btickler wrote:
tygxc wrote:
Back on topic after the spamming spree of the trolls.
...
...
5. Lol, do your math again.
...
Oh my God! Don't encourage him.
Chess will never be solved, here's why
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Oct 2, 2022
0
#4423
@MARattigan, here's another go.
A 32 piece tablebase of basic chess positions (with ply count to irreversible change for all positions that are not drawn) provides a strong solution for chess-with-an-N-move-rule-but-no-repetition-rule. (The repetition rule is only there to make games short enough for humans, not to change the results of basic positions).
Answers from this tablebase are just like the ones with a repetition rule except when you are evaluating a position that is not (basic-chess) drawn but which has been reached by repeating positions in the same reversibility-equivalance-class as the position of interest.
The point about equivalence classes is worth noting. As soon as an irreversible move occurs, previous repetitions stop being of relevance.
Further note that a basic chess tablebase provides a strong solution of chess with an N-move rule with the single constraint that for non-basic-chess-drawing positions, there has not been unnecessary faffing around in the equivalence class of the position of interest (this could have racked up the ply count so that there were not enough ply left to avoid a draw according to the N-move rule. Note also that merely replacing N by 2*N prevents this being a problem. If M is the longest ply that could ever be needed and N>M and an N-move rule is in force, then if a basic-chess-winning position has not already breached the N-move rule, another M-moves will suffice to play an irreversible move, if the position was winning in basic chess.
This shows a close relationship between the strong solutions of basic chess, N-move chess (big enough N) and chess with a repetition rule as well. They are all the same with the exception of basic-chess winning positions where there have been repetition of positions in the reversibility-equivalence-class of the position of interest.
Oct 2, 2022
0
#4424
tygxc wrote:
@5808
"how to compute tromps number"
++ Here is the explanation
https://github.com/tromp/ChessPositionRanking
Tromp counted exactly 8726713169886222032347729969256422370854716254 positions with his Haskell program.
Then he randomly sampled 1,000,000 of those positions and he found 56011 of these legal.
Thus he arrived at
8726713169886222032347729969256422370854716254 * 56011 / 1000000 = 4.82 * 10^44.
As said this should be divided by 4 to account for up / down and left / right symmetrical positions: mirror images with exactly the same game state and exactly the same mirror image moves.
Of course you understand that this is a statistical estimate. As it stands, from the evidence we can deduce the number of legal positions lies between:
56011
and
8726713169886222032347729969256422370854716254 - 1,000,000 + 56011
and there is a calculable non-zero probability of each being true.
The probability of the number being enormously greater than Tromp's number is low. But non-zero.
[Of course other directly observed and verified positions boost the lower bound, but the upper bound can only be reduced statistically (i.e. with some uncertainty]
i had come to a conclusion that i had make a mistake. I thank tygxc for showing it to me.
@5832
Tromp uses the binomial distribution to arrive at 4.82 +- 0.03 * 10^44.
You can see that the probability of either 56011, or 8726713169886222032347729969256422370854716254 - 1,000,000 + 56011
are vanishingly small.
Oct 3, 2022
0
#4427
tygxc wrote:
@5832
Tromp uses the binomial distribution to arrive at 4.82 +- 0.03 * 10^44.
You can see that the probability of either 56011, or 8726713169886222032347729969256422370854716254 - 1,000,000 + 56011
are vanishingly small.
They are very small. Smaller divergences from the mean prediction less so, of course.
The probability of a possible hypothesis is non-zero whenever the probability of the evidence conditional on the hypothesis is non-zero. That is certainly true here (the actual number can be calculated easily).
[To be absolutely precise, I am assuming discrete events here, which is always the case in the topic in question].
Oct 3, 2022
0
#4428
ChessFlair01 wrote: (topic "Chess Will Never Be Solved. Why?", but I can't post there.)
This forum is kinda stuck at 500, ey :/
Why
Could it be because you've blocked everybody?
Or maybe people think that by calling your thread "Chess Will Never Be Solved. Why?" when there's already a thread "Chess will never be solved, here's why" going, you're just trying to confuse everybody.
Oct 3, 2022
0
#4429
Elroch wrote:
@MARattigan, here's another go.
A 32 piece tablebase of basic chess positions (with ply count to irreversible change for all positions that are not drawn) provides a strong solution for chess-with-an-N-move-rule-but-no-repetition-rule. (The repetition rule is only there to make games short enough for humans, not to change the results of basic positions).
Agreed.
