Download - Signed edge domination numbers of complete tripartite graphs

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}

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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

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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.

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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

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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


+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


+1

-1

k=5, γ′s5(K8)=20-8=12

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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.

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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

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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.

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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)).

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Alex J. Carney and Abdollah Khodkar (2010)Calculated the signed edge k-domination number for Kn and Km,n .

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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).

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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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m=6

p=12

n=8

An example

-2

Assume f is a SEDF of K6,8,12 such that f(w) = -2.

w

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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

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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

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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.

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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

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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, 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.

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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.

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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

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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.

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When 2k≤m-2

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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)

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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).

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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

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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

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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.

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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.

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Thank You

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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?