by Big Blue » Wed Aug 31, 2005 10:51 am
...where the trilocation graph method actually does work. (Although only in retrospect, since I know Nick70's solution)
All you need is two triangles and two additional edges from the bilocation graph.
I'll try to draw it in ASCII
o--9--o-----o
|.........\. 3.. |.\
|..........\.6.. |...\
9...........\.9.|..3.\
|...............o...6.\
|.................\..9.|
o..................\ o
The dots should actually be just spaces.
Since this looks very ugly let me explain:
the vertex in the upper left corner is just the cell r5c2
the two edges are from the bilocation graph, with label 9
the two, er, triangles are from the trilocation graph, meaning that at the three vertices only the numbers in its area may be placed (in the particular case always the three numbers 369)
now you have an example for conflicting paths:
start in r5c2 - if 9 is NOT placed there the bilocation path that leads downwards pushes the 9 into r8; the other path pushes the 9 to the tip of the triangle - but that means its other two vertices must be 3 and 6; but these two vertices are also vertcies of the second triangle; therefore, the third vertex of the second triangle must be a 9 - but this 9 is also located in r8, which establishes conflicting paths, and thus the 9 must be placed in r5c2
but I assume that I am just rephrasing geometrically in a roundabout manner what Nick70's multiple forcing chain establishes... or the mysterious colouring method
still, for solving by hand (and if you are good with pictures both in terms of drawing and looking at them) the trilocation methods might be useful
but this is where I draw the line - no quadrilocation graphs with tetraeders for me, please - I am already struggling with 2 dimensions...