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Asked: 2026-05-23T01:35:40+00:00

I’m struggling with the negamax-algorithm for 3-in-a-row game on a 3×4 (rows x columns)

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I’m struggling with the negamax-algorithm for 3-in-a-row game on a 3×4 (rows x columns) grid. It is played like the well known 4-in-a-row, i.e. the pieces are dropped (NOT like TicTacToe). Let’s call the players R and B. R had the first move, B’s moves are controlled by negamax. Possible moves are 1, 2, 3, or 4. This the position in question which was reached after R: move 3 –> B: move 1 –> R: move 3:

  1   2   3   4

|   |   |   |   |

|   |   | R |   |

| B |   | R |   |

Now, in order to defend against move 3 by R, B has to play move 3 itself, but it refuses to do so. Instead it plays move 1 and the game is over after R’s next move.

I spent the whole day looking for an error in my negamax implementation which works perfectly for a 3×3 grid, by the way, but I couldn’t find any.

Then I started thinking: Another explanation for the behavior of the negamax-algorithm would be that B is lost in all variations no matter what, after R starts the game with move 3 on a 3×4-grid.

Can anybody refute this proposition or point me to a proof (which I would prefer ;-))?

Thanks, RSel

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1 Answer

  1. Editorial Team

    Proof that B3 loses as well:

    B3: R(1,2,4)->R1; B(1,2,4)->B2 (loses), so B1; R(2,4)->R2 Loses, so R4; B(2,4)->B2 loses, so B4; R loses on either choice now
    …so R1 will lose for R – so R won’t choose it.

    B3: R(1,2,4)->R2 loses since B2, so R won’t choose it
    B3: R(1,2,4)->R4; B2 (forced); R2 (forced); B loses on R’s next move

    …so, B3 loses for B as well as B1…so B has lost in this situation.

    EDIT: Just in case anyone is wondering about the other B options (2,4) at the end of “B3: R(1,2,4)->R1; B(1,2,4)->B2 (loses), so B1“…they are irrelevant, since as soon as Red chooses R1, this scenario shows that B (computer) can choose B1 and win. It doesn’t really matter what happens with B’s other choices – this choice will win, so Red can’t choose R1 or he will lose.

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