掌握编程过程中遇到的思路方面和技术方面的问题. 内容包括以下几个方面:
反向逻辑思维能力的培养;
建立清晰的编程逻辑思维能力;
GH 的程序设计理念;
并行数据结构深入理解和控制.
Grasshopper course of McNeel Asia focus on the cultivation of students flexible use of programming techniques, the ability to solve practical problems. Our course deep into the whole process of programming, from programming thinking model, the components principle to usage details do detailed explanation, help students complete mastery programming encountered in the process of thinking and technical aspects, include the following content:
Ability of reverse logical thinking;
Establishment of clear programming logical thinking ability;
The program design concept of Grasshopper;
Understanding parallel data tree structure and how to control it.
更多详细内容... More details…
授课讲师 Instructor 课程由Grasshopper原厂McNeel公司在中国地区的两位 Rhino 原厂技术推广工程师 – Dixon、Jessesn联合授课。课程结束后对达到授课预定目标的学员颁发唯一由Grasshopper原厂认证的结业证书.
Dixon & Jessesn, McNeel Asia Support engineer, by the end of course student who achieve the intended target will get the authentication certificate from McNeel Asia.
课程报名 Register this course 课程即日开始报名, 开课一周前停止报名, 名额满提前报名结束. This course begin to sign up, stop sign up a week ago, with the quota ahead over.
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课程日期 Schedule 7/15-7/20 Beijing 北京 7/26-7/31 Shanghai 上海 7/07-7/12 Shenzhen 深圳
课程范例演示 Samples of Grasshopper course demo
Note: pls follow below comments by Jessesn to see the samples…
number of divisions on that curve as in the defintion (i.e. by 4). The offset in the def is slightly different and should cull two or three more curves as in the lists that show my aim below.
Basically I want to look into each branch of the groups of points from each closed curve . Marking in a list whether it contains a one or a zero (0= outside 1 = coincidents).
{0;0}0. 21. 22. 23. 2 {0;1} 0. 01. 22. 03. 2 {0;2}0. 01. 02. 03. 0 {0;3}0. 21. 22. 23. 2 {0;4}0. 21. 22. 23. 2 {0;5}0. 21. 22. 23. 2 {0;6}0. 01. 22. 23. 1 {0;7}0. 21. 22. 03. 0 {0;8}0. 21. 22. 23. 2 {0;9}0. 21. 22. 23. 2 {0;10}0. 21. 22. 23. 2 {0;11}0. 21. 22. 23. 2 {0;12}0. 21. 22. 23. 2 {0;13}0. 01. 22. 23. 0 {0;14}0. 21. 22. 23. 2
I want to create a list from these points. That marks each curve that pokes out, in a cull pattern as such:
20022210222202
Using a 1 where there are co-incidents in the curve points and the boundary. A 2 for true (outside points) and a 0 for containment. So I might be able to use the 1 in future developments - however if a true false list is easiest I can live with that.
So could I use F(x) function? - to look for 0 or 1's in each bunch of points and thus list as such for a cull pattern? or will Path mapper help me here? Or can I rely on simply grafting and splitting??
I am usure of the neatest solution and would love to learn. Hope you can direct me.rgrds
J.…
output will show a tree with 3 branches of 4 integers each that I can pass on to other components. What is the best way to do it?
I have tried creating a tree and using a for loop to do so, but it didn't work.
Thank you for your help.
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etting when I merge the three trees, but what I would like to get is:
essentially a tree with 27 branches, each with a single list of either 11 or 21 points.
{0} (N=11)
{1} (N=11)
...
{10} (N=21)
{11} (N=21)
...
{17} (N=11)
{18) (N=11)
{27} (N=11)
Any help would be greatly appreciated.
All the best,
Matt
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Added by Matt Schmid at 3:06pm on December 4, 2010