#2384
You should write a letter to van den Herik.
No, you should stop talking nonsense. Are you incapable of thinking for yourself?
There's no point in offering to weakly solve chess for us if you don't even know what "weakly solve" means.
#2387
I am perfectly capable of thinking for myself and of seeing that you talk nonsense.
I adopt a generally accepted definition by one of the leading authorities in the world.
You then come up with a nonsense refutation, as usual.
It's not a generally accepted definition. I think you are possibly the only person posting on the the thread that would accept it.
But then you also accepted the Wikipaedia definition with which it is incompatible.
Like the White queen, you find it easy to believe six impossible things before breakfast.
#2389
"It's not a generally accepted definition." ++ Sorry, but it is in the world of game theory.
"I think you are possibly the only person posting on the the thread that would accept it."
++ If that were true, then I would be the only sensible person on the thread.
"But then you also accepted the Wikipaedia definition with which it is incompatible."
++ I was inclined towards accepting the Wikipedia definition, but I got criticised for that, as people did not accept Wikipedia as a reliable source. Anyway there already is enough confusion about many things, so we should at least agree to accept the same definitions as given by reputable sources. If we start to assign different meanings to the same word, then the confusion can only grow. You cannot refute a definition. You can only accept it.
https://reader.elsevier.com/reader/sd/pii/S0004370201001527?token=5EA63AC2A4C94EC9333C82E5E39A3A25210F8FC184F9458F8141645F002E759F7CEABB7B09D64FA30FBEB25ABD8CB6E2&originRegion=eu-west-1&originCreation=20220317135832
#2383
"how do you know that the probability to make a blunder is much less than the probability to make an error?"
++ Because a blunder is a double error. Example: 1 e4 (draw) e5 (draw) 2 Ba6? (lost) Nxa6 (won) 3 Nf3 (lost) Qh4?? (lost). A blunder (??) does in one move what two errors do in two moves. That is why the probability of 1 blunder = probability of 2 (consecutive) errors.
To me this is hypothesis 7. The effect of a blunder is that of two consecutive errors; that does not mean it has the same probability to occur as two consecutive errors. Besides, you are assuming that the probability to make an error is little, so the probability to make two consecutive errors can be neglected. But you infer that the error rate per move is low, by the fact that many games are drawn, as if they contained only a few errors, while you have admitted that:
We do not know exactly how many [ errors and blunders they made ], but we know if it is odd or even. We do not know which moves, but we know their parity.
So you assume that the number of errors per game is low, from that deduce the error rate per move is low, from that a much lower probability of two consecutive errors, and from that you assume that the probability to make a blunder is as low as two consecutive errors, because it has the same effect. If the error rate per move, instead, was not so low, the engine would not really know what it's doing (more on that later), and even if the last assumption was correct, the probability of two consecutive errors or a blunder could not be neglected.
#2389
"It's not a generally accepted definition." ++ Sorry, but it is in the world of game theory.
Er, no it isn't.
"I think you are possibly the only person posting on the the thread that would accept it."
++ If that were true, then I would be the only sensible person on the thread.
I wouldn't call accepting the strategy I gave for White here as a weak solution of the position sensible myself. You've obviously got a funny definition of "sensible" too.
"But then you also accepted the Wikipaedia definition with which it is incompatible."
++ I was inclined towards accepting the Wikipedia definition, but I got criticised for that, as people did not accept Wikipedia as a reliable source. Anyway there already is enough confusion about many things, so we should at least agree to accept the same definitions as given by reputable sources. If we start to assign different meanings to the same word, then the confusion can only grow. You cannot refute a definition. You can only accept it.
https://reader.elsevier.com/reader/sd/pii/S0004370201001527?token=5EA63AC2A4C94EC9333C82E5E39A3A25210F8FC184F9458F8141645F002E759F7CEABB7B09D64FA30FBEB25ABD8CB6E2&originRegion=eu-west-1&originCreation=20220317135832
No point in accepting an obviously flawed definition.
I gave a reasonably succinct definition here when the topic previously came up. I reproduce it:
weakly solved means that for the initial position either a timely strategy has been determined for one player that achieves a win against any opposition, or a timely strategy has been determined for both players that avoids a loss against any opposition.
Perhaps we could agree to assign that meaning.
If so please acknowledge your agreement. You won't get general agreement with either of the definitions you keep posting.
#2391
"The effect of a blunder is that of two consecutive errors; that does not mean it has the same probability to occur as two consecutive errors."
++ Can you agree that the probalitity of an error on move 8 and an error on move 12 is the same as the probability of an error on move 8 and an error on move 9? If not, then would you think it is higher or lower? If yes, then why would it not be the same as a double error on move 8?
"you are assuming that the probability to make an error is little, so the probability to make two consecutive errors can be neglected."
++ I constatate from ICCF results that the vast majority of games end in draws (by agreement, repetition, or table base), and thus many games have an even number of errors and very few have an odd number of errors. The only logical explanation for this fact is that there are more games with 0 errors, less with 1 error, less with 2 errors, less with 3 errors... From that I conclude that the probability of an error in the ICCF World Championship is little indeed. I do not assume, I conclude from the data. You can even replicate that with a previous ICCF World Championship, or with the Yekaterinburg Candidates' Tournament.
