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Revisiting the Crooked Bishop. Revisiting the Crooked Bishop.[All Comments] [Add Comment or Rating]
Bn Em wrote on Sat, Nov 26, 2022 04:29 PM UTC:

Fergus asked some time ago whether a piece covering the same squares as a zB but alternating between the move's arms (for which he suggested the name ‘Helical Bishop’ on account of the path's resemblance to a DNA‐style Double Helix) had already been invented. It seems we can now answer that in the affirmative: it's mentioned twice on this page, as the Zigzag Bishop. Betza also posits what Gilman would go on to call a Bruegel (t[Wzt[FAA]]; for which I initially mistook Fergus' description), as well as a piece (the t[FzDD]) ‘dual’ (in the Gilman sense) to the Harvestman of Seenschach which seems to go curiously unnamed in M&B.


Jeremy Lennert wrote on Mon, Mar 28, 2011 10:02 PM UTC:Excellent ★★★★★
I also get 9.7 for the mobility of a crooked bishop when using .7 as the magic number for crowding, and 16.9 for the empty-board mobility, both of which are about 1.2 times the mobility of a rook.

Peter Hatch wrote on Wed, Nov 27, 2002 04:08 PM UTC:
>>And there's another detail that all of us forgot!  When the crooked
bishop is on the edge of the board, one of its paths in the (0,2)
direction is blocked, even though the (0,2) square might be on the
board.<<

It wasn't forgotten - Betza mentions it on the page, and my second comment
gives what I think is the correct formula for figuring it.

Robert Shimmin wrote on Tue, Nov 26, 2002 09:10 PM UTC:
And there's another detail that all of us forgot!  When the crooked bishop
is on the edge of the board, one of its paths in the (0,2) direction is
blocked, even though the (0,2) square might be on the board.

When I included this in my evaluation, I got the result that the crooked
bishop has about 1.2 times the mobility of a rook for a broad range of
reasonable values of the magic number.

gnohmon wrote on Sun, Oct 20, 2002 03:45 AM UTC:
Another vote! It's enough to convince me. I still can't wrap my mind around
it, though I believe you now.

I'm not a mathematician. For me to pretend to be one, I had to be in the
mood. I had to be concentrating on the problem for weeks, and then I could
do it. I have not been able to get back into that state of mind. I'm not a
mathematician even though my SAT score in 1959 was 754! 

All I ever dreamt of was to get a real mathematician, or several, to
follow up on my feeble attempts. There is gold there for a real mather,
maybe a thesis or dissertation. By all means, write it up.

I pass the torch, mostly because I don't seem to be able to pick it up
anymore.

Robert Shimmin wrote on Sat, Oct 19, 2002 05:32 PM UTC:
I'm going to weigh in on Peter's side.  The probability of getting to (0,4)
should be 0.51793.

There are two ways to get there.  One requires (1,1) and (1,3) open.  The
other requires (-1,1) and (-1,3) open.  Both require (0,2) open.  

The probability of (1,1) and (1,3) open is 0.49.  (Same for the other
route.)  So the probability of this part of each route being closed is
0.51.  So the probability of both being closed is 0.51^2 = .2601, and the
probability of at least one of them being open is 1-.2601 = .7399.

Of course, in either case (0,2) must be open, so the final probability of
getting to (0,4) is .51793.

This generalizes to the probability of getting to (0,2n) being 
(1-(1-p^n)^2)*p^(n-1), where p is the magic number.  This is equivalent to
Peter's original formula, although a more compact writing of the formula
is (2-p^n)*p^(2n-1) or 2*p^(2n-1) - p^(3n-1).

If you need help convincing yourself your original statement

> 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'. 

is wrong, try thinking about it this way: suppose both d2 and f2 are
empty.  Then the crooked bishop may pass through either d4 or f4 on its
way to e5.  And the probability of being blocked on BOTH squares is 0.09,
not 0.3.  So the difference between your calculation and the answer Peter
and I get lies entirely in the cases where d2 and f2 are both empty.  In
these cases, it doesn't matter which square (d2 or f2) it passed through
to get to e3, so no binding decision about which path was taken has been
made yet.  In fact, no such decision has to be made until the first d-file
or f-file blockade is encountered.  Your method of calculation forces the
decision to be made at the very first step, even when both paths are
open.

