Python代码：将标准SOP形式转换为标准POS形式

编写一个Python程序，将标准SOP（积之和，Sum of Products）形式转换为标准POS（和之积，Product of Sums）形式。

假设：

输入的SOP表达式已经是标准形式。表达式中的变量是连续的，即如果表达式包含变量A，那么它将依次拥有变量B、C等，并且每一个乘积项中的字母都按字母顺序排列（例如ABC，而不是BAC）。

示例：

输入 : ABC‘+A‘BC+ABC+AB‘C 
输出 : (A+B+C).(A+B+C‘).(A+B‘+C).(A‘+B+C)

输入 : A‘B+AB‘
输出 : (A+B).(A‘+B‘)

解决思路：

• 首先，将每个乘积项转换为等效的二进制形式（例如：如果是ABC‘，则非补变量（A, B）取1，补变量（C）取0，因此二进制转换为110），最后将其转换为十进制形式（例如：110 = 6），并存储在一个列表中。
• 现在，为了得到POS形式，我们要选取所有不在第一步列表中出现的项，然后将每个项转换为二进制，进而转换为SOP形式。例如：假设5不在列表中，那么

5 ==> 101 (二进制)

现在，将1替换为补变量（A, C）

将0替换为非补变量（B）

并在变量之间使用 ‘+‘

101 ==> A‘+B+C‘

在每个单独的和项之后使用 ‘.‘

为了更清晰，在每个单独的项之间使用括号。

例如： (A‘+B+C‘).(A+B+C‘)

Python 代码

# Python code to convert standard SOP form 
# to standard POS form 

# function to calculate no. of variables 
# used in SOP expression 
def count_no_alphabets(SOP): 
    i = 0
    no_var = 0

    # As expression is standard so total no. 
    # of alphabets will be equal 
    # to alphabets before first ‘+‘ character 
    while (SOP[i]!=‘+‘): 

        # checking if character is alphabet             
        if (SOP[i].isalpha()):     
            no_var+= 1
        i+= 1
    return no_var 

# function to calculate the min terms in integers 
def Cal_Min_terms(Min_terms, SOP): 
    a ="" 
    i = 0
    while (i<len(SOP)): 
        if (SOP[i]=='+'): 

            # converting binary to decimal                 
            b = int(a, 2) 

            # insertion of each min term(integer) into the list                 
            Min_terms.append(b) 

            # empty the string         
            a =""                         
            i+= 1
        else: 

            # checking whether variable is complemented or not 
            if(i + 1 != len(SOP) and SOP[i + 1]=="'"): 

                # concatenating the string with '0' 
                a+='0' 

                # incrementing by 2 because 1 for alphabet and 
                # another for "'"                        
                i+= 2                            
            else: 

                # concatenating the string with '1' 
                a+='1'                        
                i+= 1

    # insertion of last min term(integer) into the list     
    Min_terms.append(int(a, 2))             

# function to calculate the max terms in binary then 
# calculate POS form of SOP 
def Cal_Max_terms(Min_terms, no_var, start_alphabet): 

    # declaration of the list 
    Max_terms =[] 

    # calculation of total no. of terms that can be 
    # formed by no_var variables                     
    max = 2**no_var                 
    for i in range(0, max): 

        # checking whether the term is not 
        # present in the min terms 
        if (Min_terms.count(i)== 0): 

            # converting integer to binary and then 
            # taking the value from 2nd index as 1st 
            # two index contains '0b' 
            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 
            Max_terms.append(b)     

    POS ="" 

    # loop till there are max terms                         
    for i in Max_terms: 

        # before every sum term append POS by '('         
        POS = POS+"("

        # acquire the starting variable came from 
        # main function in every sum term             
        value = start_alphabet 

        # loop till there are 0's or 1's in each max term     
        for j in i: 

            # checking for complement variable to be used                 
            if (j =='1'): 

                # concatenating value, ' and + in string POS                 
                POS = POS + value+"'+"

            # checking for uncomplement variable to be used     
            else: 

                # concatenating value and + in string POS                     
                POS = POS + value+"+"

            # increment the alphabet by 1     
            value = chr(ord(value)+1) 

        # for discarding the extra '+' in the last     
        POS = POS[:-1] 

        # appending the POS string by ')' 
        POS = POS+")" 

    return POS 

# Driver Code 
SOP ="ABC'+A'BC+ABC+AB'C" 
Min_terms =[] 

# to count no. of variables in the SOP expression 
no_var = count_no_alphabets(SOP) 

# function to calculate min terms(integer) from the SOP expression 
Cal_Min_terms(Min_terms, SOP) 

# we are taking A as a start_alphabet, we can ignore 
# this by passing this value through function 
start_alphabet = 'A' 

# conversion of SOP form to POS form 
POS = Cal_Max_terms(Min_terms, no_var, start_alphabet) 

print(POS) 

输出：

(A+B+C).(A+B+C‘).(A+B‘+C).(A‘+B+C)