基于python 凸包问题的解决

# -*- coding: utf-8 -*-
import turtle
import random
import time
f=open('point.txt','w')
for i in range(100):
 x=random.randrange(-250,250,10)
 y=random.randrange(-200,200,10)
 f.write(str(x)+','+str(y)+'\n')
f.close()
point=[]
 
f=open('point.txt')
for i in f:
 try:
  temp=i.split(',')
  x=float(temp[0]); y=float(temp[1])
  point.append((x,y))
 except :
  print 'a err'
f.close()
 
point=list(set(point))#去除重复的点
 
n=0
for i in range(len(point)):
 if point[n][1]>point[i][1]:
  n=i
 
p0=point[n]
point.pop(n)
def compare(a,b):
 x=[a[0]-p0[0],a[1]-p0[1]]
 y=[b[0]-p0[0],b[1]-p0[1]]
 dx=(x[0]**2+x[1]**2)**0.5
 dy=(y[0]**2+y[1]**2)**0.5
 cosa=x[0]/dx
 cosb=y[0]/dy
 if cosa < cosb:
  return 1
 elif cosa > cosb:
  return -1
 else:
  if dx<dy:
   return -1
  elif dx>dy:
   return 1
  else:
   return 0
 
point.sort(compare)
point.insert(0,p0)
ep=point[:]#复制元素，操作ep不会对point产生影响
tag=0
while tag==0:
 tag=1
 l=len(ep)
 for i in range(l):
  i1,i2,i3=(i,(i+1)%l,(i+2)%l)
  x,y,z=(ep[i1],ep[i2],ep[i3])
  a1,a2=((y[0]-x[0],y[1]-x[1]),(z[0]-y[0],z[1]-y[1]))
  if a1[0]*a2[1]-a1[1]*a2[0] < 0:
   tag=0
   ep.pop(i2)
   break
  elif a1[0]*a2[1]-a1[1]*a2[0]==0 and a1[0]*a2[0]<0:
   #==0应改写,360度的情况
   tag=0
   ep.pop(i2)
   break
 
 
def drawpoint(point,color,line):
 p=turtle.getturtle()
 p.hideturtle()
 turtle.delay(1)
 if(line=='p'):
  p.speed(0)
  for i in point:
   p.pu()
   p.color(color)
   p.goto(i)
   p.pd()
   p.dot()
 elif(line=='l'):
  p.pu()
  p.speed(1)
  for i in point:
   p.color(color)
   p.goto(i)
   p.pd()
   p.dot()
  p.goto(point[0])
 
drawpoint(point,'black','p')
drawpoint(ep,'red','l')
time.sleep(1)

import random
import itertools

def generate_num(n):
  random_list = list(itertools.product(range(1, 100), range(1, 100)))
  lists=random.sample(random_list, n)
  return lists

#判断pi是否位于pj,pk,p0组成三角形内，返回t1,t2,t3三个变量
def isin(pi,pj,pk,p0):
 x1 = float(p0[0])
 x2 = float(pj[0])
 x3 = float(pi[0])
 x4 = float(pk[0])
 y1 = float(p0[1])
 y2 = float(pj[1])
 y3 = float(pi[1])
 y4 = float(pk[1])

 k_j0=0
 b_j0=0
 k_k0=0
 b_k0=0
 k_jk=0
 b_jk=0
 perpendicular1=False
 perpendicular2 = False
 perpendicular3 = False
 #pj,p0组成的直线，看pi,pk是否位于直线同一侧

 if x2 - x1 == 0:
 #pj,p0组成直线垂直于X轴时
  t1=(x3-x2)*(x4-x2)
  perpendicular1=True
 else:
  k_j0 = (y2 - y1) / (x2 - x1)
  b_j0 = y1 - k_j0 * x1
  t1 = (k_j0 * x3 - y3 + b_j0) * (k_j0 * x4 - y4 + b_j0)

 #pk,p0组成的直线，看pi,pj是否位于直线同一侧

 if x4 - x1 == 0:
 #pk,p0组成直线垂直于X轴时
  t2=(x3-x1)*(x2-x1)
  perpendicular2=True
 else:
  k_k0 = (y4 - y1) / (x4 - x1)
  b_k0 = y1 - k_k0 * x1
  t2 = (k_k0 * x3 - y3 + b_k0) * (k_k0 * x2 - y2 + b_k0)

 # pj,pk组成的直线，看pi,p0是否位于直线同一侧

 if x4 - x2 == 0:
 # pj,pk组成直线垂直于X轴时
  t3=(x3-x2)*(x1-x2)
  perpendicular3 = True
 else:
  k_jk = (y4 - y2) / (x4 - x2)
  b_jk = y2 - k_jk * x2
  t3 = (k_jk * x3 - y3 + b_jk) * (k_jk * x1 - y1 + b_jk)
 #如果pk，p0,pj，三点位于同一直线时，不能将点删除
 if (k_j0 * x4 - y4 + b_j0)==0 and (k_k0 * x2 - y2 + b_k0)==0 and (k_jk * x1 - y1 + b_jk)==0 :
   t1=-1
 #如果pk，p0,pj，三点位于同一直线时且垂直于X轴，不能将点删除
 if perpendicular1 and perpendicular2 and perpendicular3:
   t1=-1

 return t1,t2,t3

import isintriangle

def force(lis,n):
 #集合S中点个数为3时，集合本身即为凸包集
 if n==3:
  return lis
 else:
  #集合按纵坐标排序，找出y最小的点p0
  lis.sort(key=lambda x: x[1])
  p0=lis[0]
  #除去p0的其余点集合lis_brute
  lis_brute=lis[1:]
  #temp是用来存放集合需要删除的点在lis_brute内的索引，四个点中如果有一个点在其余三个点组成的三角形内部，则该点一定不是凸包上的点
  temp=[]
  #三重循环找到所有这样在三角形内的点
  for i in range(len(lis_brute)-2):
   pi=lis_brute[i]
   #如果索引i已经在temp中，即pi一定不是凸包上的点，跳过这次循环
   if i in temp:
    continue
   for j in range(i+1,len(lis_brute) - 1):
    pj=lis_brute[j]
    #如果索引j已经在temp中，即pj一定不是凸包上的点，跳过这次循环
    if j in temp:
     continue
    for k in range(j + 1, len(lis_brute)):
     pk=lis_brute[k]

     #如果索引k已经在temp中，即pk一定不是凸包上的点，跳过这次循环
     if k in temp:
      continue
     #判断pi是否在pj,pk,p0构成的三角形内
     (it1,it2,it3)=isintriangle.isin(pi,pj,pk,p0)
     if it1>=0 and it2>=0 and it3>=0:
      if i not in temp:
       temp.append(i) 
     # 判断pj是否在pi,pk,p0构成的三角形内
     (jt1,jt2,jt3)=isintriangle.isin(pj,pi,pk,p0)
     if jt1>=0 and jt2>=0 and jt3>=0:

      if j not in temp:
       temp.append(j)

     # 判断pk是否在pj,pi,p0构成的三角形内
     (kt1, kt2, kt3) = isintriangle.isin(pk, pi, pj, p0)
     if kt1 >= 0 and kt2 >= 0 and kt3 >= 0:

      if k not in temp:
       temp.append(k)
  #listlast是最终选出的凸包集合
  lislast=[]
  for coor in lis_brute:
   loc = [i for i, x in enumerate(lis_brute) if x == coor]
   for x in loc:
    ploc = x
   if ploc not in temp:
    lislast.append(coor)
  #将p0加入凸包集合
  lislast.append(p0)
  return lislast