Python SymPy Permutation.cyclic_form()方法

Permutation.cyclic_form() : cyclic_form()是一个Sympy Python库函数，通过省略单子，从经典符号中返回循环符号。

语法 : sympy.combinatorics.permutations.Permutation.cyclic_form()

返回 :
循环记号来自于典型记号

代码 #1 : cyclic_form() 示例

# Python code explaining
# SymPy.Permutation.cyclic_form()
  
# importing SymPy libraries
from sympy.combinatorics.partitions import Partition
from sympy.combinatorics.permutations import Permutation
  
# Using from sympy.combinatorics.permutations.Permutation.cyclic_form() method 
  
# creating Permutation
a = Permutation([2, 0, 3, 1, 5, 4])
  
b = Permutation([3, 1, 2, 5, 4, 0])
  
  
print ("Permutation a - cyclic_form form : ", a.cyclic_form)
print ("Permutation b - cyclic_form form : ", b.cyclic_form)

输出 :

Permutation a – cyclic_form form : [[0, 2, 3, 1], [4, 5]]
Permutation b – cyclic_form form : [[0, 3, 5]]

代码 #2 : cyclic_form() 示例 – 二维互换

# Python code explaining
# SymPy.Permutation.cyclic_form()
  
# importing SymPy libraries
from sympy.combinatorics.partitions import Partition
from sympy.combinatorics.permutations import Permutation
  
# Using from 
# sympy.combinatorics.permutations.Permutation.cyclic_form() method 
  
# creating Permutation
a = Permutation([[2, 4, 0], 
                 [3, 1, 2],
                 [1, 5, 6]])
  
# SELF COMMUTATING    
print ("Permutation a - cyclic_form form : ", a.cyclic_form)

输出 :

Permutation a – cyclic_form form : [[0, 3, 5, 6, 1, 2, 4]]