Untitpoi's answer to the first problem covered most of the detail of what I ended up doing.
In terms of my approach to reach the same solution as Untitpoi's answer, this was a little different - rather than working directly with the graphs as given, I instead started drawing a topologically equivalent graph on paper without the mess of crossing over lines. In particular, I started by drawing
the nodes that are connected in triangles on both graphs. For the first one there is only a single triangle, and for the second there is a single triangle, and a chain of 3 triangles each sharing a common vertex.
I then connected further nodes onto that separate graph, looking for unique patterns. In particular, on the first chart, it was easy to notice, as Untitpoi did, that
only node K is connected to two different nodes (D and H) which are themselves connected to the FLI triangle. This breaks the symmetry of the triangle, uniquely identifying node C too.
The rest follows quite simply by labelling the target graph with the possible letters for each node that remains ambiguous.
For the second puzzle, note that
AJF form a triangle, and there are another chain of triangles: BCH-HGN-NED
The second chain uniquely identifies node G without even drawing anything else.
Having identified the first node for certain,
the connected nodes must be H and N, and the nodes connected to those can be differentiated because C and E are not connected to anything else, whereas B and D are both connected to two other nodes.
We can then observe that
B is connected to J and M, which includes one of the nodes of the isolated triangle, whereas D is connected to K and L neither of which is part of a triangle. This uniquely identifies all of J, M, B, C, D, E, H, N and reduces the possibilities for the others with no other observations.
By this point, my graph on the separate piece of paper looked something like this:
A M\_ G _/K
/ \ B\_ / \ _/D_
J---F _/| _H---N_ | \L
\____/ C/ \E
note that extending the (seemingly symmetric) links from B and D broke the symmetry, as one of them connected to triangle AJF.
I expect finishing off from there will be simple... and indeed it was, resulting in the following labelling (starting clockwise from the node labelled with an arrow):
C, M, I, K, H, F, D, J, N, L, B, E, G, A
In order to slightly generalise the technique which worked for both of these:
- Look for triangles on the source graph.
- Draw a new graph where these triangles are connected in a more obvious way, but without the other nodes for now.
- Extend this graph using the nodes connected to the nodes that form part of the triangle, until the symmetry is broken, so that at least some nodes can be uniquely identified (or at least narrowed down to two possibilities).
- Draw circles representing the solution on a new piece of paper, and initally draw just the lines that represent the triangles.
- Add lines to the copy of the solution graph this equivalent to the partial graph from step 3.
- Label the nodes that were identified in step 3.
- Continue to extend both charts as new nodes are identified.
- When you run out of lines to add, and/or nodes to label, you're done!
For a more general case (where we may have a graph consisting entirely of triangles, or with no triangles) step 1 etc. will need to be rephrased to look for some other identifiable characteristic that can be easily recognised.