Ralf Gering
Germany Germany

An Analysis of Byte
===================
Byte is a rather unique stacking game, although it reminds me a little bit of Yinsh (not a stacking game): http://www.gipf.com/yinsh/index.html
In (8x8)Byte you win if you complete at least two of three stacks of size eight with one of your stones on the top. In (10x10)Byte you win if you complete at least three out of five stacks of size eight.
In Yinsh you win if you're the first player to complete three rows of five stones.
Both games could be called NinaRow Games: The rows are in Byte vertical, in other words, they are stacks of height 8. In both games you must complete more than one row in order to win.
Otherwise, Yinsh and Byte are very different games.
Now let's take a closer look at Byte:
(10x10)Byte ("The International Variant" because it is played on the International Draughts Board) has 40 stones initially. Stack after stack of height 8 is formed and then removed. So you have later 32, then 24, then 16 and eventually 8 stones on the board. During game play the maximum height of a stack is at most 7. As soon as a stack grows to eight stones, it is removed from the board and no stack can grow higher than 8 stones. There is never a board position with 10 or 20 stones. The number of stones on the Byte board is always a multiple of 8.
There are only two kinds of stones: white (occassionally called "red" because in America red pieces are used in Draughts instead of white ones) and black.
A general observation is that there are a lot of moves to choose from in the opening (>60), but the number of choices a player has is soon rapidly diminishing.
If we consider these limitations, the number of legal board positions can't be that large. I suppose that stateoftheart computers could calculate the gametheoretic values of all board positions.
As I don't have such a computer, I tried other methods to analyse Byte.
As described in the previous post only eight stones remain on the board at the end of the game. This is called the "8Stone Endgame". There can be still more than two stacks, but the remaining stacks will slowly merge until only two stacks are left. Then we reach a stage of the endgame that can be called the "final two stack endgame".
This endgame position is defined by
(1) the height of the remaining two stacks (2) the composition of the remaining two stacks (3) the distance between the two remaining stacks (4) the "move"
(1) The combined height of both stacks must be 8. So if the lower stack has just one stone, the larger stack must be 7 stones high. (2) The stacks can be composed of white or black stones.
There can be 128 different stacks of height 7 (2^7), 64 different stacks of height 6 (2^6), 32 different stacks of height 5 (2^5), 32 different stacks of height 5 (2^5), 16 different stacks of height 4 (2^4), 8 different stacks of height 3 (2^3), 4 different stacks of height 2 (2^2) and two different stacks of height one (2^1). If we combine two stacks, we get every time 256 different combinations. There can be a 1stack combined with a 7stack (2x128), a 2stack with a 6stack (4x64), a 3stack with a 5stack (8x32) or two 4 stacks (16x16). There are 4x256 combinations altogether, that is 1,024.
(3) the distance between two stacks (empty spaces between) can be "odd" or "even".
(4) White can have the "move" (must move first) or Black can have the "move".
We have only 1,024 x 2 x 2 = 4,096 different twostack endgame positions. Sounds much, but, as has been pointed out before, usually the owner of the bottom stone of the lower stack wins. There are a few exceptions, however.
The exact results are:
TwoStack Final Endgame =======================
(a) Stacks of different Size:
The player who owns the bottom stone of the smaller stack always wins except
 his opponent owns the two upper stones of the smaller stack (the generalized case of "Paul's Position")  the smaller stack is a mixed stack of size 2 and both players can move stacks. Then the first player to move wins if the distance is odd, the second player wins, if the distance is even.
(b) Stacks of the same Size:
 If both players can move, the first player to move wins if the distance is even, the second player wins if the distance is odd. Exception: The top three stones of at least one stack are owned by the opponent and are not neutralized by two consecutive stones of your own in the center or bottom of the other stack. Then the opponent wins.
 If only one player can move, the moving player wins. Exception: The central two stones of at least one stack are owned by the opponent. Then the opponent wins.
This sounds far more complicated than it is. Draughts endgames tend to be much more complicated (think of the very common "First Position" which can easily take more than 60 moves to win). Try the different situations on a board with real pieces, and you'll see that you will soon understand which positions are favourable and which are not.
It shouldn't be too difficult to program a computer so that he recognizes these "winning lines".
Now you know for what you should strive in an Byte endgame.
Ralf Gering

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