Revisiting the Crooked Bishop

By Ralph Betza

In my original page about the Crooked Bishop, I introduced a new piece and said a few things about it. However, even though it is a natural piece in many ways, it is also in many ways a strange piece.

In order to be able to use it well, one must know a great deal more about it, and the purpose of this text is to investigate and understand the zFF as thoroughly as possible.

Value of the Crooked Bishop

Fortunately, you can bore some of the people some of the time, but you can't -- no, that's wrong, isn't it?

Self-Fulfilling Prophecies

Part of the problem was that I hadn't learned how to use it yet; and part of the problem is that you can't know how to use a piece until you know its value.

If you think the piece is worth a Bishop, you will trade it for a Bishop; and in that game, it will only have been worth that much. This is why playtesting without the theory of piece values could never succeed in establishing values well enough to make chess armies that were equal in strength, yet different. (And when I realized in 1976 that playtesting alone, would not do the job, I set out to create the theory!)

Common Sense Attempts

Even more startling, it has 8 distinct paths, so that in a sense it could be twice as valuable as a Rook! However, what looks like half the possible destinations on each path are shared by some other path, so the common sense approach says it covers between 1.5 and 2 Rooks worth of squares.

Compared to the "Rook and a half", the shared squares give the zFF an advantage -- when the R wants to go from e1 to e3, it is blocked if e2 is occupied, but to block the zFF both d2 and f2 must be occupied.

Flaws in Common Sense

However, another peculiarity is that a zFF has only one path from a1 to g1, because there is no such square as b0 ("b zero" would be the square South of b1.)

Gotta Calculate

However, it's tougher than calculating a Rook or Bishop, and that excuses some of my laziness. In addition, before I wrote the series of pages on Ideal and Practical Values (which isn't finished yet, by the way), I was in a crisis of doubt about the predictive value of the average mobility calculation; why do all that work to arrive at a number whose significance was dubious?

Now that my faith in the basic value of this calculation has been renewed, there's no excuse not to try the calculation.

The method is: take the sum of the average mobility contributed by each step of the piece's path, where the average mobility is the product of three things:

• the probability that the piece can move that far without being obstructed by some other piece
• the number of directions the piece moves in
• the probability that the destination square is on the board

Probability that zFF can move that far

zFF magic zMagic 

(1,1) 1.0 1.0 
(0,2) 0.7 0.91 
(1,3) 0.49 0.49[1] 
(0,4) 0.343 0.4459
(1,5) 0.2401 0.2401 
(0,6) 0.16807 0.218491 
(1,7) 0.117649 0.117649 

[1] Not 0.637, but it occurs twice as often!

The probability of zFF moving (0,2) is 0.91 because for this move to be blocked by some other piece two different (1,1) squares must both be occupied. This is exactly like finding the probability of not rolling double sixes -- if you ever had a "probability and statistics" course, this was covered in the first week.

The probability of Rook moving (0,2) is only 0.7, but the probability of zFF doing so is 0.91; right now we are only worrying about the chance of being blocked by some other piece.

The probability of zFF making a three step move is exactly the same as for the Rook. Remember, the zFF cannot choose at will: it cannot get from e1 to d4 by passing through f2! (The Zigzag Bishop can do so, but can't go e1-d2-e3-d4; calculating the value of a free-form crooked B is left as an exercize for the demented.)

The probability of zFF making a three step move is exactly the same as for the Rook, but the zFF has eight possible three step moves, while the Rook only has four. This will be taken care of in the next step of our calculation.

The probability of the zFF making a four step move was originally misstated as 0.91 times 0.7 times 0.91 -- but is that correct? When the zFF goes from e1 to e3, it has a choice of two paths, but when it goes from e1 to e5, it chooses its path when it makes its first step and then has no choice when continuing from e3 (if it got to e3 via d2, it must continue via d4). I think that's wrong.

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".

zFF Number of Directions

zFF nD

(1,1) 4
(0,2) 4
(1,3) 8
(0,4) 4
(1,5) 8
(0,6) 4
(1,7) 8

General On-Board Probability

This formula is my own invention, and it makes things a lot easier than the old method of putting the piece down on each square of the board, counting its legal moves, and dividing the sum by the number of squares on the board! Not only that, but some clever person can earn everybody's gratitude by posting a reference page giving solutions for all 28 possibilities on an 8x8 board, all 36 on a 9x9, all 45 on a 10x10. (Euler's Number when w equals h; when not equal, too tough for me.)

