algorithmic modeling for Rhino

- Comment on: Topic 'Split Tree Sequence Error'
*thanks for the correction but is there a way to have a sequence like this list? {0}[0] {1}[1] {2}[2] {3}[0] {4}[1] {5}[2] {6}[0] {7}[1] . . .*- Added by Zadig at 7:15am on May 6, 2017

- Comment on: Topic 'combining lists'
*you want each "element" to be a single Item or a single item for ALL elements. See Below 0. 20 1. 30 2. 59 3. 60 4. {9,45,29} 5. 0.0 6. 3.0 7. 6.0 Or 0. 20 30 59 60 {9,45,29} 0.0 3.0 6.0 …*- Added by Danny Boyes at 3:13am on October 29, 2013

- Topic: weaving indices
- n =5; {0;b} n=3; {0;d} n =5.can anyone tell me how to do it? thanks!…
- Added by Clemens Klein at 5:09am on September 24, 2010

- Comment on: Topic 'Path Mapper'
*--> {0;0;0} add to the heap 3 {0;0;4}(0) --> {0;0;0} add to the heap 4 {0;0;5}(0) --> {0;0;0} add to the heap 5 {0;0;6}(0) --> {0;0;0} add to the heap 6 {0;0;7}(0) --> {0;0;0} add to the heap 7 {0;0;8}(0) --> {0;0;0} add to the heap 8 {0;0;9}(0) --> {0;0;0} add to the heap 9 {A;B;C}(i) --> {0;0;0}(i) ------------------------------- {0;0;0}(0) --> {0;0;0}(0) contains 0 {0;0;1}(0) --> {0;0;0}(0) now contains 1 {0;0;2}(0) --> {0;0;0}(0) now contains 2 {0;0;3}(0) --> {0;0;0}(0) now contains 3 {0;0;4}(0) --> {0;0;0}(0) now contains 4 {0;0;5}(0) --> {0;0;0}(0) now contains 5 {0;0;6}(0) --> {0;0;0}(0) now contains 6 {0;0;7}(0) --> {0;0;0}(0) now contains 7 {0;0;8}(0) --> {0;0;0}(0) now contains 8 {0;0;9}(0) --> {0;0;0}(0) now contains 9 …*- Added by Danny Boyes at 8:04am on May 24, 2013

- Topic: Splitting a tree and reversing the order for alternate branches
- hieve 1 2 3 4 5 6 7 8 Branch 1 14 13 12 11 10 9 Branch 2 15 16 17 18 19 Branch 3 27 26 25 24 23 22 21 20 Branch 4 Many thanks William…
- Added by WT4 at 11:56am on August 16, 2016

- Comment on: Topic 'Break single path into multiple paths at specific indices.'
*0. 11. {0;3} 1. 32. {1;1} {1}3. {1;2} 0. 14. {1;4} 1. 25. {2;1} 2. 46. {2;2} {2} 0. 1 1. 2…*- Added by Jaime Sanchez-Alvarez at 8:53am on February 21, 2012

- 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

- Topic: culling, shifting, and other list manipulations on super branches...
- branches in each A's list of B's, or remove its ends etcso that if I want to remove the last B in every A{0;1},{0;2},{0;3},{0;4},{0;5},{0;6}{1;1},{1;2},{1;3},{1;4}{2;1},{2;2},{2;3},{2;4},{2;5}would become{0;1},{0;2},{0;3},{0;4},{0;5} {1;1},{1;2},{1;3} {2;1},{2;2},{2;3},{2;4}I guess the question is do I need to figure out the cull pattern- each B may have different lengths...…
- Added by Gabriel at 5:09pm on March 15, 2010

- Comment on: Topic 'Formulas x y z for mapping surfaces in grashopper...is it possible?'
*e are some fun implicit equations I found when making this file: x^2 + y^2 - 4 = 0(x+2)^2 - y^2 - 3 = 0x^3 - 2*x+1 - y^2 = 03*x * y^2 - x - 1 - x^3 - y^3 = 0y^3 + x^3 - 3*x*y = 0sin(x) + cos(y) = 0sin(x * sin(y)) - cos(y * cos(x)) = 0-x^3 + 2*x*y + y^5 - y^3 = 0(x^2 + y^2)^2 - x^2 + y^3 = 0x^3 + 3*x^2 - y^2 = 0x^4 + 20*y^3 + y^6 - 12*x*y + 12*y + 55/(2*y^2 + x^2) = C (interesting values for C=55, 30, 8)…*- Added by David Rutten at 2:23pm on January 16, 2015

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