MathLabs

Worked solution: Euler's degree argument (1736)

Step 5 of 6: From the pairing argument to Euler's exact criterion
In plain words

The pairing argument just given proves one direction outright: an Eulerian trail can never have more than two odd-degree landmasses, since only its start and finish can end up with a leftover bridge. What takes more work — and what Carl Hierholzer nailed down rigorously in 1873 — is the reverse direction: that having 00 or 22 odd-degree vertices is not just necessary but actually enough to guarantee such a trail exists, with a construction that never leaves a bridge stranded.

#{v∈V:deg⁡(v) mod 2=1}∈{0,2}\#\{v \in V : \deg(v) \bmod 2 = 1\} \in \{0, 2\}
Detailed analysis

The previous step already establishes the 'only if' half of Euler's criterion directly: if a connected graph admits a walk crossing every edge exactly once, then the number of vertices with odd degree is either 00 (a closed circuit, if the walk returns to its start) or exactly 22 (an open trail, whose two odd-degree vertices must be its start and its finish).

The converse — that 00 or 22 odd-degree vertices is also sufficient, for a connected graph — is the harder half. Euler asserted it in his 1736 paper without a complete proof; Carl Hierholzer supplied a full, constructive argument in 1873, showing how to build such a walk by stitching together closed loops (Biggs, Lloyd & Wilson 1976, Ch. 2).

Together the two directions give an exact test — #{v∈V:deg⁡(v) mod 2=1}∈{0,2}\#\{v \in V : \deg(v) \bmod 2 = 1\} \in \{0, 2\} — that applies to any connected graph whatsoever, which is exactly why this is a genuine theorem settling every possible bridge layout, and not merely an observation about one city.

Knowledge used in this step
Common mistake. An even total-degree sum (the handshaking lemma from two steps ago) holds automatically for every graph and proves nothing by itself; the sharp criterion needs the actual count of individually odd-degree vertices, which is a much stronger requirement than the sum merely being even.