[Computer-go] Alphago and solving Go

Gunnar Farnebäck gunnar at lysator.liu.se
Wed Aug 9 05:34:56 PDT 2017

```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 about how the game’s rules affect the calculation) is outdated since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
>
> ddoshay at mac.com <mailto:ddoshay at mac.com>
>
>
>
>
>
>> On 6, Aug 2017, at 3:17 PM, Gunnar Farnebäck <gunnar at lysator.liu.se
>> <mailto:gunnar at 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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```