Chess will never be solved, here's why

Let's imagine what a solution of chess requires. First it requires a big tablebase (in all seriousness, one that is too big to be infeasible. Let's pretend not and say we have a puny 12-piece tablebase. This requires about 10^24 bytes.  That's about 10000 times bigger than the total storage capacity of all computers on the planet. Ridiculous, but technologically plausible. If you have a few trillion dollars)

Next you need two strategies. The first forces a drawn position in the tablebase with white, the second forces a drawn position in the tablebase with black.

The problem is that it seem rather unlikely you could avoid needing to rely on drawn positions not in the tablebase, if the opponent fails to co-operate in exchanging pieces (eg he blocks the position and shuffles pieces).

But can you somehow get round the need to use a large state space by considering that only opposing moves that produce new positions matter. If the opponent against a drawing strategy moves to a position previously reached he has achieved nothing.

So you don't need the full state for the positions, rather you check whether new positions are already in your strategy.

Thus I think @tygxc is correct to believe that FEN states are adequate for a solution of chess, even if he is woefully wrong on the computing power needed to solve this rigourously, as achieved for checkers.

Any flaws in that reasoning?

@4951

"By which time SF will have used positions  rather than equivalence classes"
++ Yes, the 3-fold repetition rule is about repetition of a position, i.e. the same FEN without move#.

"Somebody has to move the pieces."
++ You cannot transpose the 2 mirror images @4953 into each other by moving the pieces.
Only 1 of the 4 mirror images can be relevant.

"In which case the Tromp figures are not too high" ++ Tromp gives the number of legal positions, but in weakly solving chess only 1 of the 4 mirror images can occur.

"The Gurion count is not fit for purpose" ++ The Gourion number is a better estimate, as the Tromp positions cannot result from optimal play. Look at the 3 random samples.
https://github.com/tromp/ChessPositionRanking
They all have underpromotions to rook or bishop on both sides. The only reason to underpromote to rook or bishop instead of queen is to avoid stalemate i.e. avoid a draw. It makes no sense for both sides to avoid a draw. So at least one side made a mistake by underpromoting instead of queening. So these positions cannot result from optimal play.

"solving competition rules chess"
++ The plan is to solve chess with the 3-fold repetition rule (or even simplify to 2-fold) and without the 50-moves rule, which plays no role.
The solution for chess with the 50-moves rule is the same as without.

@4952

"First it requires a big tablebase"
++ No, that is not true. There are much more chess positions around 26 men than around 12 men. A 7-men endgame table base is enough.
In TCEC the engines frequently hit their 7-men endgame table base as early as move 10.

"Next you need two strategies."
++ No, you only need 1 strategy for black to achieve a draw against all white opposition.

"The first forces a drawn position in the tablebase with white" ++ No, unneeded.

"the second forces a drawn position in the tablebase with black"
++ Can also be a prior 3-fold repetition, e.g. perpetual check, or a known draw.

"if the opponent fails to co-operate in exchanging pieces"
++ There is always compelling reason to move pawns or exchange pieces.
In ICCF WC Finals the 50-moves rule is never invoked.
Such games are > 99% sure to be perfect games with optimal play and no errors.

Intuitively solving chess with 2-fold repetition draw is equivalent to solving it with 3-fold repetition.

The solution for chess with the 50-move rule is not the same as a solution of chess without this rule. A strategy for the former could happily reach a tablebase position where the opponent couldn't mate quickly enough, so a 50-move draw was inevitable. This strategy would fail without the 50-move rule. This is trivial.

Fortunately the 50 move rule is not difficult to deal with and FEN does so.

You're speaking for yourself. I find it quite easy.

@4955

"solving chess with 2-fold repetition draw is equivalent to solving it with 3-fold repetition."
++ Yes, that is correct.

"The solution for chess with the 50-move rule is not the same as a solution of chess without"
++ Yes they are not necessarily the same.
However, if chess is solved to be a draw without the 50-moves rule, then that same solution also applies with the 50-moves rule. The reverse is not necessarily true.

"A strategy for the former could happily reach a tablebase position where the opponent couldn't mate quickly enough, so a 50-move draw was inevitable."
++ It is not necessary. In ICCF they can claim 7-men endgame table base wins that exceed 50 moves without capture or pawn move.
Such endgame table base win claims do not occur.
Endgame table base draw claims are common in 10% of games.

