More Gnomes

I recently posted a cute Shapovalov’s puzzle about gnomes. Here is another easier gnome puzzle, also by Alexander Shapovalov.

Puzzle. All gnomes are knights or knaves: knights always tell the truth, and knaves always lie. There is a gnome on every cell of a 4 by 4 chessboard. It is known that both knights and knaves are present in this group. Every gnome states, “Among my neighbors, the number of knaves is the same as the number of knights”. How many knaves are there, if by neighbors they mean orthogonally adjacent gnomes?

The next gnome puzzle has a different author, Alexander Gribalko. Gnomes in this puzzle are not knights or knaves but rather friendly and polite beings.

Puzzle. Nine gnomes repeated the following procedure three times. They arranged themselves on a 3 by 3 chessboard with one gnome per cell and greeted all their orthogonal neighbors. Prove that not all pairs of gnomes greeted each other.


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3 Comments

  1. Mathematical Gnome problem – The nth Root:

    […] I recently posted a cute Shapovalov’s puzzle about gnomes. Here is another easier gnome puzzle, also by Alexander Shapovalov. … (Tanya Khovanova’s Math Blog) […]

  2. Ivan:

    With respect to the first puzzle, I think a more interesting question is “What is the minimal number of knaves on a N x N chessboard, where N>3?” Answering it will not only give an answer to the puzzle, but will also lead to a new OEIS sequence.

  3. JBL:

    So I thought the second one would be a long case-bash, and was pleasantly surprised. Oddly, the fact that it was a chessboard and not any old 3-by-3 grid was helpful. The numbers work out nicely for an n-by-n grid whenever n is 0 or 3 modulo 4 (then n(n + 1)/4 boards gives you a total of exactly n^2 choose 2 pairs of adjacent gnomes, so you are looking to get each pair adjacent exactly once). I would guess that this can be done for n > 3, but finding an example for 16 gnomes and five 4-by-4 boards seems hard by hand.

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