# Finding a winning strategy $4\times 4$ grid game

Can someone please help me to come up with a winning strategy for a 4 x 4 grid game with the use of 16 counters.

Players take turns to place either 1,2 or 3 counters on the grid (one counter per square). If 2 or 3 counters are put on any one turn, then they must form an unbroken line (horizontally, vertically or diagonally). The loser is the player who places the last counter on the grid.

2 players

• This is an impartial game, which will yield to the Sprague-Grundy theorem. With only $2^{16}$ states (less if you exploit symmetry) a computer could make a game tree easily. Whether there is a human learnable strategy is hard to guess. – Ross Millikan Oct 24 '16 at 18:26

Well, the added rule of the unbroken line could make it more difficult. Normally, you would always have an advantage of going first because you will always be able to force taking four in total, and then win in the end (if you take three in the beginning, there are 13 squares left, so no matter how the opponent plays you can always sum up to four-> e.j., 1+3, 2+2, 3+1). If the board is laid out like so:

    ABCD
EFGH
IJKL
MNOP


You can clearly see there are at most 5 configurations for the 3-in-a-row, so it would be very difficult for your opponent to take them all. The only problem is in the worst case scenario: if the opponent tries to seal you off, you might not be able to complete three in a row. Example: You play first:

    AXXX
EFGH
IJKL
MNOP


For ease of operation opponent will always play 3 except the last round (he is "Z")

    AXXX
ZZZH
IJKL
MNOP


your turn again [take corners to prevent being forced to play only 1]

    AXXX
ZZZH
IJKL
XMOP

XXXX
ZZZZ
IJKZ
XMOZ


Z tries to prevent you from winning [But it's obvious at this point who's going to win: you]

    XXXX
ZZZZ
ZZZZ
XMOZ


As you can see the "worst case scenario" won't happen because you're taking the corners. When going first, you can take one corner from the get start, and even though Z can attempt to "seal off" a corner like so (your turn):

   AXXX
EFZZ
IJZZ
XNZP


He won't succeed because there are 7 squares left and you can force Z into a situation where he can only play two or one (and you can thus always force 5 with this situation):

   XXXX
XFZZ
IJZZ
XNZP


So in conclusion: -You win if you go first -Play 3 on the get go -Always take the corners if possible -Sum up with opponent to 4 (unless you can only play 2 or 1) -If Z tries to force you into being only able to play 2 or 1, it won't change a thing since you can still force a sum of 3 and win with the rest. (He can't force you into this situation if he only plays one at a time)