As far as I understood those pieces are "close" types. Thus by SMIRF's method their third value element is always zero because both first elements are equal. It results (please verify this) in 8x8 values: Minister 6+5/7, High Priestess 6+1/28, in 10x10 values: Minister 6+44/45, High Priestess 6+19/45. Thus a Minister seems to be about 1/2 Pawn unit more valued than a High Priestess.
Hi, Reinhard. I get slightly different numbers:
But the results are about the same.
High Priestess
4 6 8 8 8 8 6 4
6 8 12 12 12 12 8 6
8 12 16 16 16 16 12 8
8 12 16 16 16 16 12 8
8 12 16 16 16 16 12 8
8 12 16 16 16 16 12 8
6 8 12 12 12 12 8 6
4 6 8 8 8 8 6 4
52 76 104 104 104 104 76 52 672
10.5 + 10.5 = 21 21/3.5 = 6.0
Minister
6 8 8 8 8 8 8 6
8 6 13 13 13 13 10 8
10 13 16 16 16 16 13 8
10 13 16 16 16 16 13 8
10 13 16 16 16 16 13 8
10 13 16 16 16 16 13 8
8 10 13 13 13 13 10 8
6 8 8 8 8 8 8 6
68 84 106 106 106 106 88 60 = 724
11.3 + 11.3 = 22.6 22.6/3.5 = 6.5
Logically, for 10x8, for the HP, we'd add 104+104 to 672, then divide by 80, giving 11.0 then x2, then divided by 3.5= 6.3
For the Min, it's 106x2 + 724 = 936 divided by 80 = 11.7 times 2 = 23.4 divided by 3.5 = 6.7
There is a minor thing that puzzles me about these numbers. That is the value of the knight, minister and high priestess all go up a slight amount on a larger board when logically these values should decrease slightly. It certainly doesn't appear to be very significant when dealing with chessboards of around 100 squares. The error is no greater than maybe half a pawn. But it's interesting that the one odd FIDE piece, the knight, is the one to give apparently contrary results. It's a fair bet the 2 things are related.
Well, here are my two 8x8 tables: (slightly different, please compare)
High Priestess:
04 06 08 08 08 08 06 04
06 09 12 12 12 12 09 06
08 12 16 16 16 16 12 08
08 12 16 16 16 16 12 08
08 12 16 16 16 16 12 08
08 12 16 16 16 16 12 08
06 09 12 12 12 12 09 06
04 06 08 08 08 08 06 04 (Sum = 676)
Minister:
06 08 10 10 10 10 08 06
08 10 13 13 13 13 10 08
10 13 16 16 16 16 13 10
10 13 16 16 16 16 13 10
10 13 16 16 16 16 13 10
10 13 16 16 16 16 13 10
08 10 13 13 13 13 10 08
06 08 10 10 10 10 08 06 (Sum = 752)
Those pieces are weakened at the corner area. But because that part of the board is becoming less important in bigger boards those pieces increase a little bit in their value at a 10x10 board.
Reinhard, [Graeme, and anybody else who may be interested]:
I've been thinking about your last statement:
"Those pieces are weakened at the corner area. But because that part of the board is becoming less important in bigger boards those pieces increase a little bit in their value at a 10x10 board."
At first, I didn't buy this statement, but then I started thinking about just how big the edge is in a chess variant. On an 8x8 board, 28 of 64 squares are edge squares, or just under about 45% of all board positions; on a 10x10, it's 36, both squares and percent. This obviously has a massive effect on pieces that are crippled at the edges, and among the sliders, this includes bishops. The 12x12 gives 44 of 144 squares, or about 28% [please forgive minor inaccuracies in the numbers; being a technological Neanderthal, I do them all quickly, and roughly, in my head]; for 16x12, it's 52 of 192 squares, around 26%. Even on the largest boards I've used, 20x30 and 24x24, it comes in at about 16%, giving a still very large effect.
On a 100x100, it drops to 4%. So piece value calculations on this size board *should* give one very close to "true" values. I'm not ready to do all the calcs, but I would like to speculate a little. The shortrange piece values should "flatten out" and approach a "true calculated value" [if that phrase has any meaning].