Answers from this tablebase are just like the ones with a repetition rule except when you are evaluating a position that is not (basic-chess) drawn but which has been reached by repeating positions in the same reversibility-equivalance-class as the position of interest.
More or less agreed.
It depends on what you mean by "answer". The Syzygy tablebase takes only basic rules positions and those don't determine a result with an N-move rule in effect. The game-theoretic result shown on the Syzygy site may say "win" in drawn positions, but the moves will still reach the actual game-theoretic result in the absence of an n-fold repetition rule (not necessarily otherwise).
The point about equivalence classes is worth noting. As soon as an irreversible move occurs, previous repetitions stop being of relevance.
Also agreed.
Further note that a basic chess tablebase provides a strong solution of chess with an N-move rule with the single constraint that for non-basic-chess-drawing positions, there has not been unnecessary faffing around in the equivalence class of the position of interest (this could have racked up the ply count so that there were not enough ply left to avoid a draw according to the N-move rule.
Not agreed for two reasons.
Black to move, ply count 0
Nalimov (basic rules chess tablebase) gives this as mate in 85.
With a 50 move rule in place (N=50) it needs 108 moves. (Syzygy will take at least 128 against most obstinate defence).
However you define "faffing around" there can have been none in the equivalence class of the position, because it's ply count 0, so there hasn't been no nothing.
The basic rules tablebase in that case doesn't even provide a weak solution of the position.
Secondly a strong solution takes a position as a parameter. (Position in my sense meaning game state, not @tygxc's which only represents game state under basic rules.)
For a strong solution of chess that parameter is the starting position. For a strong solution with parameter the given position, a weak solution of all positions reachable from that position must be given. That is it must provide for any amount of faffing about (so long as it's legal) after the position occurs.
Note also that merely replacing N by 2*N prevents this being a problem. If M is the longest ply that could ever be needed and N>M and an N-move rule is in force, then if a basic-chess-winning position has not already breached the N-move rule, another M-moves will suffice to play an irreversible move, if the position was winning in basic chess.
Are you not assuming that the number of moves to mate is independent of the N in the N move rule there?
This shows a close relationship between the strong solutions of basic chess, N-move chess (big enough N) and chess with a repetition rule as well.
Obviously with big enough N the first two are the same. Don't follow the bit about repetition. Syzygy for example will give wrong moves as well as wrong evaluations with the triple repetition rule in effect.
They are all the same with the exception of basic-chess winning positions where there have been repetition of positions in the reversibility-equivalence-class of the position of interest.
As mentioned above; for a strong solution of the position of interest you have to take account not just of repetitions that have occurred, but all possible repetitions that could subsequently legally occur, whether or not in the same equivalence class (which set of positions will also depend on the n in the n-fold repetition rule, should there be one).
@5844
"Position in my sense meaning game state, not @tygxc's"
++ Diagram = location of all men on the board
Position = Diagram + side to move, castling rights, en passant possibility
'9.2.2 Positions are considered the same if and only if the same player has the move, pieces of the same kind and colour occupy the same squares and the possible moves of all the pieces of both players are the same. Thus positions are not the same if:
9.2.2.1 at the start of the sequence a pawn could have been captured en passant
9.2.2.2 a king had castling rights with a rook that has not been moved, but forfeited these after moving. The castling rights are lost only after the king or rook is moved.'
https://handbook.fide.com/chapter/E012018
Node = position with its evaluation and history, i.e. castling rights, repetition of moves, move turn
https://chessify.me/blog/nps-what-are-the-nodes-per-second-in-chess-engine-analysis
Oct 3, 2022
0
#4431
@tygxc
The fact that 9.2.2 defines when two positions are regarded as equal for the purposes of article 9 doesn't alter the fact that the equivalence class of positions so defined doesn't determine if a member of the class (one of your "positions") is won, drawn or lost under competition rules.
Your definition is useless for any solution of chess under competition rules. But it is adequate for chess under basic rules. You have many times made it clear that you cannot distinguish between the two games.
Incidentally no show yet for your calculation of the theoretical result and error rates in my games here. Are you still working on it?
Will it take more than 5 years?
You just need to do that then we can ignore your "calculations". I think your "calculations" are out by several orders of magnitude in those games. See what you make it.
By the way, "node" is just a vertex of a graph. It's meaning depends on what graph you're talking about. That could be the game tree graph, a reduced game tree graph where nodes with the same possible forward actions are merged into one (when a full history is unnecessary), the graph of game states that your solving procedure will construct (by the way when do we get a proper description of that) a tree of pairs (b,n) where b represents a basic rules position and n a distance to an objective in a tablebase construction and so on.
