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Signed edge domination numbers
of complete tripartite graphs
Abdollah Khodkar
Department of Mathematics
University of West Georgia
www.westga.edu/~akhodkar
Joint work with Arezoo N. Ghameshlou, University of Tehran
Overview
1. Signed edge domination
2. Previous Results
3. New Result
2
Graph G = (V(G), E(G))
e1
e2
e3
e6
e4
e5
Closed neighborhood of e1 = N[e1] = {e1, e2, e3, e6}
Closed neighborhood of e5 = N[e5] = {e4, e5, e6}
3
Signed Edge Dominating Functions
B. Xu (2001):
f : E(G) → {-1, 1}
∑y in N[x] f(y) ≥ 1,
1
1
1
Weight of f = w(f) = 1+1+1=3
for every edge x in E(G).
1
1
-1
Weight of f = w(f) = 1+1+(-1)=1
γ′s (G) = Minimum weight for a signed edge dominating function
4
Signed Edge Domination Number of
Complete Graph of Order 8
+1
-1
Max number of -1 edges:
⌊(2n-2)/4⌋=
⌊(2(8)-2)/4⌋ =3
γ′s1(K8)=16-12=4
Best Lower Bound
B. Xu (2005)
Let G be a graph with δ(G) ≥ 1, then
γ′s (G) ≥ |V(G)| - |E(G)| = n - m.
This bound is sharp.
Problem: (B. Xu (2005))
Classify all graphs G with
γ′s (G) = n - m.
6
Karami, Khodkar, Sheikholeslami (2006)
Let G be a graph of order n ≥ 2 with m edges.
Then γ′s (G) = n - m if and only if
1. The degree of each vertex is odd;
2. The number of leaves at vertex v = L(v) ≥ (deg(v) - 1 )/2.
n = 22
m = 24
-1
γ′s (G) = -2
-1
1
-1
1
-1
1
1
-1
1
-1
1
1
-1
-1
1
1
-1 -1
-1
1
1
-1
-1
7
Signed Edge k-Dominating Functions
A.J. Carney and A. Khodkar (2009):
f : E(G) → {-1, 1}, k is a positive integer
∑y in N[x] f(y) ≥ k, for every edge x in E(G).
1
1
1
1
1
-1
Weight of f = w(f) = 1+1+1=3
Weight of f = w(f) = 1+1+(-1)=1
k=3, γ′s3(K3)=3
k=1, γ′s1(K3)=1
8
Signed Edge 1-Domination Numbers
+1
-1
Max number of -1 edges:
⌊(2n-1-k)/4⌋
⌊(2(8)-1-1)/4⌋ =3
k=1, γ′s1(K8)=16-12=4
Signed Edge 3-Domination Numbers
+1
-1
Max number of -1 edges:
⌊(2n-1-k)/4⌋
⌊(2(8)-1-3)/4⌋ =3
k=3, γ′s3(K8)=18-10=8
Signed Edge 5-Domination Numbers
+1
-1
k=5, γ′s5(K8)=20-8=12
11
A sharp lower bound for signed edge
k-domination number
B. Xu (2005)
Let G be a simple graph with no isolated vertices. Then
γ′s (G) ≥ |V (G)| − |E(G)|
A. J. Carney and A. Khodkar (2009)
Let G be a simple graph with no isolated vertices and let G
admit a SEkDF. Then
γ′sk (G) ≥ |V (G)| − |E(G)| + k -1
When k ≥ 2 the equality holds if and only if G is
a star with k + b vertices, where b is a positive odd integer.
12
Upper Bounds
Trivial upper bound
γ′s (G) ≤ m,
where m is the number of edges
Conjecture: (B. Xu (2005))
γ′s (G) ≤ |V(G)| - 1 = n – 1,
where n is the number of vertices.
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The conjecture is true for trees, because
γ′s (G) ≤ m=n-1.
B. Xu (2003)
Let n ≥ 2 be an integer. Then
γ′s (Kn) = n/2
if n is even
and
γ′s (Kn) = (n − 1)/2 if n is odd.
