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Subject: An Analysis of Byte - The Final Two-Stack Endgame rss

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Ralf Gering
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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 N-in-a-Row 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 state-of-the-art computers could calculate the game-theoretic 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 "8-Stone 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 1-stack combined with a 7-stack (2x128), a 2-stack
with a 6-stack (4x64), a 3-stack with a 5-stack (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 two-stack 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:


Two-Stack 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

-------------

To find out more about Byte and other stacking games, come and join our free research forum "True-Stacking-Games":

http://games.groups.yahoo.com/group/true-stacking-games/

 
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