#1219
No, I did not ignore anything.
I just calculated the number of positions needed to visit.
sqrt (10^36) * 50 / 500
Number of positions to visit / nodes per second = time in seconds
Chess will never be solved, here's why

Theory: whoever or whatever eventually 'solves' chess -
won't get paid for it.
Additional obstacle besides billions of millenia needed:
Space required for the supercomputers doing it.
Both parallel processing 'processor' space ...
and - memory space.
Perhaps a system will be needed that's bigger than Greenland?
Bigger than Brazil?
Bigger than earth?
Cost: thousands of trillions of dollars ?
Return on investment: dubious.

You might get funding to solve chess as long as you said that it was a first step in research aiming to cure cancer.

You might get funding to solve chess as long as you said that it was a first step in research aiming to cure cancer.
By getting people to Not play it?
#1221
No, I keep getting it right.
Nothing was 'explained' to the contrary, just some loose negative remarks and unfounded conclusions.
#1223
Just a few millions are enough to rent the cloud engines.
There is no payback, but neither was there payback for the people who solved checkers, nine men's morris, connect four, antichess... They do it for the honor. Their institution pays them their salary anyway.

You keep getting it wrong. This has been explained to you.
Correct but its clear he will not accept that. Its a 'bounce'.

#1209
No, if black can draw against 1 e4 and 1 d4 then it is trivial to prove that black can draw against 1 a4 as well.
Great. Then prove it.
Until then, reality is not on your side of this position .

#1216
5 years on 3 cloud engines, 5000 years on 3 desktops.
See discussion: where? Who except me has calculated anything? Which post #?
Your "calculations" are garbage in/garbage out.

All the suggestions for pruning the positions or lines that need to be examined seem to be based on expert human judgement of which are "bad" and do not need to be considered in a proof. Of course positions impossible to reach under the rules of chess can be discounted. If expert human judgement is sufficient to decide what's "good" and "bad" the whole question of how much computer time and data storage space are needed becomes moot and we can just rely on "recognized authority" to reveal the truth.

The proof is
1) Black can draw against 1e4 and 1d4
2) It is always easier to draw against 1a4
3) Therefore black can draw against 1a4
Simples!
The argument is indeed simple, but not in the desired way!

Is there any hope at all of computers 'solving chess' from the openings end of the game ? Would the chances of same be worse than from the endgame end?
Apparently the chances would be even Worse.
But - we don't have to dismiss that.
Consider that even cheap chess software - running on modest desktop/laptop/phone hardware - can and does find mates many moves ahead.
Anytime such a computer can find a forced checkmate for itself whatever number of moves ahead - we can consider that postion and all the possible positions between that and that checkmate - as 'solved'.
(side issue - could computers be programmed to find all positions that are checkmate or stalemate- and secondary to that - all such positions that could arise legally?)
A very famous player named Lasker stated in his 'Manual of Chess' that 'ceteris paribus' the plus of a rook is sufficient to win.
Regarding ceteris paribus - I don't like the term and would prefer to put it as - if the side winning the rook is already at equal or better material and evaluates the rook-up position as having insufficient compensation for the other side (which the computer could evaluate as if it was a starting position - having skipped the positions where it doesn't win the rook) -
then can such a position and the positions leading to the plus of the rook be regarded as 'solved' ?
It seems a reasonable jump.
By extension - all positions leading to a Queen up - or two minor pieces up could also be regarded as 'solved' in similiar circumstances.
and all of the positions 'en route' that are in between.
What about a minor piece up plus a pawn or two up ?
Much harder. As would be three pawns up or whatever.
Maybe there's plays for the other side to knock off the extra pawns.
But would such mate-findings and material winnings positions offer any hope of 'solving' the game?
The point is that computers find such positions a lot.
They even announce 'mate in -'.
What percentage of positions could this take care of?
In the case of supercomputers ... how many moves ahead can they announce 'mate in' ?

How many years is a trillion seconds ?
Over 31,000 years.
What is 1 petaflop ? It is one trillion operations per second.
Apparently the world's fastest supercomputers run at under 1000 petaflops.
If there are ten^36th positions to analyze - and the computer can 'solve' each one with a trillion operations - then that's 1000 positions 'solved' per second - you're looking at 10^33rd seconds.
That's 31,000 years multiplied by a billion trillion.
Want to make it 10^24th positions instead?
Then you've got 31,000 years multiplied by a billion.
Want to make it that the computer only needs a billion operations to 'solve' a position ?
Then you've still got 31 Billion Years !! !!!

