Spotting the Loophole
For the three grids the coordinates of the vectors are
Solutions are:
Grid 1: Black, Dark blue, pink (only solution)
Grid 2:Black, Red, yellow, Green (only solution)
Grid 3:Black, yellow, red, pink (first
solution)
Dark blue, light blue, red, pink (second
solution)
Example visual proof of uniqueness
for grid 1
Only one arrow points downwards (the black one). Thus the black
arrow must be part of the closed loop, if such a loop exists.
The x-offset of the black arrow is greater than the x offset of all
of the other arrows, thus there must be at least two right facing
arrows in the closed loop. Thus two or three of pink, green and
dark blue must be present. Both dark blue and green togther would
be too high and just pink and green would be too low: thus the loop
contains pink and dark blue. These form a closed
loop.
I like this sort of argument. I could imagine Pythagoras talking in
such a way.