Class: PerfectShape::QuadraticBezierCurve

Inherits:
Shape
  • Object
show all
Includes:
MultiPoint
Defined in:
lib/perfect_shape/quadratic_bezier_curve.rb

Constant Summary collapse

BELOW = 
-2
LOWEDGE = 
-1
INSIDE = 
0
HIGHEDGE = 
1
ABOVE = 
2
OUTLINE_MINIMUM_DISTANCE_THRESHOLD = 
BigDecimal('0.001')

Instance Attribute Summary

Attributes included from MultiPoint

#points

Class Method Summary collapse

Instance Method Summary collapse

Methods included from MultiPoint

#first_point, #initialize, #max_x, #max_y, #min_x, #min_y, normalize_point_array

Methods inherited from Shape

#==, #bounding_box, #center_point, #center_x, #center_y, #height, #max_x, #max_y, #min_x, #min_y, #width

Class Method Details

.eqn(val, c1, cp, c2) ⇒ Object

Fill an array with the coefficients of the parametric equation in t, ready for solving against val with solve_quadratic. We currently have:

val = Py(t) = C1*(1-t)^2 + 2*CP*t*(1-t) + C2*t^2
            = C1 - 2*C1*t + C1*t^2 + 2*CP*t - 2*CP*t^2 + C2*t^2
            = C1 + (2*CP - 2*C1)*t + (C1 - 2*CP + C2)*t^2
          0 = (C1 - val) + (2*CP - 2*C1)*t + (C1 - 2*CP + C2)*t^2
          0 = C + Bt + At^2
C = C1 - val
B = 2*CP - 2*C1
A = C1 - 2*CP + C2



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 89

def eqn(val, c1, cp, c2)
  [
    c1 - val,
    cp + cp - c1 - c1,
    c1 - cp - cp + c2,
  ]
end

.eval_quadratic(vals, num, include0, include1, inflect, c1, ctrl, c2) ⇒ Object

Evaluate the t values in the first num slots of the vals[] array and place the evaluated values back into the same array. Only evaluate t values that are within the range <, >, including the 0 and 1 ends of the range iff the include0 or include1 booleans are true. If an “inflection” equation is handed in, then any points which represent a point of inflection for that quadratic equation are also ignored.



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 151

def eval_quadratic(vals, num,
                             include0,
                             include1,
                             inflect,
                             c1, ctrl, c2)
  j = -1
  i = 0
  while i < num
    t = vals[i]
    
    if ((include0 ? t >= 0 : t > 0) &&
        (include1 ? t <= 1 : t < 1) &&
        (inflect.nil? ||
         inflect[1] + 2*inflect[2]*t != 0))
      u = 1 - t
      vals[j+=1] = c1*u*u + 2*ctrl*t*u + c2*t*t
    end
    i+=1
  end
  j
end

.point_crossings(x1, y1, xc, yc, x2, y2, px, py, level = 0) ⇒ Object

Calculates the number of times the quadratic bézier curve from (x1,y1) to (x2,y2) crosses the ray extending to the right from (x,y). If the point lies on a part of the curve, then no crossings are counted for that intersection. the level parameter should be 0 at the top-level call and will count up for each recursion level to prevent infinite recursion +1 is added for each crossing where the Y coordinate is increasing -1 is added for each crossing where the Y coordinate is decreasing