Oct 3, 2022
0
#4432
MARattigan wrote:
Elroch wrote:
@MARattigan, here's another go.
A 32 piece tablebase of basic chess positions (with ply count to irreversible change for all positions that are not drawn) provides a strong solution for chess-with-an-N-move-rule-but-no-repetition-rule. (The repetition rule is only there to make games short enough for humans, not to change the results of basic positions).
Agreed.
Answers from this tablebase are just like the ones with a repetition rule except when you are evaluating a position that is not (basic-chess) drawn but which has been reached by repeating positions in the same reversibility-equivalance-class as the position of interest.
More or less agreed.
It depends on what you mean by "answer".
What different moves will achieve with optimal play.
The Syzygy tablebase takes only basic rules positions and those don't determine a result with an N-move rule in effect.
I was assuming (but failed to state on this occasion) that N is big enough. Otherwise it is trivial that a position can be a basic chess win or an N-move rule chess draw. And of course, you need a bit more leeway if you have wasted moves within the current equivalence class to get to the position when you consult the oracle!
The game-theoretic result shown on the Syzygy site may say "win" in drawn positions, but the moves will still reach the actual game-theoretic result in the absence of an n-fold repetition rule (not necessarily otherwise).
Only if there have been previous (n-1) repetitions of positions in the equivalence class of the position of interest, and the wrong path taken thereafter, such that the only right paths require revisiting that position. (You can think of this within the equivalence class as a network of legal (reversible) moves where all paths to the winning nodes (where the equivalence class is exited) have been blocked by unnecessary repetitions of positions).
The point about equivalence classes is worth noting. As soon as an irreversible move occurs, previous repetitions stop being of relevance.
Also agreed.
Further note that a basic chess tablebase provides a strong solution of chess with an N-move rule with the single constraint that for non-basic-chess-drawing positions, there has not been unnecessary faffing around in the equivalence class of the position of interest (this could have racked up the ply count so that there were not enough ply left to avoid a draw according to the N-move rule.
Not agreed for two reasons.
Black to move, ply count 0
Nalimov (basic rules chess tablebase) gives this as mate in 85.
With a 50 move rule in place (N=50) it needs 108 moves. (Syzygy will take at least 128 against most obstinate defence).
However you define "faffing around" there can have been none in the equivalence class of the position, because it's ply count 0, so there hasn't been no nothing.
The basic rules tablebase in that case doesn't even provide a weak solution of the position.
Only because N is too small. I have somewhere referred to N-move rules that are big enough not to interfere with the values of basic chess positions when drawing rules are added, except where previous play in the equivalence class has either spent too much of the ply count or blocked routes to the exit(s) by (N-1)-repetition.
Secondly a strong solution takes a position as a parameter. (Position in my sense meaning game state, not @tygxc's which only represents game state under basic rules.)
Yes, my point is that you can lump together the large class of such states where previous play has not scuppered the result. There are only two things that can go wrong - all paths to the exit(s) from the equivalence class blocked by (N-1)-repetitions or not enough ply left to get to the exit(s). [This is from the winning side - the inverse occurs from the losing side. A basic chess loss can only be converted to a win in chess with drawing rules by the opponent having scuppered their chance to exit the equivalence class].
For a strong solution of chess that parameter is the starting position. For a strong solution with parameter the given position, a weak solution of all positions reachable from that position must be given. That is it must provide for any amount of faffing about (so long as it's legal) after the position occurs.
Note also that merely replacing N by 2*N prevents this being a problem. If M is the longest ply that could ever be needed and N>M and an N-move rule is in force, then if a basic-chess-winning position has not already breached the N-move rule, another M-moves will suffice to play an irreversible move, if the position was winning in basic chess.
Are you not assuming that the number of moves to mate is independent of the N in the N move rule there?
The N-move rule is global. Moves-to-mate is not as theoretically interesting as the moves to traverse equivalence classes in a winning position. This might be good to included in a tablebase!
Interesting (confident) guess: the max N necessary to traverse a (basic chess) winning equivalence class is probably greater than the max N needed for a winning strategy for the start position.
This shows a close relationship between the strong solutions of basic chess, N-move chess (big enough N) and chess with a repetition rule as well.
Obviously with big enough N the first two are the same.