14
S. Akbari, S. Bolouki, P. Hatami and M. Siami (2009)
Let m and n be two positive integers and m ≤ n. Then
(i) If m and n are even, then γ′(Km,n) = min(2m, n),
(ii) If m and n are odd, then γ′(Km,n) = min(2m − 1, n),
(iii) If m is even and n is odd, then
γ′(Km,n) = min(3m, max(2m, n + 1)),
(iv) If m is odd and n is even, then
γ′(Km,n) = min(3m − 1, max(2m, n)).
15
Alex J. Carney and Abdollah Khodkar (2010)
Calculated the signed edge k-domination
number for Kn and Km,n .
16
Signed edge domination numbers of complete
tripartite graphs
Let f : E(G) → {-1, 1} be a SEDF of G : that is;
∑y∈ N[x] f(y) ≥ 1, for every edge x in E(G).
The weight of vertex v ∈ V (G) is defined by f(v) =Σe∈E(v) f(e),
where E(v) is the set of all edges at vertex v.
17
Our Strategy
Step 1: We find minimum weight for SEDFs of complete
tripartite graphs that produce vertices of negative weight.
There is a vertex v of the graph Km,n,p such that f(v) < 0.
Step 2: We find minimum weight for SEDFs of complete
tripartite graphs that do not produce vertices of negative
weight.
For all vertices v of the graph Km,n,p, f(v) ≥ 0.
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An example
n=8
m=6
Assume f is a SEDF of K6,8,12
such that f(w) = -2.
w
p=12
-2
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An example
n=8
m=6
p=12
-2
20
2
2
n=8
m=6
2
2
2
2
2
2
p=12
-2
21
2
2
n=8
m=6
4
4
4
4
4
4
p=12
-2 -2 -2 -2 -2 -2 -2 -2
22
2
2
n=8
m=6
2
2
2
2
2
2
p=12
-2 -2 -2 -2 -2 -2 -2 -2
23
2
2
n=8
m=6
2
2
2
2
2
2
p=12
-2 -2 -2 -2 -2 -2 -2 -2 0
24
2
2
10 12 12 12 12 12
n=8
m=6
2
2
2
2
2
2
w(f)=38
p=12
-2 -2 -2 -2 -2 -2 -2 -2 0
2 2 2
25
An example
n=8
m=6
w
-4
p=12
26
n=8
m=6
p=12
-4
27
4
4 4
n=8
m=6
4
4
4
4
4
4
p=12
-4 -4 -4 -4 -4 -4 -4 -2
28
4
4 4
8
8
10 10 10
n=8
m=6
4
4
4
4
4
4
w(f)=34
p=12
-4 -4 -4 -4 -4 -4 -4 -2 4 4 4 4
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An example
0
m=6
0 0 0
n=8
Let f be a SEDF of K6,8,12 such
that f(v)≥ 0 for every vertex v.
0
0
0
0
0 0 0 0 0
p=12
30
0
m=6
0 0 0
2 2
2
2
n=8
2
2
2
w(f)=14
0
0
0
0
0 0 0 0 0 2
p=12
2 2 2 2
4
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Lemma 1: Let m, n and p be all even and
1 ≤ m ≤ n ≤ p ≤ m+n. Let f be a SEDF of Km,n,p such
that f(a) < 0 for some vertex a ∈ V (G). Then
If m ≠ 2, then
w(f) ≥ m2 − 5m + 3n + 4
w(f) ≥ −n2+4 + mn − m + n
w(f) ≥ −n2+4 + mn − m + n + 1
if 2(m − 2) ≤ n
if 2(m − 2) ≥ n + 2
and n ≡ 0 (mod 4)
if 2(m − 2) ≥ n + 2
and n ≡ 2 (mod 4)
If m = 2, then w(f) ≥ n + 4. In addition, the lower bounds are
sharp.
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Sketch of Proof: m, n p are all even
n
m
w
-2k
p
33
n
m
There are (m+n+2k)/2 negative
one edges at vertex w.
p
w
-2k
34
v
2k
2k
n
m
2k
2k
There are (m+n+2k)/2 negative
one edges at w.
