#N canvas 706 96 611 655 10;
#X msg 20 198 5 12 13;
#X obj 20 251 #inner (3 4 5);
#X text 26 82 (3 4 5) cross (5 12 13) = (-8 -14 16);
#X obj 20 271 display;
#X obj 20 218 #inner (3 3 3 # 0 0 0 0 0 1 0 -1 0 0 0 -1 0 0 0 1 0 0
0 1 0 -1 0 0 0 0 0);
#X text 252 173 This considers the cross-product as a "trilinear" operator
\, in which you do a bilinear product (two consecutive #inner) but
the output is still vectorial.;
#X text 251 248 This can be thought of as multiplying a vector by a
matrix then by a vector \, except that instead of the matrix you have
a 3-D grid so that the end result is a vector too.;
#X obj 0 0 doc_demo;
#X obj 20 448 t a a;
#X obj 20 468 #finished;
#X obj 20 508 #store;
#X msg 20 488 3 2 2 2 # 0 1 1 2 0 2 1 1 0 2 1 0 0 0 1 2 0 0 1 1 0 1
1 0;
#X obj 20 568 display;
#X obj 20 528 #fold *;
#X obj 20 548 #fold -;
#X text 245 432 This version is much faster \, done in a quite completely
different way \, but does require that you join the inputs as a single
grid.;
#X text 3 25 First part made in déc.2006 by Mathieu Bouchard;
#X text 6 55 Several more or less twisted ways to do the common computation
:;
#X obj 4 154 doc_section bilinear_vectors;
#X text 430 41 This example is about math.;
#X text 430 55 No video or music in here.;
#X text 16 106 This is the operator for finding perpendicular directions
in 3-D space \, as taught in 13th grade where I live (Algèbre Linéaire
105 du DEC scientifique).;
#X obj 4 402 doc_section pick_values;
#X text 246 509 It picks up the values by telling their row numbers
and column numbers in the correct order so that they be multiplied
and subtracted. The sequence for computing x is 0 1 \, 1 2 \, 0 2 \,
1 1 is the basic pattern \, which is a 2-D determinant. It is then
shifted to the right \, with wraparound \, to compute y and z.;
#X text 251 293 Each of the 27 numbers in the grid determines how each
combination of x \, y \, z left input \, x \, y \, z right input and
x \, y \, z output (3*3*3=27) contributes to the final result : 1 for
add \, -1 for subtract \, 0 for discard.;
#X text 252 347 Those correspond to the rules for cross products of
the standard base : i*i=0 \, j*j=0 \, k*k=0 \, and all six remaining
combinations are things like i*j=k and j*i=-k ...;
#X obj 4 602 doc_section value_cycling;
#X msg 179 632 1 3 # 5 12 13;
#X msg 9 632 1 3 # 3 4 5;
#X obj 9 672 #convolve (1 3 # 0 0 1);
#X obj 9 652 #t;
#X obj 34 652 #convolve (1 3 # 1 0 0);
#X obj 179 652 #t;
#X obj 204 652 #convolve (1 3 # 0 0 1);
#X obj 179 672 #convolve (1 3 # 1 0 0);
#X obj 9 704 # *;
#X obj 189 705 # *;
#X obj 92 757 display;
#X obj 92 737 # -;
#X text 246 695 this one exchanges some columns in matrices that have
3 columns : the last column goes to the beginning or vice-versa. this
version has the advantage of being able to do many cross-products in
one shot.;
#X obj 4 796 doc_section C++;
#X msg 5 823 3 4 5;
#X msg 20 428 2 3 # 3 4 5 5 12 13;
#X obj 5 863 display;
#X text 242 818 This one solves the problem by writing a GridFlow external
in the C++ language. This is sometimes the easiest solution but it
depends for whom and it depends what kind of problem.;
#X text 3 39 but published and expanded in nov.2011;
#X obj 5 843 # cross (5 12 13);
#X text 242 861 In the case of GridFlow \, you don't have to write
it as a class \, you can also write it as a vecop \, which is a kind
of plugin for the [#] class \, and also works in several other classes
(#draw_polygon \, etc).;
#X text 419 100 It's also more like a tutorial.;
#X text 429 114 Other examples are demos.;
#X obj 0 929 doc_f;
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