2
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This puzzle is based on this wonderful puzzle.

A fairy chess piece is placed on an infinitely large chess board with no edges. It can only visit each square once. What is the smallest number of moves it can make that would cause it to become trapped?

I am interested in answers for the following fairy chess pieces:

  • Ferz: moves 1 square diagonally in any direction.
  • Camel: moves 1 square in a horizontal/vertical direction followed by 3 squares in an orthogonal direction.
  • Zebra: moves 2 squares in a horizontal/vertical direction followed by 3 squares in an orthogonal direction.
  • Giraffe: moves 1 square in a horizontal/vertical direction followed by 4 squares in an orthogonal direction.
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7
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I'm gonna go with

7 moves for the ferz (circle around a square and step in)

15 moves for the rest (do the exact same moves as you would with a knight.)

The latter depends on the following characteristics of the move:

  • there are 8 different squares to block
  • any square you can reach in 1 move is impossible to reach in 2 moves
  • if you can reach a square in 1 move, there are several ways to reach it in 3 moves.
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  • $\begingroup$ Wow is it really that simple? $\endgroup$ – Dmitry Kamenetsky Oct 21 at 4:20
7
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I'm pretty sure it is

7,15,15,15

Reasoning:

on an infinite board to get trapped the 4/8/8/8 squares we can legally move to must have been visited. Since all fantasy pieces move in a parity alternating way we need 6/14/14/14 moves to do so plus one last move to enter out prison.

This is indeed possible:

We can visit the neighboring squares, for example, clockwise. They are by definition connected through the prison square, and we can swap the two moves to leave the prison square untouched.

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2
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More direct, visual reasoning for all of the knight-variants, using Camel as the example:

...P....
........
.......Q
..A.B...
........
X.....C.
...O....
X.....X.
........
..X.X...

Starting from A, we need to define the moves to go from A to B, then from B to C, so that we can repeat them through the X's and finally move into O. It is easy to see that we can move from A to B and from B to C in exactly two moves each, not passing through O: by passing through P and Q respectively, which are the mirror images of O by the lines AB and BC.

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