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I remember playing this game with four rows of pearls, not three, but the solution method still seems to work.
It is a mathematical problem concerning the total number of pearls remaining in the game, combined with their divison amongst the rows. Without making it too complicated, suffice it to say that the method to win the game is to always pass your opponent a setup that screws them over.
The way to solve the puzzle is to do a little math. What you need to do is reduce the number of pearls on the board to binary form. So for the starting setup of pearls like this:
O O O
O O O O
O O O O O
Their conversion to binary is then written as:
0 1 1 = 3 pearls
1 0 0 = 4 pearls
1 0 1 = 5 pearls
Now in order to be "winning" the game you need to pass your opponent a setup such that the sum of each column, modular 2, is zero.
At the start of the game we have:
0 1 1
1 0 0
1 0 1
------
2 1 2 (mod 2) ===> 0 1 0
So the center column needs to be adjusted. The easiest way to do this is to remove 2 pearls from the top row, resulting in:
0 0 1
1 0 0
1 0 1
------
2 0 2 (mod 2) ===> 0 0 0
Which screws him over. Depending on what the AI does, and how many rows of pearls there are in the game, you simply repeat these steps a few times.
Now you have to use some common sense - at a certain point if you pass the opponent a board sum of "0 0 0" it will give it a win. You need to look for the "clincher" setup, one of which is a chance to pass the opponent a board of two rows of two:
O O
O O
Which, if you have tried the game a few times, the computer will have done to you and fucked you royally. Pass him this and you win.
_________________
Nephrenia
Red Mage: 60 Black Mage: 30
2-Leaf Mandragora Bud Finder Extraordinaire.
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