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懒得码字了,直接上代码+测试。
注:附上了完整的TCAX下的测试工程,要执行,请把tcCurve.py放到TCAX的util目录下。
实现
tcCurve.py
(10.02 KB, 下载次数: 2256)
- ###########################################################################################################################
- ###########################################################################################################################
- ###################### Uniform Cubic B-Spline Python Implementation, Object-Oriented Version With Const Speed Utility
- class UCBSpline:
- # pre-calculate coefficients
- def __init__(self, p):
- self.p = p # tuple or list of control points (x, y)
- n = len(p) - 1 # control points p0 ~ pn, i.e., p[0] ~ p[n]
- m = 3 + n + 1 # knots vector u0 ~um, i.e., u[0] ~ u[m]
- self.m = m
- step = 1 / (m - 2 * 3)
- self.u = 3 * [0] + [_u * step for _u in range(m - 2 * 3 + 1)] + [1] * 3 # knots vector
- num = n + 1 - 3 # number of curve segments
- self.num = num
- self.a1 = []
- self.a2 = []
- self.a3 = []
- self.a4 = []
- self.b1 = []
- self.b2 = []
- self.b3 = []
- self.b4 = []
- for i in range(num):
- self.a1.append((-p[i][0] + 3 * p[i + 1][0] - 3 * p[i + 2][0] + p[i + 3][0]) / 6.0)
- self.a2.append((3 * p[i][0] - 6 * p[i + 1][0] + 3 * p[i + 2][0]) / 6.0)
- self.a3.append((-p[i][0] + p[i + 2][0]) / 2.0)
- self.a4.append((p[i][0] + 4 * p[i + 1][0] + p[i + 2][0]) / 6.0)
- self.b1.append((-p[i][1] + 3 * p[i + 1][1] - 3 * p[i + 2][1] + p[i + 3][1]) / 6.0)
- self.b2.append((3 * p[i][1] - 6 * p[i + 1][1] + 3 * p[i + 2][1]) / 6.0)
- self.b3.append((-p[i][1] + p[i + 2][1]) / 2.0)
- self.b4.append((p[i][1] + 4 * p[i + 1][1] + p[i + 2][1]) / 6.0)
- self.SIMPSON = 100 # number of sub-intervals of composite simpson integration, must be even, usually 100 is good enough
- self.NEWTON = 0.0001 # epsilon, effective digits of the root calculated by Newton's method
- # calculate the value at t
- def __call__(self, t):
- if t < 0 or t >= 1:
- return None
- else: # find the curve segment
- for i in range(3, self.m - 3):
- if self.u[i] <= t and t < self.u[i + 1]:
- break
- idx = i - 3 # get coefficients
- a1 = self.a1[idx]
- a2 = self.a2[idx]
- a3 = self.a3[idx]
- a4 = self.a4[idx]
- b1 = self.b1[idx]
- b2 = self.b2[idx]
- b3 = self.b3[idx]
- b4 = self.b4[idx]
- t = (t - self.u[i]) / (self.u[i + 1] - self.u[i]) # map to the current curve segment
- return (a1 * t * t * t + a2 * t * t + a3 * t + a4,
- b1 * t * t * t + b2 * t * t + b3 * t + b4, 255) # 255 is the point's alpha value, to meet the TCAX's point structure requirement ((x, y, a), (x, y, a), ...)
- # total arc-length
- def length(self):
- from math import sqrt
- sqr = lambda x: x * x
- l = 0
- for i in range(self.num):
- a1 = self.a1[i]
- a2 = self.a2[i]
- a3 = self.a3[i]
- a4 = self.a4[i]
- b1 = self.b1[i]
- b2 = self.b2[i]
- b3 = self.b3[i]
- b4 = self.b4[i]
- f_grand = lambda t: sqrt(sqr(3 * a1 * t * t + 2 * a2 * t + a3) + sqr(3 * b1 * t * t + 2 * b2 * t + b3)) # integrand of UCB
- l += CompositeSimpson(f_grand, 0, 1, self.SIMPSON)
- return l
- # find t2 according to a specific arc-length s and a starting point t1
- def t2OfLength(self, t1, s):
- if t1 < 0 or t1 >= 1:
- return None
- else: # find the curve segment
- for i in range(3, self.m - 3):
- if self.u[i] <= t1 and t1 < self.u[i + 1]:
- break
- idx = i - 3 # get coefficients
- a1 = self.a1[idx]
- a2 = self.a2[idx]
- a3 = self.a3[idx]
- a4 = self.a4[idx]
- b1 = self.b1[idx]
- b2 = self.b2[idx]
- b3 = self.b3[idx]
- b4 = self.b4[idx]
- t1 = (t1 - self.u[i]) / (self.u[i + 1] - self.u[i]) # map to the current curve segment
- # Assumption: t1 and t are in the same piece of curve
- from math import sqrt
- sqr = lambda x: x * x
- f_grand = lambda t: sqrt(sqr(3 * a1 * t * t + 2 * a2 * t + a3) + sqr(3 * b1 * t * t + 2 * b2 * t + b3)) # integrand of UCB
- s1 = CompositeSimpson(f_grand, t1, 1, self.SIMPSON) # length from the starting point t1 to the end of the current curve segment
- if s1 >= s: # t1 and t are in the same segment of curve
- f_grand_d = lambda t: ((3 * a1 * t * t + 2 * a2 * t + a3) * (6 * a1 * t + 2 * a2) + (3 * b1 * t * t + 2 * b2 * t + b3) * (6 * b1 * t + 2 * b2 * t)) / f_grand(t) # derivative of f_grand