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 37

def point_crossings(x1, y1, xc, yc, x2, y2, px, py, level = 0)
  return 0 if (py <  y1 && py <  yc && py <  y2)
  return 0 if (py >= y1 && py >= yc && py >= y2)
  # Note y1 could equal y2...
  return 0 if (px >= x1 && px >= xc && px >= x2)
  if (px <  x1 && px <  xc && px <  x2)
    if (py >= y1)
      return 1 if (py < y2)
    else
      # py < y1
      return -1 if (py >= y2)
    end
    # py outside of y11 range, and/or y1==y2
    return 0
  end
  # double precision only has 52 bits of mantissa
  return PerfectShape::Line.point_crossings(x1, y1, x2, y2, px, py) if (level > 52)
  x1c = BigDecimal((x1 + xc).to_s) / 2
  y1c = BigDecimal((y1 + yc).to_s) / 2
  xc1 = BigDecimal((xc + x2).to_s) / 2
  yc1 = BigDecimal((yc + y2).to_s) / 2
  xc = BigDecimal((x1c + xc1).to_s) / 2
  yc = BigDecimal((y1c + yc1).to_s) / 2
  # [xy]c are NaN if any of [xy]0c or [xy]c1 are NaN
  # [xy]0c or [xy]c1 are NaN if any of [xy][0c1] are NaN
  # These values are also NaN if opposing infinities are added
  return 0 if (xc.nan? || yc.nan?)
  point_crossings(x1, y1, x1c, y1c, xc, yc, px, py, level+1) +
    point_crossings(xc, yc, xc1, yc1, x2, y2, px, py, level+1)
end

.solve_quadratic(eqn, res) ⇒ Object

Solves the quadratic whose coefficients are in the eqn array and places the non-complex roots into the res array, returning the number of roots. The quadratic solved is represented by the equation: <pre>

eqn = {C, B, A}
ax^2 + bx + c = 0

</pre> A return value of -1 is used to distinguish a constant equation, which might be always 0 or never 0, from an equation that has no zeroes.

Parameters:

  • eqn —

    the specified array of coefficients to use to solve the quadratic equation

  • res —

    the array that contains the non-complex roots resulting from the solution of the quadratic equation

Returns:

  • the number of roots, or -1 if the equation is a constant.



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 114

def solve_quadratic(eqn, res)
  a = eqn[2]
  b = eqn[1]
  c = eqn[0]
  roots = -1
  if a == 0.0
    #  The quadratic parabola has degenerated to a line.
    #  The line has degenerated to a constant.
    return -1 if b == 0.0
    res[roots += 1] = -c / b
  else
    #  From Numerical Recipes, 5.6, Quadratic and Cubic Equations
    d = b * b - 4.0 * a * c
    #  If d < 0.0, then there are no roots
    return 0 if d < 0.0
    d = BigDecimal(Math.sqrt(d).to_a)
    #  For accuracy, calculate one root using:
    #      (-b +/- d) / 2a
    #  and the other using:
    #      2c / (-b +/- d)
    #  Choose the sign of the +/- so that b+d gets larger in magnitude
    d = -d if b < 0.0
    q = (b + d) / -2.0
    #  We already tested a for being 0 above
    res[roots += 1] = q / a
    res[roots += 1] = c / q if q != 0.0
  end
  roots
end

.tag(coord, low, high) ⇒ Object

Determine where coord lies with respect to the range from low to high. It is assumed that low < high. The return value is one of the 5 values BELOW, LOWEDGE, INSIDE, HIGHEDGE, or ABOVE.



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 72

def tag(coord, low, high)
  return (coord < low ? BELOW : LOWEDGE) if coord <= low
  return (coord > high ? ABOVE : HIGHEDGE) if coord >= high
  INSIDE
end

Instance Method Details

#contain?(x_or_point, y = nil, outline: false, distance_tolerance: 0) ⇒ Boolean

Checks if quadratic bézier curve contains point (two-number Array or x, y args)

the quadratic bézier curve, false if the point lies outside of the quadratic bézier curve’s bounds.

Parameters:

  • x —

    The X coordinate of the point to test.

  • y (defaults to: nil) —

    The Y coordinate of the point to test.

