TITLE: Is Wolfram wrong about unique 3-colorability, or am I just confused?
QUESTION [34 upvotes]: The illustration on Wolfram's page claims to present a uniquely colorable, triangle-free graph. However, this seems to be blatantly false: the graph has a symmetry with respect to a reflection through the horizontal axis, and we can use this symmetry to construct a new colouring not isomorphic to the original one.
Am I missing something obvious here, or is the illustration simply wrong? If it's the latter, what is a simple example of a triangle-free, uniquely 3-colourable graph?

REPLY [45 votes]: Yes, Wolfram is wrong in this case. I just checked the archives of the Journal of Combinatorial Theory (where the erratum to the paper in question is published) and the two top vertices are supposed to be connected by an edge.
I cannot provide a link because it requires a login, and I was able to log in through my university's subscription to the journal.