I definitely prefer proof 1, it's a nice elegant counting argument, love a good bit of combinatorics. While also valid, the construction in proof 2 feels more contrived, the altering of the original graph obscures the argument a little.
Just a little extra detail that I needed to figure out to completely understand and recreate the proof: 2k + l = x + 2y = the number of times a red-blue edge touches a triangle.
At 09:55 when you 'amending' the picture by drawing the curves connecting the bottom right blue circle to two red and one blue circles (kinda random I'd say), why? Shouldn't you draw the curves connecting only the blue-red circles at the bottom edge? Why not connect the right blue circle to the second right red circle at the bottom edge? Shouldn't it be the case that one of the nodes outside the triangle should be considered as the 'starting point' so there would be an odd number of end points of paths?
I definitely prefer proof 1, it's a nice elegant counting argument, love a good bit of combinatorics. While also valid, the construction in proof 2 feels more contrived, the altering of the original graph obscures the argument a little.
This is amazing. Thanks for explaining so well !!!
Thank you so much for this, really intuitive
Thank you very much! It would be great to see how this Lemma is applied 🙂
The first proof seems to be easier to come up with
Very well explained sir, thank you for the video
Just a little extra detail that I needed to figure out to completely understand and recreate the proof: 2k + l = x + 2y = the number of times a red-blue edge touches a triangle.
At 09:55 when you 'amending' the picture by drawing the curves connecting the bottom right blue circle to two red and one blue circles (kinda random I'd say), why? Shouldn't you draw the curves connecting only the blue-red circles at the bottom edge? Why not connect the right blue circle to the second right red circle at the bottom edge? Shouldn't it be the case that one of the nodes outside the triangle should be considered as the 'starting point' so there would be an odd number of end points of paths?