"This strategy would fail without the 50-move rule."
++ Yes, but the strategy without the 50-move rule applies also with the 50-moves rule.

@4956
"I find it quite easy."
++ I am speaking of middlegame positions of around 26 men, where most positions are.
See table 3: https://arxiv.org/pdf/2112.09386.pdf

For positions without pawns there is even 8-fold symmetry:
up / down, left / right, and mirror images along a diagonal.

Positions with 7 men or less are not relevant: those are looked up in the endgame table base.

Not true.

A solution of chess with 2-fold repetition draw is equivalent to a solution with 3-fold repetition, but whether solving the two are equivalent depends on the method of solution. 

So far as @tygxc's method is concerned the latter equiivalence would appear not to hold. We have requested a detailed description of what he proposes, but I doubt if he can manage to produce one in five years.

So what are you proposing to solve? Chess or most middlegame positions?

There is no solution that involves two-fold repetitions being involved on the path to anywhere but a draw.
It is like in real play. A player fishing for a solution gets to a two-fold repetition and realises he needs to change his strategy from that point. When the strategies are final, this no longer occurs - two strategies that reach a two-fold repetition will reach an n-fold repetition because they are fixed.

@4960
"We have requested a detailed description of what he proposes"
++ I have explained this before.
Take an ICCF WC Finals drawn game e.g. this one.
https://www.iccf.com/game?id=1164259 
It begins from the initial position and it ends in a known drawn endgame.
Take the last move 35 Be3. Look at the top 3 alternatives. Do they draw too?
Now look at 34 a5. Look at the top 3 alternatives. Do they draw too?
Now look at 33 a4. Look at the top 3 alternatives. Do they draw too?
Now look at 32 Kf2. Look at the top 3 alternatives. Do they draw too?
...
Like that all the way down until another ICCF WC finals drawn game is reached.

Yes, but they're only called large because of definitions. It's still the case that the larger they get the more minuscule (got it right this time) they get compared with almost all the rest.

@4961
"So what are you proposing to solve? Chess or most middlegame positions?"
++ Chess, but the vast majority of chess positions is around 26 men.
32 men: 1.89 × 10^33
31 men: 1.71 × 10^34
30 men: 1.64 × 10^35
29 men: 1.53 × 10^36
28 men: 5.46 × 10^36
27 men: 1.05 × 10^37
26 men: 1.08 × 10^37
25 men: 6.14 × 10^36
24 men: 3.19 × 10^36
23 men: 5.66 × 10^35

Yes. But it's worth remembering that whether these cardinals exist is not a given.  You can do almost all of mathematics with ZFC set theory. The first inaccessible cardinal is bigger than all the cardinals that can be proven to exist in this model of mathematics: it has to be added via another axiom.

It is not provable that the existence of this cardinal is consistent with ZFC (but it is believed to be, like it is independently believed that ZFC is consistent).

@4966
This thread is about solving chess, not about transfinite numbers or the continuum hypothesis.

@Elroch

I know.

But none of your posts are relevant to solving chess either.

@4969
"none of your posts are relevant to solving chess either"
++ I beg your pardon. All of my posts are relevant to solving chess.
I have calculated how long it takes to weakly solve chess: 5 years. I have shown how to do it.
Others troll with off topic.

Yes. It doesn't work.

There's no output strategy for a start, which according to any of your posted definitions of "weakly solve" is the whole aim of the exercise.

When you get to "Look at the top 3 alternatives" there's nothing to say how you determine what they are.

When you look for the sub function "Do they draw too?" it isn't there.

No mention of a computer, so what are the cloud cuckoo computers all about? 

Not much point in going any further, it's a joke.

@4973

"There's no output strategy" ++ The output are the 3 alternative lines per white move.

"there's nothing to say how you determine what they are"
++ The top 4 moves of the cloud engine running for 17 s, (or a desktop running for 4.7 h). As previously shown the real good move is always among the top 4 moves with 1 error in 10^20.

"When you look for the sub function "Do they draw too?" it isn't there."
++ Either a table base draw, or a 3 fold repetition, or a humanly known draw, adjudicated by the good assistants.

"No mention of a computer" ++ Of course the 3 computers serve to do that. 1 computer looks at 1 e4, 1 computer at 1 d4, and a 3rd computer at 1 c4 and 1 Nf3 lines that do not transpose.