Αποτελέσματα Αναζήτησης
$\frac{n(n-1)}{2} = \binom{n}{2}$ is the number of ways to choose 2 unordered items from n distinct items. In your case, you actually want to count how many unordered pair of vertices you have, since every such pair can be exactly one edge (in a simple complete graph).
5 Σεπ 2015 · To make matters worse I have just noted that the determinant of the adjacency matrix of a complete graph with n vectors is $(-1)^{n-1}(n-1)$ which also mean my belief is also not correct. n = 6; i = 1; while (i <= n) i = i + 1; M = ones(i) - eye(i); [v, d] = eig(M) end
15 Νοε 2019 · $\begingroup$ I'd attack the problem in two stages. First, enumerate the orientations of the graph (assigning a direction to each edge) such that each node has the same number of edges entering and leaving; then find the number of Eulerian circuits in each of the resulting digraphs.
23 Φεβ 2019 · Because every two points are connected in a complete graph, each individual point is connected with every other point in the group of n points. There is a connection between every two points. There is a connection between every two points.
24 Οκτ 2015 · It's not true that in a regular graph, the degree is $|V| - 1$. The degree can be 1 (a bunch of isolated edges) or 2 (any cycle) etc. In a complete graph, the degree of each vertex is $|V| - 1$. Your argument is correct, assuming you are dealing with connected simple graphs (no multiple edges.)
22 Ιουν 2018 · I'm in with Mike, standard notation is rare in graph theory. People can't even agree if "graph" without additional information refers to "simple graph" (no loops and multiple edges between the same vertices) or "multi graph" (possibly multiple edges between the same vertices; loops or not is again not standard, some call the construct without loops "multi graph", some call it "pseudo graph ...
I can see why you would think that. For n=5 (say a,b,c,d,e) there are in fact n! unique permutations of those letters.
You need to consider two thinks, the first number of edges in a graph not addressed is given by this equation Combination(n,2) becuase you must combine all the nodes in couples, In addition you need two thing in the possibility to have addressed graphs, in this case the number of edges is given by the Permutation(n,2) because in this case the order is important.
16 Απρ 2014 · Every complete graph is also a simple graph. However, between any two distinct vertices of a complete graph, there is always exactly one edge; between any two distinct vertices of a simple graph, there is always at most one edge.
27 Αυγ 2017 · In a complete graph total number of paths between two nodes is equal to: $\\lfloor(P-2)!e\\rfloor$ This formula doesn't make sense for me at all, specially I don't know how ${e}$ plays a role in ...
