**Example**

**5:**Simplify

*S*= S

*m*(0, 1, 2, 4, 5, 7, 9, 10) +

*d*(3, 8, 15), where

**‘**

*d*’

**denotes don’t care terms.**

**Solution:**

**The K-map is drawn as shown in Fig. 2.23. Here, we represent don’t-care terms by Greek letter**

**f**, since

**f**may be thought of as a combination of

**0**and

**1**. We now look for the cells in which the

**f**s are entered, and see whether they can be used to eliminate any variable or not.

It can be observed that the don’t-care term in cell 3 can be grouped with entries in cells 1, 5 and 7. Hence, in this case, we treat the cell-3 entry

**f**as a valid**1**, and it forms a quad with the entries in cells 1, 5 and 7. Similarly, the entry**f**in cell 8 can be grouped with the**1**in cell 9 to form a pair. However,**f**in cell 15 cannot be combined with any**1**-entry, and hence it is discarded. The final solution now becomes:

*S*=

*a′d*+

*b′cd′*+

*ab′c′*

_{+}

*a′bc′*(2.26)

It is to be noted that

**f**in cell 8 can be grouped with the**1**in cell 10 also. However, the**1**in cell 10 is already paired with the**1**in cell 2. Hence, we discard the pairing with**f****in this case.**
not included 0 at min term 0 position.

ReplyDeleteso its solution is wrong