If e = {u,v} then HasEdge(G,e) returns true if the undirected graph G contains the (undirected) edge {u,v}, and false otherwise.

If a = [u,v], a directed edge, HasEdge(G,a) returns true if the undirected graph G has the undirected edge {u,v} in it.

If a = [u,v], HasArc(G,a) returns true if the directed graph G has the directed edge from vertex u to v in it, and false otherwise.

If e = {u,v}, HasArc(G,a) returns true if the directed graph G has both edges [u,v] and [v,u] in it, and false otherwise.

Because the data structure for a graph is an array of sets of neighbors, the test for edge membership uses binary search and hence the cost is O(log k) where k is the number of neighbors.

$\mathrm{with}\left(\mathrm{GraphTheory}\right)\:$

$G≔\mathrm{Graph}\left(\left\{\left\{1\,2\right\}\,\left\{1\,4\right\}\,\left\{2\,3\right\}\,\left\{3\,4\right\}\right\}\right)$

$G≔\mathrm{Graph\; 1:\; an\; undirected\; graph\; with\; 4\; vertices\; and\; 4\; edge(s)}$

$\mathrm{HasEdge}\left(G\,\left\{1\,2\right\}\right)$

$\mathrm{true}$

$\mathrm{HasEdge}\left(G\,\left[2\,1\right]\right)$

$\mathrm{true}$

$\mathrm{HasEdge}\left(G\,\left\{1\,3\right\}\right)$

$\mathrm{false}$

$N≔\mathrm{Graph}\left(\left\{\left[1\,2\right]\,\left[2\,3\right]\,\left[3\,4\right]\,\left[4\,1\right]\right\}\right)$

$N≔\mathrm{Graph\; 2:\; a\; directed\; graph\; with\; 4\; vertices\; and\; 4\; arc(s)}$

$\mathrm{HasArc}\left(N\,\left[1\,2\right]\right)$

$\mathrm{true}$

$\mathrm{HasArc}\left(N\,\left[2\,1\right]\right)$

$\mathrm{false}$

$\mathrm{HasEdge}\left(G\,\left[1\,3\right]\right)$

$\mathrm{false}$