I'm a lil confused still-how can we assume that the chains will always contain two colours (orange and blue, or yellow and pink for eg)? Mightn't things be different if the chains had more colours? Also, why does the cycle of blue and orange vertices (left side) act as a barrier to yellow-pink chain? Couldn't (say) a pink vertex connect to (say) a blue dot as well?
Great question! Think about it this way: Start with a blue vertex. If it touches any orange vertices, count that as part of the chain. Then if those count any blue vertices, consider that part of the chain. And so on. Now, once you have a blue/orange chain, yellow and pink vertices can connect with it, but no yellow/pink chain can pass *through* it. Does that help?
Thank you so much for such a prompt reply-that makes a lot more sense! So the diagrams you drew are a kind of simplifications that only show the colours we care about. Am I right in understanding that say I have a sequence of blue-yellow-pink-orange connected in a single line, I'd still count that as a blue-orange chain? But I don't quite get why no yellow/pink chain can pass through another chain. I don't see why the line you drew at 13:53 is invalid, because wouldn't the graph still be planar? Also, if I keep the line from 13:53, and at 13:58 I connect the newest pink dot to the blue dot with an "x" on it, since they're connected through a vertex, isn't that still planar as well? Why isn't this allowed?
Thanks 4 ur lecture video plz. Clarify if we have only one chain and except the middle vertex and the chained vertices all have degree 1 then still we can not hv any free color to color the middle one
This is the most excellent explanation of this theorem I have ever seen. Very well done!
That's it? I always thought the proof of the 5CC was more complicated. Thank you for setting me straight.
It's Idaho, not Iowa.
why cant all the countries come together, regardless of their color? :-(
Your teaching is colorful and detailed. Thank you so much for making it easier to understand.
This was amazing! Thank you for making this so easy to understand.
The proof was a really brilliant one!
How do know that when you swap colors in the connected chain you aren't undoing a prior coloring?
Thank you so much. What a clear, beautifully presented explanation.
This was extremely lovely as well as lively.
clean explanation thank youuu
I'm a lil confused still-how can we assume that the chains will always contain two colours (orange and blue, or yellow and pink for eg)? Mightn't things be different if the chains had more colours? Also, why does the cycle of blue and orange vertices (left side) act as a barrier to yellow-pink chain? Couldn't (say) a pink vertex connect to (say) a blue dot as well?
Great question! Think about it this way: Start with a blue vertex. If it touches any orange vertices, count that as part of the chain. Then if those count any blue vertices, consider that part of the chain. And so on. Now, once you have a blue/orange chain, yellow and pink vertices can connect with it, but no yellow/pink chain can pass *through* it. Does that help?
Thank you so much for such a prompt reply-that makes a lot more sense! So the diagrams you drew are a kind of simplifications that only show the colours we care about. Am I right in understanding that say I have a sequence of blue-yellow-pink-orange connected in a single line, I'd still count that as a blue-orange chain?
But I don't quite get why no yellow/pink chain can pass through another chain. I don't see why the line you drew at 13:53 is invalid, because wouldn't the graph still be planar?
Also, if I keep the line from 13:53, and at 13:58 I connect the newest pink dot to the blue dot with an "x" on it, since they're connected through a vertex, isn't that still planar as well? Why isn't this allowed?
Beautifully explained... Amazing Thank u so much ❤
Very cool and simple explanation dude
Amazing explanation
Thanks 4 ur lecture video plz. Clarify if we have only one chain and except the middle vertex and the chained vertices all have degree 1 then still we can not hv any free color to color the middle one
Thankyou for making this clear
Great explanation!
amazinggg thank you soo much
Amazing
little complicated idea. but got it at last
👍👍
Same proof as my graph theory course
Where are you studying graph theory, if I may ask?
Bologna, Italy (unibo)
same here, studying bioinformatics in Leipzig, Germany and this series helps me a lot with the upcoming exam!
thanks
Idaho 😭
its booring because need for clear concept
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