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Grandmaster Daniel "Danya" Naroditsky was well known for his excellent chess commentary, online streams, and educational content. One chess concept Danya enjoyed introducing to his viewers was the idea of a collinear move, a term coined by John Nunn. Given a pair of long-range pieces (e.g., 2 bishops or 2 rooks) of opposite color that attack each other, a collinear move moves one of them to any empty square along the line of attack, either toward or away from the enemy piece, without capturing any pieces.

In the 5 chessboards below, the arrows show the ranges of collinear moves currently available for both sides. Two players, each controlling pieces of one color, play a game using these boards. Beginning with the player with the white pieces, the players take turns making a collinear move with one of their pieces on one of the boards. A player with no collinear moves left loses. Does either player have a winning strategy?

A chessboard with a white rook on d2, a black rook on d7, white bishops on d8 and h5, and black bishops on d1 and h4.

A chessboard with white rooks on a4 and h8 and black rooks on a8 and h4.

A chessboard with white rooks on a5 and h5, black rooks on a8 and h8, a white bishop on f3, and a black bishop on h1.

A chessboard with white bishops on a8 and h8, black bishops on a1 and e4, and a black knight on f3.

A chessboard with white rooks on a1 and h4, black rooks on a4 and h1, white bishops on a5 and e8, black bishops on d8 and h5, and a white knight on e1.

Hint:

If they played 5 separate games each restricted to an individual chessboard, the player with the white pieces would win on 4 boards, and the player with the black pieces would win on 1 board.

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  • $\begingroup$ To clarify the rules — in the third board, 1 Rhh1 is illegal? $\endgroup$ Commented Oct 27 at 19:27
  • $\begingroup$ @msh210 Correct, as long as h1 is occupied, because captures are not allowed. $\endgroup$ Commented Oct 27 at 19:30
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    $\begingroup$ A bit unrelated but I never thought his tragic news would have reached the shores of PSE… this is a testament to his influence… I was a big fan of his content on YouTube and he seemed like a kind hearted humble man. Nice to see a tribute puzzle for him! $\endgroup$ Commented Oct 27 at 21:07
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    $\begingroup$ From all I can tell he was an outstanding person in every respect. Everyone dies too soon but him more than most. $\endgroup$ Commented Oct 28 at 0:24
  • $\begingroup$ Since there haven't been any answers, I'll say I've made progress in finding Nim values for many components by bypassing reversible moves to treat them as impartial. But some parts are tricky and elude me, like both A's. $\endgroup$ Commented Oct 29 at 5:21

2 Answers 2

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I believed it is a draw. (but may well be missing something)
CORRECTION: I believe white wins

Observations:

Any 2 non-interacting pairs on equal distance:
The winner should only play responsively
- if the opponent moves away: move the same distance toward the moved piece to push it back
- if the opponent moves closer: move the other pair the same amount closer

Board 2: the second player wins The winner should only play responsively on that board; it can be ignored
- if the opponent moves closer: make a rectangle again
- if the opponent moves away: treat it as 2 non-interacting pairs on equal distance

(without further proof) This makes the puzzle equal to winning on the following 3 boards:

Analysis:

reduced problem
Board 1 starts with distances {6,4,1}
- As soon as 2 distances are the same, one can win by reducing the third to 0
- A first player win is ->{5,4,1}->{3,2,1}
Board 2 wins by using the horse-block: h2, a2 (or?); b2
Board 3: black starting loses, but white starting is a draw!

CORRECTION!

I forgot a pair; the lone distances should not be {6,4,1} but {6,4,3,1}
This is a losing position:
- making first double looses (opponent makes second double and plays symmetrically)
- making 3210 wins (opponent must make double)
- making third<4 looses (opponent makes 3210)
- since >3 may not be made <=3 6431 is equivalent to 2031 and first player looses

Corrected strategy:

For white to win, since 1 and 3 are losing, it can prioritize board 2 and move the rook h4-h2 for a losing position there. The third board is a win for white if black starts; because black does not have the room to retreat (to fore a draw) like in the old strategy.
a87 h6 a6 h7 a7 a6 g7 h5! h7 h6 h8 h7 a7 b7, and black cannot move to a9 to gain initiative.

Old Strategy:

For white to force a draw:
If white prioritizes board one and black board 2 we can end up as below, with white to move. position at final repeating stage
On board 3 follows repeatedly:
hb6, a8; a6, a87; a5, a6; h6, g6; h4 h76; h5 h7; h6 a6;
(or a similar infinite sequence)

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  • $\begingroup$ Great analysis! Can you double check the bishop pairs? I think you might be missing one. $\endgroup$ Commented Nov 13 at 5:11
  • $\begingroup$ @noedne You are correct, and this changes the outcome :-). edited $\endgroup$ Commented Nov 13 at 8:11
  • $\begingroup$ Congrats, you've mastered collinear moves! An easier way to analyze the set of 4 lone distances is by recognizing that it's a game of Nim. $\endgroup$ Commented Nov 13 at 21:34
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This is a partial answer, with a few assumptions.

Black will win with perfect play. Each board is either winning or losing. A board is losing if playing first on it would mean that your opponent would win. Regardless of position.

Thus:

The starting configurations of the boards are [][][][][L] Note these boards are in no particular order and rely solely on the hint Thus white can go on board 5, so black will reply on 5 and thus it will be presented with the same position. So white must change the state to [L][][][][L]. Black goes on another board [L][L][][][L] and white must again play on a winning board or play till they can't on a losing board. So eventually white will play [L][L][L][][L]. Then black completes the set to [L][L][L][L][L]. So whenever white goes, black will go on the same board until black has won.

Still to be shown:

whether the starting assumption is true, the [L] board and the optimal strategy for each board

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  • $\begingroup$ This is an interesting try! When white plays on a winning board, will it always change to a losing board? $\endgroup$ Commented Nov 10 at 20:28
  • $\begingroup$ The assumption is that if a board isn't losing for the next player it is winning for them, so if they move correctly it becomes losing for the next player. Usually black. The major assumption is any losing position is losing for both and winning is winning for both. It only matters who plays next. $\endgroup$ Commented Nov 10 at 21:35
  • $\begingroup$ On second thoughts you could be right, probably another assumption to prove. (It more likely a case to consider) $\endgroup$ Commented Nov 10 at 21:38

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