Returns:

  • (Boolean) —

    true if the point lies within the bound of



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 193

def contain?(x_or_point, y = nil, outline: false, distance_tolerance: 0)
  x, y = Point.normalize_point(x_or_point, y)
  return unless x && y

  x1 = points[0][0]
  y1 = points[0][1]
  xc = points[1][0]
  yc = points[1][1]
  x2 = points[2][0]
  y2 = points[2][1]
  
  if outline
    distance_tolerance = BigDecimal(distance_tolerance.to_s)
    minimum_distance_threshold = OUTLINE_MINIMUM_DISTANCE_THRESHOLD + distance_tolerance
    point_distance(x, y, minimum_distance_threshold: minimum_distance_threshold) < minimum_distance_threshold
  else
    # We have a convex shape bounded by quad curve Pc(t)
    # and ine Pl(t).
    #
    #     P1 = (x1, y1) - start point of curve
    #     P2 = (x2, y2) - end point of curve
    #     Pc = (xc, yc) - control point
    #
    #     Pq(t) = P1*(1 - t)^2 + 2*Pc*t*(1 - t) + P2*t^2 =
    #           = (P1 - 2*Pc + P2)*t^2 + 2*(Pc - P1)*t + P1
    #     Pl(t) = P1*(1 - t) + P2*t
    #     t = [0:1]
    #
    #     P = (x, y) - point of interest
    #
    # Let's look at second derivative of quad curve equation:
    #
    #     Pq''(t) = 2 * (P1 - 2 * Pc + P2) = Pq''
    #     It's constant vector.
    #
    # Let's draw a line through P to be parallel to this
    # vector and find the intersection of the quad curve
    # and the line.
    #
    # Pq(t) is point of intersection if system of equations
    # below has the solution.
    #
    #     L(s) = P + Pq''*s == Pq(t)
    #     Pq''*s + (P - Pq(t)) == 0
    #
    #     | xq''*s + (x - xq(t)) == 0
    #     | yq''*s + (y - yq(t)) == 0
    #
    # This system has the solution if rank of its matrix equals to 1.
    # That is, determinant of the matrix should be zero.
    #
    #     (y - yq(t))*xq'' == (x - xq(t))*yq''
    #
    # Let's solve this equation with 't' variable.
    # Also let kx = x1 - 2*xc + x2
    #          ky = y1 - 2*yc + y2
    #
    #     t0q = (1/2)*((x - x1)*ky - (y - y1)*kx) /
    #                 ((xc - x1)*ky - (yc - y1)*kx)
    #
    # Let's do the same for our line Pl(t):
    #
    #     t0l = ((x - x1)*ky - (y - y1)*kx) /
    #           ((x2 - x1)*ky - (y2 - y1)*kx)
    #
    # It's easy to check that t0q == t0l. This fact means
    # we can compute t0 only one time.
    #
    # In case t0 < 0 or t0 > 1, we have an intersections outside
    # of shape bounds. So, P is definitely out of shape.
    #
    # In case t0 is inside [0:1], we should calculate Pq(t0)
    # and Pl(t0). We have three points for now, and all of them
    # lie on one line. So, we just need to detect, is our point
    # of interest between points of intersections or not.
    #
    # If the denominator in the t0q and t0l equations is
    # zero, then the points must be collinear and so the
    # curve is degenerate and encloses no area.  Thus the
    # result is false.
    kx = x1 - 2 * xc + x2
    ky = y1 - 2 * yc + y2
    dx = x - x1
    dy = y - y1
    dxl = x2 - x1
    dyl = y2 - y1
  
    t0 = (dx * ky - dy * kx) / (dxl * ky - dyl * kx)
    return false if (t0 < 0 || t0 > 1 || t0 != t0)
  
    xb = kx * t0 * t0 + 2 * (xc - x1) * t0 + x1
    yb = ky * t0 * t0 + 2 * (yc - y1) * t0 + y1
    xl = dxl * t0 + x1
    yl = dyl * t0 + y1
  
    (x >= xb && x < xl) ||
      (x >= xl && x < xb) ||
      (y >= yb && y < yl) ||
      (y >= yl && y < yb)
  end
end

#curve_center_point ⇒ Object

The center point on the outline of the curve in Array format as pair of (x, y) coordinates



