A well-known problem in graph theory is the Seven Bridges of Königsberg. In Leonhard Euler's day, Königsberg had seven bridges which connected two islands in the Pregel River with the mainland, laid out like this:
And Euler proved that it was impossible to find a walk through the city that would cross each bridge once and only once. And more generally, that an Eulerian path does not exist for a graph with more than two nodes of odd degree.
World War 2 changed the topology of the city by destroying two of the bridges. (It also brought other changes such as the transfer of the territory from Germany to Russia, but this is not topologically relevant.) This lead to the similar but solvable problem of the Five Bridges of Kaliningrad.
I doubt that the British and Soviet air forces made creation of an Euler walk a priority when they were conducting their attacks. But if they had, one would have to criticize their inefficiency, because they could have achieved the same objective by bombing only one bridge:
Generalizing the problem:
What is the minimum number of edges that need to be removed from a graph so that the remaining edges form an Eulerian path?
My first conjecture was that it would be half the number of “extra” odd degree nodes. However, a simple X-shaped graph provides a counterexample: There are 4 odd nodes (and 1 even one), but two edges need to be removed, not just one.