Adjusted Length Analysis

A same-colour stone in that region increases the connectivity. An opposed-colour stone decreases it. My idea is: for each subset of same-coloured stones in the region, find the best-connected path using those stones and the end points, where the connectivity is the product of connectivities between each pair. (The idea is that any of the 2^N combination of stones works, but only one path should be counted for each). Similarly do the same for paths using the two end points plus only opposite-coloured stones. The (direct) path using no stones is counted as a same-colour path.

Then, the net connectivity is the 1-(product of (1-(same-colour connectivities))) * (product of (opposite-colour connectivities). The assumption is that connecting the endpoints with opposite-colour stones has the same difficulty as preventing a same-colour connection. (Probably not a very good assumption at all!)

Finally, we define the expected area of triangles. (Maybe I should have called this "Adjusted Area Analysis"...) Heron's formula for the area of a triangle is sqrt((s)(s-a)(s-b)(s-c)), where a,b,c the side lengths and s=(a+b+c)/2. The area inside a triangle marked out with stones is going to be somewhat less than that; assuming we connect, the edges are 1 stone thick and half of that is inside the triangle, so we subtract (s) - but need to add about 1 back, to cover overlap at the corners. (This is somewhat inspired by Pick's theorem for area of shapes with lattice-point coordinates.) Finally, we multiply that by the net probability of connecting all three sides (and the assumption of independence here is probably very dangerous too. But it's a start... :/ )

So our net "size" for a move (and finally note that this doesn't count things like influence resulting from the move, etc.) is: C(a)*C(b)*C(c)*(sqrt((s)(s-a)(s-b)(s-c)) - s + 1), where for simple situations C(x) = e^-((x^2 - 1)/16) and s = (a+b+c)/2. Whew. Of course, for simple situations Heron's formula isn't needed for that part of the calculation.

Let's try it out:

Let's consider the 'area' resulting from each of the marked points. We'll take the circled stones plus the marked point as sides of a triangle, with the circled stones of course being solidly connected. (The paths are IkkenTobi for .83 , and nobi+nobi for 1.00 which makes the IkkenTobi irrelevant. So at least we get *something* right. ;) ) Without the squared stone the results would be proportional, since that just changes the connectivity of the one side. (or would it? For point a, the squared stone makes the connection from a to the far endpoint solid when it wouldn't otherwise be. Similarly for b and c, but not for d onwards. A slight change, then.) In each case, note that the Heron's formula part of the area just becomes the number of rows we extend down (i.e. one point for a and b, two for c and d etc.)

Point a gives connections nobi, (nobi+nobi+nobi) to the circled stones for a connectivity product of 1; the area is 1-(1+2+sqrt(5))/2 + 1 = -.62. (A negative result; reinforcing the idea that TheEmptyTriangleIsBad, perhaps?)

Point b is also solidly connected and the area is 1-(2+sqrt(2)+sqrt(2))/2 + 1 = -.41.

Point c connects as an IkkenTobi to the near stone and through four non-solid paths to the far stone. Those paths work out to give near-certain connection, though, so the connectivity product is .83 for the IkkenTobi. The unadjusted area is 2-(2+2+sqrt(8))/2 + 1 = -.41. The adjusted area, finally, is -.34. Still improving, you see, but still not the right move.

Point d connects as keima to both endpoints; the squared stone is outside the region of influence (for a keima, just the two 'waist' points count). So the connectivity is (.78)^2 = .61. The unadjusted area is 2-(2+sqrt(5)+sqrt(5))/2 + 1 = -.24; the adjusted area is -.14. A pattern emerges; the centre plays are better, because the same triangular area is enclosed with a shorter perimeter, conserving territory. Again, play at the point of symmetry. So we won't consider the side plays any further.

Point e connects as ogeima to both endpoints and again the squared stone doesn't matter (and won't from now on). The connectivity is .57^2 = .32. The unadjusted area is 3-(2+sqrt(10)+sqrt(10))/2 + 1 = -.16 and the adjusted area is -.05.

Point f gives connectivity .37^2 = .14. The unadjusted area is 4-(2+sqrt(17)+sqrt(17))/2 + 1 = -.12 and the adjusted area is -.02.

Another pattern emerges; some simplification shows that this series approaches zero asymptotically. The conclusion is that we should play as far away as we can get away with, which doesn't seem quite right. Oh well, I started writing this without having done any previous testing, so I figured it would be miraculous if the system predicted NikenTobi (point e) right off the bat (as I'm told is the usual extension from that line). Any suggestions are more than welcome :)

Some Lurker? This is a cool idea . . . and I think possesing of much merit. My only beef is that unless I am mistaken you are using Euclidean distance for this . . . which seems inappropiate. At large distances it is probably a good approximation but something like the city block metric is probably better. In thinking about it it seems to me that the city block metric isn't quite right either since Kosumi seems to be a stronger connection than a one point jump. Apologies in advance if I turn out to be full of it.

HolIgor: In computer go the city block metric is called the Manhattan distance. I think that an additional characteristic that is the multiplicity of paths can make the mathematical description better. Both kosumi and ikken tobi have the same Manhattan distance but kosumi has two paths while ikken tobi has only one. Kosumi is stronger.

Daniel: I have actually have been working on some odd variants of the distance metric based off of a formula for something known as the super ellipse; essentially D=(X^N+Y^N)^(1/N) (yes the nth root) as n increases beyond zero the value returned comes closer and closer to returning equipotential lines in the form of squares radiating outward (more information on super ellipses can be found at wolfram mathworld (they have their own site) essentially the important point is that using something along the lines of this as a distance metric rather than Manhattan distance or Euclidean distance would rate a kosumi as being superior to a ikken tobi while depending on the n you chose rating it as worth slightly less than a stand myself I like how n=32 works out for me with the computer go AI I am working on.