"So you assume that the number of errors per game is low, from that deduce the error per move is low, from that a much lower probability of two consecutive errors, and from that you assume that the probability to make a blunder is as low as two consecutive errors, because it has the same effect."
++ No, I count much more draws than decisive games in the ICCF World Championship. From that I deduce there are more games with an even than an odd number of errors. From that I deduce that the number of errors per game must be low.
"the probability of two consecutive errors or a blunder could not be neglected"
++ I do not neglect it. Maybe some of the draws in the ICCF World Championship contain an error (?) by white and an error (?) by black. Maybe some contain an error (?) by white, a blunder (??) by black, and an error (?) by white. However, the probability of 4 errors must be lower than the probability of 2 errors and lower than the probability of 0 errors to explain the data.
I cannot tell for sure that all the drawn games contain no errors and all the decisive games contain 1 error. There may be a few drawn games with 2 or 4 errors and there may be decisive games with 3 or 5 errors.
If you were to look at the tournament table of the championship of your local club, then you would notice far less draws, indicative of a far higher error rate.
@MARattigan
Yes, we have talked about that kind of position, with @Elroch too. Why the move 1. Qa7 is not optimal? Of course, it is not the best bet for White, if he faces a non optimal opponent, but his options are all equally losing. Or it has something to do with the 50-moves rule (but you wrote 1. Qa7)?
It is a perfect move.
But the point is, a strategy for one player to lose is just not what is meant by a weak solution (whatever Prof van den Herik and @tygxc say).
Sorry, @MARattigan, I have posted that before realizing you added another post with a definition of "weak solution"; I have deleted it to better study your previous post. I don't want to deal with both you and @tygxc at the same time, so to reduce my error rate 
.
#2392
"Perhaps we could agree to assign that meaning."
++ No, your personal alternative definition is not generally accepted.
I could propose my own personal alternative definition, which you would not accept either.
"To weakly solve chess is to prove that black can draw against all reasonable white moves."
Let us just accept the generally accepted definition by van den Herik, leading authority in that field.
Otherwise let us accept a definition in Wikipedia,
or from any reputable encyclopedia e.g. Britannica.
If we cannot agree on the meaning of a word,
then we do not start an argument about it, we look it up in a dictionary, e.g. Webster.
#2392
"Perhaps we could agree to assign that meaning."
++ No, your personal alternative definition is not generally accepted.
I could propose my own personal alternative definition, which you would not accept either.
"To weakly solve chess is to prove that black can draw against all reasonable white moves."
Let us just accept the generally accepted definition by van den Herik, leading athority in that field.
Otherwise let us accept a definition in Wikipedia or from any reputable encyclopedia e.g. Britannica.
If we cannot agree on the meaning of a word, then we do not start an argument about it, we look it up in a dictionary, e.g. Webster.
I would say my definition is generally accepted (also by van den Herik - he was just slipshod in the reference you gave).
The Wikipaedia definition is not even consistent.
You say, "If we cannot agree on the meaning of a word ...".
Is there a particular reason you cannot agree on the definition I gave? If so please say what the reason is.
By all means propose your own personal alternative definition (from any source if you can't think of one for yourself). I'm sure people will accept it if it's reasonable. The definitions you've posted to date aren't.
The high draw rate signifies that several of these games are likely ideal games with optimal moves. So when two toddlers play chess and cannot mate each other, these are optimal moves. <-- Fixed
The high draw rate signifies that several of these games are likely ideal games with optimal moves. So when two toddlers play chess and cannot mate each other, these are optimal moves. <-- Fixed
Using the dictionary as bible - is usually regressive.
Counter productive. Reference only.
Its a reference book - not a guide to life.
#2403
You find my personal alternative definition not reasonable but you give no reasons.
You find the definitions is Wikipedia not reasonable.
You find the definition by van den Herik slipshod.
I find your personal alternative definition not reasonable for several reasons.
"my definition is generally accepted (also by van den Herik"
++ Please show the letter where van den Herik accepts your definition.
This discussion about definitions is pointless.
Accept what is written by van den Herik.
Propose something else written in a reputable source.
#2405
I am talking about the World Championship Finals of the ICCF, with ICCF grandmasters with their engines and 5 days per move.
That is not the championship of your local kindergarten.
#2406
If people attach different meanings to the same word differently,
then misunderstanding and confusion results.
For the meaning of a word: look up in a dictionary.
For the meaning of a technical term: look up in an encyclopedia,
or in a publication by an authority in the field.
#2391
"The effect of a blunder is that of two consecutive errors; that does not mean it has the same probability to occur as two consecutive errors."
++ Can you agree that the probalitity of an error on move 8 and an error on move 12 is the same as the probability of an error on move 8 and an error on move 9? If not, then would you think it is higher or lower? If yes, then why would it not be the same as a double error on move 8?
They are not the same thing! You're jumping to conclusions. Engines and humans search and evaluate: the error rate per move is an error rate per search. In the first two cases you have two searches, in the latter just one.
Let's say that P(d) is the probability per day to have a car accident that causes a damage which costs d dollars. Let's say (for the moment, because it is not that simple) that the probability to have two car accidents per week, each causing a damage d, is P²(d). With obvious meaning for the symbols, it is not automagically P(2d)=P²(d)! Where do you read such things? If d=500$, P(1000$)=P²(500$)?
#2384
You should write a letter to van den Herik.