You're doing great work on this channel. I hope you continue to focus on advanced mathematics in the future because that is what's most lacking in the TH-cam-verse.
Tyrone Slothdrop I'm planning on moving into Linear Algebra, then hopefully some Group Theory, Rings, and Fields afterwards. We'll see how things go. Working on a side project at the moment off the channel. Big plans!
A Hamiltonian cycle, also called a Hamiltonian circuit, Hamilton cycle, or Hamilton circuit, is a graph cycle (i.e., closed loop) through a graph that visits each node exactly once.
For the 8:10 is it wrong for the vertex at the last go back to the a just because it is bipartite graph which cannot gone through it's own vertex again?
I'm sorry if you don't know the answer to this question but does the path for cubes have any relation to a Hilbert Curve? To me it feels like there's a lot of similarities between resolving an n-dimensional cube's hamilton cycle and the method of generating a Hilbert Curve. Cheers mate.
Hi! You say around 11:15 that the graph has a Hamilton cycle because Va = Vb. But that isn't sufficient to prove the existence of one I think. What if a bipartite graph with equal independent sets of vertices had a cut edge? I mean, I know the graph you gave has a Ham cycle, but I think it's over-generalizing to say "because Va = Vb" since that's not always true. I could be wrong though, I am new to this.
Yeah, Trev got that one part wrong. A very simple graph with 4 vertices: o -- o -- o -- o, is bipartite and Va = Vb, but the graph clearly does not have a Hamilton cycle. I only know this now because I thought Va = Vb was sufficient to prove the existence of a Hamilton cycle, and lost marks on a quiz! Damn it Trev! Haha
Hi Trev, how come you don't go into more depth for this topic? The textbook has a larger scope and depth, and the questions in MACM 201 go beyond what this intro video teaches. All your other topics were much more exhaustive.
![[Discrete Mathematics] Graph Coloring and Chromatic Polynomials](http://i.ytimg.com/vi/J7g5HMZgKMU/mqdefault.jpg)
![[Discrete Mathematics] Graph Coloring and Chromatic Polynomials](/img/tr.png)
![[Discrete Mathematics] Planar Graphs](http://i.ytimg.com/vi/ETtXKpfBWNI/mqdefault.jpg)
You're doing great work on this channel. I hope you continue to focus on advanced mathematics in the future because that is what's most lacking in the TH-cam-verse.
Tyrone Slothdrop I'm planning on moving into Linear Algebra, then hopefully some Group Theory, Rings, and Fields afterwards. We'll see how things go. Working on a side project at the moment off the channel. Big plans!
A Hamiltonian cycle, also called a Hamiltonian circuit, Hamilton cycle, or Hamilton circuit, is a graph cycle (i.e., closed loop) through a graph that visits each node exactly once.
Prakash Besra what about the initial node? it is visited twice once the cycle is created does it not?
this playlist was very helpful. Thanks a ton!!!
For the 8:10 is it wrong for the vertex at the last go back to the a just because it is bipartite graph which cannot gone through it's own vertex again?
I'm sorry if you don't know the answer to this question but does the path for cubes have any relation to a Hilbert Curve?
To me it feels like there's a lot of similarities between resolving an n-dimensional cube's hamilton cycle and the method of generating a Hilbert Curve.
Cheers mate.
Thanx sir, the bipartite graph result was very helpful..
Did u go to SFU?
Just a total coincidence that the courses match MACM 101 and MACM 201. ;P
LOL macm 201 final in 2 hrs thanks for all your hard work!!
Hi Trev, but does |V_a|=|V_b| guarantee a Hamilton cycle in a bipartite graph? I don't think so.
A bipartite with equal amount of nodes that have same degree I believe guarantee a hamilton cycle
Hi! You say around 11:15 that the graph has a Hamilton cycle because Va = Vb. But that isn't sufficient to prove the existence of one I think. What if a bipartite graph with equal independent sets of vertices had a cut edge? I mean, I know the graph you gave has a Ham cycle, but I think it's over-generalizing to say "because Va = Vb" since that's not always true. I could be wrong though, I am new to this.
Yeah, Trev got that one part wrong. A very simple graph with 4 vertices: o -- o -- o -- o, is bipartite and Va = Vb, but the graph clearly does not have a Hamilton cycle. I only know this now because I thought Va = Vb was sufficient to prove the existence of a Hamilton cycle, and lost marks on a quiz! Damn it Trev! Haha
is hamiltonian cycle and hamiltonian circuit are the same?
Yes, I think..
yup
my combinatorics class does all the proofs :( a little disappointed you didn't cover those.
are hamilton cycle and hamilton path same? plz rply
+abhishek sharma if the starting and ending point are same then it is a cycle otherwise it is a path.
Love you man
cavt tanem dzec
Hi Trev, how come you don't go into more depth for this topic? The textbook has a larger scope and depth, and the questions in MACM 201 go beyond what this intro video teaches. All your other topics were much more exhaustive.
Hi, where can I go into more depth for this graph topic on TH-cam
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