Actually no! Whatever N is used in the (global) rule, you could have wasted near that number of moves in the equivalence class before reaching a position of interest, and this could stop you having time to exit, so change a win to a draw. I think what you mean is if you could change N during the game that would repair any harm. Likewise, changing a past N-repetition rule to an (N+1) repetition rule on the fly will heal all past suboptimal repetition!
Don't follow the bit about repetition. Syzygy for example will give wrong moves as well as wrong evaluations with the triple repetition rule in effect.
They are all the same with the exception of basic-chess winning positions where there have been repetition of positions in the reversibility-equivalence-class of the position of interest.
As mentioned above; for a strong solution of the position of interest you have to take account not just of repetitions that have occurred, but all possible repetitions that could subsequently legally occur (which will also depend on the n in the n-fold repetition rule).
This is not true once you have exited the current equivalence class, since shortest winning paths obviously don't repeat positions ever. (Similar statement applies to the losing side).
I have noticed there is a poorly chosen use of N by me both for a ply-count drawing rule and for a repetition rule. I think they don't get confused though.
Oct 3, 2022
0
#4433
Elroch wrote:
For those interested and wondering about the technicolor; consecutively Elroch: white, MARattigan: orange, Elroch: blue, MARattigan: green
MARattigan wrote:
Elroch wrote:
@MARattigan, here's another go.
A 32 piece tablebase of basic chess positions (with ply count to irreversible change for all positions that are not drawn) provides a strong solution for chess-with-an-N-move-rule-but-no-repetition-rule. (The repetition rule is only there to make games short enough for humans, not to change the results of basic positions).
Agreed.
You don't need to refer to ply count in a 32 man DTM (basic rules) position to get a strong solution of the basic rules game or a game with an N move rule but no n-fold repetition rule for large enough N.
Answers from this tablebase are just like the ones with a repetition rule except when you are evaluating a position that is not (basic-chess) drawn but which has been reached by repeating positions in the same reversibility-equivalance-class as the position of interest.
More or less agreed.
It depends on what you mean by "answer".
What different moves will achieve with optimal play.
The Syzygy tablebase takes only basic rules positions and those don't determine a result with an N-move rule in effect.
I was assuming (but failed to state on this occasion) that N is big enough. Otherwise it is trivial that a position can be a basic chess win or an N-move rule chess draw. And of course, you need a bit more leeway if you have wasted moves within the current equivalence class to get to the position when you consult the oracle!
The game-theoretic result shown on the Syzygy site may say "win" in drawn positions, but the moves will still reach the actual game-theoretic result in the absence of an n-fold repetition rule (not necessarily otherwise).
Only if there have been previous (n-1) repetitions of positions in the equivalence class of the position of interest, and the wrong path taken thereafter, such that the only right paths require revisiting that position. (You can think of this within the equivalence class as a network of legal (reversible) moves where all paths to the winning nodes (where the equivalence class is exited) have been blocked by unnecessary repetitions of positions).
I think we are still only more or less in agreement here. The main point is, "what do you understand by a strong solution?".
I was a tad confused about when your N was to be regarded as big enough to be irrelevant and when not, because in some of the sequel N appears to be assumed less.
It's not so much that I disagree with your previous paragraph. Rather, I think it's incomprehensible with my understanding of "strong solution", which was what @tygxc was talking about in the post which kicked off the discussion. (To be precise I should say the term he was using; I don't think he knows what he's talking about.)
The point about equivalence classes is worth noting. As soon as an irreversible move occurs, previous repetitions stop being of relevance.
Also agreed.
Further note that a basic chess tablebase provides a strong solution of chess with an N-move rule with the single constraint that for non-basic-chess-drawing positions, there has not been unnecessary faffing around in the equivalence class of the position of interest (this could have racked up the ply count so that there were not enough ply left to avoid a draw according to the N-move rule.
Not agreed for two reasons.
Black to move, ply count 0
Nalimov (basic rules chess tablebase) gives this as mate in 85.
With a 50 move rule in place (N=50) it needs 108 moves. (Syzygy will take at least 128 against most obstinate defence).
However you define "faffing around" there can have been none in the equivalence class of the position, because it's ply count 0, so there hasn't been no nothing.
The basic rules tablebase in that case doesn't even provide a weak solution of the position.
Only because N is too small. I have somewhere referred to N-move rules that are big enough not to interfere with the values of basic chess positions when drawing rules are added, except where previous play in the equivalence class has either spent too much of the ply count or blocked routes to the exit(s) by (N-1)-repetition.
