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
;0;1;1;0}
{0;0;1;2;0} ...
{0;0;2;0;0}
{0;0;2;1;0}
{0;0;2;2;0} ...
{0;0;3;0;0}
{0;0;3;1;0}
{0;0;3;2;0} ...
...
and I would like to have this in two lists separated:
{0;0;0;0;0}
{0;0;0;1;0}
{0;0;0;2;0} ...
{0;0;2;0;0}
{0;0;2;1;0}
{0;0;2;2;0} ...
...
{0;0;1;0;0}
{0;0;1;1;0}
{0;0;1;2;0} ...
{0;0;3;0;0}
{0;0;3;1;0}
{0;0;3;2;0} ...
...
How can I do that?…
o now is select and group together list items by their index. {0:0} and {2:0} also {1:0} and {3:0}. They way I'm doing this at the moment is quite complicated and I'm sure there's an easier way to achieve the same result.
maybe someone in here knows a solution?
thanks a lot…
;1},{0;2},{0;3}, (note that the first item is NOT {0,0})
{1;1}{1;2},{1;3},
{2;1},{2;2},{2;3},
{3;1}{3;2},{3;3}...
Well I'll just upload an extract of the definition. The data path should be the same, but with the items in the last path shifted to the first path, and the items in the first path shifted to the second path and so on.
I'll keep trying. …
a follow up question... how do I wrap a list onto itself at a certain frequency?
i.e. I want the list {1;2;3;4;5;6;7;8;9}
to become {1,4,7; 2,6,8; 3,6,9} wrapped every 3rd item
Added by Joshua Jordan at 5:30pm on November 17, 2012