Page 76 - DCAP108_DIGITAL_CIRCUITS_AND_LOGIC_DESIGNS

Unit 4: Minimization of Boolean Algebra

This quad is looped because it contains two squares that have not been looped Notes
previously. Looping this quad produces BC.

Step 6. All 1s have already been looped.
Step 7. The terms generated by the three loops are ORed together to obtain the expression
for X.

Consider the K-map in Figure 4.21 (c).

Step 2. There are no isolated 1s.
Step 3. The 1 in square 2 is adjacent only to the 1 in square 6. This pair looped to produce
ACD. Similarly, square 9 is adjacent only to square 10.

Looping this pair produces ABC. Likewise, loop 7, 8 and loop 11, 15 product the terms
ABC and ACD, respectively.

Step 4. There are no octets.
Step 5. There is one quad formed by squares 6, 7, 10, and 11. This quad, however, is not
looped, because all the Is in the quad have been included other loops.

Step 6. All 1s have already been looped.
Step 7. The expression for X is shown in the Figure 4.21.

Consider the K map in Figure 4.22(a).
Step 2. There are no isolated 1s.
Step 3. There are no 1s that are adjacent to only one other 1.

Step 4. There are no octets.
Step 5. There are no quads.
Steps 6 and 7. There are many possible pairs. The looping must use the minimum number
of loops to account for all the 1s. For this map there are two possible loopings, which require
only four looped pairs. Figure 4.22 shows one solution and its resultant expression. Figure
4.22(b) shows the other. Note that both expressions are of the same complexity, and so
neither is better than the other.
Figure 4.21: The Same K-Map with Two Equally Solutions

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