Final answer:
The statement that the galactic shortest-path problem is NP-complete by performing a reduction from the partition problem is true. If every instance of the partition problem can be transformed into an instance of the galactic shortest-path problem, then the latter is confirmed as NP-complete.
Step-by-step explanation:
Proving that a computational problem such as the galactic shortest-path problem is NP-complete often involves showing that another problem that is already known to be NP-complete can be polynomial-time reduced to it.
The claim in the question is true as the partition problem, which is a well-known NP-complete problem, can serve as a starting point for such a reduction. This process is called a reduction because it effectively transforms instances of the partition problem into instances of the galactic shortest-path problem in polynomial time, such that solutions to the new problem give solutions to the original one.
For a problem to be NP-complete, it must satisfy two conditions: it must be in NP (that is, any proposed solution can be verified as correct in polynomial time), and every other problem in NP must be reducible to it in polynomial time.
The partition problem involves deciding whether a given set of integers can be partitioned into two subsets that have equal sums. This problem is already known to be NP-complete, and if one can show that every instance of the partition problem can be transformed into a corresponding instance of the galactic shortest-path problem such that the answer to one implies the answer to the other, then the galactic shortest-path problem would indeed be NP-complete.
A reduction typically constructs a graph which represents the partitions by utilizing a special configuration of nodes and edges, where the shortest path would line up with the equal-sum subsets if a solution to the partition problem exists. If such a reduction can be found and proven to work in polynomial time, then the galactic shortest-path problem would rightfully be categorised as NP-complete as well.