what is the mathematical formula to put n points in uniform distribution inside a mxm square?

Question

the n points inside the square have to be equidistant between themselves and the corners of the square

Answer 1

Think about the geometry of it. There's only one point that can be equidistant from the 4 corners and that's in the center.


Doug

Answer 2

It won't work unless n is itself a square number. Then you want the distance from the corner of the frame equal to the distance between the points. Let the distance from the corner of the frame be "d". Then the distance from the edge of the frame has to be
d/sqrt(2), and the distance between the points has to be d.

If there are n=a^2 points, then there will be a-1 spaces between the points. This means that 2d/sqrt(2)+(a-1)d=m. Therefore
d=m/(a-1+sqrt(2)). Is the distance between the points, and the distance of the corner points from the nearest corner of the frame.

Starting at the point [d/sqrt(2),d/sqrt(2)] (the lower left-hand corner is [0,0]) start filling in, "a" dots per row, for "a" rows.

Edit: Obviously, I could be misinterpreting your question--I'm assuming you meant that the nearest-neighbor distance is the same as the nearest point-to-corner distance.

Answer 3

If all points must be equidistant to another then the best you can do in the plane is the vertices of an equilateral triangle. They will not be equidistant to all the corners of a square.

Answer 4

?
That is only possible for n = 1

The only point inside the square that has same distance to all 4 squares is the middle. So put your point there.

for n>1 there is no other point than the middle of the square that has same distance to all 4 corners...

Can you reformulate your question ?

Answer 5

Uhhh, what????