Except 361! (~10^768) couldn't plausibly be an estimate of the number of legal positions, since ignoring the rules in that case gives the trivial upper bound of 3^361 (~10^172).

More likely it is a very, very bad attempt at estimating the number of games. Even with the extremely unsharp bound given in https://tromp.github.io/go/gostate.pdf


10^(10^48) < number of games < 10^(10^171)

the 361! estimate comes nowhere close to that interval.

/Gunnar

On 08/07/2017 04:14 AM, David Doshay wrote:
Yes, that zeroth order number (the one you get to without any thinking about how the game’s rules affect the calculation) is outdated since early last year when this result gave us the exact number of legal board positions:

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

So, a complete game tree for 19x19 Go would contain about 2.08 * 10^170 unique nodes (see the paper for all 171 digits) but some number of duplicates of those nodes for the different paths to each legal position.

In an unfortunate bit of timing, it seems that many people missed this result because of the Alpha Go news.

Cheers,
David G Doshay

ddos...@mac.com <mailto:ddos...@mac.com>





On 6, Aug 2017, at 3:17 PM, Gunnar Farnebäck <gun...@lysator.liu.se <mailto:gun...@lysator.liu.se>> wrote:

On 08/06/2017 04:39 PM, Vincent Richard wrote:
No, simply because there are way to many possibilities in the game, roughly (19x19)!

Can we lay this particular number to rest? Not that "possibilities in the game" is very well defined (what does it even mean?) but the number of permutations of 19x19 points has no meaningful connection to the game of go at all, not even "roughly".

/Gunnar
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