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I recently designed a combinatorial game (sequential game of perfect information) with an infinite branching factor, that is it has a game-tree of infinite width.

I'm wondering how is it possible to solve this game? How can I find the best move each turn? Which tools Game Theory has to study this type of games?

Best regards, Matteo Perlini

INFINITO

Infinito is a two-player game, played on a (for the moment) 8x8 square board . One player owns the black stones, the other player owns the white stones. Grey neutral stones are needed.

Each player has an infinite number of stones with an unique natural number printed on them: 0, 1, 2, 3, and so on...

Players move alternately, starting with the player controlling the white stones. Each turn consists of two actions, performed in this order:

  1. Optional move: you can move a stone exactly as a Queen’s moves in Chess, i.e. any number of cells horizontally, vertically or diagonally. If your stone ends its move next to one enemy stone whose value is less than your stone, and your stone wasn’t next to that enemy stone to start its move, replace any friendly stone – but the stone just moved – on the board with a neutral grey stone with “∞” printed on it. In one move you can end up close to more enemy stones whose value are less than your stone, in that case you replace your stones with neutral stones for each of those stones. When you remove your stones from the board, those stones will be available for future placements.

  2. Compulsory placement: you must place a stone onto any empty space.

The game ends when the board is full, whoever has the least sum of his values wins.

JDH
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Matteo Perlini
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  • What is the size of the board? Are you trying to determine a size of the board that will make this game interesting? I am asking because with a small size brute-force can work, but with a bigger size, monte-carlo approach might be better. – Djaian Apr 09 '13 at 09:15
  • Hi Djaian, yes, the exact board size in not determined yet, but we can assume for now a standard 8x8 board. How can you use a brute-force if the branching factor is unlimited? – Matteo Perlini Apr 09 '13 at 09:22
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    Well, technically you have infinite number of stones, but you can suppose it's only finite. After all, the goal of the game is to have the smallest possible sum. So it seems not a good idea to play a stone higher than 1 million for example. – Djaian Apr 09 '13 at 09:29
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    It depends on the opponent's choices: if he already have a 2.000.000 sum, maybe a 1 million stone is not so bad. My point is that intuition here can be unreliable, so I was searching for mathematical certainty. :-) Besides, I'm really interested to know how game theory and computer science can handle this kind of games. – Matteo Perlini Apr 09 '13 at 09:37
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    Can you argue that on a board of a given size every game is equivalent to a game in which the stones are bounded by some small polynomial of the board size with an additional constraint on which stones may be placed on a given turn? – Peter Taylor Apr 09 '13 at 09:51
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    @MatteoPerlini While that's true, there could be some arguments that say "without loss of generality", we can reduce the $k^{th}$ move to one of several finite options such that the game has the same result. Edit: For example, if there are 64 squares on the board and all the past moves have been low, is there ever a reason to play $3,000,000$ over some much smaller number? At each stage, reduce the choices of numbers one should play with to obtain all 'equivalent games' and then you are left with something finite to work with. (up to some sort of isomorphism) – muzzlator Apr 09 '13 at 10:04
  • @PeterTaylor I have no idea. – Matteo Perlini Apr 09 '13 at 11:55
  • @muzzlator This can be true, but I cannot see a way to proof that or its contrary. Regarding to your 3,000,000 question, it could be possible if I know I can replace that stone with a neutral one. – Matteo Perlini Apr 09 '13 at 11:56
  • @Djaian: I forgot to tag you earlier. – Matteo Perlini Apr 09 '13 at 12:52
  • Any solution to this problem? – Matteo Perlini Apr 10 '13 at 16:41
  • If we would discover there is not a upper limit to good moves, would it mean this game is undecidable? – Matteo Perlini Apr 15 '13 at 16:52
  • There is a small error in the question. 0 is not a natural number. – ghosts_in_the_code Nov 28 '14 at 11:35
  • Good question. This is my first bounty; I hope it will be worth it. – ghosts_in_the_code Nov 28 '14 at 11:48
  • I tried playing the game by myself. I know my moves were nowhere near optimal, but the game ending quite soon (100 moves or so), so I would also like to know if this game is finite if played using a Nash equilibrium. – ghosts_in_the_code Nov 29 '14 at 13:14
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    @user45195 That $0$ is or is not a natural number is a matter of convention. See, for example, here. – user1729 Dec 01 '14 at 09:17
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    I think it suffices to restrict to a stone greater than all stones of the opponent (one choice) or any available below this. The reasoning is that the opponent can get a greater number on the board anyways so there is no need to take a higher stone, the opponent would then just take an even higher number to place if he wants to. The only flaw may be that the number of stones below this may influence the optimal strategy, but it should be considered as a starting point I think. – AlexR Dec 01 '14 at 14:49
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    One may want to use a stone 2 pts more than the highest stone of the opponent, so that he can use the previous stone later. To counter this strategy, the opponent might take this stone and so the 1st player may have to start with greatest stone + 2 pts. Basically, in a perfect strategy, even a difference of 1 matters. So you may want 100001 rather than 100000 when the highest stone on the board is, say 100. – ghosts_in_the_code Dec 05 '14 at 10:15
  • Somebody could write atleast an approximate answer, because I have only 20 hours left to award the bounty, and I don't want to waste it. – ghosts_in_the_code Dec 05 '14 at 14:57
  • Okay, the bounty has expired. – ghosts_in_the_code Dec 06 '14 at 15:41
  • "next to another stone" does that mean it is touching it side by side or touching it corner by corner as well, or just side by side? Also you mention "natural numbers" but include "0"? This makes a difference as well, since if the sums were a draw and one spot was left to place on, if the 0 is included, then it is a draw (and placing player has it to place); but if 0 is not included, then it is a lose for the placing player as any stone will increment its sum above draw. – Vepir Nov 22 '17 at 20:41

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