这个类接受一个有限域多项式字符串,解析它,操作 (+-*/%),然后以与输入相同的格式输出。它工作正常(到目前为止)。但是,现在我试图在算术运算符上实现特殊方法,但我无法超越简单地连接字符串的地步。通常,想法是将输入初始化为类实例,但在这种情况下,输入上有一个正则表达式,这似乎使任何尝试这样做都变得复杂。我正在自学 Python,所以这对我来说是一部恐怖电影,但对于任何经验丰富的 Python 程序员来说可能只是一个玩具。
这些似乎有很多信息,但我不确定它们在这种情况下有多大帮助:
http://stackoverflow.com/questions/10842166/programmatically-create-arithmetic-special-methods-in-python-aka-factory-funct
http://rosettacode.org/wiki/S-Expressions
http://www.greenteapress.com/thinkpython/thinkCSpy/html/chap14.html
http://docs.cython.org/src/userguide/special_methods.html
这是我正在使用的课程和示例:
import re
class gf2pim:
def id(self,lst):
"""returns modulus 2 (1,0,0,1,1,....) for input lists"""
return [int(lst[i])%2 for i in range(len(lst))]
def listToInt(self,lst):
"""converts list to integer for later use"""
result = self.id(lst)
return int(''.join(map(str,result)))
def parsePolyToListInput(self,poly):
"""performs regex on raw string and converts to list"""
c = [int(i.group(0)) for i in re.finditer(r'\d+', poly)]
return [1 if x in c else 0 for x in xrange(max(c), -1, -1)]
def prepBinary(self,x,y):
"""converts to base 2; bina,binb are binary values like 110100101100....."""
x = self.parsePolyToListInput(x); y = self.parsePolyToListInput(y)
a = self.listToInt(x); b = self.listToInt(y)
bina = int(str(a),2); binb = int(str(b),2)
return bina,binb #
def add(self,a,b):
"""a,b are GF(2) polynomials like x**7 + x**3 + x**0 ...; returns binary string"""
bina,binb = self.prepBinary(a,b)
return self.outFormat(bina^binb)
def subtract(self,x,y):
"""same as addition in GF(2)"""
return self.add(x,y)
def quotient(self,a,b):
"""a,b are GF(2) polynomials like x**7 + x**3 + x**0 ...; returns quotient formatted as polynomial"""
a,b = self.prepBinary(a,b)
return self.outFormat(a/b)
def remainder(self,a,b):
"""a,b are GF(2) polynomials like x**7 + x**3 + x**0 ...; returns remainder formatted as polynomial"""
a,b = self.prepBinary(a,b)
return self.outFormat(a%b)
def outFormat(self,raw):
"""process resulting values into polynomial format"""
raw = "{0:b}".format(raw); raw = str(raw[::-1]); g = [] #reverse binary string for enumeration
g = [i for i,c in enumerate(raw) if c == '1']
processed = "x**"+" + x**".join(map(str, g[::-1]))
if len(g) == 0: return 0 #return 0 if list empty
return processed #returns result in gf(2) polynomial form
def __add__(self,other):
return gf2pim.add(self,other)
底部的最后一个示例显示了问题:
obj = gf2pim()
a = "x**14 + x**1 + x**0"; b = "x**6 + x**2 + x**1"
c = "x**2 + x**1 + x**0"; d = "x**3 + x**1 + x**0"
e = "x**3 + x**2 + x**1 + x**0"; f = "x**2"; g = "x**1 + x**0"; h = "x**3 + x**2 + x**0"
p = "x**13 + x**1 + x**0"; q = "x**12 + x**1"; j = "x**4 + x**3 + x**1 + x**0"
print "add: [%s] + [%s] = %s "%(a,b,obj.add(a,b))
print "add: [%s] + [%s] = %s "%(c,d,obj.add(c,d))
print "quotient (max(a,b)/min(a,b): [%s] / [%s] = %s "%(a,b,obj.quotient(a,b))
print "remainder (max(a,b) mod min(a,b)): [%s] mod [%s] = %s "%(a,b,obj.remainder(a,b))
print "quotient (max(a,b)/min(a,b): [%s] / [%s] = %s "%(c,d,obj.quotient(c,d))
print "remainder (max(a,b) mod min(a,b): [%s] mod [%s] = %s "%(c,d,obj.remainder(c,d))
print "quotient (max(a,b)/min(a,b): [%s] / [%s] = %s "%(q,q,obj.quotient(q,q))
print "remainder (max(a,b) mod min(a,b): [%s] mod [%s] = %s "%(q,q,obj.remainder(q,q))
print "modular_inverse: [%s] * [%s] mod [%s] = 1 [%s]"%(p,valuemi2[0],q,valuemi2[1])
sum1 = obj.add(a,b); quotient1 = obj.quotient(sum1,c)
### HERE THE PROBLEM IS CLEAR
print "[(a+b)/c] = ",quotient1
smadd1 = a+b
print "smadd1 ",smadd1
和输出:
>>>
add: [x**14 + x**1 + x**0] + [x**6 + x**2 + x**1] = x**14 + x**6 + x**2 + x**0
add: [x**2 + x**1 + x**0] + [x**3 + x**1 + x**0] = x**3 + x**2
quotient (max(a,b)/min(a,b): [x**14 + x**1 + x**0] / [x**6 + x**2 + x**1] = x**7 + x**6 + x**5 + x**3 + x**1
remainder (max(a,b) mod min(a,b)): [x**14 + x**1 + x**0] mod [x**6 + x**2 + x**1] = x**2 + x**1 + x**0
quotient (max(a,b)/min(a,b): [x**2 + x**1 + x**0] / [x**3 + x**1 + x**0] = 0
remainder (max(a,b) mod min(a,b): [x**2 + x**1 + x**0] mod [x**3 + x**1 + x**0] = x**2 + x**1 + x**0
quotient (max(a,b)/min(a,b): [x**12 + x**1] / [x**12 + x**1] = x**0
remainder (max(a,b) mod min(a,b): [x**12 + x**1] mod [x**12 + x**1] = 0
[(a+b)/c]*d = x**14 + x**12 + x**9 + x**1
smadd1 x**14 + x**1 + x**0x**6 + x**2 + x**1
>>>
因此,正如您所看到的,smadd1
我需要使用 + 添加这两个,而不仅仅是连接。另外,我想知道在这种情况下是否需要使用 S 表达式树。
编辑:
Multiply() 正在工作但现在不可用:
def __mul__(self,other):
"""
__multiply__ is the special method for overriding the - operator
returns product of 2 polynomials in gf2; self,other are values 10110011...
"""
self = int(str(self),2)
bitsa = reversed("{0:b}".format(self))
g = [(other<<i)*int(bit) for i,bit in enumerate(bitsa)]
return gf2infix(self.outFormat(reduce(lambda x,y: x^y,g)))
它的原始形式是:
def multiply(self,a,b):
"""a,b are GF(2) polynomials like x**7 + x**3 + x**0... ; returns product of 2 polynomials in gf2"""
a = self.prepBinary(a); b = self.prepBinary(b)
bitsa = reversed("{0:b}".format(a))
g = [(b<<i)*int(bit) for i,bit in enumerate(bitsa)]
return self.outFormat(reduce(lambda x,y: x^y,g))
忽略这个问题multiply()
,我修复了它。更改的行是:
bitsa = reversed("{0:b}".format(self.bin))
并且在那之前的那条线被取出了。