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 311

def curve_center_point
  subdivisions.last.points[0]
end

#curve_center_x ⇒ Object

The center point x on the outline of the curve



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 316

def curve_center_x
  subdivisions.last.points[0][0]
end

#curve_center_y ⇒ Object

The center point y on the outline of the curve



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 321

def curve_center_y
  subdivisions.last.points[0][1]
end

#intersect?(rectangle) ⇒ Boolean

Returns:

  • (Boolean) 


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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 388

def intersect?(rectangle)
  x = rectangle.x
  y = rectangle.y
  w = rectangle.width
  h = rectangle.height
  
  # Trivially reject non-existant rectangles
  return false if w <= 0 || h <= 0

  # Trivially accept if either endpoint is inside the rectangle
  # (not on its border since it may end there and not go inside)
  # Record where they lie with respect to the rectangle.
  #     -1 => left, 0 => inside, 1 => right
  x1 = points[0][0]
  y1 = points[0][1]
  x1tag = QuadraticBezierCurve.tag(x1, x, x+w)
  y1tag = QuadraticBezierCurve.tag(y1, y, y+h)
  return true if x1tag == INSIDE && y1tag == INSIDE
  x2 = points[2][0]
  y2 = points[2][1]
  x2tag = QuadraticBezierCurve.tag(x2, x, x+w)
  y2tag = QuadraticBezierCurve.tag(y2, y, y+h)
  return true if x2tag == INSIDE && y2tag == INSIDE
  ctrlx = points[1][0]
  ctrly = points[1][1]
  ctrlxtag = QuadraticBezierCurve.tag(ctrlx, x, x+w)
  ctrlytag = QuadraticBezierCurve.tag(ctrly, y, y+h)

  # Trivially reject if all points are entirely to one side of
  # the rectangle.
  # Returning false means All points left
  return false if x1tag < INSIDE && x2tag < INSIDE && ctrlxtag < INSIDE
  # Returning false means All points above
  return false if y1tag < INSIDE && y2tag < INSIDE && ctrlytag < INSIDE
  # Returning false means All points right
  return false if x1tag > INSIDE && x2tag > INSIDE && ctrlxtag > INSIDE
  # Returning false means All points below
  return false if y1tag > INSIDE && y2tag > INSIDE && ctrlytag > INSIDE

  # Test for endpoints on the edge where either the segment
  # or the curve is headed "inwards" from them
  # Note: These tests are a superset of the fast endpoint tests
  #       above and thus repeat those tests, but take more time
  #       and cover more cases
  # First endpoint on border with either edge moving inside
  return true if inwards(x1tag, x2tag, ctrlxtag) && inwards(y1tag, y2tag, ctrlytag)
  # Second endpoint on border with either edge moving inside
  return true if inwards(x2tag, x1tag, ctrlxtag) && inwards(y2tag, y1tag, ctrlytag)

  # Trivially accept if endpoints span directly across the rectangle
  xoverlap = (x1tag * x2tag <= 0)
  yoverlap = (y1tag * y2tag <= 0)
  return true if x1tag == INSIDE && x2tag == INSIDE && yoverlap
  return true if y1tag == INSIDE && y2tag == INSIDE && xoverlap

  # We now know that both endpoints are outside the rectangle
  # but the 3 points are not all on one side of the rectangle.
  # Therefore the curve cannot be contained inside the rectangle,
  # but the rectangle might be contained inside the curve, or
  # the curve might intersect the boundary of the rectangle.

