Search Results - 云南11选5开奖公告-『8TBH·COM』2019双色球预测专家最准确--2023年3月19日7时46分33秒.H5c2a3.bb5b1zbzz-cc

Topic: i need help..: does any one know how to create stream like {0} 0 {1} 1 2 {2} 3 4 5 {3} 6 7 8 {4} 9 10 11 12 then ~~more~~ thx lot...; Added by wei ting at 2:53am on December 19, 2012

Comment on: Video 'Diva for Grasshopper': g definition but in diva for grasshopper in material it just appear dusty_med and not metal_railings and metal_treads. How I should write the correct definition? void brightfunc dusty_med4 dirt dirt.cal -s 101 .25 dusty_med metal metal_railings005 .7 .7 .7 .3 .2 dusty_med metal metal_treads005 .5 .5 .5 .3 .2…; Added by Alexandre Priet at 10:10am on July 10, 2012

Comment on: Topic 'data list manipulation': ep is to understan the logics of what you want to do, in your case, build 4 point surfaces (u also need to know the right direction to build the surfaces). Then you can write an hipotetic list (by hand in a paper) of what you want. In your case the list was (0, 1, 3, 2) (2, 3, 5, 4) (4, 5, 7, 6), etc... if you can imagine building 2 lists, each one with the sequences (0, 2, 4, 6, etcc) and (1, 3, 5, 7, etc..) then you can manage with shift and graft to finally have four lists. A( 0 1 2 3 ...) B (1 3 5 etc..) C(3 5 7 etc..) D (2 4 6 etc..). And to achieve the 2 first lists, you need to get the odd and the pair numbers. The cull pattern does that amazingy well. With a pattern True-False you get de pair numbers, and with the False-True pattern you get de odd numbers. Hope it was clear enough…; Added by Pep Tornabell at 5:32am on November 19, 2009

Topic: Find subsequent values in a list: e 7. True 8. True <-- this one 9. True 10. False 11. True 12. False 13. True 14. True <-- this one 15. True 16. False 17. True 18. False 19. True 20. True <-- this one 21. True 22. False 23. True 24. False 25. True 26. True <-- this one 27. True 28. False 29. True 30. False 31. True 32. True <-- this one 33. True Any idea how I can solve this? Thanks!…; Added by Anja at 1:25am on January 5, 2012

