Thenthereare(m+n+2k)/2negativeedgesand(m+n−2k)/2
positive edges at w. Therefore the weight of (m+n+2k)/2 vertices in
U ∪V must be at least 2k+1 and the weight of (m+n−2k)/2 vertices
in U ∪V must be at least 2k +3. Since m ≤ n and f is a γ′(K )-
s m,n,p
function, we can assume f(u) = 2k + 1 for every u ∈ U. So there are
(n+p−2k−1)/2negativeedgesateveryvertexu∈U. Let U∪V , where
1
V ⊆V, consist of vertices of weight 2k+1 and let V =V \V consist of
1 2 1
verticesofweight2k+3. Since (m+n+2k)/2>m andf is a γ′(K )-
s m,n,p
function, it follows that there is a vertex v ∈V of weight 2k+1. Indeed,
1
|V | = (n−m+2k)/2. Therefore there are (m+p−2k−1)/2 negative
1
5
edges at v. Let W ⊆ W consist of vertices of weight −2k. Since every
1
vertex in V must be joined to every vertex in W with a negative edge
1 1
by (1), it follows that |W | ≤ (m +p−2k −1)/2. Hence, |W \W | ≥
1 1
(p−m+2k+1)/2 and every vertex in this set has weight −2k+2. Note
that (n+p−2k−1)/2−(m+p−2k−1)/2 = (n−m)/2. Let W be a
2
subset of W \W with (n−m)/2 vertices and the edges between W and
1 2
U are all negative edges. Let W =W \(W ∪W ). Then
3 1 2
|W |≥p−[(m+p−2k−1)/2+(n−m)/2]=(p−n+2k+1)/2≥1.
3
Now let w′ ∈ W . Then the edges between w′ and U ∪V are all positive
3 1
edges. Thereforthenumberofnegativeedgesatw′isatmost(n+m−2k)/2.
On the other hand, for every vertex w ∈ W there are (n+m +2k)/2
1
negativeedges. Since k ≥1,the differencebetweenthe numberofnegative
edges at w and at w′ is 2k ≥2, which is a contradiction.
Lemma 6. Let m be odd and n,p be even or m,p be even and n be odd,
where 1 ≤ m < n ≤ p. If f is a γ′(K )-function, then f(w) ≥ −1 for
s m,n,p
every vertex of w ∈W.
Proof. Theproofisbycontradiction. Letmbeoddandn,pbeeven. (The
case m,p even and n odd is similar.) By Corollary 3, we may assume that
thedifferencebetweenthenumberofnegativeedgesateverytwoverticesin
thesamepartitesetofK isatmost1. Assumef(w)=−2k−1,where
m,n,p
1≤k ≤(m+n−1)/2,forsomew ∈W,andf(w′)≥−2k−1forallw′ ∈W.
Then there are (m+n+2k+1)/2 negative edges and (m+n−2k−1)/2
positive edges at w. Therefore the weightof (m+n+2k+1)/2 verticesin
U∪V mustbe atleast2k+1andthe weightof(m+n−2k−1)/2vertices
in U ∪V must be at least 2k +3. Since m < n and f is a γ′(K )-
s m,n,p
function, we can assume f(u) = 2k+2 for every vertex u ∈ U. So there
are (n+p−2k−2)/2 negative edges at every vertex u ∈ U. Let V ⊂ V
1
consistofverticesofweight2k+1andletV =V \V consistofverticesof
2 1
weight2k+3. Since(m+n+2k+1)/2>mandf isaγ′(K )-function,
s m,n,p
it follows that there is a vertex v ∈ V of weight 2k+1. Indeed, |V | =
1 1
(n−m+2k+1)/2. Therefore there are (m+p−2k−1)/2 negative edges
atv. LetW ⊆W consistofverticesofweight−2k−1. Sinceeveryvertex
1
in V must be joined to every vertex in W with a negative edge by (1), it
1 1
followsthat|W |≤(m+p−2k−1)/2. Hence,|W\W |≥(p−m+2k+1)/2
1 1
and every vertex in this set has weight −2k+1.
Note that (n+p−2k−2)/2−(m+p−2k−1)/2 = (n−m−1)/2.
Let W be a subset of W \W with (n−m−1)/2 vertices and the edges
2 1
betweenW andU areallnegativeedges. LetW =W\(W ∪W ). Then
2 3 1 2
|W |≥p−[(m+p−2k−1)/2+(n−m−1)/2]=(p−n+2k+2)/2≥2.
3
6
Now let w′ ∈ W . Then the edges between w′ and U ∪V are all positive
3 1
edges. Therefor the number of negative edges at w′ is at most (n+m−
2k−1)/2. On the other hand, for everyvertex w ∈W there are (n+m+
1
2k+1)/2 negative edges. Since k ≥ 1, the difference between the number
ofnegativeedgesatw andatw′ is2k+1≥3,whichisacontradiction.
The proof of the following result is similar to the proof of Lemma 6.
Lemma 7. Let m,p be odd and n be even or m be even and n,p be odd,
where 1 ≤ m < n ≤ p. If f is a γ′(K )-function, then f(w) ≥ −1 for
s m,n,p
every vertex of w ∈W.