An observation [if I understand this correctly]: the value of the bishop should increase somewhat relative to the other sliders, or, better/more obviously, relative to the rook, as board size increases.
A question: as the size of the board approaches infinity, does the value of the rook approach infinity, or does it approach some limit?
Hello Joe, to calculate your desired infinite board "true" piece values, I suggest to switch to an 8x8 thorus board, whithout any corner at all. There you will get those "final" numbers.
When calculating numbers please do not forget, that SMIRF is digging for relative figures compared to a Pawn value always defined as 1, which is also increasing as absolute number within its limit, where colums would be added.
Nice idea, Reinhard, and a bit easier than what I was going to do. I used a 10x10 board for ease of calculation. The results are that both the Minister [NDW] and High Priestess [NAF] converge to a value of 9.11…, an interesting result. Then I did the calcs for the bishop, giving 200/50 = 4 for Wr; 650/50 = 13 for Wf
Wm = square root of (Wf cubed minus Wr cubed)/(10-1)(10-1)
= square root of 2133/81
= ~1.6 [sheer guess there]
giving about 18.6/3.5 = ~5.3 for the bishop
For the rook, the values are Wr = 4 (400/100); Wf = 14 (1400/100); Wm = ~ 1.8 [again, sheer guess], giving a rook value of about 5.6
Does this seem reasonable?
Hi Joe,
piece values …hmm? One of the too many things I'm currently thinking about - this in relation to my incomplete CAST method, where the "weights" used are meant to reflect the impact of the "edge". Just a few random thoughts:
- all values are relative and cannot be transfered between variants.-in other words I don't think there is a "true" piece value.
- board geometry is important. For instance there are many (infinitely many?) infinite boards - at one extreme I can imagine a board of infinite cells existing in infinite dimensions such that each cell is adjacent to every other cell; while at the other there might be an infinite board just 1 cell wide.
- I'm not sure that a finite toroidal board is a substitute for an infinite (or extremely large) board. For instance I calculate the Rook on a 10x10 toroidal to have a value of 5.73, and on a 15x15 toroidal to have a value of 8.19. Since the Pawn's values are unchanged by board size I would expect the value of any unlimited slider to approach infinty as the board size appoached infinity (assumming a "normal" geometry)
- the number and type of other pieces in play should, I feel, have an effect on a piece's relative value.
Cheers
Graeme
Hi, Graeme:
Not sure where to start, so I guess I'll just jump into the middle. I don't think that piece values are non-transferrable from one variant to the next. I don't fully agree with your first, second, and fourth statements, which I see as all part and parcel of the same idea. I don't believe there is a single "true" value for any piece, but I do believe that in the vast majority of circumstances we are liable to see, there is a number value for a piece which most variants [that we would see at CV, for example] that contain that piece will come close to - poorly expressed, but you get the idea. There are infinite numbers of boards, and the geometry of the board can be crucial - consider your linear infinite board in 2 versions, an orthogonal one and a diagonal one - clearly the values of bishop and rook are greatly affected by which board they're on. But you won't get many games using either of those boards among the games played online. I was thinking of an infinite 2D board, or at least a very large roughly square 2D board [which is not at all the same thing, admittedly].
I agree the type of piece affects the value of the other pieces, and vice-versa, again, poorly expressed, but, again, you get the idea. However, I don't agree that the number of pieces affects the value of a piece. The number of moves per turn maybe, but with the "FIDE Rules unless otherwise specified" stipulation, I don't see why sheer numbers of a piece will change the values of the pieces on the board.
My argument with your 1st, 2nd, and 4th statements is more in degree than in kind, though. I see what you're saying and I appreciate it, and can see where you're coming from. I just don't go as far as you do in that direction; I veer to the sides, or take a somewhat different road. But specifics here can be cheerfully argued over at another time. I'll end this here, and take up statement 3 separately.
