I tried to see this as follows: If we number our glasses 0_7, then either for the four even or for the gour odd ones, there is only one poisoned one. Assume that among {0,2,4,6} only one is poisoned.

Then the other three can be drunk and the sage can pick the central one of then. (We have to think mod 8 since they are arranged circularly)

E.g if 2 is poisoned then the sage drinks from 6.

I don’t know how to solve the extra question without direct verificstion case by case though…

]]>Five different arrangements are possible for 8 glasses.

You can always find in each arrangement 3 alternate safe glasses. A will select the middle one and B and C will select the other two.

The main presented puzzle can then be seen as the base S=2, P=1, T=3 doubled into S=2, P=2+1=3, T=3*2=6. The easier puzzle as the base S=3, P=1, T=4 doubled into S=3, P=3, T=8.

A good question is whether there is any solvable instance harder than the these sequences.

]]>http://sedcore.eu.org/tmp/dig.c

I don’t know what to think. Did I solve it or not? Is it cheating? The computer explored the state space (which is small), then I just post-process what it says. What kind of thinking is this? I wrote the program, yes. But… I don’t know. I’m puzzled.

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