Proof. Theproofisbycontradiction. Letm,pbeoddandnbeeven. (The
case m even and n,p odd is similar.) By Corollary 3, we may assume that
thedifferencebetweenthenumberofnegativeedgesateverytwoverticesin
thesamepartitesetofK isatmost1. Assumef(w)=−2k−1,where
m,n,p
1≤k ≤(m+n−1)/2,forsomew ∈W,andf(w′)≥−2k−1forallw′ ∈W.
Then there are (m+n+2k+1)/2 negative edges and (m+n−2k−1)/2
positive edges at w. Therefore the weightof (m+n+2k+1)/2 verticesin
U∪V mustbe atleast2k+1andthe weightof(m+n−2k−1)/2vertices
in U ∪V must be at least 2k +3. Since m < n and f is a γ′(K )-
s m,n,p
function, we can assume f(u) = 2k+1 for every vertex u ∈ U. So there
are (n+p−2k−1)/2 negative edges at every vertex u ∈ U. Let V ⊂ V
1
consistofverticesofweight2k+2andletV =V \V consistofverticesof
2 1
weight2k+4. Since(m+n+2k+1)/2>mandf isaγ′(K )-function,
s m,n,p
it follows that there is a vertex v ∈ V of weight 2k+2. Indeed, |V | =
1 1
(n−m+2k+1)/2. Therefore there are (m+p−2k−2)/2 negative edges
atv. LetW ⊆W consistofverticesofweight−2k−1. Sinceeveryvertex
1
in V must be joined to every vertex in W with a negative edge by (1), it
1 1
followsthat|W |≤(m+p−2k−2)/2. Hence,|W\W |≥(p−m+2k+2)/2
1 1
and every vertex in this set has weight −2k+1.
Note that (n+p−2k−1)/2−(m+p−2k−2)/2 = (n−m+1)/2.
Let W be a subset of W \W with (n−m+1)/2 vertices and the edges
2 1
betweenW andU areallnegativeedges. LetW =W\(W ∪W ). Then
2 3 1 2
|W |≥p−[(m+p−2k−2)/2+(n−m+1)/2]=(p−n+2k+1)/2≥1.
3
Now let w′ ∈ W . Then the edges between w′ and U ∪V are all positive
3 1
edges. Therefor the number of negative edges at w′ is at most (n+m−
2k−1)/2. On the other hand, for everyvertex w ∈W there are (n+m+
1
2k+1)/2 negative edges. Since k ≥ 1, the difference between the number
ofnegativeedgesatw andatw′ is2k+1≥3,whichisacontradiction.
7
4 The SEDN of K
m,n,p
Consider the complete tripartite graph K whose partite sets are U,V
m,n,p
andW. Throughoutthissectionweassume|U|=m,|V|=nand|W |=p,
wherem,nandparepositiveintegers,m≤nandp≥m+nInthissection
we compute the signed edge domination number of K , where m ≥ 2
m,n,p
and (m,n)6=(2,2) if p is odd.
Theorem 8. Let m,n and p be even and p≥m+n. Then γ′(K )=
s m,n,p
m+n.
Proof. Consider the graph K whose partite sets are U,V and W. By
m,n,p
assumption
m(n+p−2)/2+n(m+p−2)/2−p(m+n)/2=mn−m−n
is even. Firstwe label(mn−m−n)/2edges betweenU and V with −1in
the following way. Label an edge uv, where u∈U and v ∈V, with −1 if
1. the total number of negative edges between U and V is less than
(mn−m−n)/2,
2. the number of negative edges at u is less than (n+p−2)/2,
3. the number of negative edges at v is less than (m+p−2)/2,
4. the numberofnegativeedgesatuislessthanorequaltothenumber
of negative edges at u′ for every u′ ∈(U \{u}), and
5. the numberofnegativeedgesatv islessthanorequaltothe number
of negative edges at v′ for every v′ ∈(V \{v}).
Thenwelabelp(m+n)/2edgesbetweenU∪V andW with−1inasimilar
fashion described above. An edge rw, where r ∈ (U ∪V) and w ∈ W is
labelled by −1 if
1. the number of negative edges between U ∪ V and W is less than
p(m+n)/2,
2. the number of negative edges at r is less than (n+p−2)/2 if r∈U,
3. the number ofnegative edgesat r is less than(m+p−2)/2 ifr ∈V,
4. the number of negative edges at w is less than (m+n)/2,
5. the numberofnegativeedgesatr islessthanorequaltothe number
of negative edges at u for every u∈(U \{r}) if r∈U,
8
6. the numberofnegativeedgesatr islessthanorequaltothe number
of negative edges at v for every v ∈(V \{r}) if r ∈V,
7. thenumberofnegativeedgesatw islessthanorequaltothenumber
of negative edges at w′ for every w′ ∈(W \{w}).