- # using Newton's method to find the root of function f
- f = lambda t: CompositeSimpson(f_grand, t1, t, self.SIMPSON) - s
- fd = lambda t: CompositeSimpsonDerivative(f_grand, f_grand_d, t1, t, self.SIMPSON) # derivative of f
- return NewtonMethod(f, fd, t1, self.NEWTON) * (self.u[i + 1] - self.u[i]) + self.u[i] # find the t and map to the whole curve
- # otherwise, t is in next segment(s)
- else:
- if idx + 1 == self.num: # no more curve segment
- return 1
- for i in range(idx + 1, self.num):
- a1 = self.a1[i]
- a2 = self.a2[i]
- a3 = self.a3[i]
- a4 = self.a4[i]
- b1 = self.b1[i]
- b2 = self.b2[i]
- b3 = self.b3[i]
- b4 = self.b4[i]
- f_grand = lambda t: sqrt(sqr(3 * a1 * t * t + 2 * a2 * t + a3) + sqr(3 * b1 * t * t + 2 * b2 * t + b3)) # integrand of UCB
- s0 = s1
- s1 += CompositeSimpson(f_grand, 0, 1, self.SIMPSON)
- if s1 >= s: # we find at which segment the t lies
- break
- if i == self.num: # t is beyond the whole curve
- return 1
- f_grand_d = lambda t: ((3 * a1 * t * t + 2 * a2 * t + a3) * (6 * a1 * t + 2 * a2) + (3 * b1 * t * t + 2 * b2 * t + b3) * (6 * b1 * t + 2 * b2 * t)) / f_grand(t) # derivative of f_grand
- # using Newton's method to find the root of function f
- f = lambda t: CompositeSimpson(f_grand, 0, t, self.SIMPSON) - (s - s0)
- fd = lambda t: CompositeSimpsonDerivative(f_grand, f_grand_d, 0, t, self.SIMPSON) # derivative of f
- return NewtonMethod(f, fd, 0, self.NEWTON) * (self.u[i + 3 + 1] - self.u[i + 3]) + self.u[i + 3] # find the t and map to the whole curve, note, here i is actually idx, should + 3 to be real i
- # f=function, a=initial value, b=end value, n=number of intervals of size h, n must be even
- def CompositeSimpson(f, a, b, n):
- h = (b - a) / n
- S = f(a)
- for i in range(1, n, 2):
- x = a + h * i
- S += 4 * f(x)
- for i in range(2, n - 1, 2):
- x = a + h * i
- S += 2 * f(x)
- S += f(b)
- return h * S / 3
- # f=function, fd=derivative of f, t1=initial value (constant), t=end value (variable), n=number of intervals of size h, n must be even
- def CompositeSimpsonDerivative(f, fd, t1, t, n):
- h = (t - t1) / n
- hd = 1 / n
- S = f(t1)
- for i in range(1, n, 2):
- x = t1 + h * i
- S += 4 * f(x)
- for i in range(2, n - 1, 2):
- x = t1 + h * i
- S += 2 * f(x)
- S += f(t)
- part1 = hd * S / 3
- # part2
- S = 0
- for i in range(1, n, 2):
- x = t1 + h * i
- xd = hd * i
- S += 4 * fd(x) * xd
- for i in range(2, n - 1, 2):
- x = t1 + h * i
- xd = hd * i
- S += 2 * fd(x) * xd
- S += fd(t)
- part2 = h * S / 3
- return part1 + part2
- # f is the function whose root needs to be found, fd is the derivative of function f, x0 is the starting point, e is an epsilon, i.e., effective digits
- def NewtonMethod(f, fd, x0, e):
- x = x0
- while True:
- x_new = x - f(x) / fd(x)
- if abs(x_new - x) < e:
- return x_new
- x = x_new
- ###########################################################################################################################
- ###########################################################################################################################
- ###################### Uniform Cubic B-Spline Python Implementation
- def UniformCubicBSpline(p, t):
- n = len(p) - 1 # control points p0 ~ pn, i.e., p[0] ~ p[n]
- m = 3 + n + 1 # knots vector u0 ~um, i.e., u[0] ~ u[m]
- step = 1 / (m - 2 * 3)
- u = 3 * [0] + [_u * step for _u in range(m - 2 * 3 + 1)] + [1] * 3
- if t < 0 or t >= 1:
- return None
- else:
- for i in range(3, m - 3):
- if u[i] <= t and t < u[i + 1]:
- break
- return UniformCubicBSpline_Calc(p[i - 3], p[i - 2], p[i - 1], p[i], (t - u[i]) / (u[i + 1] - u[i])) # can use either calc or calc2
- def UniformCubicBSpline_Calc(p1, p2, p3, p4, t):
- a1 = (-p1[0] + 3 * p2[0] - 3 * p3[0] + p4[0]) / 6.0
- a2 = (3 * p1[0] - 6 * p2[0] + 3 * p3[0]) / 6.0
- a3 = (-p1[0] + p3[0]) / 2.0
- a4 = (p1[0] + 4 * p2[0] + p3[0]) / 6.0
- b1 = (-p1[1] + 3 * p2[1] - 3 * p3[1] + p4[1]) / 6.0
- b2 = (3 * p1[1] - 6 * p2[1] + 3 * p3[1]) / 6.0
- b3 = (-p1[1] + p3[1]) / 2.0
- b4 = (p1[1] + 4 * p2[1] + p3[1]) / 6.0
- return (a1 * t * t * t + a2 * t * t + a3 * t + a4,
- b1 * t * t * t + b2 * t * t + b3 * t + b4, 255) # 255 is the point's alpha value, to meet the TCAX's point structure requirement ((x, y, a), (x, y, a), ...)
