ow do you sort data trees with distinctive number of list items in each branch?
In this example, there are 161 branches with three distinctive number of list items, 3, 4, and 5. I want to know if there is another way to sort out this data tree into this way: 3 vertices {0;0}, {0;1}, {0;2],...4 vertices {1;0}, {1;1}, {1;2},... 5 vertices {2;0}, {2;1}, {2;2},...
2. Is there a way to sort our surfaces based on number of distinctive list items automatically? For example, if there are surfaces with 3, 4, 5, 6, and 7 vertices, I want to make five branches with the same number of vertices automatically.
I hope my questions make sense...
Thank you!
Dongyeop …
Added by Dongyeop Lee at 12:14pm on October 3, 2016
dimension of matrices must be identical) and division is the same as multiplication (dimension must be in the order of A(mxn)*\/B(nxk) where n is the common dimension): to divide one element by another you just multiply it by 1/value (part or all of the elements can multiply while part or all of the elements divide):
so for example matrix addition of matrices A(2x2): {2,-1}{1,2} and B(2,2): {3,-5}{4,-2} will result in matrix C(2x2):{5,-6}{5,0}. subtraction of those matrices will result in D(2x2): {-1,4}{-3,4}
Division of matrices A(2x2): {2,0.5}{2,4} and B(2x1) :{2}{2} will result in matrix C(2x1): {1+0.25}{1+2}={1.25,3}. Multiplication of those matrices will result in D(2x1):{4+1}{4+8}={5,12}.…
13;2} ... 20.{13;12}
21. {21;0}22. {21;1}23. {21;2} ... 41. {21;20}
42. {34;0}43. {34;1}44. {34;2} ... 75. {34;33}
76. {55;0}77. {55;1} ... ....
I want to grab the first 8 [0-7], the next 13[8-20], the next 21[21-42] etc
so i have the (known fibonacci seq) list of numbers on the left here:
C S
8 0
13 8
21 21
34 42
55 76
89 131
144 220
233 364
and i need the list on the right, so that i can select items using a Series (N=1 and S and C from the list above) and a List Item component.
the simple question is:
is there a component that can take a list and accumulate it in this way that I need?
if not, is there anyone that can point me to a simple relevant VB example so i could easily adapt it?
many thanks,
gotjosh…
i-branches is removing similar branches. This will only be removing 0's.
e.g.
{1;1;1;3;0}
{1;1;1;4;0}
{1;1;1;5;0}
would end up after Simplify as:
{3}
{4}
{5}
But the single branch (remove zeros algorithm, as summarised above) would give:
{1;1;1;3} …