Support the production of this course by joining Wrath of Math as a Channel Member for exclusive and early videos, original music, and upcoming lecture notes for the graph theory series! Plus your comments will be highlighted for me so it is more likely I'll answer your questions! th-cam.com/channels/yEKvaxi8mt9FMc62MHcliw.htmljoin Graph Theory course: th-cam.com/play/PLztBpqftvzxXBhbYxoaZJmnZF6AUQr1mH.html Graph Theory exercises: th-cam.com/play/PLztBpqftvzxXtYASoshtU3yEKqEmo1o1L.html
thank you so much for your videos! I want to kindly request a video on how to easily find (and draw) K33 or K5 configuration in the nonplanar graphs ! :) hope you have a good day!
No problem, glad to help! Thanks for watching and let me know if you ever have any questions. Check out my graph theory playlist if you haven't! th-cam.com/play/PLztBpqftvzxXBhbYxoaZJmnZF6AUQr1mH.html
Thanks for watching and if G is planar, then this result applies to each of G's components, so in fact such a graph would have at least k vertices of degree 5 or less.
suppose , i will draw a regular hexagon and a vertex in the middle and join all the vertices to the middle one , now i have a vertex with degree 6, and planar .. can u please explain this ?
Thanks for watching and for your question! We have to be careful to make sure we understand the result correctly. We haven't proved that all vertices of any planar graph have degree five or less - that is not true, as your example shows. We proved that every planar graph HAS a vertex of degree 5 or less. In the example you gave, every vertex has degree 5 or less except for the one in the middle. Does that help?
Thanks for watching and I am not sure what your question is. Are you asking for a graph whose vertices all have degree 5? A complete graph on 6 vertices fits that condition.
Thank you! Check out my graph theory playlist if you're looking for more! th-cam.com/play/PLztBpqftvzxXBhbYxoaZJmnZF6AUQr1mH.html Let me know if you ever have any video requests!
Does this just only prove that you cannot have a graph with delta>=6; but at the same time, don't prove that a graph must have a vertex of degree 5 or less. I mean, the same logic could be used to prove that you can't have a graph with delta>=7 but this dont prove that you can have a planar graph with minimun degree of 6
this is exactly what i want to ask too, @WrathOfMath can you shed some light on this? We just said graph cannot have minimum degree >=6, nowhere its said 6 is least minimum degree from which graphs become non-planar.
Thanks for watching and indeed, every planar graph has a vertex of degree 5 or less, but there is nothing stopping them from having other vertices with a degree greater than 5. Consider a star graph with one center vertex that is adjacent to 100 other vertices!
Support the production of this course by joining Wrath of Math as a Channel Member for exclusive and early videos, original music, and upcoming lecture notes for the graph theory series! Plus your comments will be highlighted for me so it is more likely I'll answer your questions!
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Graph Theory course: th-cam.com/play/PLztBpqftvzxXBhbYxoaZJmnZF6AUQr1mH.html
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Dude that shirt is absolute fire
Initially I thought the way to prove this would be with contradiction, but I see how nicely contraposition works out here!
thank you so much for your videos! I want to kindly request a video on how to easily find (and draw) K33 or K5 configuration in the nonplanar graphs ! :) hope you have a good day!
Thanks! ❤
Thank you!
literally the best math videos on youtude
Thanks so much, Shizhe - I do my best! Let me know if you ever have any questions!
Great lesson, Can we prove the same using Contrdiction
isn' t (m
Hello my friend, I need cite this result. Which article is the correct? Thank you.
It's easy to understand! Thx so much!
No problem, glad to help! Thanks for watching and let me know if you ever have any questions. Check out my graph theory playlist if you haven't! th-cam.com/play/PLztBpqftvzxXBhbYxoaZJmnZF6AUQr1mH.html
@@WrathofMath Thx a lot !
how to prove this result for if G has k connected components?
Thanks for watching and if G is planar, then this result applies to each of G's components, so in fact such a graph would have at least k vertices of degree 5 or less.
@@WrathofMath Thanks a lot.
suppose , i will draw a regular hexagon and a vertex in the middle and join all the vertices to the middle one , now i have a vertex with degree 6, and planar .. can u please explain this ?
Thanks for watching and for your question! We have to be careful to make sure we understand the result correctly. We haven't proved that all vertices of any planar graph have degree five or less - that is not true, as your example shows. We proved that every planar graph HAS a vertex of degree 5 or less. In the example you gave, every vertex has degree 5 or less except for the one in the middle. Does that help?
thanks dude ❤️
Answered my question, thank you🙏🙏
You're welcome, I am glad it helped and thanks for watching!
Sir,Can u draw a graph for the degree of vertices is 5...
Thanks for watching and I am not sure what your question is. Are you asking for a graph whose vertices all have degree 5? A complete graph on 6 vertices fits that condition.
@@WrathofMath thanks for ur reply sir.
Great video.
Thank you! Check out my graph theory playlist if you're looking for more! th-cam.com/play/PLztBpqftvzxXBhbYxoaZJmnZF6AUQr1mH.html
Let me know if you ever have any video requests!
nice brother
Thank you!
Does this just only prove that you cannot have a graph with delta>=6; but at the same time, don't prove that a graph must have a vertex of degree 5 or less. I mean, the same logic could be used to prove that you can't have a graph with delta>=7 but this dont prove that you can have a planar graph with minimun degree of 6
this is exactly what i want to ask too, @WrathOfMath can you shed some light on this? We just said graph cannot have minimum degree >=6, nowhere its said 6 is least minimum degree from which graphs become non-planar.
But there still some cases in which some vertices have degree greater than 5 and some have less than 5
Thanks for watching and indeed, every planar graph has a vertex of degree 5 or less, but there is nothing stopping them from having other vertices with a degree greater than 5. Consider a star graph with one center vertex that is adjacent to 100 other vertices!
Sean baba kral video
Legend
Thanks for watching!
draw it!!!!!!!!