Enjoy,
Joe
Hey, Graeme:
I was in a rush when I posted my initial calculations, and I screwed up the numbers, didn't error check, and omitted any caveats - all cardinal sins. First, I am also not sure the toroidal approximation is accurate or adequate - I only looked at it from the point of view of the Minister and High Priestess pieces, where it seems to be fine; 10x10 or 100x100, the numbers come out the same. I then assumed it was also good for the sliders, without running numbers. First error [and obvious]. Then I used Reinhard's numbers for the 8x8s for my 10x10 calcs for B and R - error 2. And I never stated I had doubts about the method: error 3. So, once again, I rush what I'm trying to do, then get to repent at leisure. Now I'll try to do this systematically.
10x10 toroidal calcs for the rook:
Wr = 400/100 = 4 [unchanged - the rook can attack 4 pieces from any point on a torus]
Wf for 10x10 = 1800/100 = 18 [corrected value]
Wm = (18x18x18 - 64)/81 to the 1/2 power = (5832-64)/81 to the 1/2 = 71.2 to the 1/2 =
Wm = 8.5 [corrected value]
VR = 4 + 18 + 8.5 = 30.5
VR/VP = 30.5/3.5 = 8.7 for a 10x10 torus, corrected value for Rook
10x10 toroidal calcs for the bishop:
Wr = 4
Wf = 17x50/50 = 17
Wm = (17x17x17 - 64)/81 to the 1/2 = (4913 - 64)/81 to the 1/2 = 59.9 to the 1/2 =
Wm = 7.7
VB = 4 + 17 + 7.7 = 28.7
VB/VP = 28.7/3.5 = 8.2
I like these numbers a lot better than the first ones I ran; they make more sense. They are close to the 9.1 values of the minister and high priestess also, which makes more sense that the low initial numbers. These numbers are reasonable, so the following can be said:
Clearly, as the board gets larger, the value of the bishop approaches that of the rook.
Now there's just 1 more piece of the puzzle left to do.
Hi Joe,
wrt your figures
- on a toroidal board the Pawn has, I believe, a value of 4
- I think the equation for Wm is (((Wf3 - Wr3)0.5) x (coverage))/(h-1)/(w-1)
So your figures become:
- Bishop (17 + 4 + 0.43)/4 = 5.36
- Rook (18 + 4 + 0.94)/4 = 5.73
The last piece is doing rook and bishop calcs for a 20x20 torus.
Rook - 20x20 toroidal board:
Wr = 4 (4x200/200)
Wf = 38x200/200 = 38
Wm = (38x38x38 - 64)/19x19 to the 1/2 = (54,872 - 64)/361 to the 1/2 = 151.8 to the 1/2 =
Wm = 12.3
VR = 4 + 38 + 12.3 = 54.3
VR/VP = 54.3/3.5 = 15.5 for a 20x20 torus
Bishop - 20x20 toroidal board
Wr = 4
Wf = 37
Wm = (50,653 - 64)/361 to the 1/2 = 140.1 to the 1/2 = 11.8
VB/VP = 15.1
Conclusion: a smaller toroidal board is not equivalent to a larger "flat" 2D board.
So, shortrange pieces apparently approach a maximum value asymptotically as board size increases; and unlimited sliders quickly converge toward the strongest piece value and increase to infinity as the board [2D, flat] increases to infinity.
Joe
Hi, Graeme - still screwing up my numbers, I see - but I think the general thrust is correct. I admit I do not understand exactly what Reinhard is saying with "Wm", so I'll cheerfully buy your specific figures. But over all, the general idea is as I've stated in my 3 previous posts. It's time for specific board calcs, I guess, size by size [yeow!]; how is Wm figured? Joe
Hi Joe,
just a quick note wrt to number of pieces affecting piece value.
My thinking here is that it is all very well having a move potential of say,14 cells, but it does you no good if your stuck behind a triple row of Pawns. In other words if a piece is blockable it is of less value than a similar piece that can leap. And I think the difference between the two is afected by the number of pieces in play.
Again, with CAST, I am trying to introduce a negative "blocking" value based on the initial density factor.
Cheers
Graeme
Hey, Graeme:
With sliders, both sides' pieces create blocks; with leapers, only your own side creates blocks, so the effect of density is halved.
Joe
ps: you're up early, and I'm up late…
Conclusion: a smaller toroidal board is not equivalent to a larger "flat" 2D board.