You will notice that for the general calculations of average mobility, I am forced to use extremely tedious and brutal methods because I have not yet discovered a formula to streamline the work!

zFF On-Board Probability

In fact, there is always one legal path by which the zFF can go from a1 to a3, but the fact that there is only one path and not two throws a monkeywrench into my calculations!

I'm not a mathematician, nor, as I wrote elsewhere, do I play one on TV. Therefore I had a lot of trouble with this, until I decided that for a zFF's move to the (0,2) square I should use the ob-board probability of the (1,2) square, and not only that, it's commutative and I don't have to go crazy finding the right order of operations. Thirty seconds after coming to that conclusion, I could no longer remember my reasons for deciding that way -- this is considered proof of correctness, is it not?

Warning: the previous offhand estimate that the zF2 would be "worth" F plus 0.91 D was incorrect! In fact, the zF2 "is worth" a mere F plus 0.793975 D -- an error of perhaps one sixth of a Pawn. Probably nobody was harmed by this error.

zFF (using) on-board

(1,1) (1,1) 49/64 = 0.765625 
(0,2) (1,2) 42/64 = 0.65625
(1,3) (1,3) 35/64 = 0.546875
(0,4) (1,4) 28/64 = 0.4375
(1,5) (1,5) 21/64 = 0.328125
(0,6) (1,6) 14/64 = 0.21875
(1,7) (1,7) 7/64 = 0.109375

The Griffion (see below) "is worth" an estimated 1.42 Rooks.

Summing Up

Take with a grain of salt. In my old Short Rook page I noted that the mobility calc for shorter pieces seemed to overestimate the strength and that the same calc for longer pieces underestimates the strength.

zFF zMagic nD on-board contrib subtotal piece

(1,1) 1.0 * 4 * 0.765625 = 3.0625 3.0625 zF1 == F
(0,2) 0.91 * 4 * 0.65625 = 2.38875 5.45125 zF2
(1,3) 0.49 * 8 * 0.546875 = 2.14375 7.595 zF3
(0,4) 0.4459 * 4 * 0.4375 = 0.780325 8.375325 zF4
(1,5) 0.2401 * 8 * 0.328125 = 0.6302625 9.0055875 zF5
(0,6) 0.218491 * 4 * 0.21875 = 0.191179625 9.196767125 zF6
(1,7) 0.117649 * 8 * 0.109375 = 0.102942875 9.29971 zF7

Empty-Board Mobility

zFF nD on-board contrib subtotal
(1,1) 4 * 0.765625 = 3.0625 3.0625
(0,2) 4 * 0.65625 = 2.625 5.6875 
(1,3) 8 * 0.546875 = 4.375 10.0625
(0,4) 4 * 0.4375 = 1.75 11.8125
(1,5) 8 * 0.328125 = 2.625 14.4375
(0,6) 4 * 0.21875 = 0.875 15.3125
(1,7) 8 * 0.109375 = 0.875 16.1875

Evaluating the Results

However, the Bishop and the zFF are both colorbound. The practical value of the Bishop compared to the Rook is less than the average crowded-board mobility of the B compared to that of the R. To get a feel for this, convert to Pawns: if the R is worth 5 Pawns, then the ratio of 8.1 to 5.9 would make the Bishop worth 3.64 Pawns.

Because the Bishop and the zFF are both colorbound and both long range, it is reasonable to assume that the relationship of mobility to practical value is the same for both. Depending on how you evaluate the relative values of R and B, the mobility figures seem to overvalue it by 1.1 to 1.2; and if we compromise at 1.15, then it looks like the practical value of the zFF ought to be approximately exactly a Rook.

