The Games Journal | A Magazine About Boardgames

Letters - November, 2004

Mikko Saari: The Saratoga game Sandy Psiurski was looking for [See Letters - October, 2004] sounds a lot like Liverpool aka Contract Rummy. That can be found in Pagat.

GGA - Thanks to the many people who wrote in about this.

Joe Scoleri: Actually, those "hieroglyphics" are called cuneiform, Greg. [See Babylone review] I thought it was a neat way to decorate the pieces. Here's some additional info if you are curious:

http://www.wsu.edu8080/~dee/GLOSSARY/CUNEI.HTM

GGA - Thanks for the information Joe. The Babylone puzzle was pretty easy but if anyone didn't get it the correct move is to combine the green 3-stack with either of the green 1-stacks. The remaining moves are all forced.

Honourable mention goes to Josh Adelson who noticed that the position depicted is not possible. I'll leave it as an exercise for the reader to determine why.

Mark Goadrich: I too was sucked in by the underlying puzzle of Babylone when I found it on Bruno's website (http://faidutti.free.fr/) one day a few months ago.

It is indeed solvable; with perfect play, the second player should always win, and there are six different possible end states where they can force this to happen. I believe there is an on-line program which demonstrates this, linked from Bruno's website and his forums, unfortunately, I do not speak French, and babelfish translations only get me so far. But the real puzzle for me is trying to find some simple heuristics to evaluate the current game state instead of relying on a pure look-ahead strategy.

To do this, we need to get a handle on the state space of this game. An individual state can be defined as a collection of piles, where each pile has a color and a height, and we can be more concise by including a count of identical piles. In formal math terms, a pile is a tuple of color {A,B,C,D}, height {1 to 12} and count {1 to 3}. For example (A,6,1) would show one pile six tablets high with color A on top, and (C,1,3) would show three piles one tablet high of color C. So, the initial state of the game can be written as (A,1,3),(B,1,3),(C,1,3),(D,1,3).

As for the symmetry of states, it's helpful to use Bruno's hint on his website: "In Babylone, the first player has only two possible moves—stacking two pawns of the same color, or two pawns of different colours." In general, we can cut down the number of states by always ordering the piles by height (tallest first), followed by count (smallest first), and then relabeling the colors when necessary. So, any move you make from the initial state will be symmetric to either the state:

(A,2,1),(A,1,1),(B,1,3),(C,1,3),(D,1,3) 

or 

(A,2,1),(A,1,2),(B,1,2),(C,1,3),(D,1,3) 

Using this symmetry, there are only 846 reachable game states (including the initial state), which is a much more manageable game tree. A minimax algorithm can quickly find that there is always a path for the second player to win, where the six possible winning end states for that player are

(A,6,1),(B,3,1),(C,2,1),(D,1,1)

(A,9,1),(B,3,1)

(A,8,1),(B,4,1)

(A,10,1),(B,2,1)

(A,7,1),(B,5,1)

(A,11,1),(B,1,1)

You can now take each state in the tree, assess it as being a winning state or losing state for the second player given optimal play, and try to find out why some states are good and others are bad. I have yet to find some good heuristics, even after running the state spaces through a machine learning algorithm good at detecting patterns in data, and so the game continues to intrigue me. For those interested in doing some analysis of their own, I've posted a text file that I've generated of the game state tree at http://goadrich.com/babylone4color.txt (it's a text file, since trying to drawing the actual tree turned out to be quite a mess).

Oh, and for the puzzle in your article, you should combine the green three tablet pile with either one of the green one stone piles. Your opponent can only combine the green four tablet pile with the other green one tablet pile, leaving you to make the last move of combining the green five tablet pile with the black five tablet pile. Or in the state notation above, (A,5,1), (B,3,1), (C,2,1), (B,1,2) --> (A,5,1), (B,4,1), (C,2,1), (B,1,1) --> (A,5,1), (B,5,1), (C,2,1) --> (A,10,1),(B,2,1). Your only other move is to combine the two green one tablet piles, which then they combine with the white two tablet pile and leave the white tablet on top for the last move.

GGA - Thanks for the involved explanation Mark. I should also mention that reader Mark Kalmes also sent a similar proof for the fact that Babylone is solved.

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