1
$\begingroup$

How do we show if the graphs are complete or not? We will use the cartesian product of two complete graphs. We need to show two cases: 1) the cartesian product of two complete graphs is complete, 2) the cartesian products of two complete graphs is not complete. We just need it for our research. Thank you for some help!

  • 3
    Do you know the definition of a complete graph? – 2017-02-25
  • 0
    Hi, Joel! The only thing I know about complete graphs is that all the vertices are connected. Should I simply put to my paper that its vertices are connected or not? Thank you for responding by the way! – 2017-02-25
  • 1
    And in particular the definition of "cartesian product" for graphs? – 2017-02-25
  • 0
    The Cartesian product of two graphs, G and H is the graph G□H having vertex set V(G) x V(H) and in which vertex (u1, u2) is adjacent to (v1, v2) if and only if either u1=v1 and u2 is adjacent to v2 in H, or u2=v2 and u1 is adjacent to v1 in G. – 2017-02-25
  • 1
    "all the vertices are connected." Not exactly. For example, a graph that looks like a square is connected but is not complete. – 2017-02-25
  • 1
    Note that there are two natural kinds of product of graphs: the [cartesian product](https://en.wikipedia.org/wiki/Cartesian_product_of_graphs) and the [tensor product](https://en.wikipedia.org/wiki/Tensor_product_of_graphs). One of these produces a complete graph as the product of two complete graphs; the other doesn't! – 2017-02-25
  • 1
    Henning Makholm! I think I will look into that. Which one produces complete graphs? I've notice at K1 x Kn (cartesian product) produces complete graphs by observing the graph. At Km x Kn, m=2, 3, 4 it doesn't produce complete graphs. – 2017-02-25
  • 0
    Okay @JoelReyesNoche i will look into that. Thank you! – 2017-02-25
  • 1
    @Joy: Well, there's your answer, then. – 2017-02-25
  • 0
    Yes, the Cartesian product of complete graphs yields a complete graph. – 2017-02-26
  • 0
    @JazzyMatrix, not always. (Read Joy's comment above.) – 2017-02-27
  • 0
    Ah, so the "or" in the definition of edges is xor? – 2017-02-27
  • 0
    @JazzyMatrix, the Cartesian product of complete graphs does not always yield a complete graph. For example, consider the Cartesian product of $K_2$ and $K_2$. (It is not $K_4$.) – 2017-03-01
  • 0
    So the "or" in the definition of the edges in a Cartesian product is an "xor"? – 2017-03-01
  • 0
    @JazzyMatrix, the "or" in the definition of a Cartesian product of graphs is an "xor." Note that both conditions cannot be true at the same time. (For example, if $u=v$ then $u$ cannot be adjacent to $v$.) Also, I don't understand what you mean by "edges." $(u,v)$ is a vertex of the Cartesian product, not an edge. – 2017-03-01

0 Answers 0