quick way to get all transitions of a neighborhood
Posted: September 12th, 2026, 11:43 am
o is the number of orthogonal cells, d is the number of diagonal cells, q is the number of oblique cells
Code: Select all
def D4(o,d,q):
if(o == 0):
if(d == 0):
if(q == 0):
return(1)
return(2*D4(0,0,q-1))
return(2*D4(0,d-1,q))
return(2*D4(o-1,d,q))
def C4(o,d,q):
if(o == 0):
if(d == 0):
if(q == 0):
return(0)
return(4*C4(0,0,q-1)+D4(0,0,q-1))
return(2*C4(0,d-1,q))
return(2*C4(o-1,d,q))
def D20(o,d,q):
if(o == 0):
if(d == 0):
if(q == 0):
return(0)
return(4*D20(0,0,q-1)+D4(0,0,q-1))
return(2*D20(0,d-1,q))
return(4*D20(o-1,d,q)+D4(o-1,d,q))
def D21(o,d,q):
return(D20(d,o,q))
def C2(o,d,q):
if(o == 0):
if(d == 0):
if(q == 0):
return(0)
return(16*C2(0,0,q-1)+6*D20(0,0,q-1)+6*D21(0,0,q-1)+4*C4(0,0,q-1)+2*D4(0,0,q-1))
return(4*C2(0,d-1,q)+D20(0,d-1,q)+C4(0,d-1,q))
return(4*C2(o-1,d,q)+D21(o-1,d,q)+C4(o-1,d,q))
def D10(o,d,q):
if(o == 0):
if(d == 0):
if(q == 0):
return(0)
return(16*D10(0,0,q-1)+12*D20(0,0,q-1)+6*D4(0,0,q-1))
return(4*D10(0,d-1,q)+2*D20(0,d-1,q)+D4(0,d-1,q))
return(8*D10(o-1,d,q)+4*D20(o-1,d,q)+2*D4(o-1,d,q))
def D11(o,d,q):
return(D10(d,o,q))
def C1(o,d,q):
if(o == 0):
if(d == 0):
if(q == 0):
return(0)
return(256*C1(0,0,q-1)+120*D10(0,0,q-1)+120*D11(0,0,q-1)+120*C2(0,0,q-1)+54*D20(0,0,q-1)+54*D21(0,0,q-1)+60*C4(0,0,q-1)+24*D4(0,0,q-1))
return(C1(d,0,q))
return(16*C1(o-1,d,q)+6*D11(o-1,d,q)+4*D10(o-1,d,q)+6*C2(o-1,d,q)+2*D21(o-1,d,q)+D20(o-1,d,q)+3*C4(o-1,d,q))
def N(o,d,q):
return(D4(o,d,q)+C4(o,d,q)+D20(o,d,q)+D21(o,d,q)+C2(o,d,q)+D10(o,d,q)+D11(o,d,q)+C1(o,d,q))
print("o:")
O = int(input())
print("d:")
D = int(input())
print("q:")
Q = int(input())
print(N(O,D,Q))