Chess will never be solved, here's why

I continue to say they have uniqueness from each other.
Just as positions would be unique from each other where castling rights had gone even though some pieces are in castling position - as opposed to those where castling rights persist.
By the way - positions are also unique from each other where the board setup is the same - but two different cases of whose move it is.

What isn't contributing to board uniqueness and solving is how many repetitions of position have occurred - and how many 'moves of 50' have happened.
If there's a mate available then there's a mate available.
If there isn't there isn't.  
Arbitrating a number of moves in advance doesn't change that.
Getting into repetitions and 50 move rules is getting away from positions and tablebases - and getting back into 'games' again.

Martin - we won't agree on some of these points.
Some of them are semantic points.
When people disagree - often they understand the other view - often they don't.  
As I've also pointed out - people often argue when they Agree too  !! 

Dear @tygxc,

If I understand you well, to find those 4 candidate moves for a colour and be sure the best move is among them, you would leave SF analyse 60 hours. Then, you should leave SF analyse the position for each of those candidates to find the best 4 options for the other colour, and so on. So to move from depth n to depth n+1 (in plies) you should use 4*60 hours. How many hours to reach the endgame after, let's say, 20 moves?

Has Stockfish 14 played millions of games with itself ?
Probably many projects have been done where a computer plays itself - or another computer - with all variations of c number of candidate moves throughout the game - where c is a constant in each project - but might be varied from project to project.  

Ironically - this leads to another 'candidate' for 'weak solving'.
Just let the strongest computer find one best move for each position - play itself that way - and play the game through to conclusion and see what the result is.
If it turns out to be draw - then many can raise both hands very high and proclaim "You see??  This proves chess is a Draw with best play.  It is Written !"

Then - all iterations of two candidate moves can be tried out too.
But then the number of games builds up rapidly.
Instead of one game ...
They build up in powers of two.
Many would be tempted to invoke the 50 move rule - to set an upper bound on the length of the games.
After 10 ply - you've got a 'kilo'.
After 20 ply - a 'meg'.
30 ply gives you a 'gig'.  But that's still a very short game.  15 moves.  
But look what's happened !
With just two candidate moves each time - and all of them played -
you've already got over a Billion games !
Something beginning to look Rotten in the state of Denmark  !!
(just Shakespeare I believe - not meant to be political)  

Yes I know - many of them might transpose back into each other - and then you'd get a big mathematical 'cutdown'. 

Try a little harder.

Point:
If computers isolate just two candidate best  moves at each ply (half move) - and then go ahead and play both - in separate games ...
to see where the results lead ...  then after fifteen full moves you're going to have 2 raised to the 30th power ...  games !   
Over a billion games !  
Could the computers handle it?  Most of them would not be over after 15  moves - but already you've got over a billion games going !  
Uh oh !

It just got harder.

It was always hard.

LOL

اثممخ

#1989

Dear @haiaku,

"If I understand you well, to find those 4 candidate moves for a colour and be sure the best move is among them, you would leave SF analyse 60 hours. Then, you should leave SF analyse the position for each of those candidates to find the best 4 options for the other colour, and so on." ++ No, not exactly. I do 4 candidate moves for white only, I do 1 candidate move for black. I store intermediate results for later reference so as not to duplicate the same calculation.
"So to move from depth n to depth n+1 (in plies) you should use 4*60 hours. How many hours to reach the endgame after, let's say, 20 moves?"
++ I do not start at move 1, but rather at the 26-men tabiya. 1 hour = 3600*10^9 positions. Based on ICCF games I guess that most positions terminate in a draw by 3-fold (or 2-fold) repetition before. You can see data on table base hits and nodes/s in the TCEC superfinals.

#1990
"Just let the strongest computer find one best move for each position - play itself that way - and play the game through to conclusion and see what the result is.
If it turns out to be draw - then many can raise both hands very high and proclaim "You see??  This proves chess is a Draw with best play.  It is Written !"
++ That is not proof yet, that is begin of proof. The autoplay game ending in a draw is a candidate ideal game. Black's moves are optimal, as the result is a draw. It rests to prove that white's moves are optimal too. That is done by retracting all white moves from the last to the first and replacing them with the 2nd choice, then the 3rd choice, and finally the 4th choice so as to ascertain that all white's moves are optimal too.
#1993
"If computers isolate just two candidate best  moves at each ply (half move) - and then go ahead and play both - in separate games ...
to see where the results lead ...  then after fifteen full moves you're going to have 2 raised to the 30th power ...  games !   "
++ As explained before, I only go for 4 white alternatives and 1 black alternative.
Moreover, in chess there are many transpositions: different games that lead to the same position. E.g. 1 e4 e5 2 Nf3 Nc6 = 1 e4 Nc6 2 Nf3 e5 = 1 Nf3 Nc6 2 e4 e5

I didn't imply you have to start at move 1, but from a 26-men position you would need some moves to reach the endgame, right? So, you check only one candidate move for Black... and that should be enough to find her optimal strategy?

The question still remains: even with only one option for Black and 4 for White, but 60 hours spent to determine those 4 candidates when White has to move, how many hours to build the proof tree?

In my "simplified rules", reaching a repeated position is an automatic draw. This seems not to remove anything game theoretically significant, as if you have a win there is no point in repeating a position or giving your opponent the opportunity to do so unless you have been inaccurate up to that point.

What is the point of your last caveat? You are usually crystalline in your clarity.

Nothing deep.

It simplifies the understanding of the role of 3-fold repetition in solving chess. Basically a repetition of a position is admission of a draw (or grabbing of a draw - technically the same thing). There is also some redundancy in the drawing rules for theoretical purposes. Any analog of the 50 move rule will suffice without a repetition of position rule (since any number of moves will be exceeded by optimal play that has no better than to repeat). The reverse is only true when the N-move rule does not interfere with it being possible to achieve all basic rules tablebase wins.

Just stating simple facts (I hope) out loud.

"or giving your opponent the opportunity to do so unless you have been inaccurate up to that point."

I was asking about that bit.

There are good reasons for the threefold repetition rule (as opposed to twofold): it allows the attacking side to try for a quicker win etc. Obviously, in the hypothetical situation where there's a solution, the lines are all worked out and repetition achieves nothing. The first repetition acts as a signal. Nothing is gained by losing that rule. For the purposes of "solving chess", it should be a two-fold repetition which draws, to save moves.