Ratings & Comments

I'm not really a mathematician or stastician - I merely enjoy math and am
somewhat talented at it.

I have read your general theory of piece values - in fact, I think I've
read it roughly ten times, starting back when you were still adding to it. 
I'm afraid I can't tell you how accurate it is, as I feel very much the
midget when it comes to playing chess. (I think I may have read your theory
of piece values more often than I have played chess in the last five
years.)

I've meant to e-mail you with various comments about it for many years now,
but I never got around to it.  This handy comment system makes it easy
enough that I'll finally stop procrastinating, though.  I'll start some new
threads, I think.

I hesitate to mention it because I'm currently working on the revision
(which should suck less), but Fantasy Grand Chess is my chess variant with
different armies.  I didn't analyze things very thoroughly (mostly I just
guessed at what looked right), and mostly assumed values would be the same
on an 8x8 board as 10x10, so it needs work.  (Which is what I'm currently
doing.)  I'm also making changes to help the theme, and dropping it down to
a more manageable four armies.

If there are any other numbers in particular you want me to check, let me
know.  I'm currently calculating for a Crooked Rook, which should be simple
after the Bishop, and then I'm going to do mRcpR and RcpR.

Various and sundry ideas about calculating the value of chess pieces.

First off, it is quite interesting to instead of picking a magic number as
the chance of a square being empty, calculate the value for everything
between 32 pieces on the board and 3 pieces on the board.  Currently I'm
then just averaging all the numbers, and it gives me numbers slightly
higher than using 0.7 as the magic number (for Runners - Knights and other
single step pieces are of course the same).  One advantage of it is that it
becomes easier to adjust to other starting setups - for Grand Chess I can
calculate everything between 40 pieces on the board and 3, and it should
work.  With a magic number I'd have to guess what the new value should be,
as it would probably be higher since the board starts emptier.  One
disadvantage is that I have no idea whether or not the numbers suck. :) 
Interesting embellishments could be added - social and anti-social
characteristics could modify the values before they are averaged, and
graphs of the values would be interesting.  It would be interesting to
compare the official armies from Chess with Different Armies at the final
average and at each particular value.  It might be possible to do something
besides averaging based on the shape of the graph - the simplest idea would
be if a piece declines in power, subtract a little from it's value but
ignore the ending part, assuming that it will be traded off before the
endgame.

Secondly, I'm not sure what to do with the numbers, but it is interesting
to calculate the average number of moves it takes a piece to get from one
square to another, by putting the piece on each square in turn and then
calculate the number of moves it takes to get for there to every other
square.  So for example a Rook (regardless of it's position on the board)
can get to 15 squares in 1 move, 48 squares in 2 moves, and 1 square in 0
move (which I included for simplicity, but which should probably be left
out) so the average would be 1.75.  I've got some old numbers for this on
my computer which are probably accurate, but I no longer know how I got
them.   Here's a sampling:

Knight: 2.83
Bishop: 1.66 (can't get to half the squares)
Rook: 1.75
Queen: 1.61
King: 3.69
Wazir: 5.25
Ferz: 3.65 (can't get to half the squares)

This concept seems to be directly related to distance.  Perhaps some method
of weighting the squares could make it account for forwardness as well.

Finally, on the value of Kings.  They are generally considered to have
infinite value, as losing them costs you the game.  But what if you assume
that the standard method is to lose when you have lost all your pieces, and
that kings have the special disadvantage that losing it loses you the game?
 I first assumed this would make the value fairly negative, but preliminary
testing in Zillions seems to indicate it is somewhere around zero.  If it
is zero, that would be very nifty, but I'll leave it to someone much better
than me at chess to figure out it's true value.

'Yellow is the color of mystery in Italy' is an arcane little i18n joke. A
paperback pulp mystery story is colloquially called 'un giallo' (a yellow)
because of its yellow cover. Even the publisher Mondadori uses the term, as
its series is titled 'Il Giallo Mondadori'. Number 1331, 'Quella Bomba di
Nero Wolfe' (Please Pass the Guilt) was published in 1974 and it is weekly,
therefore the series began around 1948; but it also says 'new series', so
the usage of a yellow in this sense may be older.

This is *not* the sort of color usage that can get you into i18n trouble,
though it sounds like the typical 'White is the color of death in China'
warning, and that's the little joke.

For true madness and horror, you should look into the methods of
internationalization that were used in the days before the current
standards existed....

>>   Would 0.91 times 0.7 times 0.7 be correct? Yes, this is the answer
>>   to 'it can move there if either d2 or f2 is empty AND e3 is empty
>>   AND the corresponding square (d4 if d2, or f4 if f2) is empty'.

>  This isn't right (I think). It can move there if e3 is empty and
>  either d2 and d4 are empty or f2 and f4 are empty. So that's 0.7 * (1
>  - (1 - 0.49) * (1 - 0.49) ), which works out to 0.51793, as compared
>  to 0.4459. I think the generalized equation, where X is the (always
>  even) number of squares moved, would be 0.7^(X/2 - 1) * (1 - (1 -
>  0.7^(X/2))^2)

(We're talking about the probability of the zFF being able to make a four
step 
move, for example from e1 to e5.)

My verbal description is saying that the choice between the two paths 
is made only once, and therefore the two-path probability correction should

be made only once in the calculation; this gives me a simpler formula 
for doing the calc by hand. Upon review I am even more convinced that
this is correct, but in order to feel perfectly secure I must find 
your error. 

