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Kevin Pacey wrote on Wed, Feb 1 08:56 PM UTC in reply to H. G. Muller from 07:52 PM:

Hi H.G.

For the Alibaba if I recall right I gave a considerable (x2) leaper bonus to both the A and D components, before primitively concluding AD (aka Spider)=A+D+P=2.25

Where A=D (roughly) = (N-P)/[2 times 2] (roughly) each (Why the N? - I treated an A or D leap as kind of similar enough to a N leap, except they each only go to half as many cells). Why the 'N-P'? Well, if Q=R+B+P then my imprecise way of guessing a piece half a Q's power would be (Q-P)/2, for example. Why '/[2 times 2]'? Well that's the final penalty factor, where one of the 2's means that an A (or D) only has half as many moves as a N.

If I also recall right, it's

Where A is binded 3 ways, and D is bound just 2 ways, but is often slower than an A (thus a penalty factor of /[2*8] should perhaps be used for each [rather than /4] I feel, but the leaper bonus I gave for the A and D each is a factor of x2, and I [perhaps generously] gave A and D each a x2 bonus for distance often covered faster by a series of leaps compared to a series of N leaps, so thus the final /4 penalty factor).

So, A+D+P (for AD, aka Spider) = (N-P)/4+(N-P)/4+P = 0.625+0.625+1 = 2.25

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