§ Euler tours

§ Tucker's proof: undirected graph with all even degree has an euler tour

I find this proof much more intuitive because it's extremely clear where the even condition is used.
  1. For each vertex vv, arbitrarily pair edges incident at vv to get chains u --- v --- w.
  2. Connect chains to get cycles.
  3. Find a spanning tree of cycles to get an euler tour, where we have an edge between two cycles if they share a common vertex.
It's super clear why we need all vertices to be even degree; You can't pair up vertices otherwise!

§ References