1+2+3+4+5+6 = 21
1+2+3+4+5+6+7 = 28
1+2+3+4+5+6+7+8 = 36
1+2+3+4+5+6+7+8+9 = 45
Is there a tool, that can do me that job?
How do I get this List {1,3,6,10,15,21,28,36,45}?…
Added by Ahmed Hossam at 2:19pm on September 22, 2013
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
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
ersect (2, 3, 4, 5, 6) with the line and the ones which do not intersect (0, 1, 7). Intersect is done! But how to get the non intersecting vectors (0, 1, 7)?
So I e. g. could deselect vectors 2, 3, 4, 5, 6 so I would display/use only vectors 0, 1, 7 and the bounced ones.
Appreciate your help!
Rudi…
53 → 53 → 63 → 74 → 74 → 84 → 9
As you can see from the above list the connection sequence comes in waves of three, where each group of similar indices on the left is associated with a group of three incrementing indices on the right.
Some combination of Series components will probably generate this list, but it'll only work for the first ring, the second one will need a different connection pattern. It is perhaps better to just encode the integer pairs by hand. But then you cannot change your mind about the number of sides later.…
Added by David Rutten at 10:39am on October 21, 2015