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.
在线报名参加课程...
Sign up to this course…
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…
+ 1/8 .... 1/∞
Let's reverse the sequence and use common denomitator.
1/∞ + 2/∞ + 4/∞ + 8/∞ .... ∞/∞
So 1 = ∞/∞
Also take a look at this marvelous proof for 1/∞ = 0.
I think that we can safely assume 1/∞ = 0. Let's go back to our sequence :
1/∞ = 0
2/∞ = 2*1/∞ = 2*0 = 0
4/∞ = 4*1/∞ = 4*0 = 0
.
.
.
∞/∞ = ∞*1/∞ = ∞*0 = 0
therefore
1 = 0…
Branch Address in the form {0;1;2;3;4;5;6;7;8;9;.......;N}
Try Grafting the input (I) Create Path by right clicking on it and selecting graft from the context menu…
in the desired order.
0 = 0
1 = 1
2 = 6
3 = 7
4 = 8
5 = 9
6 = 12
7 = 13
8 = 2
9 = 3
10 = 4
11 = 5
12 = 10
13 = 11
Where the first number is the index and the second number is the actual sorting key. Then you sort these keys while sorting your curves in parallel using the A input of the Sort component.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
0;5} n8 I can easily retrieve a specific tree branch, then modify the data, lets say to each have 6 items in their list instead of 8. the end resut i want is:
{0;0} n6
{0;1} n8
{0;2} n8
{0;3} n8
{0;4} n8
{0;5} n6
But i cant seem to get the "replace branch" componet to work
In reality i have a tree that consists of 108 branches, so exploding the tree and entwining (which is my normal work around) is impractical
any help is appreciated!…
the one-but-last list [4]. After running out of the n- items avalaible it should continue with the second item of list 0 and so on for all items on all the lists.
Intput, six lists of 30 items each
[0] (n=30)
[1] (n=30)
[2] (n=30)
[3] (n=30)
[4] (n=30)
[5] (n=30)
Output, 18 lists of 10 items each
[0],i=0;[5],i=4; [4],i=7;...
[0],i=1;[5],i=5; [4],i=8;...
...
[5],i=1;[4],i=5; [3],i=0;...
I thought perhaps the weave component or the relative tree item component but didn't manage to figure out how to compose the mask. I couldn't find much on how to use these. I guess it should wrap the lists, but not the items.
Any help would be greatly appreciated.…
Added by Thorsten Lang at 2:27am on January 24, 2011
the tree layout would be:
{0;0}(0) (first column, first point)
{0;0}(1) (first column, second point)
{0;0}(2) (first column, third point)
.....
{0;4}(8) (fifth column, ninth point)
.....
{0;9}(9) (last column, last point)
What if you want to connect every point in this grid with a point two levels higher up and one column over?
You can specify an offset like so:
{0;+1}(+2)
If we apply this offset to the list of items above, we'll get:
{0;1}(2) (second column, third point)
{0;1}(3) (second column, fourth point)
{0;1}(4) (second column, fifth point)
.....
{0;5}(0) (sixth column, first point)*
.....
{0;0}(1) (first column, second point)**
* If you overstep the list length and Item Wrap is set to false, you will not get any result. Otherwise you'll get the cycled item.
** If you overstep the number of paths and Path Wrap is set to false, you will not get any result.
At the moment it only supports whole integers, and the items in the offset are simply added to the original path+index. I'm not sure yet if I'll move towards expressions in the offset pattern. There are two components, one allows you to get relative items from the same data tree, the other from two distinct data trees.
--
David Rutten
david@mcneel.com
Seattle, WA…
Added by David Rutten at 11:25am on November 20, 2010