I said "approximately exactly" and I meant it. During the opening and midgame, colorbound pieces are not handicapped much by being colorbound; it is only in the endgame that it begins to be a problem. (This suggests that in FIDE Chess if you have Bishop versus Rook you want to avoid exchanges...) If the whole-game value of the B is two thirds of a Rook, and its endgame value is less, its opening/midgame value must be more -- probably 3/4 of a Rook. (That's 1.15 times 2/3, but I did not make this guess by multiplying the value times the "colorbound correction".)

I always wanted to have a colorbound correction number. In the 1970s I used 10% as a guess, which wasn't far off -- if a piece's value is 90% of its calculated mobility, its mobility overstates its value by 1.11; and the difference between 1.11 and 1.15 is much less than the uncertainty and doubt attached to all these numbers.

Practical Value of the Crooked Bishop

The difference is half a Pawn, or even three-quarters of a Pawn. This is a significant difference, and therefore if you start the game with a pair of zFF but your opponent has a pair of R, you ought to be able to either win in the midgame or to enter the endgame with a material advantage that more than compensates for the fact that the Rook is worth more than zFF in the endgame.

Note that the Pawnless endgame K + zFF versus K is only a draw, and K + zFF + wrong Rook's Pawn is a draw in most cases. Endgames with zFF versus Bishop -- or even Ferz! --of opposite color will often be drawn, but zFF versus concordant B is only sometimes a win (White, Ke6, Pe7, zFF on g8; Black Be8, Kc7. 1...Be8-c6 2. Ke6-f6 Bc6-a4 3. zFFg8-f7 Ba4-b5 4. zFF f7-e6 Bb5-a4 5. zFFe6-d5 1-0; but 2...Kc7-d6!, therefore 2 Ke6-e5 Ba4 3 zFF-f7 Bb5 4 zFF-e6 Ba4 5 zFF-d5 Ba4-d7 6 Ke5-f6 Kc7-c8 7 e7-e8/Q, the Bd7 is pinned). Not easy!

In other words, if the whole-game value of a colorbound piece is roughly 87 per cent of its average crowded-board mobility, its endgame value must be less than 87 per cent, and the zFF shows this expected theoretical weakness in specific practical positions.

Characteristics of the Crooked Bishop

The most interesting other characteristic of the zFF is the funny shape of its move. An "open line" for the zFF has a different shape than an open line for a Rook or a Bishop, which means that in order to use the zFF effectively, or to defend against it effectively, the player needs to learn a great deal: need to learn to see these open lines as easily as seeing diagonals or files, need to learn how to nurture and provoke open lines of such a shape, need to learn the tactical tricks of zFF versus R, zFF versus Knight, zFF versus Fibnif, and so on; and need to learn the tricks of the shared path.

For example, consider a position where White has B on g3, N on b3 and c7, Kd4, and a zFF on d3, while Black has Kd1 with Pawns on b4, c2, d6, and e2. Here's the UAD:

. * . * . * . *
* . N . * . * .
. * . p . * . *
* . * . * . * .
. p . K . * . *
* N * Z * . B .
. * p * p * . *
* . * k * . * .

Consider the opening of a game (named "zFF versus Rooke Test Chess") where White has zFF in place of Rooks, and of course O-O-O must put the K on b1 and move the zFF from a1 to c1 (lest it change color), but the rest is as in FIDE Chess: 1 e4 e5 2 Nf3 Nc6 3 Bc4 Nf6 4 Ng5 d5 5 ed5 Nd5? 6 N:f7! K:f7 7 Qf3+ Ke6 8 O-O; White threatens Qf3:d5 with double check (zFF from f1 via e2, f3, e4, f5, e6) and there's no defense.

Instead, 1 e4 e5 2 Nf3 Nc6 3 Bc4 Nf6 4 Ng5 d5 5 ed5 Na5 6 Bb5+ c6 7 d5:c6 b:c6 8 O-O!? threatens f7, and 8...e5-e4 9 Bb5-a4 (9 Be2 h6; 9 Bd3!?) looks good for White (9 Ba4 Bg4 10 Qe1 Qd5 11 d3; and remember b2-b4 is a threat because the zFF on a1 defends b4). Looks like the Two Knights' Defense is busted.

