Proof: Connected Graph of Order n Has at least n-1 Edges | Graph Theory
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- เผยแพร่เมื่อ 30 ก.ค. 2020
- A connected graph of order n has at least n-1 edges, in other words - tree graphs are the minimally connected graphs. We'll be proving this result in today's graph theory lesson!
We'll be using a special type of contradiction proof called a proof by minimum counterexample. If we assume that there is a graph contradicting our claim, we can consider one such graph of minimum order, then we will show that this "minimum order counterexample" is actually not a minimum order counterexample, producing a contradiction and proving our claim.
Connected graphs with one less edge than vertices NEVER have cycles, and connected graphs with no cycles are called tree graphs; check out these lesson on them...
Intro to Tree Graphs: • Intro to Tree Graphs |...
Proof that a tree graph of order n has size n-1: • Proof: Tree Graph of O...
Proof that a connected graph with n vertices and n-1 edges is a tree graph: • Proof: Graph with n Ve...
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Minimal counterexample? More like "Magnificent lectures, with quality that's more than ample!"
Didn't get the minimum order part of proof, but luckily i can understand alternate contradiction using same argument.
Thank you very much for your elegant proof.
Consider an undirected graph G that has n distinct vertices where each vertex has degree 2. Assume n≥3. What is the maximum number of circuits that G can contain? If all the vertices of G are contained in a single circuit, what is the maximum number of vertices that can be contained in an independent set?
I don't know what to say
BUT the triangel has 3 sides. Why should be just 2 edges?
Neatly explained 👌
Great proof , thanks a lot for this !!
Good explanation. Hope it saves me for the exams 😅
How do you prove that after removing an end vertex the graph still remains connected
Hello can you make a vdieo on Prufer code
what a clever neat proof this was
I am still confused lol
Hey!!could you pls make a video on the que:
Nice class
dont you think using induction is way easier than this proof?
Really appreciate your videos.