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Alice and Bob are playing the neighboring game

which is originally a single person game with the aim of getting the highest points at the end.

You start with an empty 4x4 grid. At each turn, you can choose an empty cell and place a value in it. The placed value is given by the following rules:

  • If the chosen cell has no neighboring (horizontal or vertical) cells that are already occupied then the placed value is 1.
  • Otherwise the placed value is the number of neighboring cells that are already occupied. So, for instance, if Alice chooses a cell which has 2 neighboring cells (horizontal or vertical) that are already occupied, then her placed value will be 2.

Once all the cells have been filled in then each person's score becomes the sum of the values they have put in. Alice goes first. Who will end up with a higher score at the end, Alice or Bob, if both play optimally?

Here is another version of the game: Alice and Bob version 1

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    $\begingroup$ (1) The sum of the number of neighbors: does it mean how many neighbors which are already occupied? (2) What is the higher score? Is it the last number to put, or the sum of the numbers they put? $\endgroup$
    – athin
    Nov 6 '21 at 5:23
  • $\begingroup$ @athin , the sum of the number of numbers means the number of neighboring cells that are already occupied. So, if 3 of the neighboring cells are occupied, then the value you will put is 3. Higher score is the sum of the numbers you have put. P.S : I just edited the question to make it clearer. $\endgroup$ Nov 6 '21 at 11:07
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No matter what,

Bob can always win by at least two points, by simply reflecting Alice's last move through one of the axes of symmetry of the board. Since the board will always be symmetric after Bob moves, Alice's move and Bob's response will have the same number of neighbors unless Alice plays next to the central axis. This must happen four times in a game, but up to two of those four turns can have Alice place a piece with zero neighbors; since this still awards one point, Bob's response with one neighbor won't give him a leg up.

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I haven't run any exhaustive simulations to prove it, but in my optimal game,

Bob ends up with the higher score, A12-B13

An example of this play is:

01-A1 | 02-B1 | 03-A1 | 04-B1
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14-B3 | 12-B1 | 15-A4 | 05-A1
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11-A1 | 16-B4 | 13-A2 | 06-B1
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10-B1 | 09-A1 | 08-B1 | 07-A1

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  • $\begingroup$ If B plays move 12 at the position of your move 14 then he scores 2 points on that move. Then with optimal play to the remaining four centre squares, A gets a 2, B gets a 3, A gets a 3, and B gets a 4 for the final move. So after your move 11-A1, B can force a win with score A11-B14. $\endgroup$
    – Penguino
    Nov 6 '21 at 19:56
  • $\begingroup$ I thought the goal was to get as low a score as possible; I misread. I can make much higher scores. lol $\endgroup$
    – Vaekor
    Nov 6 '21 at 20:09
  • $\begingroup$ Correction (beyond 5 minutes), score is A10-B14 in my scenario. $\endgroup$
    – Penguino
    Nov 6 '21 at 20:09

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