Five Sages
Here is a new puzzle by Nikolai Chernyatiev.
Puzzle. Five sages, who all know one another, are blindfolded, seated in a row in a dimly lit hall, and then have their blindfolds removed. Each sage can see both of their immediate neighbors, but no farther; the sages at the ends know that they are at the ends. After that, each sage writes down one of the numbers 1, 2, or 3. The complete information — who wrote which number, in seating order — is then announced to everyone.
Before being seated, the sages may agree on a rule for choosing 1, 2, or 3 based on what they see. After the five numbers are announced, each sage must reconstruct the full left-to-right order of all five sages.
I love puzzles related to information theory, and this is a lovely example. Let’s do a quick sanity check. There are 5! = 120 possible orders of the sages. The announced numbers form a ternary string of length 5, giving 35 = 243 possible announcements. That is more than enough in principle; so far, so good.
AI can produce a possible table of answers, but the resulting strategy is not very inspiring. Fortunately, there is a much more elegant solution based on the following neat fact:
Among any three distinct residues modulo 5, exactly one is the average of the other two. Equivalently, any three vertices of a regular pentagon form an isosceles triangle: one of the three vertices lies on the axis of symmetry of the other two.
But wait: Konstantin Knop proved a much tighter result. Each sage can get away with writing down only one of two numbers. Wow!
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