Chess will never be solved, here's why

Less positions, yes, never said otherwise.  10^17 positions?  No.  Just no.

To be pedantic, it's not less positions. It's fewer.

Also, the solution of checkers which has more directionality than chess (hence less transposition until the deep endgame, helping "small" solutions) only permitted a reduction by a power of around 2/3 rather than 1/2 to the state space complexity.

This point merits emphasis.

Unfortunately, it refutes the one node solution as well as others.

A discussion with Ty, who may be very similar to a participant in this group.

Me: you claim positions with underpromoted pieces cannot be part of a solution.

Ty: yes

Me: but underpromotion occasionally occurs as the only best move in the tiny sample of master games that comprises classical chess theory?

Ty: yes

Me: so positions with an underpromoted piece can't really be ignored, can they?

Ty: maybe not with one underpromoted piece.

Me: is there any reason that a position with one underpromoted piece can't lead to a position where the only best move is to underpromote?

Ty: might be, but I can't think of one. It's unlikely?

Me: right. There is no known reason. Then how about the same with positions with a couple of underpromoted pieces. Why would these not occasionally have a move where the best move is to underpromote? Even one in ten million

Ty: seems unlikely enough to ignore.

Me: so no good reason then?  And so on, so that you need to include positions with multiple underpromotions in a solution of chess, and you have a million times more work to do. Perhaps you should increase your fee?

Ty: [thinking]

what is this chaos

@8792

"positions with underpromoted pieces cannot be part of a solution"
++ The only reason to underpromote to a rook or a bishop instead of a queen is to avoid stalemate. Only the side that is winning has reason to avoid a draw resulting from stalemate.
Positions with 3 or more bishops or rooks on one side cannot be part of a solution of Chess.
Positions with 3 or more knights on one side could be part of a solution of Chess, and underpromotions to knights occur from time to time in perfect games with optimal play from both sides, but only in situations where a knight already has been captured.

The Laws of Chess include:
'3.7.3.4    The player's choice is not restricted to pieces that have been captured previously.'
If we modify this to:
'3.7.3.4modified    The player's choice is restricted to either a queen or pieces that have been captured previously.',
then all perfect games with optimal play from both sides stay exactly the same.

That is why Gourion's 10^37 is a more suitable estimate than Tromp's 10^44.

Since I said, "75% of promotions in the total might be expected to be under-promotions, I won't provide a proof. The observation follows from the fact that there are 4 possible promotions, 3 of which are under-promotions.

You don't say whether you won't ask for a proof because you think its's obvious (which it isn't, quite) or because, as I suspect, you're mixing up games, perfect games and games that @tygxc might consider sensible. The reason you do give ("because statistically it would be much more than that and realistically, far less") doesn't seem to make any sort of sense, but maybe you're talking about alligators as usual. 

Is your algorithm for deciding when a queen promotion is not appropriate (presumably meaning not perfect if you could bring yourself to write the word) coming later, or is it proving harder to formulate than you suggest?

@8802

"the fact that there are 4 possible promotions, 3 of which are under-promotions"
++ The fact that there are 4 possible promotions does not make them equally likely.
Normally promotion is to a queen.
In rare cases there is an underpromotion to a knight for its unique properties.
In nearly all of these cases a knight has been previously captured.
Underpromotions to rook or bishop only make sense to avoid stalemate.

@8804

"If we're discussing the total number of games" ++ We are not discussing the total number of games, but the total number of positions relevant to weakly solving Chess, i.e. 10^17.

"Normally in what?" ++ Query any game data base. Count how many cases of 3, 4, 5, 6 queens there are and how many with 3, 4, 5, 6 knights, how many with 3, 4, 5, 6 bishops or rooks.
3 or 4 queens occur in less than 1% of games. > 3 knights, bishops, rooks do not occur at all.

"no games are ruled out on the grounds of some ulterior purpose"
++ The subject is weakly solving Chess.

"Proof?" ++ Query any game data base and count how many cases with 3 or more knights.

"what makes sense to you" ++ Underpromoting to rook or bishop cannot be optimal play, except for the winning side in an already winning position.

@8806

"a vanishingly small proportion of the total" ++ That is correct.

"It's not relevant to what we were discussing." ++ It is. This thread is about weakly solving Chess. My point is that the Gourion number 10^37 is a better starting point to estimate the time to weakly solving Chess than the Tromp number 10^44 as Tromp allows unlimited underpromotions to pieces not previously captured and Gourion excludes promotions to pieces not previously captured.

The answer, of course, is that Tygxc cannot do that and still hold on to his Sveshnikov number...and he pas amply proven he will sacrifice anything on the altar of that number.

You don't know this, and cannot prove this.  You cannot even know or prove that a single "perfect game" has ever been played.  You cannot prove "optimal play" until you hit a forcing sequence, either.

@MARattigan has explained how my dialog only represented half of the story. My point was that multiple underpromotions for the proponent of a strategy may be rare but can't be justifiably ignored. But for the defender, all of the underpromotions have the same status - they are simply legal moves that absolutely have to be dealt with. No excuses.

Note that an interesting observation is that this means that it is only the case of multiple underpromotions by BOTH sides that can be said to be "unlikely" to occur in a solution. The only realistic way to make this definite is to exhibit such a solution (a la chequers) and say "look, we manage without positions with a lot of underpromotions by both side.

Still wouldn't be definite until you've produced all solutions and counted how many have positions with a lot of underpromotions by both sides (and, of course, set the level that would count as "a lot" and the probability that would count as "likely").

(If the statement were restricted to solutions produced by particular methods, the answers would vary depending on the method.) 

Well, exhibiting one (weak) solution which respected some constraint on the underpromotions would suffice to demonstrate that you can solve chess using only positions respecting that constraint on underpromotions.

Doing it without exhibiting a solution is not feasible.

Bro what you are literally strawmanning what I am saying.  If you continue to make bad faith arguments and claims I’m not going to respond to you as it clearly won’t be worth it.

at least Tygxc bothers to read what I’m saying and learn the context to claims.