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Aberg variation of Capablanca's Chess. Different setup and castling rules. (10x8, Cells: 80) [All Comments] [Add Comment or Rating]
H.G.Muller wrote on Sun, Apr 20, 2008 10:21 PM UTC:
'The quality of computer play correlates strongly as a function of ply
depth completion which, in turn, is a function of time where exponentially
greater time is generally required to complete each successive ply.'

Exactly. So the ply depth depends only logarithmically on search time,
which is VERY WEAKLY.

So if you had wanted to show any understanding of the matter at hand, you
should have written RIGHT! in stead of WRONG! above it...

'When you speak of what is needed to 'win the game' you are fixating
upon the mating power of pieces which translates to endgame relative piece
values- NOT opening game or midgame relative piece values.  '

Absolute nonsense. Most Capablanca Chess games are won by annihilation of
the opponents Piece army, after which the winning side can easily push as
many Pawns to promotion as he needs to perform a quick mate.
Closely-matched end-games are relatively rare, and mating power almost
plays no role at all. As long as the Pawns can promote to pieces with
mating power, like Queens.

Your gobbledygook about 'suppreme piece enhancements' seems to
completely undermine your own theory. What are you saying? That the values
you gave below should _never_ be used, because they will unavoidably get
bonuses as the other pieces are traded, so that one should include the
bonuses beforehand? That would be an admission that I correctly recognized
the numbers you gave as useless nonsense, as they apparently do not include
these always-present bonuses. In fact it would mean the table you give are
not piece values at all. Piece values are by definition additive
quantities, the difference of which for both sides tell you who is likely
winning (all positional factors being equal). Non-linear corrections to
that should average out to zero over all piece combinations, or you would
be hiding part of the piece values in these bonuses.

But lets cut the beating around the bush, and give us a clear statement
about the following:
1) If I delete, from the Capablanca opening setup, Ac8, Nb1 and Bc3. You
are now claiming that this gives a winning advantage to white, and that
apparently the values of A, B and N you give below do not apply (as they
would sum up to approximate equality)? So what are the total piece values
of A, B and N in that position (including all bonuses)?
2) Give us a position where the values given below (without any bonuses)
would apply, in a situation where they do not automatically cancel because
both sides have equal material of that type.

As to the alleged similarity of your model to worthless nonsense, the
litmus test is if that model can (statistically) predict results of games
(like the Elo system can for having different players, rather than
different material). So it is really very simple:

Take the position:
rncbqkbcnr/pppppp1ppp/10/10/10/10/PPPPPPPPPP/RNABQKBANR w KQkq - 0 1
Now which side does your piece-value model predict has the advantage here,
and how big is it, compared to an advantage consisting only of Pawns (and,
just to be sure, does that 'advantage' mean he will win more often, like
the advantage of the first move will make white win more often than black,
or do you consider it an 'advantage' that he loses more often?) Then
play that position between equally strong opponents of your own choice
(i.e. opponents that score 50-50 from the normal Capablanca opening
setup), with a time control of your choice, for as many games as needed to
get a statistically significant measure for the deviation of the score
percentage from 50%, and see if this matches the prediction of your
theory.

You see, that is the nice thing about science. People don't have to take
my or your word for it, to know if your theory is any good. Even if you
avoid doing such tests out of fear for being utterly falsified, anyone
else can do it too. And when your theory predicts that the Chancellors
have the advantage here, because it is 'ridiculous' to assume A and C
are approximately equally strong, they will all discover soon enough that
it is actually your theory which is ridiculous nonsense.

I have done that test. So I know the answer. You, apparently, have not...