  eqn = nil
  res = []
  if !yoverlap
      # Both Y coordinates for the closing segment are above or
      # below the rectangle which means that we can only intersect
      # if the curve crosses the top (or bottom) of the rectangle
      # in more than one place and if those crossing locations
      # span the horizontal range of the rectangle.
      eqn = QuadraticBezierCurve.eqn((y1tag < INSIDE ? y : y+h), y1, ctrly, y2)
      return (QuadraticBezierCurve.solve_quadratic(eqn, res) == 2 &&
              QuadraticBezierCurve.eval_quadratic(res, 2, true, true, nil,
                            x1, ctrlx, x2) == 2 &&
              QuadraticBezierCurve.tag(res[0], x, x+w) * QuadraticBezierCurve.tag(res[1], x, x+w) <= 0)
  end

  # Y ranges overlap.  Now we examine the X ranges
  if !xoverlap
      # Both X coordinates for the closing segment are left of
      # or right of the rectangle which means that we can only
      # intersect if the curve crosses the left (or right) edge
      # of the rectangle in more than one place and if those
      # crossing locations span the vertical range of the rectangle.
      eqn = QuadraticBezierCurve.eqn((x1tag < INSIDE ? x : x+w), x1, ctrlx, x2)
      return (QuadraticBezierCurve.solve_quadratic(eqn, res) == 2 &&
              QuadraticBezierCurve.eval_quadratic(res, 2, true, true, nil,
                            y1, ctrly, y2) == 2 &&
              QuadraticBezierCurve.tag(res[0], y, y+h) * QuadraticBezierCurve.tag(res[1], y, y+h) <= 0)
  end

  # The X and Y ranges of the endpoints overlap the X and Y
  # ranges of the rectangle, now find out how the endpoint
  # line segment intersects the Y range of the rectangle
  dx = x2 - x1
  dy = y2 - y1
  k = y2 * x1 - x2 * y1
  c1tag = c2tag = nil
  if y1tag == INSIDE
    c1tag = x1tag
  else
    c1tag = QuadraticBezierCurve.tag((k + dx * (y1tag < INSIDE ? y : y+h)) / dy, x, x+w)
  end
  if y2tag == INSIDE
    c2tag = x2tag
  else
    c2tag = QuadraticBezierCurve.tag((k + dx * (y2tag < INSIDE ? y : y+h)) / dy, x, x+w)
  end
  # If the part of the line segment that intersects the Y range
  # of the rectangle crosses it horizontally - trivially accept
  return true if c1tag * c2tag <= 0

  # Now we know that both the X and Y ranges intersect and that
  # the endpoint line segment does not directly cross the rectangle.
  #
  # We can almost treat this case like one of the cases above
  # where both endpoints are to one side, except that we will
  # only get one intersection of the curve with the vertical
  # side of the rectangle.  This is because the endpoint segment
  # accounts for the other intersection.
  #
  # (Remember there is overlap in both the X and Y ranges which
  #  means that the segment must cross at least one vertical edge
  #  of the rectangle - in particular, the "near vertical side" -
  #  leaving only one intersection for the curve.)
  #
  # Now we calculate the y tags of the two intersections on the
  # "near vertical side" of the rectangle.  We will have one with
  # the endpoint segment, and one with the curve.  If those two
  # vertical intersections overlap the Y range of the rectangle,
  # we have an intersection.  Otherwise, we don't.

  # c1tag = vertical intersection class of the endpoint segment
  #
  # Choose the y tag of the endpoint that was not on the same
  # side of the rectangle as the subsegment calculated above.
  # Note that we can "steal" the existing Y tag of that endpoint
  # since it will be provably the same as the vertical intersection.
  c1tag = ((c1tag * x1tag <= 0) ? y1tag : y2tag)

  # c2tag = vertical intersection class of the curve
  #
  # We have to calculate this one the straightforward way.
  # Note that the c2tag can still tell us which vertical edge
  # to test against.
  eqn = QuadraticBezierCurve.eqn((c2tag < INSIDE ? x : x+w), x1, ctrlx, x2)
  num = QuadraticBezierCurve.solve_quadratic(eqn, res)

  # Note: We should be able to assert(num == 2) since the
  # X range "crosses" (not touches) the vertical boundary,
  # but we pass num to QuadraticBezierCurve.eval_quadratic for completeness.
  QuadraticBezierCurve.eval_quadratic(res, num, true, true, nil, y1, ctrly, y2)