Make any sense?

gnohmon wrote on Sat, May 11, 2002 01:43 AM UTC:
You just about have me convinced, but I can't admit that you're right until
I find the error in my way of doing it. This will have to wait until I'm
both well-rested *and* in a proper frame of mind. 

Presuming you're right, you have my deepest thanks. 

My confirmation of my error and your correctness is on my to-do list.

Once it's confirmed, I'll have to rewrite the 'Revisiting the zFF' page
because the error is large enough to make a difference in the evaluation of
the piece. My evil twin would curse you for it, but I am above all that. 

Evil twin. Hmmm. Evil Twin Chess. James T. Kirk's evil twin was from an
alternate universe, so a two-board setup reminiscent of Alice's Chess seems
natural. Same move on both boards, all is well. But the N and its evil twin
must have different powers of capture or movement; when you make differnt
moves, what's the rule to make it a playable game? Why of course! you have
to move both the Q and her evil twin, unless one of the two has been
captured! Mate on either board wins. The army: N twins with Fibnif, B twins
with fBbN, R twins with mRcHcWcD, Q twins with Chancellor, K twins with K,
P twins with berolina P. No playtest but I'll bet it works with at most one
more rule needed somewhere.

Peter Hatch wrote on Fri, May 10, 2002 09:05 PM UTC:
> When you're really stuck with probabilities, you can use the
> laborious case-by-case analysis.

Indeed.  Instead of checking each case for two-path bonus, it's easier to
just check each case to see whether it is blocked, and just add up the
probabilites.  With 5 squares that can be blocked (e3, d2, d4, f2, and f4)
there will be 2^5 = 32 total possibilities.  1 possibilty has no squares
blocked, 5 have one square blocked, 10 have two squares blocked, 10 have
three squares blocked, 5 have four squares blocked, and 1 has all five
squares blocked.  The probability of each possibility is 0.7^(number of
unblocked square) * 0.3^(number of blocked squares).

The no squares blocked possibility lets us reach the destination square, as
do 4 of the one square blocked possibilities (all but e3), and 2 of the two
squares blocked possibilities (d2 and d4, f2 and f4).  None of the rest do.
 So the total probability is 0.7^5 + 4 * 0.7^4 * 0.3 + 2 * 0.7^3 * 0.3^2 =
0.51793, which agrees with my formula.  So I'm confident I'm right again.
:)

Peter Hatch wrote on Sun, Apr 14, 2002 06:15 AM UTC:
I still think I'm right, but I'm not as sure anymore...

I think I'm not fully applying the two-path for the f4/d4 - getting the
full two path bonus twice would be 0.91 * 0.7 * 0.91 = 0.57967, just
getting it once is 0.91 * 0.7 * 0.7 = 0.4459, and my method is in my first
comment as 0.51793.  So it is between the values of one and two two-path
bonuses.

Mostly I think it is right because the verbal description seems to be
right, and you agreed with my math on turning that description into a
formula.  I think that formula is properly figuring out how much of a
two-path bonus to give.

Does this seem reasonable to you?

gnohmon wrote on Wed, Jan 30, 2002 12:00 AM UTC:
Thanks for cooking my problem. I never claimed to be a good composer.

Now, if you'll pardon me, I'll go sit in a corner and cry.

JorgKnappen wrote on Wed, Dec 5, 2001 12:00 AM UTC:Excellent ★★★★★
Have I overlooked something in the first version of the problem? 1. Ke3 d5 2. zBe2 gives mate in 2 moves. The second version looks fine (but does not exhibit the ability of the zB to pin two pieces at the same time). --J'org Knappen

gnohmon wrote on Sun, Apr 14, 2002 04:26 AM UTC:
'Why is it that when I encounter an Ultima variant, it inevitably seems
more complex than Ultima, not less?'

What defines an Ultima variant? Is it possible that no game simpler than
Ultima fits your definition of 'an Ultima variant'?

At some unspecified time (1970s most likely) I collaborated with John
Ishkanian on an Ultima variant named 'Ultimate Ultima'; I still have a copy
and have seen it within the last few weeks but it would take me years to
find it again (Phil Cohen probably still has a copy and can find it
quickly). The premise was based on my idea that the duration of a game
depends on the ratio of power to space; and we tried to create a playable
game with so much power to space that games would rarely last longer than 4
moves -- this would be a great game for playing at lightspeed radio against
an opponent in another star system!

