Comments by Tony Hecker

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Tony Hecker wrote on Sun, May 25, 2008 01:47 PM UTC:

'Actually the chance for twice the same flip in a row is 1/2.'

H.G. is correct here.
- The probability of two heads in a row is 1/4.
- The probability of two tails in a row is 1/4.
- The probability of two same flips in a row is the sum of these two
outcomes: 1/4 + 1/4 = 1/2.

Another way to think about it:
With two coin flips, there are 4 equally likely outcomes: HH, HT, TH, TT.
In 2 of the 4 (equally likely) outcomes, the same flip result occurs twice
in a row.

Tony Hecker wrote on Tue, May 27, 2008 04:29 AM UTC:

I'm not very familiar with H.G.'s randomization technique, so I really
have no idea how well it works.  It sounds like he adds small random
values to leaf node evaluations, which is of course different than
selecting a random 'good' move from the root of the search.

Note that it is definitely true that randomness can be helpful for a chess
engine, even though it might seem counter-intuitive.

For example, basically all strong chess engines (as far as I know) use
random (pseudo-random) Zobrist keys for hashing.  The random keys may be
generated at run-time, or pre-generated, but they are random either way. 
Using different random keys will cause the engine to give slightly
different results without necessarily changing the engine's overall
strength.

Obviously, if used incorrectly, randomness could severely hurt an
engine's strength as well.  For example, if an engine just plays random
moves.  :)

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Tony Hecker wrote on Sun, Jul 13, 2008 01:58 PM UTC:

TJchess10x8 currently uses these values, which are just guesses.  I know
that they aren't optimal.  I think they should be revised in later
versions.

pawn        1
knight      2.6
bishop      3
rook        5
archbishop  7
chancellor  8.5
queen       9

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Tony Hecker wrote on Sat, Aug 9, 2008 03:07 AM UTC:

I think I see what happened.  The FEN for the starting position is this:
1nbackqbn1/pppppppppp/10/10/10/10/PPPPPPPPPP/RNBACK1BNR w KQkq - 0 1

Black doesn't have any rooks, but according to the FEN, both players
still have castling rights (KQkq).  I think that makes it an illegal FEN,
and it probably should have been this:
1nbackqbn1/pppppppppp/10/10/10/10/PPPPPPPPPP/RNBACK1BNR w KQ - 0 1

I see 3 ways an engine could handle this:
a) It could reject the FEN.
b) It could accept the FEN but not grant any illegal castling rights (if
the rook is missing or the king is in the wrong place).
c) It could accept the FEN and always grant the castling rights.

Sounds like Joker80 does c), although I'd tend to prefer b).

Tony Hecker wrote on Sat, Aug 9, 2008 09:16 PM UTC:

Here is the pgn:

[Event 'Computer Chess Game']
[Site 'DEREK-UFTPLMLF1']
[Date '2008.08.07']
[Round '-']
[White 'Joker80.np']
[Black 'Joker80.np']
[Result '1-0']
[TimeControl '1/1500']
[Variant 'capablanca']
[FEN '1nbackqbn1/pppppppppp/10/10/10/10/PPPPPPPPPP/RNBACK1BNR w KQkq - 0
1']
[SetUp '1']

{--------------
. n b a c k q b n .
p p p p p p p p p p
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
P P P P P P P P P P
R N B A C K . B N R
white to play
--------------}
1. d3 {+0.04/12 0} d6 {-0.03/12} 2. Nh3 {+0.02/12 0} Nh6 {-0.03/12} 3.
c4 {+0.02/13 0} g6 {+0.00/14} 4. g3 {+0.00/13 0} Ng4 {+0.00/13} 5.
Kg1 {-0.13/12 0} Bg7 {+0.35/14} 6. Ng5 {-0.22/9 0} Bj4 {+0.66/14} 7.
Nf3 {-0.60/12 0} Qj8 {+0.64/14} 8. Nc3 {-0.71/14 12:15}
{Xboard: Forfeit due to invalid move: f8i8 (f8i8) res=0} 1-0

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