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Similar to How Many Knights on a Chess Board? however speedy knights move the way the knight moves, then the way the knight moves AGAIN. However they cannot backtrack to their starting position. How many can you place on a board so to as none are being threatened.

For example, this 1 speedy knight, attacks all these kings:

enter image description here

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  • $\begingroup$ Do they threaten the square they move to on their first move, or only the one they move to on their second move? $\endgroup$
    – Miniman
    Commented Oct 13, 2014 at 22:14
  • $\begingroup$ @Mini I misunderstood. They do not threaten their "first move" they basically move twice the distance of knights (4 up/down, 2 right/left, or, 4 right/left, 2 up/down) $\endgroup$
    – warspyking
    Commented Oct 14, 2014 at 0:01
  • $\begingroup$ @Mini But they also move according to the first knight move... $\endgroup$
    – warspyking
    Commented Oct 14, 2014 at 0:04
  • $\begingroup$ @Mini Hold on I'll add a picture to help explain. $\endgroup$
    – warspyking
    Commented Oct 14, 2014 at 0:04
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    $\begingroup$ A regular knight cannot move to a square with a same coloured piece in it, so you need to explicitly state that the half-way square doesn't have to be empty. Otherwise, you could fit 64 non-threatening speedy knights pretty trivially. $\endgroup$
    – Bass
    Commented Oct 18, 2018 at 6:32

3 Answers 3

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I managed to fit 12 speedy knights onto the board. These 12 attack every other square on the board, so I think it's unlikely that more than 12 will fit.

12 Speedy Knights

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8
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I believe I can fit 8 speedy-knights on a board. Indeed, obviously you'd have at least of one colour. Then you'll notice that there's always that pesky diagonal that a knight has trouble getting to:

enter image description here

Let me know if this is what you meant.

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  • $\begingroup$ It works! I'll give others a while to get 9 or more, if they can't you get Best Answer/Accepted Answer! $\endgroup$
    – warspyking
    Commented Oct 14, 2014 at 0:43
  • $\begingroup$ No problem! I really enjoy these puzzles. $\endgroup$
    – warspyking
    Commented Oct 14, 2014 at 0:53
  • $\begingroup$ @warspyking how about ticking the answer :p $\endgroup$
    – d'alar'cop
    Commented Oct 18, 2014 at 14:14
  • $\begingroup$ Done! Obviously nobody could beat 8. I have expected someone to post 2, one on each colour XD $\endgroup$
    – warspyking
    Commented Oct 18, 2014 at 14:21
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I can prove that 12 is the best you can do:

First, let's agree to call a move of form $(a, b)$ a move that is $a$ vertical steps and $b$ horizontal steps, or vice-versa. So, for example, a standard knight move is $(2, 1)$.

Step 1: Notice that a speedy knight can make all moves $$(2, 0), (3, 1), (3, 3), (4, 0), (4, 2),$$ anywhere on the board. Also, a speedy knight can make any move $$(1, 1)$$ unless one endpoint of the move is in the corner -- I'll call this the corner exception.

In other words, if a move $(a, b)$ has $a$ and $b$ at most $4$, and same parity, then it is legal unless it is $(2, 2)$, $(4, 4)$ or the corner exception.

Step 2: By symmetry it's enough to show that there can be at most $3$ knights on light-colored squares in the bottom half of the board. (Notice light- and dark- colored squares don't interact.) Let's agree to number the columns 1 through 8.

Step 3: Consider two knights on light-colored squares in the bottom half of the board. The vertical distance between them is at most 3. I claim the horizontal distance between them can't be 0 (same column), 1 (adjacent columns), 3, or 4, except for the corner exception. For example if the horizontal distance was 3 then it would either be a $(3, 1)$-move or a $(3, 3)$-move between them, and either way they are attacking each other. The other cases are easy to check.

Step 4: Note that the corner exception only occurs in columns 7 and 8. (Remember, we're looking at light-colored squares on the lower half of the board.)

Step 5: We're almost done! Notice there can be at most: -one knight in columns 1, 2, 5 -one knight in columns 3, 6, 7 -one knight in columns 4, 8. Hence, at most three knights total. The proof is complete.

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