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Wazir. Moves one square orthogonally.[All Comments] [Add Comment or Rating]
H. G. Muller wrote on Wed, Nov 1, 2017 08:52 AM UTC:

@Kevin: That is a pretty thorough analysis, about as good as can be expected without computer studies. From the latter I can add the following: A Knight is slightly stronger than a pair of Ferzes (on opposite square shade), but as the Kaufman value of the Knight is actually 3.25 Pawns, this agrees perfectly with the Ferz value being 1.5. One assumes that there is a pair bonus involved here, like with Bishops, because of the color binding; the value itself is so low, however, that it would be pretty hard to actually measure how large this bonus is. (Plus the problem that Fairy-Max isn't really the ideal tool to measure pair bonuses.) My guess based purely on intuition is that a lone Ferz might be worth 1.4, and an oppositely shaded pair 3 (so pair bonus = 0.2). It is also completely unknown how the value of 'inhomogeneous pairs' such as Bishop + Ferz would depend on their square shade. (I.e. whether the Ferz is worth more when on the other shade than the Bishop, rather than on the same.)

As to the Wazir: the computer studies of opening values reveal a phenomenon that is common amongst all purely orthogonal movers (including Rook): when they start behind a closed rank of Pawns, they seem to lose 0.25 Pawns in value, presumably due to the difficulty of developing them. A Wazir in such a situation tests as about equal to a Pawn. While starting it before the Pawns makes it worth ~1.25. That is still a bit lower than the conjectured value 1.4 of a lone Ferz, probably due to the 'forwardness' effect that you mention. It is also true that Wazirs are much less effective in stopping passers; instead of the 'rule of the square' that you have for Kings and Ferzes you would have a 'rule of the triagle'. Diagonal steps are not only geometrically longer than orthogonal ones: they really can get you somewhere faster (at the price of skipping half the squares).

As to cooperative bonuses for combining pieces: the formula for short-range leapers with N moves, value = 1.1*N*(30+5/8*N), already accounts for cooperativity through the quadratic term. For N=4 it gives 1.43, for N=8 it gives 3.08, so a cooperativity bonus of 0.22. It seems natural that as the piece values themselves go up, the bonuses would too. Based on their conventional values the Bishop moves would have an ' effective leap count' of N=8.5 (330cP), and the Rook N=12 (496cP). Together that gives N=20.5 for their compound, the Queen, which results in 9.79cP. So the cooperativity bonus in a Queen is about what you expect. For Chancellor you would have N=20, resulting in 9.35, which reproduces the empirical value difference with the Queen. Only the Archbishop turns out to be worth much more than can be reasonably expected from the sum of its parts.

All this is for 8x8 boards. It would be a very interesting question to see how the relative importance of leaper and slider moves depends on board size. This has not been investigated numerically at all. It seems reasonable that the 'effective leap count' of diagonal and orthogonal slides would go up on larger boards. It would probably not be strictly proportional, though, but saturate at some point. Very long slides are hardly useful on a crowded board, as you would in practice hardly have any room to make them. And even on a 12x12 board KBNK remains just as won as KBBK. Also, even on a quite empty 100x100 board extra W steps would remain important on a Bishop (and F steps on a Rook), because they help you to aim the long move more precisely at a distant target. Perhaps the value of sliders should be calculated incorporating a single 'can slide' bonus (which would go up with board size) due to it being able to quickly go to where the action is, plus individual 'manoeuvrability' contributions for each slide, which would have an effective leap count independent of board size.