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192 CHAPTER 4/LOGIC FUNCTION REPRESENTATION AND MINIMIZATION

4.12 The two four-variable functions shown are presented in canonical POS form. Follow
the discussion in Section 4.5 and find the optimized SOP and POS minima for the two
functions taken as a system. Use the gate/input tally, including inverters, to determine
which is simpler, the SOP result or the POS result. Implement the simpler of the two
forms in either NAND/INV or NOR/INV logic. Assume that the inputs and outputs
are all active high.
gi(A, B, C, D} = Y\M(0, 3, 4,11, 12, 13, 15) • 0(2, 5, 6)

g 2(A, B, C,D) = Y\ M(0, 1, 9, 12, 13) • 0(2, 3, 4, 10)

4.13 Given below is a set of three functions, each of four variables. Follow the discussion
in Section 4.5 and find the optimized SOP and POS minima for the three functions
taken as a system. Use the gate/input tally, excluding inverters, to determine which is
simpler, the SOP result or the POS result. [Hint: In determining the shared Pis, don't
forget to include the ANDed and ORed functions (y\ • y^ • ys) and (yi + y 2 + ys).]

yi(a, b, c,d) = ^m(0, 1,2,5,1, 8, 10, 14, 15)

y 2(a, b, c, d) = £/n(0, 2, 4, 5, 6,1, 10, 12)
2 3 4 6 8 9 10
y 3(a, b, c,d) = J2™(°' !> > ' > > > > > !!)
4.14 Extract minimum SOP and POS expressions (cover) from the K-maps shown in
Fig. P4.2. Where appropriate, application of the loop-out protocol discussed in Section
4.4 will help to avoid redundancy.
4.15 Following the discussion in Section 4.6, compress each function in Problem 4.2 into
a second-order K-map (Map Key = 2) and extract minimum SOP and POS cover. Use
the LSB variable as the entered variable (EV).
4.16 Following the discussion in Section 4.6, compress each function in Problem 4.3 into
a third-order K-map (Map Key = 2) and extract minimum SOP and POS cover. Use
the LSB variable as the entered variable (EV).

4.17 Following the discussion in Section 4.6, compress each function in Problem 4.7 into
a second-order K-map (Map Key = 2) and extract minimum SOP and POS cover. Use
the LSB variable as the entered variable (EV).

AB\ oo 01 11 10
\ B 1 \BC 01 11 10 00 E 0 1 0
A\ 0 A\ °°
A
I
0 1 0 1 X 0 D 1 D 1 01 E 0 E E
1 B 1 X 0 1 D 1 0 D 11
/_ 1 0 0 1
h h h
' 1 ' 2 ' 3
10 1 E 0 0
(a) (b) (c) (d)

FIGURE P.4.2