By David Joyner, Minh Van Nguyen, Nathann Cohen

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Sage : G = Graph ({1:[2 ,4] , 2:[1 ,4] , 3:[2 ,6] , 4:[1 ,3] , 5:[4 ,2] , 6:[3 ,1]}) sage : G . vertices () [1 , 2 , 3 , 4 , 5 , 6] sage : E1 = Set ( G . edges ( labels = False )); E1 {(1 , 2) , (4 , 5) , (1 , 4) , (2 , 3) , (3 , 6) , (1 , 6) , (2 , 5) , (3 , 4) , (2 , 4)} sage : E4 = Set ( G . edges_incident ( vertices =[4] , labels = False )); E4 {(4 , 5) , (3 , 4) , (2 , 4) , (1 , 4)} sage : G . delete_vertex (4) sage : G . vertices () [1 , 2 , 3 , 5 , 6] sage : E2 = Set ( G . edges ( labels = False )); E2 {(1 , 2) , (1 , 6) , (2 , 5) , (2 , 3) , (3 , 6)} sage : E1 .

In the case where V1 = V2 , then G1 ∆G2 is simply the empty graph. 21 for an illustration of the symmetric difference of two graphs. The join of two disjoint graphs G1 and G2 , denoted G1 +G2 , is their graph union, with each vertex of one graph connecting to each vertex of the other graph. For example, the join of the cycle graph Cn−1 with a single vertex graph is the wheel graph Wn . 22 shows various wheel graphs. 5. 21: The symmetric difference of graphs. 22: The wheel graphs Wn for n = 4, .

Find matrix theoretic criteria on J1 and J2 which hold if and only if G1 ∼ = G2 . 22 for incidence matrices. 13. Show that the complement of an edgeless graph is a complete graph. 14. Let G H be the Cartesian product of two graphs G and H. Show that |E(G H)| = |V (G)| · |E(H)| + |E(G)| · |V (H)|. 15. In 1751, Leonhard Euler posed a problem to Christian Goldbach, a problem that now bears the name “Euler’s polygon division problem”. Given a plane convex polygon having n sides, how many ways are there to divide the polygon into triangles using only diagonals?

### Algorithmic Graph Theory by David Joyner, Minh Van Nguyen, Nathann Cohen

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