How to fit closest point to an equation? Python

Question

Basically I have some code for which it will find the equation of a plane and then attempt to place a 1 in a list if a point satisfies this equation of a line, otherwise a 0 is put in for the list. Unfortunately, there has to be an increment amount, so how would you get a point that is closest to the equation so that an approximate plane can be made without a bunch of empty spaces?

Here is the code so far:

def plane(self):
    p1 = self.first_pos
    p2 = self.second_pos
    p3 = self.third_pos
    x1,y1,z1 = self.fourth_pos
    x2,y2,z2 = self.fifth_pos
    a = (p2[0] - p1[0],p2[1] - p1[1],p2[2] - p1[2])
    b = (p3[0] - p1[0],p3[1] - p1[1],p3[2] - p1[2])
    abc = ((a[1] * b[2]) - (a[2] * b[1]),(a[2] * b[0]) - (a[0] * b[2]), (a[0] * b[1]) - (a[1] * b[0]))
    constant = (p1[0] *abc[0] * -1) - (p1[1] * abc[1]) - (p1[2] * abc[2])
    lx = []
    lxy = []
    axyz = []
    if x1 > x2 : x1, x2 = x2, x1
    if y1 > y2 : y1, y2 = y2, y1
    if z1 > z2 : z1, z2 = z2, z1
    for z in range(z1, z2+1):
        for y in range(y1,y2+1):
            for x in range(x1,x2+1):
                if int(round(((abc[1] *y) + (abc[2] *z) + constant + 0.6 ) / (-1 * abc[0]))) == x:
                    lx.append(1)
                else:
                    lx.append(0)
                if x == x2:
                    lxy.append(lx)
                    lx = []
            if y == y2:
                axyz.append(lxy)
                lxy = []
    self.first_pos = self.fourth_pos
    self.second_pos = self.fifth_pos
    self.buildMatrix(axyz)
    self.BuildCuboid(axyz)

Here is an example code for drawing a line that works with the closest points to the actual line being used:

def DrawLine(self):
    self.bot.sendMessage("Drawing line.",ignorable=True)
    fp = self.first_pos
    sp = self.second_pos
    ## This is the vector from pt 1 to pt 2
    x,y,z = sp[0] - fp[0], sp[1] - fp[1], sp[2] - fp[2]

    ## magnitude of that vector
    dist = self.bot.dist3d(fp[0], fp[1], fp[2], sp[0], sp[1], sp[2] )

    ## unit vector
    n_x, n_y, n_z = x/dist, y/dist, z/dist

    ## stepping a dist of 1 in the direction of the unit vector, find the
    ## whole coordinate and place a block at that location
    coords = []
    for d in xrange(0, int(dist)):
        self.blocks.append( (
                       self.block_type,
                       int(round(fp[0] + (n_x * d))),
                       int(round(fp[1] + (n_y * d))),
                       int(round(fp[2] + (n_z * d)))
                       ) )
    self.DrawBlocks()

Answer 1

If I'm understanding your intent correctly, you have a plane defined by three points (p0, p1, p2) and then want to evaluate whether some other point lies in that plane (or very nearly so).

This is most easily expressed using matrices rather than by manipulating the individual coordinate components in the code snippet listed above. Here's a link showing how to use matrices to solve this and related problems: http://paulbourke.net/geometry/planeeq/ .

It looks like your code is already close to representing the equation of the plane (you can verify it by substituting in the original points to see if it evaluates to zero or near zero).

Next, substitute the candidate point to see whether it evaluates to zero or near zero.

Answer 2

Your problem is a 3-dimensional version of the 2-dimensional line-plotting done by Bresenham's Algorithm. B-A plots a line using cells in a grid, including connecting cells so that you get a continuous line. That is, instead of just indexing column by column and computing the proper cell for the equivalent x-value (which would leave spaces between points if the line is not exactly vertical, horizontal, or 45degrees), B-A moves from cell to cell, determining which adjacent cell is best to fit to the linear equation being plotted.

Adapting this to 3-dimensions might be done by plotting each linear slice in x, and then again for each slice in y. Implementation left as an exercise for the OP. The Wikipedia page includes some links for n-dimensional treatments.