  # Note: We can assert(num evals == 1) since one of the
  # 2 crossings will be out of the [0,1] range.
  c2tag = QuadraticBezierCurve.tag(res[0], y, y+h)

  # Finally, we have an intersection if the two crossings
  # overlap the Y range of the rectangle.
  c1tag * c2tag <= 0
end

#point_crossings(x_or_point, y = nil, level = 0) ⇒ Object

Calculates the number of times the quad crosses the ray extending to the right from (x,y). If the point lies on a part of the curve, then no crossings are counted for that intersection. the level parameter should be 0 at the top-level call and will count up for each recursion level to prevent infinite recursion +1 is added for each crossing where the Y coordinate is increasing -1 is added for each crossing where the Y coordinate is decreasing



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 303

def point_crossings(x_or_point, y = nil, level = 0)
  x, y = Point.normalize_point(x_or_point, y)
  return unless x && y
  QuadraticBezierCurve.point_crossings(points[0][0], points[0][1], points[1][0], points[1][1], points[2][0], points[2][1], x, y, level)
end

#point_distance(x_or_point, y = nil, minimum_distance_threshold: OUTLINE_MINIMUM_DISTANCE_THRESHOLD) ⇒ Object 



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 360

def point_distance(x_or_point, y = nil, minimum_distance_threshold: OUTLINE_MINIMUM_DISTANCE_THRESHOLD)
  x, y = Point.normalize_point(x_or_point, y)
  return unless x && y
  
  point = Point.new(x, y)
  current_curve = self
  minimum_distance = point.point_distance(curve_center_point)
  last_minimum_distance = minimum_distance + 1 # start bigger to ensure going through loop once at least
  while minimum_distance >= minimum_distance_threshold && minimum_distance < last_minimum_distance
    curve1, curve2 = current_curve.subdivisions
    distance1 = point.point_distance(curve1.curve_center_point)
    distance2 = point.point_distance(curve2.curve_center_point)
    last_minimum_distance = minimum_distance
    if distance1 < distance2
      minimum_distance = distance1
      current_curve = curve1
    else
      minimum_distance = distance2
      current_curve = curve2
    end
  end
  if minimum_distance < minimum_distance_threshold
    minimum_distance
  else
    last_minimum_distance
  end
end

#rect_crossings(rxmin, rymin, rxmax, rymax, level, crossings = 0) ⇒ Object

Accumulate the number of times the quad crosses the shadow extending to the right of the rectangle. See the comment for the RECT_INTERSECTS constant for more complete details.

crossings arg is the initial crossings value to add to (useful in cases where you want to accumulate crossings from multiple shapes)



