Signed edge domination numbers of complete tripartite graphs

1. 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

2. Overview 1. Signed edge domination 2. Previous Results 3. New Result

3. GraphG = (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}

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 1 1 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

5. 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

6. Best Lower Bound B. Xu (2005) Let G be a graph withδ(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 ordern ≥ 2withmedges. 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 -1 -1 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

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 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

9. 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

10. 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

11. Signed Edge 5-Domination Numbers +1 -1 k=5, γ′s5(K8)=20-8=12 11

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 is a star with k + b vertices, where b is a positive odd integer. 12

13. 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.

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. 14

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)). 15

16. Alex J. Carney and Abdollah Khodkar (2010) Calculated the signed edge k-domination number for Kn and Km,n . 16

17. 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

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. 18

19. An example n=8 m=6 Assume f is a SEDF of K6,8,12 such that f(w) = -2. w p=12 -2 19

20. An example n=8 m=6 p=12 -2 20

21. 2 2 n=8 m=6 2 2 2 2 2 2 p=12 -2 21

22. 2 2 n=8 m=6 4 4 4 4 4 4 p=12 -2 -2 -2 -2 -2 -2 -2 -2 22

23. 2 2 n=8 m=6 2 2 2 2 2 2 p=12 -2 -2 -2 -2 -2 -2 -2 -2 23

24. 2 2 n=8 m=6 2 2 2 2 2 2 p=12 -2 -2 -2 -2 -2 -2 -2 -2 0 24

25. 12 10 2 2 12 12 12 12 n=8 m=6 2 2 2 2 w(f)=38 2 2 p=12 -2 -2 -2 2 2 -2 -2 -2 2 -2 -2 0 25

26. An example n=8 m=6 w p=12 -4 26

27. n=8 m=6 p=12 -4 27

28. 4 4 4 n=8 m=6 4 4 4 4 4 4 p=12 -4 -4 -4 -4 -4 -2 -4 -4 28

29. 10 4 4 4 10 8 10 8 n=8 m=6 4 4 4 4 w(f)=34 4 4 p=12 -4 -4 -4 -4 -4 -2 -4 -4 4 4 4 4 29

30. An example 0 0 0 0 m=6 n=8 Let f be a SEDF of K6,8,12 such that f(v)≥ 0 for every vertex v. 0 0 0 p=12 0 0 0 0 0 0 30

31. 2 0 0 0 m=6 2 0 2 2 n=8 2 2 2 w(f)=14 0 0 0 p=12 0 0 0 2 0 0 2 0 4 2 2 2 31

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,psuch that f(a) < 0 for some vertex a ∈ V (G). Then If m ≠ 2, then w(f) ≥m2− 5m + 3n + 4 if 2(m − 2) ≤ n w(f) ≥−n2+4 + mn − m + nif 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. 32

33. Sketch of Proof: m, n p are all even n m w p -2k 33

34. n m There are (m+n+2k)/2 negative one edges at vertex w. p w -2k 34

35. 2k 2k v n m 2k There are (m+n+2k)/2 negative one edges at w. 2k 2k There are at most (n+p-2k)/2 negative one edges at u. 2k 2k There are at most (m+p-2k)/2 negative one edges at v. 2k u p w -2k 35

36. 2k 2k n m There are (m+n+2k)/2 negative one edges at w. 2k 2k There are at most (n+p-2k)/2 negative one edges at u. 2k 2k There are at most (m+p-2k)/2 negative one edges at v. 2k 2k p -2k -2k -2k -2k -2k -2k -2k -2k 36

37. 2k 2k n m 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 2k 2k 2k w1 p -2k -2k -2k -2k -2k -2k -2k -2k -2k+2 37

38. When 2k≤m-2 38

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) ≤ n w(f) ≥−n2+4 + mn − m + nif 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) 39

40. 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,psuch 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

41. m, n and p are all even and m + n + p ≡ 0 (mod 4) n 0 0 0 2 2 2 0 2 m 2 2 w(f)=(m+n+p)/2 2 0 0 0 p 2 0 2 2 0 2 0 0 2 2 41

42. m, n and p are all even and m + n + p ≡ 2 (mod 4) n 0 0 0 2 2 2 0 2 m 2 2 w(f)=(m+n+p+2)/2 2 0 0 0 p 2 0 2 4 0 2 0 0 2 2 42

43. 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

44. 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

45. Thank You 45

46. Example : People A B : A and B are working on a task Proposal -1 -1 Votes: Yes = 1 No = -1 1 1 -1 -1 1 1 Should the proposal be accepted? -1 -1