Secondly a strong solution takes a position as a parameter. (Position in my sense meaning game state, not @tygxc's which only represents game state under basic rules.)
Yes, my point is that you can lump together the large class of such states where previous play has not scuppered the result. There are only two things that can go wrong - all paths to the exit(s) from the equivalence class blocked by (N-1)-repetitions or not enough ply left to get to the exit(s). [This is from the winning side - the inverse occurs from the losing side. A basic chess loss can only be converted to a win in chess with drawing rules by the opponent having scuppered their chance to exit the equivalence class].
And my point is you don't seem to be talking about a strong solution. A strong solution of a position that is not scuppered must include a weak solution of all scuppered positions that can be reached from it.
The only rôle, when you talk about a strong solution of a position, that the position itself has, is to determine the positions that can be legally reached from it (including positions outside of its equivalence class), so I found it difficult to interpret some of what you said.
The "meat" of my objection is my following paragraph, not my previous one.
For a strong solution of chess that parameter is the starting position. For a strong solution with parameter the given position, a weak solution of all positions reachable from that position must be given. That is it must provide for any amount of faffing about (so long as it's legal) after the position occurs.
Note also that merely replacing N by 2*N prevents this being a problem. If M is the longest ply that could ever be needed and N>M and an N-move rule is in force, then if a basic-chess-winning position has not already breached the N-move rule, another M-moves will suffice to play an irreversible move, if the position was winning in basic chess.
Are you not assuming that the number of moves to mate is independent of the N in the N move rule there?
The N-move rule is global. Moves-to-mate is not as theoretically interesting as the moves to traverse equivalence classes in a winning position. This might be good to included in a tablebase!
See Haworth's paper on DTR (distance to the rule) tablebases. General consensus was to go for same again with more men given limited resources, so none have been produced.
But I should have said, "Are you not assuming that the number of moves to traverse equivalence classes in a winning position is independent of the N in the N move rule there?". Which brings us back to the KNNKP position I posted where the questions are essentially the same so far as White wins go.
What would you say would be the number of moves required to mate would be with a 51 move rule?
Interesting (confident) guess: the max N necessary to traverse a (basic chess) winning equivalence class is probably greater than the max N needed for a winning strategy for the start position.
Assuming there is a winning strategy from the start position for the N you have in mind. (E.g. If n=1 I think you would struggle to find one or maybe even with N=50.)
But a winning strategy for which version of chess? I would guess (not confident) the maximum length mate from the starting position under competition rules is already less than the number of moves required to traverse some 7 man positions, I'd also guess (even less confident) the same applies under basic rules chess.
This shows a close relationship between the strong solutions of basic chess, N-move chess (big enough N) and chess with a repetition rule as well.
Obviously with big enough N the first two are the same.
Actually no! Whatever N is used in the (global) rule, you could have wasted near that number of moves in the equivalence class before reaching a position of interest, and this could stop you having time to exit, so change a win to a draw. I think what you mean is if you could change N during the game that would repair any harm. Likewise, changing a past N-repetition rule to an (N+1) repetition rule on the fly will heal all past suboptimal repetition!
No I didn't mean that. For large enough N a DTM (basic rules) and DTMN (N move rule) generation would result in identical tablebases and be strong solutions of both with no n-fold repetition rule.
I still think you're talking at cross purposes regarding the term "strong solution".
Don't follow the bit about repetition. Syzygy for example will give wrong moves as well as wrong evaluations with the triple repetition rule in effect.
They are all the same with the exception of basic-chess winning positions where there have been repetition of positions in the reversibility-equivalence-class of the position of interest.
As mentioned above; for a strong solution of the position of interest you have to take account not just of repetitions that have occurred, but all possible repetitions that could subsequently legally occur (which will also depend on the n in the n-fold repetition rule).
This is not true once you have exited the current equivalence class, since shortest winning paths obviously don't repeat positions ever. (Similar statement applies to the losing side).
No, but a strong solution has to work when the paths are not shortest (read your Wikipedia article).
I have noticed there is a poorly chosen use of N by me both for a ply-count drawing rule and for a repetition rule. I think they don't get confused though.
Oct 3, 2022
0
#4434
To tone down the classic confusing depth of discussion
MARattigan wrote:
And my point is you don't seem to be talking about a strong solution. A strong solution of a position that is not scuppered must include a weak solution of all scuppered positions that can be reached from it.
I have not seen anyone (except for in this discussion) refer to strong solutions of states. But it would be reasonable to define it as "strong solution of the game like chess but with specified starting state". I think we can agree on that.
