I'm glad it was helpful to you! When I was reading through the textbook that I work with, it wasn't too clear to me either - hence the video! Hopefully the other videos that I am making on Decision Maths are helpful to you too. The series is not complete yet, but soonish it will have a video for all of the algorithms from the Edexcel Decision Maths course. See the vidoe description for the (work in progress) playlist
Thx, that's a model of clarity. I taught the previous Decision Maths syllabus several years ago, but this was new to me - a 'redraw-the-network-with-no-intersections' conundrum (if I've understood it aright).
I'm glad it was helpful. It's not a straightforward algorithm so this seemed the best way of presenting it for my students. Yes, you have understood it correctly.
If there are no cycles in your graph, then it is a tree and is thus automatically planar! Drawing the version of the garph in a plane should be quite straightforward.
@@mathshelpwithmrorys8555 No cycle is different to no hamiltonian cycle. Do you mean one should only consider the biggest subgraph where all nodes belong to a hamiltonian cycle?
The Hamiltonian Cycle requirement is for it to be a subgraph (I possibly should have made this clearer within the video). If you cannot find a cycle as a subgraph, you have a tree, and all trees are planar!
Bearing in mind that this is the demonstration of an algorithm for an exam, it is safe to say that the exam paper will not contain a graph that doesn't have a cycle including every vertex! Assuming that you are not doing this for an exam, there may well be other algorithms that are better suited, but my first impression would be to first isolate all vertices where a cycle can be found, determine if that section is planar. If it is not, job done! If it is, carefully consider adding other vertices in, ensuring planarity. If you reach a point which causes the graph to be non-planar when adding vertices back in, check carefully whether another arrangement would solve it. It may be necessary to perform the planarity algorithm on a subgraph as a part of this. Good luck!
Planarity? More like “plain and clear to see”! Thanks for another truly illuminating lecture.
Amazingly clear, thank you. Much better than the textbook I'm going through.
I'm glad it was helpful to you! When I was reading through the textbook that I work with, it wasn't too clear to me either - hence the video! Hopefully the other videos that I am making on Decision Maths are helpful to you too. The series is not complete yet, but soonish it will have a video for all of the algorithms from the Edexcel Decision Maths course. See the vidoe description for the (work in progress) playlist
How did the rest of your class go?
Very clear explanation and easy to follow
Thx, that's a model of clarity. I taught the previous Decision Maths syllabus several years ago, but this was new to me - a 'redraw-the-network-with-no-intersections' conundrum (if I've understood it aright).
I'm glad it was helpful. It's not a straightforward algorithm so this seemed the best way of presenting it for my students. Yes, you have understood it correctly.
Even maths teachers need help :D
Absolute legend
Thank you. I am glad it was helpful to you.
Brilliant. Absolutely brilliant.
Thank you!
Brilliant explanation!
What if there is not Hamiltonian cycle?
If there are no cycles in your graph, then it is a tree and is thus automatically planar! Drawing the version of the garph in a plane should be quite straightforward.
@@mathshelpwithmrorys8555 No cycle is different to no hamiltonian cycle. Do you mean one should only consider the biggest subgraph where all nodes belong to a hamiltonian cycle?
Great video! thank you
You're welcome. I'm glad it was useful for you.
can i ask a question ?
what if the graph don't have a hamaltonian cycle ?
The Hamiltonian Cycle requirement is for it to be a subgraph (I possibly should have made this clearer within the video). If you cannot find a cycle as a subgraph, you have a tree, and all trees are planar!
Could you send to me a search paper about it
And if there isn't a cycle that includes every vertex of the graph/
Bearing in mind that this is the demonstration of an algorithm for an exam, it is safe to say that the exam paper will not contain a graph that doesn't have a cycle including every vertex! Assuming that you are not doing this for an exam, there may well be other algorithms that are better suited, but my first impression would be to first isolate all vertices where a cycle can be found, determine if that section is planar. If it is not, job done! If it is, carefully consider adding other vertices in, ensuring planarity. If you reach a point which causes the graph to be non-planar when adding vertices back in, check carefully whether another arrangement would solve it. It may be necessary to perform the planarity algorithm on a subgraph as a part of this. Good luck!
@@mathshelpwithmrorys8555 Thank you I found a method like this which uses bridges and the same idea and worked out. Thanks!
Could you please tell me the Name of this algorithm ?
The planarity algorithm.
@@FarmYardGaming u doing some last minute A level revision too? 😁
@@christianwilliams2248 For AS exams, but yep, haha
Amazing video
Thank you!
first i thought it would be hard but found it easy