Chess will never be solved, here's why

Imagine a chess position of X paradigms.

Now, a chess computer rated 3000 solves that position. All well and good.

Could another computer rated a zillion solve that position better than Rybka?

No, because not even chess computer zillion could solve the Ruy Lopez better than a sad FIDE master could.

the point is, there's chess positions with exact solutions. Either e4, or d4, or c4, etc.

nothing in the world can change that.

So if you are talking about chess as a competitive sport, then chess has already been solved by kasparov, heck, by capablanca.

If you are talking chess as a meaningless sequence of algorithms, where solving chess equates not to logical solutions of positional and tactical prowess, but as 'how many chess positions could ensure from this one?'' type of solutions, then, the solutions are infinite.

So can chess be solved? If it is as a competitive sport where one side must, win, then it has already been solved. Every possible BEST move in chess has been deduced long ago.

If chess is a meaningless set of moves, with no goal in sight, then sure, chess will never be solved.

@12003

"every position reached generates around 35 positions"
"Show where it has been disproved"

++ No, not every position reached generates around 35 positions.
It generates massive transpositions to positions previously generated.
10^38 = 3^80
Thus 3 non-transposing positions per ply generate all 10^38 reasonable chess positions in 40 moves.

Your reasoning here is appalling. It assumes that the rate at which new positions is added is constant, picks a random number of moves (40) and then you imply that you believe the number of moves goes up by a factor of 3 all the way to 40 moves then stops dead. Obviously, those assumptions are simply wrong.

In addition, you use an undefined term (intuitively based on weak chess play) and make a statement which is meaningless because the term is undefined.

According to Tromp there are positions that take over 60 moves to reach (the max is unknown), and no-one cares whether a weak player thinks they are "reasonable".

If there were 35 non-transposing moves per position,
then that would generate all 10^38 reasonable chess positions in 25 ply, i.e. 12.5 moves.
35^25 = 10^38
An average ICCF WC Finals draw lasts 40 moves.
12.5 moves is clearly too small, thus 35 non-transposing choices is clearly too many.

"each of the two strategies will generate at least 10^19 positions after a mere 40 moves" ++ OK

"Everyone knows positions after 40 moves are rarely in a tablebase"
++ They usually are. Games in the ICCF WC Finals last average 40 moves.

This is a lie. You know perfectly well that 40 move positions in ICCF games usually have more than 8 pieces on the board. Games often last more than 40 moves. Here is an ICCF game with 9 pieces on the board on move 77 (found by just randomly looking at a handful of games).

They end in a 7-men endgame table base draw (10 draws), or a prior 3-fold repetition (37 draws), or a draw by agreement in a position where neither side has any hope to win (57 draws), or 10 games where Dronov passed away in otherwise drawn positions.

Here you are making a new version of one of your habitual mistakes, confusing positions on the board and positions in the (almost always non-rigorous) analysis from a position on the board. Only the first is relevant to the point being made (simply because the positions many ply deeper in the analysis are those in hypothetical games with much more than 40 moves.)

"ignoring (huge numbers of) positions based on an evaluation is invalid"
++ Yes, that is what I say all the time.

If only what you said was consistent as well as being occasionally correct (such as here). It is not.

For the black side all moves but one can be ignored as long as the final result is a certain draw. It does not matter how that one move was generated, maybe Caïssa appeared in a dream and whispered the move in the (grand)master's ear.
As for the white moves, all reasonable white moves need consideration.

Not sure if you are blundering here or using loose language (probably the former). If you believe chess is a draw, the status of the two sides is equal. You need to exhibit 2 strategies, one to draw with white, one to draw with black. No excuses!

This is a red herring, as a weak solution of basic chess with an added n-move rule automatically provides a weak solution of chess with an additional repeated position rule (the one that greatly increases the state space). This is true whether chess is a win or a draw.

You know this already, so why ignore it again?

@tygxc's positions don't transpose mainly because of the added n-move rule. I didn't mention the triple repetition rule.

The fact is that when @tygxc asserts that his "positions" transpose, they don't. We might as well be accurate.

It also seems to me that you're also ignoring the fact that the triple move rule can be ignored in solving only with a procedure correctly designed to do that. Apart from the fact that @tygxc's procedure isn't actually designed to solve chess anyway it doesn't appear to me to conform with that. He's happy to accept a draw by SF under the triple move rule as proving a draw even when the initial occurrence was actually a win.