Here's another: 1 e4 e5 2 Nf3 Nc6 3 Bc4 Nf6 4 Nc3 (inferior because 4 Ng5 seems to win) N:e4 5 B:f7+ K:f7 6 N:e4 d5 7 Nf3-g5+ Ke8 (Kg8? 8 Qf3 as in FIDE Chess) 8 O-O!? d5:e4 9 zFF f1:e4+ (a sort of double check) Ke7 10 Nf7 Qd4 11 d3 h6 (Bg5 checkmate must be prevented) 12 N:h8 and W is winning.

Thus, in this game where in theory the opening/midgame superiority of the zFF should be expected to win, in practice we can see concrete examples; but the examples were chosen so that Black walks right into the buzz saw.

Another opening: 1 e4 e5 2 Nf3 d6 3 d4 Nf6 4 Nc3 Nbd7 5 Bc4 Be7 6 B:f7+ and so forth; because the zFF on h1 defends g2, Black's normal counterplay fails, and instead of winning, he loses.

It hardly seems fair to show a variation where W wins because the zFF defends g2 without also showing one where he loses because zFF does not defend h2, so 1 e4 e5 2 f4? Qh4+ 3 g3 (3 Ke2 Q:h2 wins the trapped zFF) Q:h2 4 zFF h1-g2 Q:g3+ 5 Ke2 and so forth.

In the interest of fairness, we should also consider the zFF that doesn't Castle, usually the one that starts on a1. It has absolutely no move until you move the b-Pawn, which would in many positions be a waste of tempo, a creation of weakness, or both.

The bad placement on a1 might be enough to compensate for the zFF's midgame strength.

Would a real problemist prefer the following version?

. * . * . * . * 
* . * . * . N . 
. * . * . * . * 
* . * . * p * .
. * . * . * . K 
* . * . * N * Z 
. * . * . * p * 
* . * . * . * k

Combinations with the Crooked Bishop

When used in combination with something that removes its colorboundness, the zFF contributes its full ideal value of 1.15 Rooks to the combined piece; therefore, a zFFN (I'll name it the "Crooked Cardinal") should be worth enough more than the Queen to make a difference.

That extra 0.15 R makes it slightly difficult to use the zFF in combined pieces that have just the right value; if a Rook is three standard atoms, the zFF is nearly three and a half. The zFF also overlaps with Ferz (F) and Dabbabah (D), and is too strong for H (with the (0,3) leap added, the resulting piece would win material in the opening position against most armies), combining with Alfil (A) would be unbalanced toward colorboundness, and all that's left is Wazir (W) and Knight (N).

The obvious choice is fbWffNsbNzFF, that is, Crooked Bishop plus vertical Wazir plus Crab. I think it's fully worth a Queen. I guess it's the "Crooked Crabinal".

A colorbound Queen might be zFFAADD, Crooked Bishop plus AlfilRider plus DabbabahRider; note that the DD power overlaps existing zFF power and becomes nearly worthless in the late endgame.

My favorite combination is zFF plus Griffion. The Griffion, if I have the name correct, makes a Ferz move, then if that square was empty it turns and continues outwards as a Rook. Its power on an 8x8 board should be about 2 Rooks times 7/8, and subtract a third of a R from that: 1.42 Rooks. But when you add it to the zFF, there is considerable overlap! They both start with an F move, and both movement paths include the (1,3) and (1,5) and (1,7) squares. This piece is worth more than a Queen, probably about equal to Raven, I'd guess.

Variations of the Crooked Bishop

The Zigzag Bishop goes from e1 to d2, e3, f4, e5, d6, e7, f8; or from e1 to f2, e3, d4, e5, f6, e7, d8. It covers the same squares as the zFF but follows a different route, and its value is slightly less than zFF. (This article tells you how to calculate its average mobility.)

An interesting piece I thought of while writing this would move as F, then if that square was empty it would turn 90 degrees and continue as a Crooked Alfil-Rider: from e1 to d2, f4, d6, f8; or from e1 to f2, d4, f6, d8.

Even more interesting, F then zDD.

The End

White zFFd5 and Pd6, Black Kd8: stalemate. White zFFe4 and Pe6, Black Ne5: N is trapped, zugzwang. And so on.

However, I hope that I have given enough info to make the zFF useful. It is an attractive piece, an excellent substitute for the Rook, and deserves to be used in many games.