ve a Vertex [V] connected to four other Vertexs [N1-N4].
Each of the has a Value:
V ... 1
N1 ... 5
N2 ... 3
N3 ... 8
N4 ... 11
The Average Filter would set the Value of [V] to
(1+5+3+8+11)/5 = 5,6
The Median Filter would Sort Values and pick the middle one
1,3, [5], 8, 11
Hope that helped...…
ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009
ards to the number before the start number...
i.e. 9, 0, 1, 2, 3, 4, 5, 6, 7, 8
then it will need to repeat this pattern (continuing to count upwards) and the repeat number is based on a slider (for example 3 in the case illustrated below):
9, 0, 1, 2, 3, 4, 5, 6, 7, 8
19, 10, 11, 12, 13, 14, 15, 16, 17, 18,
29, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
if anyone has any ideas on how to create this series it would be great
M.…
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
…
This must be a bug because its true for dividing by all odd numbers:
(i-1)/3
(i-2)/5
(i-3)/7
(i-4)/9
(i-5)/11
....
(i-n)/2n+1
And you can't make it work for even numbers
Added by Danny Boyes at 5:06pm on January 13, 2010