The number of potential positions on a Go board

   

There's a reference to research done by Tromp and Taylor at Position that may be relevant. Their estimate is higher by many orders of magnitude. --Hyperpapeterie


Theorem: There are more than one googol (10^100) potential positions on a 19x19 Go board, allowing technically illegal positions in which a group of stones has no liberties.


Proof: Each point on the board either contains a black stone, a white stone, or no stone. There are 19*19=361 points on a Go board. Therefore the number of combinations is 3^361 > 3^360 = 3^(3*180) = (3^3)^180 = 27^180 > 10^180 > 10^100


Lemma: There are more than eight googol (8*10^100) legal board positions on a Go board. Why do I say eight googol instead of one googol? See my next theorem after this lemma.


Proof: Suppose you allow board positions with only black stones and no white stones. Each point either contains a black stone or no stone. The number of combinations is 2^361. However, exactly one of these board positions is illegal:

[Diagram]

The only illegal board position with no white stones

As you can see, this is illegal, because the huge black group has no liberties. But any other combination of black stones and empty points is legal. Therefore, there are 2^361 - 1 such combinations.

Let S be the set of legal positions allowing black and white stones. Let n be the number of elements in S. The set of legal positions allowing only black stones is a subset of S. Therefore, n > 2^361 - 1

-1 > -2^360

n > 2^361 - 1 > 2^361 - 2^360 = 2*2^360 - 2^360 = (2 - 1)*2^360 = 1*2^360 = 2^360 = 2^(10*36) = (2^10)^36 = 1024^36 > 1000^36 = (10^3)^36 = 10^(3*36) = 10^108 > 10^101 = 10*10^100 > 8*10^100


Theorem: There are more than one googol legal board positions that are unique after taking transformations into account. The following are examples of transformations:

[Diagram]

Original position

[Diagram]

Rotation

[Diagram]

Reflection


Proof: Let T be a set of legal board positions that are unique after taking transformations into account. T is not the only such set, but it contains the maximum number of elements, which we'll call m. By replacing each element in T by eight elements that are the transformations of that element, the resulting set is identical to S. However, Some board positions are rotationally and/or reflectively symmetrical, meaning that they transform into themselves, Here are some examples:

[Diagram]

Rotational symmetry

[Diagram]

Reflective symmetry

Therefore, 8m > n > 8*10^100

m > 10^100


This is a copy of the living page "The number of potential positions on a Go board" at Sensei's Library.
(OC) 2012 the Authors, published under the OpenContent License V1.0.
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