Arithmagons are polygons with a hidden number on each vertex and a given number on each side such that each given number is the sum of the two hidden numbers adjacent to it. These are well-known tasks for primary school.
It is known that for simple (isomorphic to regular) polygons there is always a unique solution when the number of sides is odd. If the number of sides is even, there are infinitely many solutions provided that the alternating sum of the edges is 0; otherwise there are no solutions.
In the book "Thinking mathematically" authors suggest to consider polygons with "a more general arrangement of vertices", in fact a networks or graphs:
Is there really a finite algorithm that allows any such graph problem to answer whether a solution exists or not. And if there is, determine one of them. It is clear from the context that the analysis of system of the linear equations is not assumed.
To begin with, we can assume that the graph is connected; if it is not, we should try to solve each connected component separately. With that in mind, the biggest distinction is already there in the polygon case. We want to know: does the graph contain an odd cycle?
If so, then there is a unique solution for the vertices of that cycle, taking just the cycle's edges into account. Once we have that solution, we can propagate it: for as long as there's an edge $uv$ where $u$ has been given a number but $v$ has not, we can identify a unique value for $v$ to satisfy $uv$'s sum constraint. Finally, once all vertices are labeled, we can check whether all the edge constraints we haven't used yet are satisfied. If they are, we've found the unique soltuion; if they are not, no solution exists.
On the other hand, if the graph has no odd cycles, then it is bipartite: we can find a partition $A \cup B$ of its vertices such that all edges join a vertex in $A$ to a vertex in $B$. In this case, if we have one solution, we get an infinite family of solutions: we get more by increasing the label of every vertex in $A$ by a constant, then decreasing the label of every vertex in $B$ by the same constant.
So in the bipartite case, we are looking for an infinite family of solutions. To find it, pick an arbitrary vertex, and label it with a variable $x$. Then, propagate as before, except that the label we give every other vertex will be a function of $x$. (Specifically, every other vertex will get a label of the form $a+x$ or $a-x$ for some constant $a$.) Again, once we have labeled every vertex, check all the unused edge constraints to make sure that they're satisfied: this will tell us if we have infinitely many solutions or none.