On Thu, Jun 19, 2008 at 1:59 PM, Roy Stogner <[EMAIL PROTECTED]> wrote: > > We might want to split off a QNegativeGauss quadrature class, which > would be non-default. Some applications are going to be more tolerant > of negative weights than others, and for those applications people > could have the option of choosing quadrature rules with fewer total > points.
Yeah, it turns out that Walkington's seventh-order rule with 35 points that I mentioned earlier has negative weights also. This one-point-less fifth-order rule may just be an anomaly. The separate class idea is an interesting one, though there would be a lot of empty cases. Another possibility might be a bit in QBase which you could flip to allow computation with potentially more efficient quadrature rules that also just happen to have negative weights. > However, this rule doesn't just have negative weights, it has some > negative barycentric coordinates! That's seriously wrong. Is > "QFuckedUpGauss" too politically incorrect a class name for libMesh > use? As far as I can tell, the barycentric coords of x=(x_1, x_2, ..., x_d) are (x_0, x_1, ..., x_d) where the "zeroth" coordinate is given by x_0 = 1 - x_1 - x_2 - ... - x_d so, I think it's possible for it to be negative. > >> Did you happen to catch the meaning of "class" C_0, C_1, etc.? I'm >> not exactly sure how to get the 87 points out of the tabulated data. > > A "class" is set of points in R^3 s.t. for some symmetry s in G_3 (the > symmetry group of the simplex) the barycentric coordinates > representation of s(x) has the form > lambda(s(x)) = (a0,...,a0,a1,...,a1,a2...) > where a0 appears m0 times, a1 appears m1 times, etc. > > The specific classes here are: > C0: m = [4] > C1: m = [3, 1] > C2: m = [2, 2] > C3: m = [2, 1, 1] > C4: m = [1, 1, 1, 1] > > So I think C0 is the tet center, C1 is along the symmetry lines from > each vertex to the opposite face's centroid, etc. Each C1 point has > four unique permutations, each of which is also a point in the > quadrature rule. Each C2 point has 6 permutations, each C3 point has > 12 permutations. So the 1 + 5 + 1 + 5 lines in that data table become > 1*1 + 5*4 + 1*6 + 5*12 = 87 points. Excellent. Walkington's paper uses similar permutation groups but I just wasn't so sure about this one's notation. -- John ------------------------------------------------------------------------- Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://sourceforge.net/services/buy/index.php _______________________________________________ Libmesh-devel mailing list Libmesh-devel@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/libmesh-devel