Comment on: Topic 'recursion in c#': n; n = (int) (Math.Log10((double) nfaces / 20.0) / Math.Log10(4.0)); double t = (1.0 + Math.Sqrt(5.0)) / 2.0; double c = Math.Sqrt(1 + (1.0 + Math.Sqrt(5.0)) * (1.0 + Math.Sqrt(5.0)) / 4.0); //Icosaedron Rhino.Geometry.Mesh mesh = new Rhino.Geometry.Mesh(); mesh.Vertices.Add(-1 / c, t / c, 0); mesh.Vertices.Add(1 / c, t / c, 0); mesh.Vertices.Add(-1 / c, -t / c, 0); mesh.Vertices.Add(1 / c, -t / c, 0); mesh.Vertices.Add(0, -1 / c, t / c); mesh.Vertices.Add(0, 1 / c, t / c); mesh.Vertices.Add(0, -1 / c, -t / c); mesh.Vertices.Add(0, 1 / c, -t / c); mesh.Vertices.Add(t / c, 0, -1 / c); mesh.Vertices.Add(t / c, 0, 1 / c); mesh.Vertices.Add(-t / c, 0, -1 / c); mesh.Vertices.Add(-t / c, 0, 1 / c); mesh.Faces.AddFace(0, 11, 5); mesh.Faces.AddFace(0, 5, 1); mesh.Faces.AddFace(0, 1, 7); mesh.Faces.AddFace(0, 7, 10); mesh.Faces.AddFace(0, 10, 11); mesh.Faces.AddFace(1, 5, 9); mesh.Faces.AddFace(5, 11, 4); mesh.Faces.AddFace(11, 10, 2); mesh.Faces.AddFace(10, 7, 6); mesh.Faces.AddFace(7, 1, 8); // 5 faces around point 3 mesh.Faces.AddFace(3, 9, 4); mesh.Faces.AddFace(3, 4, 2); mesh.Faces.AddFace(3, 2, 6); mesh.Faces.AddFace(3, 6, 8); mesh.Faces.AddFace(3, 8, 9); // 5 adjacent faces mesh.Faces.AddFace(4, 9, 5); mesh.Faces.AddFace(2, 4, 11); mesh.Faces.AddFace(6, 2, 10); mesh.Faces.AddFace(8, 6, 7); mesh.Faces.AddFace(9, 8, 1); for (int i = 0; i < n; i++) { Rhino.Geometry.Mesh mesh_rec = new Rhino.Geometry.Mesh(); recursiveSubdivision(mesh, ref mesh_rec); mesh = mesh_rec; } mesh.Vertices.CombineIdentical(true, true); mesh.Normals.ComputeNormals(); mesh.Compact(); A = mesh; } // <Custom additional code> public void recursiveSubdivision(Mesh mesh, ref Mesh mesh_rec) { int ind = 0; for (int i = 0; i < mesh.Faces.Count; i++) { Point3f A = new Point3f(mesh.Vertices[mesh.Faces[i].A].X, mesh.Vertices[mesh.Faces[i].A].Y, mesh.Vertices[mesh.Faces[i].A].Z); Point3f B = new Point3f(mesh.Vertices[mesh.Faces[i].B].X, mesh.Vertices[mesh.Faces[i].B].Y, mesh.Vertices[mesh.Faces[i].B].Z); Point3f C = new Point3f(mesh.Vertices[mesh.Faces[i].C].X, mesh.Vertices[mesh.Faces[i].C].Y, mesh.Vertices[mesh.Faces[i].C].Z); ind = mesh_rec.Vertices.Count; mesh_rec.Vertices.Add(A);//0 mesh_rec.Vertices.Add(middle(A, B)); mesh_rec.Vertices.Add(B); mesh_rec.Vertices.Add(middle(B, C)); mesh_rec.Vertices.Add(C); mesh_rec.Vertices.Add(middle(C, A)); mesh_rec.Faces.AddFace(ind + 0, ind + 1, ind + 5); mesh_rec.Faces.AddFace(ind + 1, ind + 3, ind + 5); mesh_rec.Faces.AddFace(ind + 1, ind + 2, ind + 3); mesh_rec.Faces.AddFace(ind + 5, ind + 3, ind + 4); } } public Point3f middle(Point3f A, Point3f B) { Point3f AB = new Point3f(); float length; AB.X = (A.X + B.X) / 2; AB.Y = (A.Y + B.Y) / 2; AB.Z = (A.Z + B.Z) / 2; length = (float) Math.Sqrt(AB.X * AB.X + AB.Y * AB.Y + AB.Z * AB.Z); AB.X = AB.X / length; AB.Y = AB.Y / length; AB.Z = AB.Z / length; return AB; }…; Added by Laurent DELRIEU at 12:58pm on August 6, 2016

Comment on: Topic 'i need help..': Hey, {0} 0 {1} 1 2 {2} 3 4 5 {3} 6 7 8 9 {4} 9 10 11 12 13 14 ???; Added by Rémy Maurcot at 3:03am on December 19, 2012

Comment on: Topic 'DataTree selection rules': [1,4,...,10] = 1 to 10 by 3 = 1 4 7 10 [1,6,...,10] = 1 to 10 by 5 = 1 6 [1,2,...,10] = 1 to 10 by 1 = 1 to 10 = 1 2 3 4 5 6 7 8 9 10; Added by Danny Boyes at 3:37pm on November 11, 2013

Comment on: Topic 'culling similar points': erations, is not it? This is what I finally want and how I plan to do it: In the starting tree, points are listed accordingly to the (11) 4-side-panels they belong to. I need to do a tree where each of the 11 lists contains not the points composing the panels but the points code of the points composing the panels. the point code is the number in the flattened list that refers to the points (like, panel 1 is made of points n 0,1,4,18, and so on). To do this, I suppose that I will use my tree of 11 lists of 33 true-false values, and apply it to cull 11 times a series of numbers from 0 to 32. I ll post it if it works!…; Added by Vale at 10:35am on September 15, 2015

Topic: Get all vertices in each subgraph (for disjoint graph): {4}-0;3 {5}-6;7 {6}-5;7 {7}-5;6 Here it can be shown that there are two subgraphs containing 0,1,2,3,4 and 5,6,7. How can I use spiderweb (either using scripting or the components) to give me this result when I have many more vertices?? Thanks, Sam…; Added by Sam Gregson to SpiderWeb at 3:49am on November 24, 2015