def both_connections(con_A,con_B):
l1 = []
l2 = []
#get the values for linekrs which have connectison
#in an agreeable format
for k in range(len(con_A)):
a = []
b = []
for i in range(len(con_A[k])):
a.append(con_A[k][i][1])
for i in range(len(con_B[k])):
b.append(con_B[k][i][1])
l1.append(a)
l2.append(b)
#count the number that are unique and the same
one=[]
two=[]
two_list=[]
one_list=[]
for k in range(len(l1)):
two.append(len(points.intersection(l1[k],l2[k])))
one.append(len(points.difference(l1[k],l2[k])))
for k in range(len(l1)):
two_list.append((points.intersection(l1[k],l2[k])))
one_list.append((points.difference(l1[k],l2[k])))
return one,two,one_list, two_list
def search(connected,A,B):
add=False
for i in range(len(connected)):
#is it in there already
if connected[i].count(A)==1:
if connected[i].count(B)==0:
connected[i].append(B)
if connected[i].count(B)==1:
if connected[i].count(A)==0:
connected[i].append(A)
#look through connectd if there is an intersection, take the union and
#delete what was there
for i in range(len(connected)):
for j in range(len(connected)):
if intersection(connected[i],connected[j])!=[]:
connected[i]=union(connected[i],connected[j])
connected[j]=connected[i]
#Removed any doubles
connected=unique(connected)
return connected
