Seriously, no-one is going to buy this as any kind of a solution for chess...The entire, artificial idea of "weakly solving", "strongly solving" etc is complete nonsense because in practice, they overlap considerably.
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Weakly solving overlaps strongly solving in exactly the same way that vegetables overlaps cabbages. That doesn't mean that no-one is going to buy any cabbages.
There are (64!)/(32!) positions just with all the prices on the board...
Er, you sure about that? I think I prefer the gibberish.
So long as we're talking about basic rules positions at any rate (@tygxc obviously isn't).
No I ment positions in total not legal positions, that is the number if you just put every piece in a random position.
Only if you label the pieces, e.g. calling your white horses Prancer and Charger.
one of the things they check when they check if the position is legal or not is stuff like bishops not on the same colour, pawns not on first and last row, kings not near each other ….
Last is normally king of side not to move not in check, but you haven't mentioned side to move. Some tablebases have extra constraints like no triple checks (but unless you can get hold of the generating code you'll probably need to find out by trial and error). The upshot is to significantly reduce the number.
The best available estimate of legal positions under basic rules to date is the paper you didn't like, as far as I know. And it's not 10^44 as @tygxc keeps trying to tell everyone, but (4.82 +- 0.03) * 10^44.
If you want to talk about legal positions under competition rules and equate them to game states (as @tygxc does) then you have to increase that number by a ridiculously huge unknown factor (as @tygxc doesn't - he instead divides it by something rather less huge but more ridiculous).
(! Is factorial, it is the number times every integer smaller, so 5! = 5•4•3•2•1=120, 6! = 6•5•4•3•2•1=720, 10! = 10•9•8•7•…•2•1=3,628,800)
I assumed so.