Python Sympy Polyhedron.pgroup()用法及代碼示例

Polyhedron.pgroup()：pgroup()是一個sympy Python庫函數，它返回多麵體的排列。

用法： sympy.combinatorics.Polyhedrons.Polyhedron.pgroup()

代碼1：pgroup()示例–四麵體

# Python code explaining 
# SymPy.Polyhedron.pgroup() 
  
# importing SymPy libraries 
from sympy.combinatorics import Permutation, Cycle 
from sympy.combinatorics.polyhedron import tetrahedron, octahedron 
  
# Using from  
# sympy.combinatorics.polyhedron.Polyhedron.pgroup() 
  
# Creating Polyhedron 
a = tetrahedron.copy() 
  
print ("Polyhedron - pgroup form : ", a.pgroup) 
  
a.rotate(0) 
print ("\nPolyhedron - pgroup form : ", a.pgroup)

輸出：

Polyhedron – pgroup form : PermutationGroup([
(1 2 3),
(3)(0 1 2),
(0 3 2),
(0 3 1),
(0 1)(2 3),
(0 2)(1 3),
(0 3)(1 2)])

Polyhedron – pgroup form : PermutationGroup([
(1 2 3),
(3)(0 1 2),
(0 3 2),
(0 3 1),
(0 1)(2 3),
(0 2)(1 3),
(0 3)(1 2)])

代碼2：pgroup()示例–八麵體

# Python code explaining 
# SymPy.Polyhedron.pgroup() 
  
# importing SymPy libraries 
from sympy.combinatorics import Permutation, Cycle 
from sympy.combinatorics.polyhedron import tetrahedron, octahedron 
  
# Using from  
# sympy.combinatorics.polyhedron.Polyhedron.pgroup() 
  
# Creating Polyhedron 
a = octahedron.copy() 
  
print ("Polyhedron - pgroup form : ", a.pgroup) 
  
a.rotate(0) 
print ("\nPolyhedron - pgroup form : ", a.pgroup)

輸出：

Polyhedron – pgroup form : PermutationGroup([
(5)(1 2 3 4),
(0 4 5 2),
(0 1 5 3),
(0 1)(2 4)(3 5),
(0 2)(1 3)(4 5),
(0 3)(1 5)(2 4),
(0 4)(1 3)(2 5),
(0 5)(1 4)(2 3),
(0 5)(1 2)(3 4),
(0 4 1)(2 3 5),
(0 1 2)(3 4 5),
(0 2 3)(1 5 4),
(0 4 3)(1 5 2)])

Polyhedron – pgroup form : PermutationGroup([
(5)(1 2 3 4),
(0 4 5 2),
(0 1 5 3),
(0 1)(2 4)(3 5),
(0 2)(1 3)(4 5),
(0 3)(1 5)(2 4),
(0 4)(1 3)(2 5),
(0 5)(1 4)(2 3),
(0 5)(1 2)(3 4),
(0 4 1)(2 3 5),
(0 1 2)(3 4 5),
(0 2 3)(1 5 4),
(0 4 3)(1 5 2)])