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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. - PowerPoint PPT Presentation
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Abdollah KhodkarDepartment of Mathematics
University of West Georgia
www.westga.edu/~akhodkar Joint work with Arezoo N. Ghameshlou, University of Tehran
Signed edge domination numbers of complete tripartite graphs
2
Overview
2. Previous Results
1. Signed edge domination
3. New Result
3
e1 e2
e3
e6 e4
e5
Graph G = (V(G), E(G))
Closed neighborhood of e1 = N[e1] = {e1, e2, e3, e6}
Closed neighborhood of e5 = N[e5] = {e4, e5, e6}
4
Signed Edge Dominating Functions
B. Xu (2001): f : E(G) → {-1, 1}
∑y in N[x] f(y) ≥ 1, for every edge x in E(G).
1 1 11
1 -1
Weight of f = w(f) = 1+1+1=3 Weight of f = w(f) = 1+1+(-1)=1
γ′s (G) = Minimum weight for a signed edge dominating function
Signed Edge Domination Number of Complete Graph of Order 8
+1
-1
γ′s1(K8)=16-12=4
Max number of -1 edges:⌊(2n-2)/4 =⌋ ⌊(2(8)-2)/4⌋ =3
6
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.
7
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
γ′s (G) = -2 1
1
-1
-1
-1
1
-1
-11
1
1 -1
1
1
1
-1-1-1
1
-1-11
-1-1
8
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 11
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
Signed Edge 1-Domination Numbers
+1
-1
k=1, γ′s1(K8)=16-12=4
Max number of -1 edges:⌊(2n-1-k)/4⌋
⌊(2(8)-1-1)/4⌋ =3
Signed Edge 3-Domination Numbers
+1
-1
k=3, γ′s3(K8)=18-10=8
Max number of -1 edges:⌊(2n-1-k)/4⌋
⌊(2(8)-1-3)/4⌋ =3
11
Signed Edge 5-Domination Numbers
+1
-1
k=5, γ′s5(K8)=20-8=12
12
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 isa star with k + b vertices, where b is a positive odd integer.
13
Upper Bounds
Conjecture: (B. Xu (2005))
γ′s (G) ≤ |V(G)| - 1 = n – 1,
where n is the number of vertices.
Trivial upper bound
γ′s (G) ≤ m,
where m is the number of edges
14
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.
15
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)).
16
Alex J. Carney and Abdollah Khodkar (2010)Calculated the signed edge k-domination number for Kn and Km,n .
17
Signed edge domination numbers of complete tripartite graphs
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.
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).
18
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.
19
m=6
p=12
n=8
An example
-2
Assume f is a SEDF of K6,8,12 such that f(w) = -2.
w
20
m=6
p=12
n=8
An example
-2
21
m=6
p=12
n=8
-2
2 2
2 2
2
2
2
2
22
m=6
p=12
n=8
-2
2 2
4
4
4
4
4
4
-2 -2-2-2-2-2-2
23
m=6
p=12
n=8
-2
2 2
2
2
2
2
2
2
-2 -2-2-2-2-2-2
24
m=6
p=12
n=8
-2
2 2
2
2
2
2
2
2
-2 -2-2-2-2-2-2 0
25
m=6
p=12
n=8
-2
2 2
2
2
2
2
2
2
-2 -2-2-2-2-2-2 0
12 12 12 12 12 10
2 2 2
w(f)=38
26
m=6
p=12
n=8
An example
-4
w
27
m=6
p=12
n=8
-4
28
m=6
p=12
n=8
-4
4 4
4 4
4
4
4
4
-4 -2-4-4-4-4-4
4
29
m=6
p=12
n=8
-4
4 4
4 4
4
4
4
4
-4 -2-4-4-4-4-4
4
4 4 4 4
8 8 10 10 10
w(f)=34
30
m=6
p=12
n=8
An example
0
0
0
0 0
0
0
0 0 0 0 0 0
Let f be a SEDF of K6,8,12 such that f(v)≥ 0 for every vertex v.
31
m=6
p=12
n=8
0
0
0
0 0
0
0
0 0 0 0 0 0 2
2
2 4 2 2 2
2
2
2
2
2 2
w(f)=14
32
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, thenw(f) ≥ m2 − 5m + 3n + 4 if 2(m − 2) ≤ nw(f) ≥ −n2+4 + mn − m + n if 2(m − 2) ≥ n + 2 and n ≡ 0 (mod 4)w(f) ≥ −n2+4 + mn − m + n + 1 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.
33
m
p
n
Sketch of Proof: m, n p are all even
-2k
w
34
m
p
n
-2k
There are (m+n+2k)/2 negative one edges at vertex w.
w
35
m
p
n
-2k
2k 2k
2k 2k
2k
2k
2k
2k
v
w
u
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.
There are (m+n+2k)/2 negative one edges at w.
36
m
p
n
-2k
2k 2k
2k 2k
2k
2k
2k
2k
-2k-2k -2k-2k-2k-2k
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
37
m
p
n
-2k
2k 2k
2k 2k
2k
2k
2k
2k
-2k-2k -2k-2k-2k-2k -2k
w1
-2k+2
If 2k≤m-2, then (n-m)/2 vertices in W can have weight -2k+2 andthe remaining vertices in W can be joined to the remaining vertices in V.
38
When 2k≤m-2
39
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 if 2(m − 2) ≤ nw(f) ≥ −n2+4 + mn − m + n if 2(m − 2) ≥ n + 2 and n ≡ 0 (mod 4)w(f) ≥ −n2+4 + mn − m + n + 1 if 2(m − 2) ≥ n + 2 and n ≡ 2 (mod 4)
40
If m ≠ 1, then
if 2(m − 1) ≤ n − 1, then w(f) ≥ m2 − 3m + 2n + 1
if 2(m − 1) ≥ n + 1, thenw(f) ≥ (−n2 + 1)/4 + mn − m + n.
If m = 1, then w(f) ≥ 2n + 1.In addition, the lower bounds are sharp.
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).
41
m
p
n
m, n and p are all even and m + n + p ≡ 0 (mod 4)
0
0
0
0 2
0
0
0 0 0 0 2
0
2 2 2 2 2
2
2
2
2
2 2
w(f)=(m+n+p)/2
42
m
p
n
m, n and p are all even and m + n + p ≡ 2 (mod 4)
0
0
0
0 2
0
0
0 0 0 0 2
0
2 4 2 2 2
2
2
2
2
2 2
w(f)=(m+n+p+2)/2
43
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.
Main Theorem
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.
Let m, n and p be positive integers and m ≤ n ≤ p ≤ m+ n.
44
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.
Main Theorem (Continued)
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.
45
Thank You
46
Example
: People
: A and B are working on a taskA B
Proposal
-1
11
-1
-1
-1
-1
-1
1 1
Votes: Yes = 1 No = -1
Should the proposalbe accepted?
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