Class: PerfectShape::QuadraticBezierCurve

Inherits:

Object
Shape
PerfectShape::QuadraticBezierCurve

Includes:: MultiPoint

Defined in:: lib/perfect_shape/quadratic_bezier_curve.rb

Constant Summary collapse

BELOW =

-2

LOWEDGE =

-1

INSIDE =

HIGHEDGE =

ABOVE =

OUTLINE_MINIMUM_DISTANCE_THRESHOLD =

BigDecimal('0.001')

Instance Attribute Summary

Attributes included from MultiPoint

#points

Class Method Summary collapse

.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.
.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.
.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).
.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.
.tag(coord, low, high) ⇒ Object

Determine where coord lies with respect to the range from low to high.

Instance Method Summary collapse

#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).
#curve_center_point ⇒ Object

The center point on the outline of the curve in Array format as pair of (x, y) coordinates.
#curve_center_x ⇒ Object

The center point x on the outline of the curve.
#curve_center_y ⇒ Object

The center point y on the outline of the curve.
#intersect?(rectangle) ⇒ Boolean
#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).
#point_distance(x_or_point, y = nil, minimum_distance_threshold: OUTLINE_MINIMUM_DISTANCE_THRESHOLD) ⇒ Object
#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.
#subdivisions(level = 1) ⇒ Object

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

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

# 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.

# 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

# 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.

# 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.

# 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

# 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



311
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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



316
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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



321
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323

# 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)

# 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

# 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`

# 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)

# 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)

# 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

Class: PerfectShape::QuadraticBezierCurve

Constant Summary collapse

Instance Attribute Summary

Attributes included from MultiPoint

Class Method Summary collapse

Instance Method Summary collapse

Methods included from MultiPoint

Methods inherited from Shape

Class Method Details

.eqn(val, c1, cp, c2) ⇒ Object

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

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

.solve_quadratic(eqn, res) ⇒ Object

.tag(coord, low, high) ⇒ Object

Instance Method Details

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

#curve_center_point ⇒ Object

#curve_center_x ⇒ Object

#curve_center_y ⇒ Object

#intersect?(rectangle) ⇒ Boolean

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

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

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

#subdivisions(level = 1) ⇒ Object

.eqn(val, c1, cp, c2) ⇒ `Object`

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

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

.solve_quadratic(eqn, res) ⇒ `Object`

.tag(coord, low, high) ⇒ `Object`

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

#curve_center_point ⇒ `Object`

#curve_center_x ⇒ `Object`

#curve_center_y ⇒ `Object`

#intersect?(rectangle) ⇒ `Boolean`

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

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

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

#subdivisions(level = 1) ⇒ `Object`