The value of Cylindrical Pieces?

[hgm]

As to the value of B*: First I tried to play B*+B* versus B+B, in a full opening, but giving the B*+B* side Pawn odds. It still seemed to crush the B+B side with more than 70%, though, in a small number of games. So I switched to playing B*+B-P vs B+B. This is now running at 47.5% after 121 games, so close to equality. So tentatively B* = B + 100 cP.

Note that there are several kinds of B pairs now: B+B, B+B* and B*+B*, which might all have different bonuses. The easiest way to measure the value of a Bishop pair is to play both sides with two Bishops, but put them on equally colored squares for one side, and on differently colored squares for the other. For the B pair this give a 50 cP advantage (both in 8x8 and 10x8 Chess), i.e. the pair wins by about half the Pawn odds score. I should try this now for a B* pair, to see if the advantage is bigger.

[hgm]

B* vs B+P ended in 49% for the B* after 290 games. So the conclusion B* - B = 100 cP stands.

I am now doing R*+R* vs R+R+P; after 66 games this was at 47% (but standard error is still ~6% after 66 games). It thus seems that R* - R ~ 50 cP. (Note that the R* could not be used in castling, which might suppress their value a little. I guess I should have played this with castling rights disabled for the normal Rooks as well. But castling rights do not amount to vey much in Fairy-Max.)

Note that, when I run such measurements for normal Chess (using Joker) I do not find the classical 1:3:3:5:9 piece values. In stead I find:

Q - (R+R) = 100 cP
(R+B) - (B+B+P) = 50 cP (scorewise)
(B+B) - (N+N) = 50 cP
B - N = 0
(N+N) - (R+P) = 100 cP

This suggests 50 cP for the B-pair bonus, and thus R-B = R-N = 200 cP. The last measurement then suggest (after putting R = N+200 cP) than B = N = 400 cP, and thus R = 600 cP, Q = 1300 cP. For better comparison with classical values it is better to rescale this by 75%, so that we get

P = 75
N = 300
B = 300 (but B+B = 650)
R = 450
Q = 975

For cylinder pieces we have then:

N* = 330
B* = 400 (B*+B* still unknown, but perhaps > 850)
R* = 550
Q* = 1175

This is all in the context of cylinder pieces embedded in a largely normal context. In cylinder Chess, however, they face a K* in stead of a K, and this might reflect back on the piece values. In particular, since a K* cannot be mated by K*+R*, one might expect the value of the R* to suffer from this compared to B* (or R vs B, for that matter).

[Ovyron]

N* = 330
B* = 400 (B*+B* still unknown, but perhaps > 850)
R* = 550
Q* = 1175

Thank you very much, this is quite satisfactory

In cylinder Chess, however, they face a K* in stead of a K

I was more interested in the value of the pieces* in the normal context than in the cylinder variation, as I dislike the K* as royal piece, but I would still be interested on the value of the K and that of a K* (On a variation in where all pieces must me captured, perhaps?)

Also, I take that the value of the P* is negligible compared to the P as to not needing to be tested.

[hgm]

It depends perhaps on if you let P* promote to Q or to Q*.

Small correction on the values above: After 183 games the R*+R* vs R+R+P score had dropped to 43.8%, i.e. about 1/3 of the Pawn-odds score in favor of the R+R+P. So we have to devaluate the R* a little bit, to

R* = R + 35 cP

I also tested the B*+B* (pair) against B*+B* (anti-pair, i.e. on same color). The pair won this by 55.5%, i.e. about 1/3 of a Pawn. Now it is not absolutely sure that this represents the value of the B*-pair, as I can imagine that a B* anti-pair is also worth some bonus, because of the batteries it can form. For a B anti-pair this is not so easy as for a B* anti-pair, as most diagonals are pretty short. So if you need two squares on the diagonal to build the battery, there is on the average not much that you can attack with it. Due to the wrap around, the diagonals for B* are all 8 squares long.

[mjlef]

For cylinder pieces we have then:

N* = 330
B* = 400 (B*+B* still unknown, but perhaps > 850)
R* = 550
Q* = 1175

I am rather stunned how they almost exactly agree (except the queen) with the Zillions values, computed on the fly. For refernce, those ratios were:

q 9.46
r 5.5
b 4.15
n 3.3
p 1

Mark

[Ovyron]

Yes, I only had the chance to play 1 Zillions-Cylmax game but it seems Zillions plays a very decent Cylinder Chess, despite Cylmax searching much deeper in some positions.

[Tony]

It depends perhaps on if you let P* promote to Q or to Q*.

Small correction on the values above: After 183 games the R*+R* vs R+R+P score had dropped to 43.8%, i.e. about 1/3 of the Pawn-odds score in favor of the R+R+P. So we have to devaluate the R* a little bit, to

R* = R + 35 cP

I also tested the B*+B* (pair) against B*+B* (anti-pair, i.e. on same color). The pair won this by 55.5%, i.e. about 1/3 of a Pawn. Now it is not absolutely sure that this represents the value of the B*-pair, as I can imagine that a B* anti-pair is also worth some bonus, because of the batteries it can form. For a B anti-pair this is not so easy as for a B* anti-pair, as most diagonals are pretty short. So if you need two squares on the diagonal to build the battery, there is on the average not much that you can attack with it. Due to the wrap around, the diagonals for B* are all 8 squares long.

Hi Harm Geert,

I was wondering about your test setup. Can it prove/disprove the folowing assumptions ?

1) The value of a cylindrical piece depends on the amount of cylindrical pieces my opponent has.

2) The value of a cylindrical piece depends on the game stage.

Tony

[hgm]

In principle, yes to both. But it is all extremely time consuming. It is just a matter of playing a few hundred games with all combinations of unequal material.

For instance, we could measure the B*-B difference by playing B*+B against B+B, or B*+B* against B*+B. And then repeat that eith all R of both sides replaced by R*, and the Q by Q*. And measure it accurate enough that you could see the difference above the statistical noise.

I do nor really expect a strong difference, btw. Certainly not if you replace R by R*. But even when you replace diagonally moving pieces by cylindrical ones, so that they 'fight back' over the board edge, the value of the B* would probably decrease, as it can no longer attack such pieces with impunity. But the value of normal B would also suffer, because now it can be attacked by the cylindrical diagonal movers, without having any defense. So the B*-B difference would stay largely constant.

Note that I did see a similar effect in 10x8 Chess on the value of the Knight: normally an opening setup without Knights but the Bishops on the same color (so no pair) can give additional Pawn odds to a setup without Bishops, to achieve almost perfact balance. But if you now delete all Archbishops and Chancellors as well, the Knights win by 56%. If you replace A&C by Queens, the Knights still win by 53%. So it seems the fact that other pieces fight back the way you attack depresses the value of a piece.

For the game stage, if you define that as the total material on the board, you could simply setup "openings" with more material deleted from both sides. I tried this to test the hypothesis if the Archbishop value would decrease during the game. So I set up non-tactical positions of A+5P against R+N+5P, with symmetrical pawn structures and King fortresses to eliminate any bias other than the material balance under test. The score was the same as when I deleted A vs R+N from a full opening, however.