My computer was behaving slightly strangely today, and I just realised I had left Leela Zero analysing 1. e4 a5 for six hours.
Anyhow after that time, the winning probability (for 2. d4) got nudged up from 63.8% to 64.2%.
My computer was behaving slightly strangely today, and I just realised I had left Leela Zero analysing 1. e4 a5 for six hours.
Anyhow after that time, the winning probability (for 2. d4) got nudged up from 63.8% to 64.2%.
Look at Leela in post #20 here https://www.chess.com/forum/view/general/chess-will-never-be-solved-heres-why?page=1
SF8-12 (I don't have SF13) also lost from the same drawing position at first attempt (SF14 took me to the 55 move rule) but none of them get themselves mated on h1 before reaching h3.
How can this be surprising? If Leela plays a million games against itself how many will finish up in a KNNKP ending?
Troitzky said of the White to win positions in the endgame that there were 20 different endgames depending on the position of the pawn. How many games that Leela plays will finish up in a KNNKP endgame with the pawn on a2 or h2?
So what's it going to "learn" about the endgame?
I've spent a completely deranged amount of time looking at this endgame and undoubtedly played thousands of White mates against Nalimov with the pawns in those positions. (Practically all KNNKP mates with rook's pawns finish up that way).
So I don't expect Leela to be a reasonable opponent in that particular endgame anytime soon.
On the other hand I did have a version of Rybka nearly 20 years ago running on slow kit that I couldn't have beaten from any drawn KNNKP position. (It had an "e" for "endgame" on the end of it's version number, meaning that it's evaluations were designed to play endgames well with no tablebase - I got the last version - Rybka can't now manage even a bishop and knight checkmate.)
The fact is that the positional evaluation function is a lot more important than the machine speed. Leela and AZ will presumably be good at that for positions near the start of the game, but always crap at positions with only a few men (usually called endgames, but all positions are endgames to a perfect player - the tag is only relevant to players with a limited look ahead).
#1227
"I did a simple calculation. I assumed a 60 move game with 5 reasonable variations at each ply, That would be 5 raised to the power of 120 permutations but obviously"
++ As pointed out you arrive at a number of positions higher than the number of legal and sensible chess positions, so your number is too high.
a) 60 moves is too much. In most human and engine games they reach a draw by repetition or a table base position before move 40.
b) It is not necessary to include alternatives for black moves as long as the candidate ideal game ends in a draw: that retroactively validates all black moves as good enough to draw.
c) 5 variations per white move are too much: as calculated above 4 suffice i.e. 3 validation passes to prove all moves of a candidate ideal game are optimal to achieve less than 1 error per the total number of relevant chess positions. Maybe 3 alternatives (Carlsen) or 2 alternatives suffice. 4 alternatives i.e. 3 verification passes is at the safe side.
"some moves can transpose to positions that could occur with a different move order"
++ That is correct and another reason why your count is too high. There are many, many transpositions i.e. different games that lead to the same position.
E.g. 1 e4 c5 2 Nf3 = 1 Nf3 c5 2 e4
"I'm not sure how to adjust for that"
++ If you assume that moves are interchangeable, then for p possibilities e.g. p = 4 and m moves e.g. m = 30 instead of
p^m
you would have
p^m / m!
where
m! = m * (m - 1) * ... * 3 * 2 * 1
That however is still not the right approach.
The right approach is to split the problem in two.
1) How long does it take to strongly or weakly solve chess?
2) How to do it?
1A) To strongly solve chess i.e. a 32-men table base: 10^36 positions, that is unfeasible
1B) To weakly solve chess i.e. provide an ideal game with proof the moves are optimal:
sqrt (10^36) * 50 / 500 positions, that is feasible in 5 years
2A) Let Stockfish play against itself at 60 h / move, verify it ends in a table base draw or a repetition of moves, this gives a candidate ideal game
2B) All black moves are good enough to draw, so can be considered optimal and need no takeback.
To prove all white moves are optimal: take back all white moves starting at the last one and replace them by the 2nd choice. This is the 1st verification pass.
Now a 2nd verification pass, now a 3rd verification pass. As calculated above 3 verification passes suffice to make less than 1 error over the number of legal and sensible chess positions.
Your calculation was based on ignoring vast numbers of variations, some of which are correct (i.e. reach the same result with perfect play) and some of which are bad, and ALL of which need to be solved.
All you have offered to deal with such variations is "they are trivial", which is (to be more blunt than MARattigan) false, unsupported and unsupportable.
Leela Zero estimates the probability of white winning after 1. e4 a5 or 1. d4 a5 to be 63.8%, which indicates this defense is more likely to be sound than to be unsound (it is nearer to 50% than to 100%). Thus solving it is probably no more trivial than another other probably sound line.