Alice and Bob are playing the following game: They have a 4x4 empty grid and take turns coloring one square each, starting with Alice. Assume that both players will colour the squares in with the same colour. Whoever completes any $2\times2$ area on the grid after having made their move, is the loser.

Is there any winning strategy for either of the two players?

  • $\begingroup$ This was flagged as off-topic; however I think adding the above clarity will resolve this issue. I have therefore edited the question per @philip's comments below to reflect that he intended both players to colour in squares with the same colour. $\endgroup$
    – El-Guest
    Commented Aug 16, 2018 at 14:52
  • 3
    $\begingroup$ The same question was posted on a different stack by (apparently) a different user one hour earlier: math.stackexchange.com/q/2884764/5676 $\endgroup$ Commented Aug 16, 2018 at 17:37
  • 2
    $\begingroup$ You also posted the answer seen on the math.se question here. I pointed out a mistake, then you posted that exact same comment on the math.se answer, then deleted your answer here. That makes three things you've plagiarized so far. $\endgroup$
    – Riley
    Commented Aug 16, 2018 at 23:22

2 Answers 2


First of all, I'm assuming that Bob and Alice are painting with different colors and that we are only concerned with a $2\times 2$ square of the same color. Now I realize that I may have interpreted the problem incorrectly.


can always force a win or draw with the following strategy:

Whatever square Alice colors, Bob should color the square symmetric with respect to the origin (rotate 180 degrees). This way, Bob can never complete a $2\times 2$ square unless Alice has just completed a rotated version of that square. In fact, Bob can win/draw this way on any $2n\times 2n$ grid.

  • $\begingroup$ this is a draw, not a win $\endgroup$
    – JMP
    Commented Aug 16, 2018 at 14:02
  • $\begingroup$ This is wrong; What if Alice does the middle 2 squares of the second row? Then Bob would be the one to finish it $\endgroup$ Commented Aug 16, 2018 at 14:04
  • $\begingroup$ @PotatoLatte; so they use the same colour then? $\endgroup$
    – JMP
    Commented Aug 16, 2018 at 14:06
  • $\begingroup$ I misunderstood the problem, assuming that Alice and Bob are painting with different colors, and we are only concerned with $2\times 2$ squares of the same color. $\endgroup$
    – Riley
    Commented Aug 16, 2018 at 14:09
  • $\begingroup$ Well I didn't write the problem; So this may be right $\endgroup$ Commented Aug 16, 2018 at 14:11

Bob wins if

both Alice and Bob only colour the edges. Then Alice is eventually forced to play in the middle.

If Alice colours an inner cell for the first time:

Bob colours the diagonally opposite, and reverts to the edge strategy ignoring the extremal corners of the extremal sub-grids with inner cells. We either end up with a saturated grid (Bob wins), or Alice colours one, and so Bob colours the opposite corner with the same effect.

Alice colours a second inner cell:

Bob colours the opposite extremal corner, and the same strategy gives Bob the win.


Bob always wins.


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