我一直在尝试用Python实现礼品包装算法,我目前有以下代码:
def createIslandPolygon(particleCoords):
startPoint = min(particleCoords.iteritems(),key = lambda x: x[1][1])[1]
check = 1
islandPolygon = []
particleList = []
for key in particleCoords:
particleList.append(particleCoords[key])
currentPoint = startPoint
while(currentPoint != startPoint or check == 1):
islandPolygon.append(currentPoint)
check = 0
angleDict = {}
angleList = []
for point in particleList:
if point != currentPoint:
angleDict[(angleBetweenTwoPoints(currentPoint, point))] = point
angleList.append(angleBetweenTwoPoints(currentPoint, point))
smallestAngle = min(angleList)
currentPoint = angleDict[smallestAngle]
return islandPolygon
用于计算极坐标:
def angleBetweenTwoPoints(p1, p2):
p3 = (p1[0], p1[1] + 2)
a = (p1[0] - p2[0], p1[1] - p2[1])
b = (p1[0] - p3[0], p1[1] - p3[1])
theta = ((a[0]*b[0]) + (a[1]*b[1]))
theta = theta / (sqrt((a[0]*a[0]) + (a[1]*a[1])) * sqrt((b[0]*b[0]) + (b[1]*b[1])))
theta = math.acos(theta)
return theta
问题是代码似乎永远不会离开while循环,我不知道为什么。有没有人有任何想法?
感谢。
(是的,代码非常粗制滥造,我只是把它快速扔到一起)
编辑:打印出坐标似乎表明它们只在两个坐标之间跳跃。
答案 0 :(得分:1)
根据http://en.wikipedia.org/wiki/Gift_wrapping_algorithm,你需要这样做:
pointOnHull = leftmost point in S
i = 0
repeat
P[i] = pointOnHull
endpoint = S[0] // initial endpoint for a candidate edge on the hull
for j from 1 to |S|-1
if (endpoint == pointOnHull) or (S[j] is on left of line from P[i] to endpoint)
endpoint = S[j] // found greater left turn, update endpoint
i = i+1
pointOnHull = endpoint
until endpoint == P[0] // wrapped around to first hull point
你有这个正确的:
pointOnHull = leftmost point in S
而且:
P[i] = pointOnHull
但这是我不确定的部分:
(S[j] is on left of line from P[i] to endpoint)
相反,您可以在所有其他点的角度中找到最小角度。但根据维基百科你想要的是最左边的角度,它与所有其他点的角度。我有一些处理角度的代码:
def normalizeangle(radians):
return divmod(radians, math.pi*2)[1]
def arclength(radians1, radians2 = 0):
radians1, radians2 = normalizeangle(radians1), normalizeangle(radians2)
return min(normalizeangle(radians1 - radians2), normalizeangle(radians2 - radians1))
def arcdir(radians1, radians2 = 0):
radians1, radians2 = normalizeangle(radians1), normalizeangle(radians2)
return cmp(normalizeangle(radians1 - radians2), normalizeangle(radians2 - radians1))
arcdir
会告诉您角度是在另一个角度的左侧还是右侧,因此您可以使用它来查看角度是否更左,因此应该使用。
如果沿着点移动总是将最左边的角度移动到下一个点,你将绕过多边形的周边并再次到达开头(因为你选择了最左边的点,你知道它必须在周长,将再次到达)