So, shortrange pieces apparently approach a maximum value asymptotically as board size increases; and unlimited sliders quickly converge toward the strongest piece value and increase to infinity as the board [2D, flat] increases to infinity.
Agreed.
Also, I think the maximum value, under this method of evalaution, for a shortrange piece is given by the toroidal board. Thus a Pawn always has an absolute value of 4, while a Knight has 16 - or a maximum relative value 4.
Cheers
Graeme
The obverse of density and negative blocking arises with those pieces that need a screen in order to effect a capture. Here a higher density has a positive effect - but I'm still struggling to quantify either.
Too many questions…, I should have stuck with playing Bridge or Poker!
Cheers
Graeme
I have been applying the 'contest' method for determining empirical piece values for some time, now, and it works quite well. It turns out that, to a high degree of accurracy, an imbalance in material corresponds to a winning probability for the given posiiton that is additive. So, for instance, if being a Pawn ahead in the opening produces a score of 62% in a match of several hundred games (excess score 12%), and having a Chancellor in stead of a Queen (again in the openening) a score of 45% (excess score -5%), then having a Chancellor plus a Pawn against a Queen produces an excess score 12%-5%=7% (i.e. total score 57%).
This obviously only works for scores close enough to 50%; if Knight odds from the opening would score 85% (excess 35%), being two Knights ahead cannot produce a score of 120%. For material imbalance so big that the score gets outside the range 30-70% there no longer is a linear relation between imbalance and excess score, as the latter tails off to +/- 50%.
This makes it possible to predict scores (for games played from the early opening phase) from a set of piece values, simply adding the piece values for each side, and taking the difference. Conversely, by playing a large number of matches from opening positions with a material imbalance, you can derive a set of piece values that optimally predict the scores. For instance, on an 10x8 board with Capablanca pieces, a Pawn advantage typically translates to a 12% score advantage. Playing an unpaired Bishop against a Knight (i.e. deleting B+N for one side, and B+B for the other from the opening array) gives a 6% excess in fafour of the Bishop. Playing a paired B vs N (so deleting B on one side, and N on the other) gives an excess of 12% for the Bishop. Deleting B on one side and N+P for the other, the score becomes equal.
So it makes sense to assign a base value to a Bishop that is half a Pawn higher than that of a Knight, and give a bonus of another half Pawn for possession of the Bishop pair. (This unlike normal Chess: if you play unpaired B vs N there, it scores 5%, and the base values of B and N would be equal.)
Applying this method systematically to all pieces from Capablanca Chess, will give the following piece values that best explain all match results:
P=85
N=300
B=350 (pair bonus = 40)
R=475
A=875
C=900
Q=950
Note in particular how strong an Archbishop is in real play; the given value is based on the observation that A hbeats R+B and R+N+P, that A+P beats Q or C, and that even a pair of Archbishops plus a Pawn have no trouble in beating a pair of Chancellors.
Well… the archbishop is obviously very strong, but it really has a major disadvantage against chancellors and even commoners and other shortrange pieces. maybe that might be why the commoner, although slow and shortranged, is stronger than a bishop or a knight. Here is my list: P=1.2
N=3.3
B=3.2-3.4
R=4.6
Lightning war machine=4.9
A=7.2
HP=7.7
M=7.9
Q=9.2
C=9.4
You did not mention what you base these claims on, or even what size board they apply to. But they seem very far from realistic. A Commoner is very similar in value to a Knight, as might be expected, as they are both short-range pieces with 8 moves. As to global properties, the Knight has more 'speed', but the Commoner has mating potential. Mating potential in general isn't worth very much, as most games are decided in the presence of Pawns. Perhaps the strongest advantage of the Commoner is that it is a 'strong defender', able to salvage a draw against a super-piece like Queen, which even a Rook cannot.
Even tests with engines that take all that into account, the end-game value of the Commoner tests hardly better than that of a Knight, perhaps 0.2 Pawn (e.g. by playing one side with 1 or 2 Knights and the other with 1 or 2 Commoners, in the presence of 3-5 Pawns each). For opening value, the Commoner tests slightly weaker than Knight (e.g. replacing one side's Knights by Commoners in the FIDE setup).