A basic chess tablebase plus "ply to exit equivalence class" provides every winning strategy that avoids increasing the ply to exit equivalence class.
You can't have a (deterministic) winning strategy with a loop in basic chess. However, the above paragraph seems clearly not to be all winning strategies. I assert that there are some other winning basic chess strategies that in some positions unnecessarily increase the ply to exit the equivalence class but do so in a way that does not change the result (because the new position has a forced exit that never returns to the position where the ply was increased).
So I am not disagreeing with you (using the reasonable definition of strong solution of a state), but pointing out that the basic-chess-tablebase-with-ply-count-to-irreversible-move (for winning or losing moves) provides something which is very substantial for versions of chess with additional drawing rules, and shows they are very closely related games.
Question: is ply to mate good enough for this purpose instead of the apparently superior ply to irreversible move?
@5840
"is ply to mate good enough instead of the apparently superior ply to irreversible move?"
++ Just draw / no draw is good enough. It is obvious whether a no draw is a win or a loss.
Solving chess is hopping from one draw to another and avoiding the pitfalls to losses.
A win is a win. There are no bonus points for a faster checkmate.
Oct 4, 2022
0
#4436
Your answer is incorrect with respect to the moves in a strategy, because it does not guarantee achieving the best result.
Consider a rook ending. Every move that does not lose the rook or stalemate is winning. So just picking one of these moves without consideration of the ply count to an irreversible move (mate in this case) can lead to infinite play without reaching mate. This either runs into a draw rule like 3-fold repetition or 50-move or (in basic chess rules) an infinite game (which is outside of combinatorial game theory but is clearly not a win).
[Note that a slightly subtle possibility MARattigan and I have been discussing is that playing inefficiently can lead to a position where no draw rule has been triggered but it is no longer possible to win without triggering one. This can occur either because the number of moves to mate with optimal play (in the sense that the winner aims to mate quickly and the loser aims to lose slowly) exceeds the number of moves left until the 50 move rule is triggered, or because all paths to mate pass through positions that have already been visited twice].
@5842
"it does not guarantee achieving the best result"
++ It does guarantee that. The best result is a draw.
Even after a mistake, e.g. 1 g4? black is winning.
That means in every position there is at least 1 move that keeps the win.
Even if a position is repeated once or twice, there is a way out with a winning move.
"Consider a rook ending. Every move that does not lose the rook or stalemate is winning.
So just picking one of these moves without consideration of the ply count to an irreversible move (mate in this case) can lead to infinite play without reaching mate."
++ Maybe some position gets repeated twice,
but then in a winning position there exists at least 1 move that wins and avoids the repetition.
"a draw rule like 3-fold repetition" ++ You can avoid the 3rd repetition in a won position
"or 50-move" ++ 50-moves rule plays no role in solving chess
"an infinite game (which is outside of combinatorial game theory but is clearly not a win"
++ The 3-fold repetition rule is essential
"all paths to mate pass through positions that have already been visited twice"
++ Then it was an error to visit the pivotal position twice.
Oct 4, 2022
0
#4438
tygxc wrote:
@5842
"it does not guarantee achieving the best result"
++ It does guarantee that. The best result is a draw.
Sigh.
The discussion related to positions that were basic chess winning. Your own post said:
"A win is a win. There are no bonus points for a faster checkmate."!
And your claim about the best result is an over-confident belief, not a confirmed fact (the difference between high confidence and certainty is HUGE). I have no expectation that you will ever understand that!
@5844
Sigh.
"The discussion related to positions that were basic chess winning."
++ There is no such thing as basic chess. The 3-fold repetition rule is essential in solving chess. The 50-moves rule plays no role in solving chess.
In a position that is winning like 1 g4? or 1 e4 e5 2 Ba6? there is always a winning move that avoids the 3-fold repetition.
"the difference between high confidence and certainty is HUGE" ++ Chess is a draw is certain.
@5808
"how to compute tromps number"
++ Here is the explanation
https://github.com/tromp/ChessPositionRanking
Tromp counted exactly 8726713169886222032347729969256422370854716254 positions with his Haskell program.
Then he randomly sampled 1,000,000 of those positions and he found 56011 of these legal.
Thus he arrived at
8726713169886222032347729969256422370854716254 * 56011 / 1000000 = 4.82 * 10^44.
As said this should be divided by 4 to account for up / down and left / right symmetrical positions: mirror images with exactly the same game state and exactly the same mirror image moves.