Worked solution: Euler's degree argument (1736)

Step 6 of 6: Königsberg has four odd-degree vertices, so no such walk exists

In plain words

Plug Königsberg's own bridge-counts into the rule just proved: check how many of its four landmasses have an odd number of bridges, and compare that count against the only allowed values, $0$ or $2$ .

\#\{v \in V : \deg(v) \bmod 2 = 1\} = 4 \notin \{0, 2\}

Detailed analysis

All four vertices $A, B, C, D$ of the Königsberg graph have odd degree ( $5, 3, 3, 3$ are all odd), so exactly $4$ vertices are odd — neither $0$ nor $2$ .

By the criterion of the previous step, this means no walk, starting anywhere and ending anywhere, can cross each of the seven bridges of Königsberg exactly once. Euler's 1736 paper, which settled the question negatively by exactly this kind of counting argument rather than by exhaustively trying routes, is usually credited as the founding argument of graph theory.