The essence of metastrategy

A recent post on Less Wrong, Levels of Action, reminded me of a game I created whose dynamics I wanted to explore. I still have not explored the dynamics to a great level of depth, but I thought it would be interesting to the nerdy community that reads my blog.

The idea came after playing Castle Wars 2. In that game you try to build your castle as tall as possible while keeping your opponents castle as short as possible. The basic game dynamic is an action/meta-action trade off: (oversimplifying) you can play a card to gain 10 bricks, or you can play a card to gain one brick per turn for the rest of the game. I was surprised by the amount of subtlety derived from such a simple dynamic, and I recommend the game to anyone wanting to kill an hour. It’s not the best game ever, but it’s not as trivial as it at first seems.

I wondered what would happen if I removed the cards, the weapons, the defense from that game and replaced them with more levels of this same dynamic. Here’s what I came up with.

You can play it with a chessboard and poker chips (my old game design standby). You don’t need 6 of the rows of the board. Each player plays on a side of the board, and has eight squares which we will label, from left to right, 1 to 8. Each square can have up to eight chips in it. The goal of the game is to get eight chips in the eighth square. Here is how play proceeds:

On your turn, place a single chip in any of your squares. Your opponent does the same. Before each turn, “cancel out” any chips that both players have on corresponding squares. That is, if you have 4 chips on the 5th square, and your opponent has 5 chips on the 5th square, remove 4 chips from both, so that you have none and your opponent has one. Then (still before your turn) duplicate each square to the next higher position and truncate down to 8. So before this action if your eight squares had these values:

```0 0 1 4 5 2 0 3
```

Then after this action, the state of your board will be:

```0 0 1 5 8 7 2 3
```

Another way to think about it is that you slide a copy of your board one position to the right and add (then truncate).

Then place your new chip, and your opponent takes his turn. That’s it. The first player to eight in the eighth square wins.

Despite this game’s simplicity, I have been unable to devise a good strategy for it. The strategy for the game seems to revolve around estimating the length of the game. If you know how many turns the game will last, it is fairly easy to determine how to play optimally. But knowing how long the game will last is not so easy to determine.

Try it out, think about it. Let me know if you discover anything.

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3 thoughts on “The essence of metastrategy”

1. Exasperation says:

It strikes me that this game suffers from the tic-tac-toe problem; Player 1 can adopt a strategy that guarantees either a stalemate or a victory. Consider: P1 plays on square 7 on the first turn. P2 can either play opposite that piece, or allow P1 to get a free piece on square 8. If P2 plays on square 7, the situation is reduced to the initial state. If P2 plays on square 8, the free piece P1 got on square 8 will cancel P2’s piece before it can duplicate, leaving P1 to place another piece on square 7. If P2 plays anywhere else, P1 simply matches P2’s move, again leaving P1 up a piece. If both players always play on square 7 (really any square other than 8 as long as P2 always matches P1’s move) the game is a stalemate. If P2 ever plays on any square other than the one P1 just played on, P1’s victory becomes inevitable as long as P1 maintains the strategy of always playing so that there is at least one piece opposite P2’s last move.

2. Luke says:

@Exasperation, that’s a good observation. I was a bit sloppy with the technical details, and figured some degenerate case might be hiding in there. Perhaps we can patch it by making the plays simultaneous (which is complicated to carry out in practice), or perhaps just simultaneously adjudicated (probably giving P2 the advantage from the additional information).

3. Mathematically speaking you don’t need to count the two players separately. One player can play the positive numbers, the other negative. That removes the need for the canceling out step.

Obviously a stalemate is always possible by player 2 matching (and thereby canceling out) the opponents move, at least if he gets to see player 1 play first. Another variation would be to disallow canceling the opponents previous move. That at least rules out many trivial strategies.