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 556

def rect_crossings(rxmin, rymin, rxmax, rymax, level, crossings = 0)
  x0 = points[0][0]
  y0 = points[0][1]
  xc = points[1][0]
  yc = points[1][1]
  x1 = points[2][0]
  y1 = points[2][1]
  return crossings if y0 >= rymax && yc >= rymax && y1 >= rymax
  return crossings if y0 <= rymin && yc <= rymin && y1 <= rymin
  return crossings if x0 <= rxmin && xc <= rxmin && x1 <= rxmin
  if x0 >= rxmax && xc >= rxmax && x1 >= rxmax
    # Quad is entirely to the right of the rect
    # and the vertical range of the 3 Y coordinates of the quad
    # overlaps the vertical range of the rect by a non-empty amount
    # We now judge the crossings solely based on the line segment
    # connecting the endpoints of the quad.
    # Note that we may have 0, 1, or 2 crossings as the control
    # point may be causing the Y range intersection while the
    # two endpoints are entirely above or below.
    if y0 < y1
      # y-increasing line segment...
      crossings += 1 if y0 <= rymin && y1 >  rymin
      crossings += 1 if y0 <  rymax && y1 >= rymax
    elsif y1 < y0
      # y-decreasing line segment...
      crossings -= 1 if y1 <= rymin && y0 >  rymin
      crossings -= 1 if y1 <  rymax && y0 >= rymax
    end
    return crossings
  end
  # The intersection of ranges is more complicated
  # First do trivial INTERSECTS rejection of the cases
  # where one of the endpoints is inside the rectangle.
  return PerfectShape::Rectangle::RECT_INTERSECTS if (x0 < rxmax && x0 > rxmin && y0 < rymax && y0 > rymin) ||
    (x1 < rxmax && x1 > rxmin && y1 < rymax && y1 > rymin)
  # Otherwise, subdivide and look for one of the cases above.
  # double precision only has 52 bits of mantissa
  if level > 52
    line = PerfectShape::Line.new(points: [x0, y0, x1, y1])
    return line.rect_crossings(rxmin, rymin, rxmax, rymax, crossings)
  end
  x0c = BigDecimal((x0 + xc).to_s) / 2
  y0c = BigDecimal((y0 + yc).to_s) / 2
  xc1 = BigDecimal((xc + x1).to_s) / 2
  yc1 = BigDecimal((yc + y1).to_s) / 2
  xc = BigDecimal((x0c + xc1).to_s) / 2
  yc = BigDecimal((y0c + yc1).to_s) / 2
  # [xy]c are NaN if any of [xy]0c or [xy]c1 are NaN
  # [xy]0c or [xy]c1 are NaN if any of [xy][0c1] are NaN
  # These values are also NaN if opposing infinities are added
  return 0 if xc.nan? || yc.nan?
  quad1 = QuadraticBezierCurve.new(points: [x0, y0, x0c, y0c, xc, yc])
  crossings = quad1.rect_crossings(rxmin, rymin, rxmax, rymax, level+1, crossings)
  if crossings != PerfectShape::Rectangle::RECT_INTERSECTS
    quad2 = QuadraticBezierCurve.new(points: [xc, yc, xc1, yc1, x1, y1])
    crossings = quad2.rect_crossings(rxmin, rymin, rxmax, rymax, level+1, crossings)
  end
  crossings
end

#subdivisions(level = 1) ⇒ Object

Subdivides QuadraticBezierCurve exactly at its curve center returning 2 QuadraticBezierCurve’s as a two-element Array by default

Optional ‘level` parameter specifies the level of recursions to perform to get more subdivisions. The number of resulting subdivisions is 2 to the power of `level` (e.g. 2 subdivisions for level=1, 4 subdivisions for level=2, and 8 subdivisions for level=3)



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# File 'lib/perfect_shape/quadratic_bezier_curve.rb', line 332

def subdivisions(level = 1)
  level -= 1 # consume 1 level
  
  x1 = points[0][0]
  y1 = points[0][1]
  ctrlx = points[1][0]
  ctrly = points[1][1]
  x2 = points[2][0]
  y2 = points[2][1]
  ctrlx1 = BigDecimal((x1 + ctrlx).to_s) / 2
  ctrly1 = BigDecimal((y1 + ctrly).to_s) / 2
  ctrlx2 = BigDecimal((x2 + ctrlx).to_s) / 2
  ctrly2 = BigDecimal((y2 + ctrly).to_s) / 2
  centerx = BigDecimal((ctrlx1 + ctrlx2).to_s) / 2
  centery = BigDecimal((ctrly1 + ctrly2).to_s) / 2
  
  default_subdivisions = [
    QuadraticBezierCurve.new(points: [x1, y1, ctrlx1, ctrly1, centerx, centery]),
    QuadraticBezierCurve.new(points: [centerx, centery, ctrlx2, ctrly2, x2, y2])
  ]
  
  if level == 0
    default_subdivisions
  else
    default_subdivisions.map { |curve| curve.subdivisions(level) }.flatten
  end
end