이 기사에서는 아래 주어진 문제 설명에 대한 솔루션에 대해 알아볼 것입니다.
문제 설명 − 동등한 sop 형식으로 변환하는 데 필요한 pos 형식이 제공됩니다.
변환은 먼저 pos 형식의 알파벳 수를 세고 모든 최대값과 최소값을 계산하여 수행할 수 있습니다.
이제 아래 구현에서 개념을 살펴보겠습니다-
예시
# 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()
출력
Standard SOP form of (A'+B'+C).(A+B+C').(A+B'+C).(A'+B+C) ==> A'B'C'+A'BC+AB'C+ABC
모든 변수는 로컬 범위에서 선언되며 해당 참조는 위 그림과 같습니다.
결론
이 기사에서는 pos를 sop 형식으로 변환하는 방법에 대해 배웠습니다.