Two types of designations of the 4-variable K-map are shown in Figs. 2.20(a) and (b), respectively.

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)

**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*= Group 1 + Group 2 + Group 3 + Group 4

=

*a'b'*+ bcd +*abc*+ abd (2.24)**Example 4:**

*S*= S(

*m*

_{0},

*m*

_{1},

*m*

_{4},

*m*,

_{5}*m*

_{6},

*m*

_{7},

*m*

_{8},

*m*

_{9},

*m*

_{10},

*m*

_{11},

*m*

_{12},

*m*

_{13},

*m*

_{14},

*m*

_{15})

**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)*
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