Page 42 - Computer science 868 Class 12
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Example 2: Minimise a Boolean expression F(A, B, C) = π(0, 1, 4, 5)
              Solution: First, draw a POS K-map with A at the left and B, C at the top.

              Then plot 0 in the grids mentioned and 1 in the remaining boxes.

                                                        B+C     B+C'   B'+C'    B'+C
                                                  A    0 0     0 1      1 3     1 2
                                                 A'    0 4     0 5      1 7     1 6

              A quad is formed containing 4 0’s placed together horizontally or vertically.
                                                    A quad with squares M , M , M , M
                                                                          1
                                                                      0
                                                                                5
                                                                             4
                                                    is formed.
                                                    M  = A+B+C
                                                     0
                                                    M  = A+B+C'
                                                     1
                                                    M  = A'+B+C
                                                     4
                                                    M  = A'+B+C'
                                                     5
              We find that literals A and C are complemented as well as uncomplemented. So A and C are cancelled and the minimised
              POS expression is B.

              Example 3: Minimise a Boolean expression F(A, B, C) = Σ(0, 1, 2, 3, 4, 5, 6, 7)
              Solution: First draw the SOP K-map properly mentioning the literals and cardinal   An octet with the blocks m , m , m ,
                                                                                                              0
              numbers and place 1 in the required grids.                                m , m , m , m , m  is formed.  1  2
                                                                                                    6
                                                                                                5
                                                                                          3
                                                                                             4
                                                                                                       7
                                                                                        m  = A'.B'.C'
              We find that all the grids are filled with 1 in the K-map.                  0
                                                                                        m  = A'.B'.C
                                                                                          1
                                                       B'.C'    B'.C    B.C     B.C'    m  = A'.B.C'
                                                                                          2
                                                 A'    1 0     1 1     1 3     1 2      m  = A'.B.C
                                                                                          3
                                                 A     1 4     1 5     1 7     1 6      m  = A. B'.C'
                                                                                          4
                                                                                        m  = A.B'.C
              Example 4: Minimise a Boolean expression F(A, B, C, D) = Σ(0, 1, 2, 3, 13, 15)  5
                                                                                        m  = A.B.C'
                                                                                          6
              Solution: First, draw the SOP K-map. Fill all the mentioned squares with 1 and   m  = A.B.C
                                                                                          7
              the rest with 0.                                                          All  literals  in  this  expression  have
                                                       C'.D'   C'.D     C.D     C.D'    changed their signs.  Hence,  the
                                                A'.B'  1 0     1 1     1 3     1 2      reduced SOP expression is 1.
                                                A'.B   0 4     0 5     0 7     0 6
                                                A.B     0    12   1    13   1    15   0    14
                                                A.B'   0 8     0 9      0    11   0    10
              The following grouping can be made:
                                  Quad 1 m , m , m , m 3             Pair 1 m , m 15
                                                 2
                                              1
                                           0
                                                                           13
                                  m  = A'.B'.C'.D'                   m  = A.B.C'.D
                                                                      13
                                    0
                                  m  = A'.B'.C'.D                    m  = A.B.C.D
                                    1
                                                                      15
                                  m  = A'.B'.C.D                     Literal C in  the expression  changes
                                    2
                                  m  = A'.B'.C.D'                    sign while A, B and D remain same.
                                    3
                                  Literals C and D in the expression are   So, the reduced form = A.B.D
                                  both  in  complemented and  normal
                                  form while A' and B' remain same.
                                  So, the reduced form = A'.B'
              Hence, the reduced expression is A'.B' + A.B.D
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