@montanarealtyc1@skdh Empathy make us humans, and when all orientation appears erased, remember the empathy bring us all together.
Look inside to your self, and will recognize your essence present inside others.
We are never alone.
We can always choice true love and hope, and the life will embrace us.
@_Mira___Mira_ Well, from a conformal point of view, it is just the result of a Cl(1,3)~SL(2,C)xU(1) symmetry, together with a category of noncommutative dualities compositions, where one can swap time like <-> space like intervals, just like a BIon.
Let's do a mental exercise.
@grok lets start with N distinct points in plane.
Label the points from 1 to N in any initial order.
What is the complexity to find the matrice where, each line and column has, at most, zero at diagonal elements, the value of the two nearest euclidean distance to point in the line or column to the point correspondent to diagonal label, and the logical value "false" to remain cases ?
@grok@grok , if the N points were equally spaced points of a circl3s in the plane , what would be the sum of sequencial distance of the points following the sequence of secondary labels of the previous algorithm?
Now lets keep the two matrices, the two nearest neighboor matrices and the full geodesic matrice, with the same label points identifications.
Now start the following algorithm:
1-Starting from the H point of label P=1, and let M=floor(N/2)+1. Put a secondary label to point H as 1 too.
2- Find the label of one point T with the largest geodesic distance to the point with secondary label P and atribute to it a secondary label equals to M.
3-Find one smalest distance neighboor point NT from point T of neighboor matrice and put it's secondary label as M+1, and update M to M+1.
4-Discarding the points H and T, find the new point NH as a point with the largest geodesic distance of the point NT and put it's secondary label as P+1, and update P to P+1.
-discard all points with secondary label less then P , and all points with secondary label greater than floor(N/2) and less than M , let H to be the point with secondary label equals to P and loops to step 2 while there are any point without a secondary label.
@grok what would be the complexity?