Description: The Königsberg Bridge problem. If G is the Königsberg
graph, i.e. a graph on four vertices 0 , 1 , 2 , 3 , with edges
{ 0 , 1 } , { 0 , 2 } , { 0 , 3 } , { 1 , 2 } , { 1 , 2 } ,{ 2 , 3 } , { 2 , 3 } , then vertices 0 , 1 , 3 each have degree
three, and 2 has degree five, so there are four vertices of odd
degree and thus by eulerpath the graph cannot have an Eulerian path.
It is sufficient to show that there are 3 vertices of odd degree, since
a graph having an Eulerian path can only have 0 or 2 vertices of odd
degree. This is Metamath 100 proof #54. (Contributed by Mario
Carneiro, 11-Mar-2015)(Revised by Mario Carneiro, 28-Feb-2016)(Revised by AV, 9-Mar-2021)