0, 5, 10, 15, 20
1, 6, 11, 16, 21
2, 7, 12, 17, 22
3, 8, 13, 18, 23
4, 9, 14, 19, 24
and if i'm here is because i'm not able... :)
can you help me?
thank you
…
dont get you, i am saying sleect numbers in range 1 to 10, starting from 1 with a step of 2.
1 to 10 by 3 = 1 4 7 10
1 to 10 by 5 = 1 6
1 to 10 by 1 = 1 to 10 = 1 2 3 4 5 6 7 8 9 10
Added by Steve Lewis at 3:15pm on November 11, 2013
now I want to combine some branches together ,the rule is : For path{2} contain number 2 and 5, then conbine the two paths together ,and for path{5} includes only 2&5,no other number ,so it's end .For path{3}, includes number 3&6 ,so we go to path{6}, path{6} includes 3&6&18, then wo go to path{18} , path{18} contains a new number 27, so we check path{27} ,path{27} includes only 27&18, no new numbers ,so it is end.
With this logic, path{2}&{5} become one tree finally , the contains is 2&5 ,and so path{3}&{6} &{18} &{27}(the contents is 3,6,18,27), and so others .
so what I want is:
{2}(2,5)+{5}(2,5)={2/5/anything}(2,5) ## the new path index doesnot matter{3}(3,6)+{6}(3,6,18)+{18}(18,27)+{27}(27,18)={3/6/18/27/?}(3,6,18,27) ``````etc
I tried path mapper, but I donot think it can do the trick this time. may be I just miss something very visible?? Awaiting for your kind help~Thanks in advance.…
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
ok, thanks, Isaw on that page before and from the begining I had a problem, because I tough these points which are like this one 3√3 / 7, the number 7 divide all the equation.
Thaks.
identical knots on end ends of the curve, so the knot vector will look something like:
0,0,0,1,2,2,2
So one at each end and one in the middle.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 3:47am on January 7, 2013
13 5 15 6 17 7 ... …
But it seems I have not been clear:
the input values (3,5,7,9,11,13) should be presented in the way of "number slider" parameter. So by increasing the value of the slider from minimum 3, to 5, 7, 9, 13, 15, 17... I should get an output numbers of 0,1,3,4,5,6,7...
So I if choose value 3 on the input "number slider", on the other end, I want 0 as an output.
Or if I choose value 5 on the input "number slider", I want 1 as an output, and so on.
Is this possible?…