Equation of a Square?
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This is going to be quite a strange question...
Does a _square_ have an equation?
You know, an equation for the points on a square?
For a circle there's r^2 = dx^2 + dy^2, but what would a square equation look like?
disjointed
a set of separate equations with some arbitrary semantics that indicate when you need to switch from one to the next
a set of separate equations with some arbitrary semantics that indicate when you need to switch from one to the next
It's discontinuous (it has corners) and isn't a function, so it's not as straight forward, but thinking about it, this seems like it might work:
x = |d|, y = |d|, for all x,y ≤ |d|
x = |d|, y = |d|, for all x,y ≤ |d|
If you want to draw the square I guess a useful way would be to describe it as a linear combination of two perpendicular unit vectors.
For example, with the two vectors
A = (0, -1, 0)
B = (1, 0, 0)
Any point in the square is given by
P = Ar + Bs
Where r, s <= l (and l = length of the sides of the square).
If you want to describe the edge you just fix either r or s at 0 or l and you have a line equation.
A and B doesn't have to be unit vectors though, but it makes it easier to set the size of the square if they are.
For example, with the two vectors
A = (0, -1, 0)
B = (1, 0, 0)
Any point in the square is given by
P = Ar + Bs
Where r, s <= l (and l = length of the sides of the square).
If you want to describe the edge you just fix either r or s at 0 or l and you have a line equation.
A and B doesn't have to be unit vectors though, but it makes it easier to set the size of the square if they are.
Equation of a triangle
P in triangle (A, B, C) :
P = A + t * (B - A) + u * (C - A)
0 <= t <= 1,
0 <= u <= 1,
0 <= (t+u) <= 1,
let's say a parallelogram, which is like, two triangles... I am not sure waht would be the constraints on that.
0 <= (t+u) <= 2 maybe?
P in triangle (A, B, C) :
P = A + t * (B - A) + u * (C - A)
0 <= t <= 1,
0 <= u <= 1,
0 <= (t+u) <= 1,
let's say a parallelogram, which is like, two triangles... I am not sure waht would be the constraints on that.
0 <= (t+u) <= 2 maybe?
another one :)
say you have square (x0, y0)-(x1, y1) (x0 < x1, y0 < y1)
x = min(max(x, x0), x1)
y = max(min(y, y0), y1)
say you have square (x0, y0)-(x1, y1) (x0 < x1, y0 < y1)
x = min(max(x, x0), x1)
y = max(min(y, y0), y1)
Actually, you just remove the the sum restriction u + t ≤ 1 completely to get the full quadrilateral (keep 0 ≤ u/t ≤ 1). But it describes the entire inner area. You would have more logic to get just the edges.
As nmi said, the equation for a zero-centered axis-aligned square of edge 2r is:
max(|x|,|y|) = r
You may apply a transform first to rotate, move or scale the square.
max(|x|,|y|) = r
You may apply a transform first to rotate, move or scale the square.
Quote:Original post by erissian
It's discontinuous (it has corners)
Not quite. A square is continuous (there are certainly no gaps), but it isn't smooth. If one were to describe it piecewise-implicitly or parametrically, then it's the first derivatives (and hence all that follow) that would be discontinuous. </unnecessary aside>
Admiral
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