In this article, we will learn about the solution to the problem statement given below.
Problem statement − We are given pos form we need to convert it into its equivalent sop form
The conversion can be done by first counting the number of alphabets in the pos form and then calculating all the max and the minterms.
Now let’s observe the concept in the implementation below−
# 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) print("Standard SOP form of " + POS_expr + " ==> " + SOP_expr) # Driver code if __name__=="__main__": main()
Standard SOP form of (A'+B'+C).(A+B+C').(A+B'+C).(A'+B+C) ==> A'B'C'+A'BC+AB'C+ABC
All the variables are declared in the local scope and their references are seen in the figure above.
In this article, we have learned about how we can convert pos to sop form