- def UniformCubicBSpline_Calc2(p1, p2, p3, p4, t):
- A1 = (1 - t) * (1 - t) * (1 - t) / 6.0
- A2 = (3 * t * t * t - 6 * t * t + 4) / 6.0
- A3 = (-3 * t * t * t + 3 * t * t + 3 * t + 1) / 6.0
- A4 = t * t * t / 6.0
- return (A1 * p1[0] + A2 * p2[0] + A3 * p3[0] + A4 * p4[0],
- A1 * p1[1] + A2 * p2[1] + A3 * p3[1] + A4 * p4[1], 255) # 255 is the point's alpha value, to meet the TCAX's point structure requirement ((x, y, a), (x, y, a), ...)
复制代码 测试用例
test_curves.rar
(1.48 KB, 下载次数: 7491)
- from tcaxPy import *
- from util.cairo import *
- from util.tcCurve import *
- def tcaxPy_User():
- surface = ImageSurface(FORMAT_ARGB32, GetVal(val_ResolutionX), GetVal(val_ResolutionY))
- ctx = Context(surface)
- # 控制点列表1
- P = [(200, 500), (300, 100), (700, 100), (400, 500), (600, 700), (800, 400), (700, 200), (900, 300), (660, 400)]
- # 控制点列表2
- # P = [(640, 0)]
- # for i in range(10):
- # P.extend([(240 + randint(-40, 40), 200 + randint(-50, 50)), (1040 + randint(-40, 40), 200 + randint(-50, 50)), (640, 0)])
- # P.extend([(340 + randint(-60, 60), 600 + randint(-50, 50)), (940 + randint(-60, 60), 600 + randint(-50, 50)), (640, 0)])
- # 控制点列表3
- # P = [(0, 100), (100, 0), (200, 0), (300, 100), (400, 200), (500, 200), (600, 100), (400, 400), (700, 50), (800, 200)]
- # 控制点列表4
- # P = [(300, 100), (400, 200), (500, 200), (600, 100)]
- # 控制点列表5
- # P = [(400, 100), (400, 500), (500, 500), (500, 100)]
- # 绘制控制点
- ctx.set_source_rgb(0, 0, 0)
- num = len(P)
- for i in range(num):
- ctx.arc(P[i][0], P[i][1], 2, 0, 2 * pi)
- ctx.fill()
- # 计算曲线
- ucb = UCBSpline(P)
- L = ucb.length() # 曲线总长度
- STEP_N = int(L / 10 + 0.5) # 取样点数
- step_size = 1 / STEP_N
- print('控制点: {0} 曲线总长度: {1:.02f} 所取点数: {2}'.format(len(P), L, STEP_N))
- t1 = 0
- points = []
- for i in range(STEP_N):
- # 普通1, 使用UniformCubicBSpline函数
- #points.append(UniformCubicBSpline(P, i * step_size))
- # 普通2
- #points.append(ucb(i * step_size))
- # 匀速版本
- points.append(ucb(t1))
- t1 = ucb.t2OfLength(t1, step_size * L)
- # 绘制独立点
- ctx.set_source_rgb(0, 1, 0)
- num = len(points)
- for i in range(num):
- ctx.arc(points[i][0], points[i][1], 2, 0, 2 * pi)
- ctx.fill()
- # 绘制连续曲线
- # ctx.move_to(points[0][0], points[0][1])
- # for i in range(1, num):
- # ctx.line_to(points[i][0], points[i][1])
- # ctx.stroke()
- # save the result
- surface.write_to_png(abspath('1.png'))
- PIX = surface.get_pix()
- PIX = PixBlur(PIX, 10)
- SavePix(abspath('2.png'), PIX)
- #sys.exit()
复制代码 若干效果截图
计算效率初步测试
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ミルク 发表于12楼
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根据这篇论文,重新理解了下UCBS,并简化了代码(更好理解)。
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