@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.
@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.
If we're discussing the total number of games, assigning different probabilities to the promotions makes no sense.
Normally promotion is to a queen.
Normally in what? In the set of all games?
In rare cases there is an underpromotion to a knight for its unique properties.
If we're talking about the set of all games no games are ruled out on the grounds of some ulterior purpose.
In nearly all of these cases a knight has been previously captured.
Proof?
Underpromotions to rook or bishop only make sense to avoid stalemate.
If we're talking about the total number of games, what makes sense to you is quite irrelevant. Or what makes sense to Syzygy, which is different from what makes sense to you, as I've pointed out before.
All the moves in the above are optimal according to the Syzygy tablebase. All the moves are perfect. all the moves are in accordance with a weak solution of the initial position.
@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.
@8804
"If we're discussing the total number of games" ++ We are not discussing total number of games, but total number of positions relevant to weakly solving Chess, i.e. 10^17.
You may be talking about that - but you talk almost 100% nonsense.
We are not.
@Optimissed's post to which I responded referred specifically to, "a vanishingly small proportion of the total". (My italics.)
"Normally in what?" ++ Take 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.
No.
It's not relevant to what we were talking about (or what you are talking about for that matter).
You can do it if you like.
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.
The subject of the posts on which you are commenting is whether games with more than two under-promotions constitute a vanishingly small proportion of the total.
"Proof?" ++ Check any game data base and count how many cases with 3 or more knights.
I seem to remember somebody already pointed out you have no concept of "proof", so I'll refrain from pointing it out again. It gets a bit repetitive .
"what makes sense to you" ++ Underpromoting to rook or bishop cannot be optimal play, except for the winning side in an already winning position.
Do you ever think before you type? Try again.
@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.
@8806
"[Reinserted for context: In reality, games with more than two under-promotions will be constitute [sic]] a vanishingly small proportion of the total" ++ That is correct.
I gave an argument in #8797 that would suggest it's very far from the truth. What would be your counter-argument?
"It's not relevant to what we were discussing [Reinserted for context: (or what you are talking about for that matter)}." ++ It is. This thread is about weakly solving Chess.
Obviously you have reading difficulties. You can't have overlooked my post #8806; it's the one you're responding to.
It was made abundantly clear that, "what we were discussing", referred specifically to what @Optimissed and I were discussing not what the thread is about in general. (Why you conveniently dropped the section I reinserted above?)
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.
Pretty Polly, Pretty Polly, but not relevant to what @Optimissed and I were discussing.
In reality, games with more than two under-promotions will be constitute a vanishingly small proportion of the total. However, there really doesn't seem to be any point in ignoring it. Sure, it's an addition to the calculations that have to be done; but not exponentially so. So why is he resisting? This normal continuation is something that has to be analysed. Is it because he didn't think of it first?
It really shouldn't be hard to teach an algorithm to recognise situations where Queen promotion is not appropriate. It would just be a quick check on all promotions and all prospective promotions.
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.
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.
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.
You also still haven’t addressed the fact that I have objectively proven that the strategy stealing method cannot work.
I wish you'd concentrate and stop introducing irrelevancies, supported by your bad thinking. I already told you that you have not objectively proven anything. You made an assertion that "it's what other people think", which is not a proof.
You need to learn to argue honestly, as a precursor to learning to argue correctly. Correctness doesn't stand alone and all you're doing is ganging up against a person who doesn't really know what he's doing, with other people who don't pick you up on your errors.
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.
@8778
"alpha zero game database used human-set openings"
++ Where in Figure 2 did you read that?
“Figure 2. AlphaZero self-play game outcomes under different time controls. As moves are determined in a deterministic fashion given the same conditions, diversity was enforced by sampling the first 20 plies in each game proportional to their MCTS visit counts.”
@8821
++ Yes, the openings were indeed enforced for diversity. It makes sense.
If they had 10,000 games with say the Berlin Defense of the Ruy Lopez, then somebody could object that the Sicilian or the Queen's Gambit would yield entirely different results.
If it doesn’t matter then why did you object so vehemently?
Again, where does it show the amount of errors per game?
@8823
The errors per game follow from the number of decisive games.
10,000 games at 1 s/move: 11.8% decisive games.
Thus the probability of an odd number of errors must be 11.8%.
Assume a Poisson distribution of the errors / game.
That is a plausible assumption also used for radioactive decay, phone calls etc.
0.118 = Poisson (1; lambda; 0) + Poisson (3; lambda, 0) + Poisson (5; lambda; 0) + ...
Solution: lambda = 0.134 errors / game average
Decisive games with 1 error: 1176
Decisive games with 3 errors: 4
Decisive games with 5 or more errors: 0
1,000 games at 1 min/move: 2.1% decisive games.
0.021 = Poisson (1; lambda; 0) + Poisson (3; lambda, 0) + Poisson (5; lambda; 0) + ...
Solution: lambda = 0.021 errors / game average
Decisive games with 1 error: 21
Decisive games with 3 or more errors: 0
@8823
The errors per game . . .
You can't reliably count errors per game without having solved chess first.
“Thus the probability of an odd number of errors must be 11.8%. “
It could be even too
“Assume a Poisson distribution of the errors / game.
That is a plausible assumption…”
it actually isn’t a plausible assumption. It assumes rare events that aren’t interconnected. Errors are rare and connected
.
In reality, games with more than two under-promotions will be constitute a vanishingly small proportion of the total.
75% of promotions in the total might be expected to be under-promotions. More than 40% of games with three promotions can be expected to have have more than two under-promotions and the percentage will rise very quickly for games with a higher number of promotions.
Are you not mixing up games, perfect games and games that @tygxc might consider sensible?
...
It really shouldn't be hard to teach an algorithm to recognise situations where Queen promotion is not appropriate. It would just be a quick check on all promotions and all prospective promotions.
Please give your detailed description of such an algorithm. It is likely to be very interesting (and apparently really not hard).
Although you have previous expertise, all your criticisms are facetious, to draw attention away from your incapability of thinking well on the subject. I won't ask for a proof of 75% of promotions in the total might be expected to be under-promotions, because statistically it would be much more than that and realistically, far less. Maybe you guessed at a median between one extreme and reality. You're talking rubbish, as usual. All you want to do is draw attention towards yourself and away from correct appraisals, which, obviously, you cannot make. That's why you and ty are almost identical.
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?