Chess will never be solved, here's why

Interesting @MARattigan: do you have a link for the 595 move mate?

@11111

"Why not the cube root"
For strongly solving: N white moves * N black replies = N² positions
For weakly solving: N white moves * 1 black reply = N positions = Sqrt (N²) positions

@11112

"Its already two years since his initial claim of 'five years'."
++ We are now about there: 106 draws out of 106 games in the ICCF WC Finals.
Besides, it was not my claim, but that of GM Sveshnikov.

My comment about cube root was a joke.
But the square root idea looks like its a joke too.

Here.

yeah i ran into the square root stuff on my own in other situations, it cant be traced back to a singular person

@Elroch and MEGA
E - you commented that the square root idea is 'intuitively trivial'.
And you made an excellent point that the square root idea more pertains to number of games as opposed to positions - if I've got that right.
Yes - number of games.
Which accounts for that divider I mentioned of 30 to the 40th power - 
a bigger number than number of atoms in the part of the big bang we can observe from here?
but maybe square root fans get help there.
They could argue that since it pertains to number of games - that number of positions is much smaller - helping out their square root argument.
Will tygxc try to jump on that one?

@11114

"5 x 10^44 versus ~10^120"

++ Wrong on both counts.
There are 10^44 legal positions, but they cannot result from optimal pley by both sides because of multiple underpromotions to pieces not previously captured and by both sides.

A better figure is 10^37 An upper bound for the number of chess diagrams without promotion, but it is a bit too restrictive, as in perfect games with optimal play from both sides positions with 3 or 4 queens do occur. Hence multiply by 10 to include positions with 3 or 4 queens, which leaves 10^38 positions without underpromotions to pieces not previously captured.

Inspection of a random sample of 10,000 positions shows none can result from optimal play by both sides either, that leaves 10^38 / 10,000 = 10^34 positions.

For weakly solving Chess only the square root of these i.e. Sqrt (10^34) = 10^17 are relevant.

The number of possible Chess games lies between 10^29241 and 10^34082.
The longest possible chess game, and bounds on the number of possible chess games

People play around with what it is.
Including figuratively. Here.
tygxc has already conceded that chess can't be solved with today's technology.
kv - there's a 75% chance that tygxc will soon assert something like this:
'No! said it can't be Strongly solved ...'

@11115

"there aren't even that many chess positions at all"
++ 10^38 positions without underpromotions to pieces not previously captured is an enormous number. However, 10^38 = 3^80.
This means 10^38 is reached after 40 moves with 3 choices per move that do not transpose.

Why? You give no reasoning. Do you have a big red telephone too?

Did you read https://www.researchgate.net/publication/231216842_Checkers_Is_Solved? I think @tygxc posted a link earlier. It doesn't manage square root but it gives some idea of the reasoning.

Actually @tygxc now appears to have changed his plan and takes the definition

weakly solved means that for the initial position a strategy has been determined to achieve the game-theoretic value against any opposition,

that he quotes to mean collect all the ICCF games so long as they draw. So he's now reduced the number of positions to around 10^5. (But he doesn't seem to have changed his cut and paste files.)

Hey - maybe 'number of games not number of positions' will put the square root trick in Trouble instead of 'fanning it'.

Martin - hi.
I gave various reasonings.
No red telephone.
You want a reduction of 10 to the 17th power?
Really?
You think that could be valid?
Hey - you got a right.
We've got rights.

I like my spaghetti shredded cheese to be a provolone-parmesan-mozzarella-asiago combo too.

Martin you want to defend the square root trick?
Hey maybe we'll still be talking about it another 10,000 posts and two years from now.
And interesting that tygxc changed his mind about 'weakly solving'.
Which you just pointed out.
'Multiple definitions'.
tygxc changing his mind?
Hey - he's an intelligent man.
And under-rated here.

But with far less frequency in FIDE competition rules chess and ICCF chess if you count a position as reappearing only if the possible forward play is the same in both cases.

And where on Earth does 10^120 come from?

Martin - 'evidence' evolves.
I asked just now in different words whether you think the gigantic reduction of square-rooting is valid.
You don't want to answer that?
That's fine.
Questions aren't thunder and don't have lightning coming out of them.

The answer is no it's not valid, but it does, as @MEGACHE3SE said at least have some merit unlike @tygxc's other ridiculous reductions which have none.

Try reading https://www.researchgate.net/publication/231216842_Checkers_Is_Solved as I suggested and then posting.

@11141

"where on Earth does 10^120 come from?"
++ It is the erroneous Shannon number

"count a position as reappearing only if the possible forward play is the same in both cases"
A diagram = place of men on the board
A position = diagram + side to move, castling flags, en passant flag
A node = position + history + provisional evaluation

'9.2.3 Positions are considered the same if and only if the same player has the move, pieces of the same kind and colour occupy the same squares and the possible moves of all the pieces of both players are the same. Thus positions are not the same if:
9.2.3.1 at the start of the sequence a pawn could have been captured en passant.
9.2.3.2 a king had castling rights with a rook that has not been moved, but forfeited these after moving. The castling rights are lost only after the king or rook is moved.'
Laws of Chess

@optumissed

You keep skipping the question: what is your argument against Cantor?

@tygxc you keep skiping this question: Because a supercomputer calculating x number of moves fail to find a win, how does that assue that a maskine calculating x +1 moves would fail to find a win?

@11145

"calculating x number of moves fail to find a win, how does that assue that a maskine calculating x +1 moves would fail to find a win?"
++ It is the other way around. As in 106 out of 106 games the ICCF World Championship Finalists with their computers managed to find a draw: either a 7-men endgame table base draw, or a 3-fold repetition, or a transition to a known drawn endgame.
Moreover, they did it in a redundant way: they found not the required 1, but several different ways to draw for black against whatever white tried.