Four pages of dense and terse single-spaced typewriter text with characters
out to the narrowest margins. Rules not all that complex, but interactions
beyond belief. The Carrier (is this the right name?) could move like Q, but
at each single step could pick up or drop pieces; and if you drop a Mixer
it can rearrange all adjacent pieces (all of this happening within the
context of the single=move multi-square move of the Carrier) and by
rearranging a Transporter it could cause pieces to teleport to other
squares, and if you teleported a Converter it could make enemy pieces yours
and so on.)

We spent probably 3 months hashing out the rules and then played 2 games,
which lasted maybe 5 plies between them (I won both).

That's my idea of complex.

Compared to that, the Game of Nemoroth is so simple!

gnohmon wrote on Sun, Apr 14, 2002 04:01 AM UTC:
Thank you for finding my error. I'm wrong, so I take back my nyaah nyaaah.
But you're also wrong.

When you're really stuck with probabilities, you can use the laborious
case-by-case analysis. With a zFF going from e1 to e5, if e3 is occupied it
can't get there and f4/d4 deserves no two-path bonus. If e3 empty and both
f2/d2 occupied, no twopath. If e3 empty and one of f2/d2 occupied, no
two-path. If e3 empty and both f2/d2 empty, I was in error, twopath
applies. I haven't worked out the correct number, but it's higher than
mine, lower than yours, closer to mine than yours. You applied twopath in
too many of these cases.

I'm not sure to write this in a more general manner. (for example, zFF
going from e1 to e9 on a larger board).

I haven't worked out the number the laborious way, pending your agreement
to this.

You found an error, I think, but I think you were also wrong. 

I know you'll reply!

Peter Hatch wrote on Fri, Apr 12, 2002 05:33 PM UTC:
>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! :)

gnohmon wrote on Fri, Apr 12, 2002 01:54 PM UTC:
>>   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.

Peter Hatch wrote on Fri, Apr 12, 2002 05:51 AM UTC:
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.

gnohmon wrote on Fri, Apr 12, 2002 04:23 AM UTC:Excellent ★★★★★
Excellent for the feedback, that is.

You have no idea how hungry I have been for so many years to find a
mathematician or statistician who would be in the mood to criticize my
numbers or my methods and point out the errors that must be there.

With all due respect, I give you this instant reply, but I do not examine
the specifics of what you said nor do I respond to them. I am in the midst
of other things and not in condition to reply.

I give you my double-barrelled platinum promise that the specific numeric
algorithmic probabilistic things you said will be closely and extensively
examined by me and that a serious reply will be forthcoming.

Meanwhile, literary criticism of your reply suggests that you agree with my
basic method but merely cavil at a few of my specific applications. Is this
right? If so, I celebrate. If not, I cerebrate.

If you haven't read my general 'theory of piece values', please please do
and if you can (though I hope you can't) tell me I'm full of it.

The general public here believes in my numbers more than I believe in my
numbers. Perhaps you can have the deciding vote, since paolo has declined
to speak up.

Did you know that a giant standing on a midget's shoulders can see further?
Well, in doing this math stuff about piece values let me tell you I've
always felt like a midget. But right now I can only write silly answers. I
just spent a few hours writing serious. The promises I made in previous
paragraphs are serious, though.

Peter Hatch wrote on Fri, Apr 12, 2002 03:43 AM UTC:
If I'm right in the previous two comments (and if I've done the calculations right), the mobility is 9.7.

Peter Hatch wrote on Fri, Apr 12, 2002 03:18 AM UTC:Excellent ★★★★★
I think the way to find the on-board probability is to divide it into two parts. The on-board probablity for having two paths on a (X,0) (where X is any even number) move would be (X,2). The probability for having just one path on a (X,0) move would be (X,6) (on a 8x8 board, generally (X,board size - 2)). I think this works - moving two squares up the board can be done on all but the last two rows, and has two paths on all but the outer two columns.

Peter Hatch wrote on Fri, Apr 12, 2002 02:47 AM UTC:Excellent ★★★★★
<blockquote>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'.</blockquote> 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)

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