Past exam of the short options course of the University of Oxford / 2021 / 7 / c / ii / Solution

By (i), two simple graph paths with different edge sets have different lengths. In particular two distinct simple graph paths from the fixed root $v$ to the same graph vertex $u$ have different edge sets, so there is a unique shortest graph path from $v$ to each $u$. Positive weights ensure that a shortest walk has no repeated graph vertex: removing a graph cycle strictly reduces its length.

For each $u\ne v$, take the final edge of its unique shortest graph path. Every initial segment of a shortest graph path is itself shortest, since replacing it by a shorter one would shorten the whole route. Thus these edges agree with the graph paths' unique predecessors. Following a predecessor strictly decreases $d_G(v,u)$, so no predecessor graph cycle is possible, and all graph vertices eventually connect to $v$. They form a shortest-path tree. Every other such tree would have to use the same unique root-to-vertex graph paths, provingLiteral wording qualification. “Any two paths” needs a convention. Reversing an undirected graph path preserves its edge set and length; distinct zero-edge graph paths at different graph vertices also have length zero. Thus the unrestricted statement is false if these count as different graph paths. The proved edge-set assertion, and in particular uniqueness for graph paths with the same ordered endpoints, is exactly what is needed for the deduction. Arbitrary walks are not included.