Then there are exactly ((p+n)/2)−1, ((p+m)/2)−1 and (m+n)/2
negative edges at every vertex in U,V and W, respectively. Label the
remaining edges of K by +1. Then all vertices in U ∪V have weight
m,n,p
2 and all the vertices in W have weight zero. Hence, this labeling defines
a SEDF f of K by (1), andω(f)=m+n. Note that since the weight
m,n,p
of every vertex of W is zero, no vertices in U ∪V can have weight zero by
(1). Now by Lemma 4 and the facts that f(W)=0 and f(r)=2 for every
r ∈U ∪V, it follows that γ′(K )=m+n.
s m,n,p
Theorem 9. Let m,n and p be odd, m,n ≥ 3 and p ≥ m+n+1. Then
γ′(K )=m+n+1.
s m,n,p
Proof. Consider the graph K whose partite sets are U,V and W. By
m,n,p
assumption,
m(n+p−2)/2+n(m+p−2)/2−p(m+n)/2=mn−m−n
is odd. In addition, (n+p−2)/2, (m+p−2)/2 and (m+n)/2 are all
odd, or two are even and one is odd. Hence, there is no graph whose m
verticeshavedegree(n+p−2)/2, n verticeshavedegree(n+p−2)/2and
p vertices have degree (m+n)/2.
On the other hand,
m(n+p−2)/2+n(m+p−2)/2−(p−1)(m+n)/2
−(m+n−2)/2=mn−m−n+1,
is an even number. We label (mn−m−n+1)/2 edges between U and V
and(p−1)(m+n)/2+(m+n−2)/2edgesbetweenU∪V andW with−1
in a similar fashion described in Theorem 8. Then there are (n+p−2)/2
negative edges at each vertex of U, (m+p−2)/2 negative edges at each
vertex of V and (m+n)/2 negative edges at each vertex of W except one
vertex which is incident with (m+n−2)/2 negative edges. We label the
remaining edges of K with +1. Then the weight of vertices in U ∪V
m,n,p
are all 2 and the weight of vertices in W are all zero except one vertex of
W whose weight is 2. Hence, this labeling defines a SEDF f of K by
m,n,p
(1), and ω(f)=m+n+1.
Note that since the weight of every vertex of W is zero, no vertices in
U ∪V can have weight zero by (1). Now by Lemma 4 and the facts that
f(W) = 2 and f(r) = 2 for every r ∈ U ∪V, it follows that γ′(K ) =
s m,n,p
m+n+1.
9
Theorem 10. Let m,n≥3 be odd, p be even and p≥m+n. Then
3m+3n+2
1. γ′(K )= if m+n≡0 (mod 4),
s m,n,p 2
3m+3n
2. γ′(K )= if m+n≡2 (mod 4).
s m,n,p 2
Proof. Consider the graph K whose partite sets are U,V and W.
m,n,p
Case 1. m+n≡0 (mod 4).
By assumption,
(m−1)(n+p−3)/2+(n+p−5)/2+n(m+p−3)/2
−p(m+n)/2=mn−1−(3m+3n)/2
is even. Label (1/2)[mn−1−(3m+3n)/2] edges between U and V and
p(m + n)/2 edges between U ∪ V and W with −1 in a similar fashion
described in the proof of Theorem 8. Then every vertex in U is incident
with (n+p−3)/2 negative edges except one vertex which is incident with
(n+p−5)/2negativeedges. EveryverticesinV isincidentwith(m+p−3)/2
negative edges and every vertex in W is incident with (m+n)/2 negative
edges. Label the remaining edges of K by +1. Then the weight of
n,m,p
vertices in U ∪V are all 3 except one vertex of U whose weight is 5, and
theweightofverticesinW areallzero. Hence,thislabelingdefinesaSEDF
f with w(f)=(3m+3n+2)/2.
Note that since the weight of every vertex of W is zero, no vertices in
U ∪V can have weight one by (1). Now by Lemma 5 and the facts that
f(W) = 0 and f(r) = 3 for every r ∈ U ∪V except one vertex which has
weight 5, it follows that γ′(K )=(3m+3n+2)/2.
s m,n,p
Case 2. m+n≡2 (mod 4).
By assumption,
m(p+n−3)/2+n(m+p−3)/2−p(m+n)/2=mn−(3m+3n)/2
iseven. Label(1/2)[mn−(3m+3n)/2]edgesbetweenU andV andp(m+
n)/2edgesbetweenU∪V andW with−1inasimilarfashiondescribedin
theproofofTheorem8. TheneveryvertexinU isincidentwith(n+p−3)/2
negative edges, every vertex in V is incident with (m+p−3)/2 negative
edges and every vertex in W is incident with (m+n)/2 negative edges.
Label the remaining edges of K with +1. Then the weight of vertices
n,m,p
in U ∪V are all 3 and the weight of vertices in W are all zero. Hence, this
labeling defines a SEDF f with w(f)=(3m+3n)/2.
Note that since the weight of every vertex of W is zero, no vertices in
U ∪V can have weight one by (1). Now by Lemma 5 and the facts that
f(W) = 0 and f(r) = 3 for every r ∈ U ∪V, it follows that γ′(K ) =
s m,n,p
(3m+3n)/2.
10