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}.…
s suivantes
{0;0} {0;0}
0=longueur (300) 0=longueur (400)
1=largeur (8) 1=largeur (6)
2=hauteur (20) 2=hauteur (8)
{0;1} {0;1}
0=longueur (250) 0=longueur (250)
1=largeur (8) 1=largeur (8)
2=hauteur (12) 2=hauteur (12)
{0;2} {0;2}
0=longueur (400) 0=longueur (300)
1=largeur (6) 1=largeur (8)
2=hauteur (8) 2=hauteur (20)
je souhaite réorganiser mes listes en fonction de la croissance de l'item 1 (soit la largeur) puis en fonction de l'item 2 (la hauteur) néanmoins pour chacune de ces listes les items sont indissociables .
dans l'exemple de gauche la liste de rhino et dans l'exemple de droite la liste que je souhaite obtenir c'est d'abord la largeur qui va croissante puis ,lorsque les largeurs sont identiques ce sont les hauteurs qui vont dans l'ordre croissant .Les longueurs restent assignées à la liste.…
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} …
TREE B
{0} n=1 {0;1} n=4
{1} n=1 {0;4} n=4
{2} n=1 {1;1} n=4
{1;2} n=4
{1;3} n=4
{1;4} n=4
{2;1} n=2
{2;2} n=4
{2;3} n=4
{2;4} n=4
Both trees are generated from sliders, so could have any number of branches, although they are tied together. Tree A is a set of division points on a line, Tree B is a set of intersections from lines generated radially from the first (in this case three) points. I am trying to perform a "closest point" operation between the first tree and the second tree-- only, I do not want them to cross list, or long or short list. I want the {0} point to operate with those entries in the 2nd tree that start with {0,x}. So it would look like
{0} --> closest point with {0;1},{0;4}
{1} --> closest point with {1;1},{1,2},{1,3},{1,4} etc
I cannot figure out how this works. What I am visually trying to do is cast rays from a string of points so that they stop when they encounter another curve. I am having trouble picking through the intersection events to get what I want. Check the attached files for some clarity. THANK YOU…
Added by Joshua Jordan at 12:06am on February 5, 2012