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法律研究Python程序——将POS转换为SOP
在本文中，我们将会了解如何解决以下给出的问题陈述。
问题陈述 − 给定POS形式，我们需要将其转换成等效的SOP形式
转换可以通过首先计算POS形式中字母的数量，然后计算所有最大项和最小项来完成。
现在让我们观察以下实现中的概念−
范例
# Python code to convert standard POS form
# to standard SOP form
# to calculate number of variables
def count_no_alphabets(POS):
i = 0
no_var = 0
# total no. of alphabets before will be equal to alphabets before first '.' character
while (POS[i]!='.'):
# character is a alphabet or not
if (POS[i].isalpha()):
no_var+= 1
i+= 1
return no_var
# maximum terms in an integer
def Cal_Max_terms(Max_terms, POS):
a = ""
i = 0
while (i<len(POS)):
if (POS[i]=='.'):
# binary to decimal conversion
b = int(a, 2)
# append each min term(integer type) into the list
Max_terms.append(b)
# assign empty strings
a =""
i+= 1
elif(POS[i].isalpha()):
# checking whether variable is having complement as superscript
if(i + 1 != len(POS) and POS[i + 1]=="'"):
# concatenating the string with '1'
a += '1'
# incrementing by 2 because 1 for alphabet and another for a symbol "'"
i += 2
else:
# concatenating the string with '0'
a += '0'
i += 1
else:
i+= 1
# append last min term(integer type) into the list
Max_terms.append(int(a, 2))
# conversion of minterms in binary and finally converting it to SOP
def Cal_Min_terms(Max_terms, no_var, start_alphabet):
# declaration of the list
Min_terms =[]
# calculation of total no. of terms formed by all variables max = 2**no_var
for i in range(0, max):
# is current term present in max_terms or not
if (Max_terms.count(i)== 0):
# converting integer to binary
b = bin(i)[2:]
# loop used for inserting 0's before the
# binary value so that its length will be
# equal to no. of variables present in
# each product term
while(len(b)!= no_var):
b ='0'+b
# appending the max terms(integer) in the list
Min_terms.append(b)
SOP = ""
# iterated untill minterms are available
for i in Min_terms:
# fetching the variable
value = start_alphabet
# iterate till there are 0's and 1's
for j in i:
# check whether the varailble is complemented or not
if (j =='0'):
# concatenating vaue and complement operator
SOP = SOP + value+ "'"
# check the non complement variable
else:
# concatenating value
SOP = SOP + value
# increment the alphabet by the next adjacent alaphabet
value = chr(ord(value)+1)
# concatenating the "+" operator
SOP = SOP+ "+"
# for discarding the extra '+'
SOP = SOP[:-1]
return SOP
# main function
def main():
# input
POS_expr ="(A'+B'+C).(A+B+C').(A+B'+C).(A'+B+C)"
Max_terms = []
no_var = count_no_alphabets(POS_expr)
Cal_Max_terms(Max_terms, POS_expr)
SOP_expr = Cal_Min_terms(Max_terms, no_var, POS_expr[1])
print("Standard SOP form of " + POS_expr + " ==> " + SOP_expr)
# Driver code
if __name__=="__main__":
main()输出
Standard SOP form of (A'+B'+C).(A+B+C').(A+B'+C).(A'+B+C) ==> A'B'C'+A'BC+AB'C+ABC
所有变量都声明在局部范围内，其引用关系如图所示。
结论
在本文中，我们已经了解了如何将POS转换为SOP形式
更新于： 2019-12-24
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