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Inspired by this question

Here's a quick Knight Checker (a chess variant) puzzle. In case you haven't read the rules to it in the above link, I'll repost them here:

  1. Each player begins with 1 knight on each square on their back rank (the knights move and capture as normal)
  2. To win, a player must move a knight to the other player's back rank (a touchdown) or capture all of the other player's knights.
  3. If a player moves a knight to the other player's back rank, and the other player has a knight guarding that square, then the other player can capture the knight on their next move, preventing the touchdown.

Here's a Knight Checker puzzle I came up with. (It's probably a little easy, because it's my first Knight Checker puzzle. I may do more in the future.)

enter image description here

White to move and touchdown in 3.

Good luck!

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2 Answers 2

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I think this works:

  1. White: Ng6. Black: Nf7 (to protect against touchdown on h8).

  2. White: Nb8. Black: Nxb8.

  3. White: Nf8, touchdown!

The key realisation is that

White must sacrifice one knight in order to draw Black's guarding knight (the one at d7) away from the position it needs to be in to stop White's second knight.

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    $\begingroup$ (+1, and accepted once I can) This was the intended answer, but switching some of the moves is possible. $\endgroup$
    – Excited Raichu
    Oct 29, 2018 at 12:27
  • $\begingroup$ @ExcitedRaichu Yep, after the 1st pair of moves the two white knights are essentially interchangeable: it doesn't matter which of them moves first. $\endgroup$
    – Rand al'Thor
    Oct 29, 2018 at 12:28
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Another solution, besides Rand al'Thor's accepted one, involves

Zugzwang

White can play

1.Ng6! Nf7 (otherwise 2.Nh8 touchdowns) 2.Nc7! , not sacrificing anything.

After which

Black, on move, is zugzwanged: each of his knights must stay in place to prevent touchdown.

Indeed

If Nb6 moves, then 3.Na8 ; If Nd7 moves, then 3.Nf8 ; If Nf6 moves, then 3.Ne8 ; If Nf7 moves, then 3.Nh8

With touchdown in each case.

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    $\begingroup$ Nice answer, this makes me think there's a harder variant puzzle $\endgroup$
    – theREALyumdub
    Oct 29, 2018 at 18:00

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