You are saying 'if e3 empty and ((d2 empty and d4 empty) or (f2 empty and
f4 empty))'. The verbal description is clearly correct, although it
makes things more complicated when you extend to 4 step and 6 step moves.
The probability that d2 empty and d4 empty is 0.49; the probability that
p or q is (1 - ((1 - p) * (1 - q))). Ouch, that's convincing.

Wouldn't another fair way of stating it be 
'(d2 empty and e3 empty and d4 empty) or (f2 and e3 and f4)'?
But that gives me a completely different number, even higher.

Aha! '(d2 and e3 and d4) and (f2 and e3 and f4)' is incorrect because 
in effect it applies the two-path correction to e3, but e3 non-empty
blocks both paths!

But then by the same token, your 'e3 and ((d2 and d4) or (f2 and f4))'
must apply the two-path correction twice!! 

I'm right, you're wrong. Nyaah, nyaah! (If I were a licensed
mathematician
I would be able to say Q.E.D., but since I'm not I can only say nyaah
nyaah.)

That was difficult. My head hurts.

'First off, it is quite interesting to instead of picking a magic number
as
the chance of a square being empty, calculate the value for everything
between 32 pieces on the board and 3 pieces on the board.  Currently I'm
then just averaging all the numbers,'

I've done that, too. The problem is, if the only reason you accept the
results is because they are similar to the results given by the 
magic number, then the results have no special validity, they mean
nothing more than the magic results. So why add the extra computational
burden?

If, on the other hand, you had a sound and convincing theory of why 
averaging the results was correct, that would be a different story.

'This concept seems to be directly related to distance.' Actually, I
think
I'd call it 'speed'. I'm pretty sure that I've played with those numbers
but gave up because I couldn't figure out what to do with them. 
Maybe you can; I encourage you to try.

>My verbal description is saying that the choice between the two paths is
made
>only once, and therefore the two-path probability correction should be
made 
>only once in the calculation

This works fine if the first step forces you to make a choice, but
sometimes both directions are unblocked after the first step, so you
still
have a choice of which way to go when you get to the third step.  A piece
that moves 2 squares as a Crooked Bishop then started moving as a Rook
would be easier to block than a Crooked Bishop is, as it would only get
the
two-path correction once.  Likewise, a piece that made bigger zig-zags,
going to c3 or g3 instead of e3, would get the higher number.

Nyaah, nyaah! :)

'Is The Game of Nemoroth a Chess Variant?' I believe it is, though it
stretches the boundaries. For me, the telling point is that there's a kind
of checkmate (provided by compulsion).

Because the basic condition of victory is stalemate, and because the pieces
all have different moves, it would also stretch the boundaries to call it
an ultima variant.

The complexity of interactions of the pieces feels a bit Ultima-ish,
though.

I have rarely seen so much chatter as for this game. (N.B. there is
significant commentary on Nemoroth in the Yellow Journalism thread.)

A couple of points:

Is Nemoroth a chess variant?  If gnohmon says it is, who am I to gainsay
him?  I am an 'inclusionist' when it comes to chess variants, anyway.  It
actually seems more like an Amazons variant, and there are other more
chess-like games that make use of the 'shrinking board' mechanism, but what
the heck.  (Bob Abbott, who invented Ultima, did not think it was chess,
because it did not use replacement captures.  He was an 'exclusionist'.)

When Nemoroth is refined, and the rules settle down, may we expect pages on
'The Value of the Nemoroth Pieces' and 'Nemoroth with Different Armies'? 
Should we reserve the name www.nemorothvariants.com?

If interest remains high, how about the CVP sponsor a contest in Nemoroth
problem composition?

Hi,

I have worked out a slightly different method of setting up Fischer random
chess positions with a single six-sided die.   It's fairly easy to
memorizem because it follows logically from the positional rules of the
game.   As far as I can tell it will create all possible positions. Here it
is:

All die rolls are counted from the left side of the board from white's
point of view and apply to remaining empty and 'legal' squares only.

Because the king must be between both rooks, it can only occupy the central
six squares on each side.  Roll a die and place the king on one of the six
'central' squares.

Now place the rooks.   Roll a die for the left rook.  If the number exceeds
the number of squares on the left side of the king, roll again.    Repeat
for the right rook.  If there is only one square to the right or left of
the king, skip the rolls and simply place the rook.

Now place the Bishops.   Place the first bishop based on a die roll.   If
the roll value exceeds the number of remaining squares, roll again.   Place
the second bishop in a similar manner counting only the available squares
of the opposite color of the already placed bishop.

Place the queen with a die roll.  If the die number is 4-6 then subtract 3
from its value (to minimize the number of rolls necessary.)

Place the two knights on the last two squares.

I have yet to study this method in detail to determine if it favors certain
positions.


A modification of the die roll procedure to minimize re-rolls is as
follows:   If there are 2-3 'legal' squares for the rooks or the second
bishop take the remainder of the die in the 'modula' of the number of
remaining squares.  For example, if there are two legal squares for the
left rook, and one rolls a 5, one counts this as a '1', as 1 is the
remainder when one divides 5 by 2.  If the roll had been a '4' one would
count this as a '2'.   In the case of 3 empty squares, one a '5' would
count as a '2'.  A '6' would count as a '3' and a '4' would count  as a '1'
(as in the queen roll, which will always have 3).   

This method will not work without bias when there are 4-6 legal squares
remaining, and re-rolls must be employed.  However, statistically speaking,
fewer rolls will be necessary in such a case anyway.   

It is possible, though highly improbable, that one might require a very
large number of rolls to finally 'nail down' a position for the rooks and
bishops.  But once they are placed, only 1 roll remains.

What do you think?

Brad Hoehne- Columbus, Ohio.