2k
There are at most (n+p-2k)/2
negative one edges at u.
2k
2k
2k
There are at most (m+p-2k)/2
negative one edges at v.
u
p
w
-2k
35
2k
m
2k
2k
2k
2k
2k
2k
2k
n
There are (m+n+2k)/2 negative
one edges at w.
There are at most (n+p-2k)/2
negative one edges at u.
There are at most (m+p-2k)/2
negative one edges at v.
-2k -2k -2k -2k -2k -2k -2k -2k
p
36
2k
2k
n
m
2k
2k
2k
If 2k≤m-2, then (n-m)/2 vertices
in W can have weight -2k+2 and
the remaining vertices in W can
be joined to the remaining
vertices in V.
2k
2k
2k
w1
-2k -2k -2k -2k -2k -2k -2k -2k
-2k+2
p
37
When 2k≤m-2
38
Hence,
w(f) ≥ mn + mp + np
- 2 [m (n + p - 2k)/2
+ ((n - m + 2k)/2) (m + p - 2k)/2
+ ((n - m)/2) (n - m + 2k-2)/2
+ ((p - n + 2k)/2) (m + n - 2k)/2]
= 4k2 - 2nk + mn – m + n
We minimize 4k2 -2nk + mn-m+n subject to m ≤ n and 2 ≤ 2k ≤ m-2.
w(f) ≥ m2 − 5m + 3n + 4
w(f) ≥ −n2+4 + mn − m + n
w(f) ≥ −n2+4 + mn − m + n + 1
if 2(m − 2) ≤ n
if 2(m − 2) ≥ n + 2
and n ≡ 0 (mod 4)
if 2(m − 2) ≥ n + 2
and n ≡ 2 (mod 4)
39
Lemma 2: Let m, n and p be all odd and
1 ≤ m ≤ n ≤ p ≤m+n.
Let f be a SEDF of Km,n,p such that f(a) < 0 for some
vertex a ∈ V (G).
If m ≠ 1, then
if 2(m − 1) ≤ n − 1, then w(f) ≥ m2 − 3m + 2n + 1
if 2(m − 1) ≥ n + 1, then
w(f) ≥ (−n2 + 1)/4 + mn − m + n.
If m = 1, then w(f) ≥ 2n + 1.
In addition, the lower bounds are sharp.
40
m, n and p are all even and m + n + p ≡ 0 (mod 4)
m
0
0
0
0
2
2
2
2
n
2
2
w(f)=(m+n+p)/2
2
0
0
0
p
0
0
0 0
2
2
2
2
2
2
41
m, n and p are all even and m + n + p ≡ 2 (mod 4)
m
0
0
0
0
2
2
2
2
n
2
2
w(f)=(m+n+p+2)/2
2
0
0
0
p
0
0
0 0
2
2
2
2
2
4
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Main Theorem
Let m, n and p be positive integers and m ≤ n ≤ p ≤ m+ n.
A. Let m, n and p be even.
1. If m + n + p ≡ 0 (mod 4), then γ′s (Km,n,p) = (m + n + p)/2.
2. If m + n + p ≡ 2 (mod 4), then γ′s (Km,n,p) = (m + n + p+ 2)/2.
B. Let m, n and p be odd.
1. If m + n + p ≡ 1 (mod 4), then γ′s (Km,n,p) = (m + n + p + 1)/2.
2. If m + n+ p ≡ 3 (mod 4), then γ′s (Km,n,p) = (m + n + p + 3)/2.
43
Main Theorem (Continued)
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′s (Km,n,p) = (m + n)/2 + p + 1.
2. If (m + n) ≡ 2 (mod 4), then γ′s (Km,n,p) = (m + n)/2 + 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′s (Km,n,p) = (m + p)/2 + n + 1.
2. If m + p ≡ 2 (mod 4), then γ′s (Km,n,p) = (m + p)/2 + n.
44
Thank You
45
Example
: People
A
B
: A and B are working on a task
Proposal
Votes: Yes = 1
No = -1
-1
1
-1
1
-1
-1
1
Should the proposal
be accepted?
1
-1
-1
46