In the euclidean plane an common heuristic for the TSP is to start with the convex hull of the point set and then successively integrate as the next point and insertion position the combination that incurs least tour expansion.

If all points except one are in convex configuration then the greedy insertion strategy yields the optimal tour; even more is true: if we remove (i.e. shortcut) a single vertex and then reinsert it with the greedy insertion strategy, the resulting tour will also be the optimal one.

Let's now interpret the points of a planar euclidean TSP instance as days, expanding a tour via according to the greedy insertion strategy as the passing of the day corresponding to the integrated point, then let your survival depend on *being able* to arrive at the optimal tour with the insertion of the last vertex.

Question:does the above resemble the Unexpected Hanging Paradox because the reasoning is analogous in that we base our induction hypothesis on not having made any mistakes in any prior tour expansion as the analogue of not having been killed till the end of Thursday. The backward induction to the first tour expansion, resp. surviving Monday also seems to follow the same "logic".

And finally isn't the disappointment with the reliability of that kind of backward induction is comparable in both cases?

It should be noted that greedy tour expansion also meets the aspect of unexpected failure, independent of whether or not we make the flawed backward analysis:
suppose we tell someone familiar with heuristic and its bad performance that we are about to solve a very large planar euclidean tsp instance with the greedy tour-expansion heuristic and start with the convex hull, just to hear "*you will surely make a wrong insertion, but you will not know when*"