Every tree is a connected directed acylic graph.
Tree (since tree is connected acyclic graph)
A graph becomes a tree when it is connected and acyclic, meaning there are no loops or cycles present. Additionally, for a graph with ( n ) vertices to be a tree, it must contain exactly ( n-1 ) edges. This structure ensures that there is exactly one path between any two vertices, fulfilling the properties of a tree.
n * (n - 1) / 2 That would ignore the "acyclic" part of the question. An acyclic graph with the maximum number of edges is a tree. The correct answer is n-1 edges.
dichotomous key can be improved when by changing the tree structure into a directed acyclic graph
A tree is a connected graph in which only 1 path exist between any two vertices of the graph i.e. if the graph has no cycles. A spanning tree of a connected graph G is a tree which includes all the vertices of the graph G.There can be more than one spanning tree for a connected graph G.
In a Directed Acyclic Graph (DAG), the longest path is the path with the greatest number of edges between two vertices, without forming a cycle.
Tree is directed, cycle-less, connected graph.
Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!
Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!
The longest path in a directed acyclic graph is the path with the greatest total weight or distance between two vertices, without repeating any vertices or going in a cycle.
n - 1
n-1