Your argument for case 1 is not true. After alternating the path starting at vi and ending at u the color red is seen by u, but now the blue color is not seen, and hence e1 can now be colored blue and eo be colored orange. Hence you are done.
+Tilde My Larsen Thank you for this comment, I got really hung up when he said u doesn't have an incident red edge when he literally just drew an incident red edge. Otherwise great presentation. Well aside from the fact two vertices are labeled v2 by the end.
In the final case, what if there was some edge from v3 which had the color blue. you cannot recolor the red edge ending at v3 with blue !! If v3 did not have a blue edge, it would have been the vi vertex !
+mohit bhura Then v3 is not the end of the red-blue path and a different case arises, where the end vertex is not v3 (can be either one of three cases.) It only ends in v3 when v3 does not have color blue.
graph theory can frustrate even the best brains. some problems are so tough. worst thing is doing a course in graph theory when the instructor gives tough problems in exams.
Your argument for case 1 is not true. After alternating the path starting at vi and ending at u the color red is seen by u, but now the blue color is not seen, and hence e1 can now be colored blue and eo be colored orange. Hence you are done.
+Tilde My Larsen Thank you for this comment, I got really hung up when he said u doesn't have an incident red edge when he literally just drew an incident red edge. Otherwise great presentation. Well aside from the fact two vertices are labeled v2 by the end.
Thank you for your comment, you literally saved the proof.
@@HHHH-vy5zv no prob. It is always Nice to contrubute
Thank you very much for the clear proof! Much easier to understand than the Diestel's one.
In the final case, what if there was some edge from v3 which had the color blue. you cannot recolor the red edge ending at v3 with blue !! If v3 did not have a blue edge, it would have been the vi vertex !
+mohit bhura Then v3 is not the end of the red-blue path and a different case arises, where the end vertex is not v3 (can be either one of three cases.) It only ends in v3 when v3 does not have color blue.
I love your presentation. thanks sir.
graph theory can frustrate even the best brains. some problems are so tough. worst thing is doing a course in graph theory when the instructor gives tough problems in exams.