Proof: Forest Graphs have n-k Edges | Graph Theory

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A forest of order n with k components has size n-k. This is a generalization of the result that tree graphs with n vertices have n-1 edges. We prove this generalization in today's graph theory lesson!

A forest is an acyclic graph. It is like a tree graph except it does not need to be connected, thus its components are trees.

We rely on the obvious fact that the order of a graph is the sum of the orders of its components (recall that the order of a graph is the number of vertices it has), similarly the size of a graph (its number of edges) is the sum of the sizes of its components. We also use the result that a tree of order n has size n-1.

I hope you find this video helpful, and be sure to ask any questions down in the comments!
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n-k? More like "Magnificent lectures that are much more than okay!" 👍

PunmasterSTP

Sir, you just saved me. great explanation!!

JustXPriest

Great explanations. Do you have a video on "a simple graph with n vertices and k components can have atmost (n-k)(n-k+1) edges".

abrarahmadwalicsek

Please can do a video on the question below for me. I appreciate and thanks

Given a connected weighted graph G and a vertex u, Prim’s algorithm runs as follows. Initially
let S = {u}. Let F = {u}. As long as S ̸= V (G), we do the following: among all edges in G
that have one endpoint in S and the other endpoint in V (G) \ S, we choose one, say e = xy with
smallest weight, where x ∈ S, y /∈ S. Let S = S ∪{y} and F = F ∪e. Prove that the algorithm does
terminate with S = V (G) and that the final F obtained at the end of the algorithm is a minimum
weight spanning tree of G. (comment: so, first you will need to argue that F is a spanning tree of
G and second show it has smallest weight among all spanning trees of G.)

rexfordboakye

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