[harryjo]

I don't think anyone believes that Go is somehow more complex than the universe it is a subset of. The point is that enumerating all cases of Go is impossible and always will be, so more sophisticated analysis is required.

[tromp]

Indeed, enumerating all

208168199381979984699478633344862770286522453884530548425 639456820927419612738015378525648451698519643907259916015 628128546089888314427129715319317557736620397247064840935

positions in Go is impossible.

[ChristianBundy]

That's called "counting". Enumeration is iterative.

[dogecoinbase]

In your enumeration, what's the board look like at position 348277381979984699478633344862652779770286522453884530548425639456820927419612?

[tromp]

I can't say, because I didn't enumerate them. I only counted them. See

http://tromp.github.io/go/legal.html

for the method used, which is a form of dynamic programming.

[nhaehnle]

Though if it is dynamic programming, then it should be possible for you to answer dogecoinbase's question using not much more computational power than you used to count them in the first place, right?

If you think of dynamic programming as counting the number of paths in a directed graph (in this case, from skimming the paper, the nodes correspond to border states), then given a path number, you can trace the path backwards through the graph, as long as you remember the number of paths ending in every vertex.

[tromp]

Yes, you could if you preserved all intermediate counts. But the graph I used has 362 layers each of which can have up to 363 billion nodes, so I had to recycle the space used for the counts (4TB per layer). Also, I didn't even compute with full counts. I reconstructed them using the Chinese Remaineder theorem from 9 separate modular counts. So, yes it's possible, but highly impractical...

[Jyaif]

1/ Convert the number to base 3.

2/ Each digit represents an intersection on the goban, assuming the following mapping: 0 = no stone, 1 = black stone, 2 = white stone.

[tromp]

That would work for all positions regardless of legality, which are in 1-1 correspondence to {0,1,2}^(19*19).

The count above is for legal positions only, i.e. those where every connected group of stones is adjacent to an empty point.

[vacri]

Sametmax's point is that you're comparing a simple count (atoms) to a factorial (combinations of pieces). For example, it's hardly surprising that 6! is larger than 6 - factorials grow much faster than simple counts.