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Abdollah Khodkar Department 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

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

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3 e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 Graph G = (V(G), E(G)) Closed neighborhood of e 1 = N[e 1 ] = {e 1, e 2, e 3, e 6 } Closed neighborhood of e 5 = N[e 5 ] = {e 4, e 5, e 6 }

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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). 1 111 1 Weight of f = w(f) = 1+1+1=3Weight of f = w(f) = 1+1+(-1)=1 γ′ s (G) = Minimum weight for a signed edge dominating function

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Signed Edge Domination Number of Complete Graph of Order 8 +1 γ′ s1 (K 8 )=16-12=4 Max number of -1 edges: ⌊ (2n-2)/4 ⌋ = ⌊ (2(8)-2)/4 ⌋ =3

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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.

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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 1 1 1 1

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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). 1 111 1 Weight of f = w(f) = 1+1+1=3Weight of f = w(f) = 1+1+(-1)=1 k=3, γ′ s3 (K 3 )=3k=1, γ′ s1 (K 3 )=1

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Signed Edge 1-Domination Numbers +1 k=1, γ′ s1 (K 8 )=16-12=4 Max number of -1 edges: ⌊ (2n-1-k)/4 ⌋ ⌊ (2(8)-1-1)/4 ⌋ =3

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Signed Edge 3-Domination Numbers +1 k=3, γ′ s3 (K 8 )=18-10=8 Max number of -1 edges: ⌊ (2n-1-k)/4 ⌋ ⌊ (2(8)-1-3)/4 ⌋ =3

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

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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.

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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

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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 (K n ) = n/2 if n is even and γ′ s (K n ) = (n − 1)/2 if n is odd.

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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′(K m,n ) = min(2m, n), (ii) If m and n are odd, then γ′(K m,n ) = min(2m − 1, n), (iii) If m is even and n is odd, then γ′(K m,n ) = min(3m, max(2m, n + 1)), (iv) If m is odd and n is even, then γ′(K m,n ) = min(3m − 1, max(2m, n)).

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

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17 Signed edge domination numbers of complete tripartite graphs The weight of vertex v ∈ V (G) is defined by f(v) =Σ e ∈ 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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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 K m,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 K m,n,p, f(v) ≥ 0.

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

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

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

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

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

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

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

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

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

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

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29 m=6 p=12 n=8 -4 4 4 4 4 4 4 4 4 -2 -4 4 4 4 44 8 8 10 w(f)=34

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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 K 6,8,12 such that f(v)≥ 0 for every vertex v.

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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

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32 Lemma 1: Let m, n and p be all even and 1 ≤ m ≤ n ≤ p ≤ m+n. Let f be a SEDF of K m,n,p such that f(a) < 0 for some vertex a ∈ V (G). Then If m ≠ 2, then w(f) ≥ m 2 − 5m + 3n + 4 if 2(m − 2) ≤ n w(f) ≥ −n 2 +4 + mn − m + n if 2(m − 2) ≥ n + 2 and n ≡ 0 (mod 4) w(f) ≥ −n 2 +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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33 m p n Sketch of Proof: m, n p are all even -2k w

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

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35 m p n -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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36 m p n -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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37 m p n -2k 2k -2k w1w1 -2k+2 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.

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38 When 2k≤m-2

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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] = 4k 2 - 2nk + mn – m + n We minimize 4k 2 -2nk + mn-m+n subject to m ≤ n and 2 ≤ 2k ≤ m-2. w(f) ≥ m 2 − 5m + 3n + 4 if 2(m − 2) ≤ n w(f) ≥ −n 2 +4 + mn − m + n if 2(m − 2) ≥ n + 2 and n ≡ 0 (mod 4) w(f) ≥ −n 2 +4 + mn − m + n + 1 if 2(m − 2) ≥ n + 2 and n ≡ 2 (mod 4)

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40 If m ≠ 1, then if 2(m − 1) ≤ n − 1, then w(f) ≥ m 2 − 3m + 2n + 1 if 2(m − 1) ≥ n + 1, then w(f) ≥ (−n 2 + 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 K m,n,p such that f(a) < 0 for some vertex a ∈ V (G).

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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

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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

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43 A. Let m, n and p be even. 1. If m + n + p ≡ 0 (mod 4), then γ′ s (K m,n,p ) = (m + n + p)/2. 2. If m + n + p ≡ 2 (mod 4), then γ′ s (K m,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 (K m,n,p ) = (m + n + p + 1)/2. 2. If m + n+ p ≡ 3 (mod 4), then γ′ s (K m,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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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 (K m,n,p ) = (m + n)/2 + p + 1. 2. If (m + n) ≡ 2 (mod 4), then γ′ s (K m,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 (K m,n,p ) = (m + p)/2 + n + 1. 2. If m + p ≡ 2 (mod 4), then γ′ s (K m,n,p ) = (m + p)/2 + n.

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45 Thank You

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46 Example : People : A and B are working on a task AB Proposal 11 11 Votes: Yes = 1 No = -1 Should the proposal be accepted?

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PP Test Review Sections 6-1 to 6-6

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