Two types of designations of the 4-variable K-map are shown in Figs. 2.20(a) and (b), respectively.
Solution: The desired four-variable K-map is drawn as shown in Fig. 2.21.
From the figure, we find that there are 5 groups in this example. Of these, Group 1 is a quad and Group 2 to 5 are pairs. We find that the members of Group 5 have already been selected by Groups 1 and 2; therefore it (Group 5) is a redundant group and can be discarded from the final expression. The desired answer is:
S = Octet 1 + Octet 2+ Quad
It can be seen that the four-variable K-map may be regarded as a combination of a three-variable horizontal K-map and a three-variable vertical K-map. The cells are designated as shown in the figures. In the Format-1 scheme shown in Fig. 2.20(a), which we follow in this text, the left topmost cell is designated as 0 and this represents cell 0000. The right cell adjacent to this is designated as 4 and this represents cell 0100. Similarly, the bottom cell adjacent to cell 0 is designated as 1 and this represents cell 0001. The other designations follow this pattern.
In the Format-2 scheme, cell 0 is the same as in our system. But cell 1 in the new scheme is our cell 4. Similarly, cell 4 in the new scheme is cell 1 in our scheme. The Type-2 designations are illustrated in Fig. 2.20(b).
EXAMPLE 3: Simplify S = S (0, 1, 2, 3, 7, 13, 14, 15)
From the figure, we find that there are 5 groups in this example. Of these, Group 1 is a quad and Group 2 to 5 are pairs. We find that the members of Group 5 have already been selected by Groups 1 and 2; therefore it (Group 5) is a redundant group and can be discarded from the final expression. The desired answer is:
S = Group 1 + Group 2 + Group 3 + Group 4
= a'b' + bcd + abc + abd (2.24)
Example 4: S = S(m0, m1, m4, m5, m6, m7, m8, m9, m10, m11, m12, m13, m14, m15)
Solution: The K-map for Example 4 is shown in Fig. 2.22. We find that we can combine eight members horizontally, and eight members vertically. A grouping of eight members (called as an octet) eliminates three variables. Thus the vertical group Octet 1 can be seen to eliminate the variables a, c, and d, as they appear in the complemented and uncomplemented forms. This leaves b as one of the variable left uneliminated in the final output. Similarly, the horizontal group Octet 2 can be seen to eliminate the variables a, b, and d, as they also appear in the complemented and uncomplemented forms. This grouping leaves c′ as the uneliminated variable. Finally, we notice that a quad group also exists in this case. The quad produces the product term ab′. The reduced sum S can now be obtained as the sum of the two uneliminated varables b and c′ and the term ab′. Thus we obtain:
S = Octet 1 + Octet 2+ Quad
= b + c′+ ab′ (2.25)
