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8/14/2019 Chapter 4 Problem Solutions 4.1. w 0 0 0 0
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- 4.1 -
Chapter 4
Problem Solutions
4.1.
w x y z f (a) (b) (c) (d) (e) (f) (g) (h)
0 0 0 0 0 0 0 0 0 0 1 1 0
0 0 0 1 0 0 1 0 0 0 1 1 0
0 0 1 0 0 0 1 0 0 0 1 1 0
0 0 1 1 0 0 1 0 0 0 0 1 0
0 1 0 0 0 0 0 1 0 0 1 1 0
0 1 0 1 1 0 1 1 0 0 1 1 1
0 1 1 0 0 0 1 1 0 1 1 1 0
0 1 1 1 0 0 1 1 0 0 0 1 0
1 0 0 0 1 1 0 1 0 0 1 1 0
1 0 0 1 1 0 1 1 1 0 1 1 0
1 0 1 0 1 1 1 1 0 0 1 1 0
1 0 1 1 1 0 1 1 1 0 1 1 0
1 1 0 0 1 1 0 1 0 0 1 1 0
1 1 0 1 0 0 1 1 0 0 1 0 0
1 1 1 0 1 1 1 1 0 1 1 1 0
1 1 1 1 0 0 1 1 0 0 1 0 0
(a) wz-is an implicant since it implies f. However, neither
the term w nor the term z-are implicants. Therefore, wz
-
is a prime implicant (1).
(b) y+z is neither an implicant or implicate (6).
(c) w+x is an implicate since it is implied by f. However,
nether the term w nor the term x are implicates.
Therefore, w+x is a prime implicate (3).
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4.1. (continued)
(d) wx-z is an implicant since it implies f. However, wx
-z
subsumes wx-which is also an implicant. Therefore, wx
-z
is an implicant, but not prime (2).
(e) xyz-is neither an implicant or implicate (6).
(f) w-+y-+z-is an implicate since it is implied by f.
However, w-+y-+z-subsumes w+y
-which is also an implicate.
Therefore, w-+y-+z-
is an implicate, but not prime (4).
(g) w-+x-+z-is an implicate since it is implied by f.
Furthermore, none of the sum terms w-+x-, w-+z-, and x-+z- are
implicates. Therefore, w-+x-+z-is a prime implicate (3).
(h) w-xy-z is an implicant since it implies f. However, none
of the product terms w-xy-, w-xz, w
-y-z, and xy
-z are
implicants. Therefore, w-xy-z is a prime implicant (1).
4.2. (a) (b)
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4.2. (continued)
(c) (d)
(e) (f)
4.3. The Karnaugh map is
8/14/2019 Chapter 4 Problem Solutions 4.1. w 0 0 0 0
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4.3. (continued)
The implicants correspond to all possible subcubes of 1-
cells.
one-cell subcubes
(minterms)
two-cell subcubes four-cell subcubes
w-x-y-z-
w-x-y-
wy
w-x-y-z w
-x-z-
w-x-yz-
w-y-z
w-xy-z wxy
wx-yz-
wx-y
wx-yz x
-yz-
wxyz- wyz
wxyz wyz-
The prime implicants correspond to those subcubes which are
not properly contained in some other subcube of 1-cells.
The prime implicants are shown on the map. They are wy,w-x-y-, w
-x-z-, w-y-z, and x
-yz-.
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4.4. (a)
The prime implicants are xz, xy, y-z, yz
-, x-z-, and x
-y-.
There are no essential prime implicants.
(b)
The prime implicants are w-x, xz, w
-y-z, and wyz. All four
prime implicants are essential as a result of the
essential 1-cells indicated by asterisks.
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4.4. (continued)
(c)
The prime implicants are w-x-, w-z__, x-z-__, and x-y __. The
essential prime implicants are underlined.
(d)
The prime implicants are y-z-, xy
-__, x
-z-
__, and w-xz ___. The
essential prime implicants are underlined.
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4.5. (a)
The prime implicates are x-+y+z and x+y
-+z-. Both prime
implicates are essential.
(b)
The prime implicates are w-+z, x+z, w+x+y
-, and w
-+x+y.
All three prime implicates are essential.
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4.5. (continued)
(c)
The prime implicates are w-+x-, x-+z, and w-+y+z-. All three
prime implicates are essential.
(d)
The prime implicates are x+z-
___, x-+y-+z_____, w
-+x-+y-, and w
-+y-+z-.
The essential prime implicates are underlined.
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4.6. (continued)
(c)
Minimal sum:
f = x-y-+ x-z + y
-z + xyz
-
Minimal product:
f = (x+y-+z)(x-+y+z)(x-+y-+z-)
(d)
Minimal sum:
f = y-z-+ x-yz
Minimal products:
f = (y+z-)(y-+z)(x
-+z-)
f = (y+z-)(y-+z)(x
-+y-)
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4.7. (a)
Minimal sum:
f = x + yz-
Minimal product
f = (x+y)(x+z-)
(b)
Minimal sum:
f = x-+ y + z
-
Minimal product:
f = x-+ y + z
-
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4.7. (continued)
(c)
Minimal sum:
f = x-z-+ yz
Minimal product:
f = (y+z-)(x-+z)
(d)
Minimal sum:
f = y + x-z-
Minimal product:
f = (x-+y)(y+z
-)
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4.8. (a)
Minimal sum:
f = xy + w-x-y- + x-y-z-
Minimal products:
f = (x-+y)(x+y
-)(w-+y+z
-)
f = (x-+y)(x+y
-)(w-+x+z
-)
(b)
Minimal sum:
f = xz-+ wz + x
-yz
Minimal products:
f = (x+z)(w+x-+z-)(w+x+y)
f = (x+z)(w+x-+z-)(w+y+z
-)
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4.8. (continued)
(c)
Minimal sum:
f = wz + x-z + yz + w
-xz-
Minimal product:
f = (w-+z)(x+z)(w+x
-+y+z
-)
(d)
Minimal sums:
f = w-x-z-+ w-xy + wxz + wx
-y-
f = x-y-z-+ wy
-z + xyz + w
-yz-
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4.8. (continued)
Minimal products:
f = (w+x+z-)(w+x-+y)(w-+x-+z)(w-+x+y-)
f = (x-+y+z)(w+y+z
-)(x+y
-+z-)(w-+y-+z)
(e)
Minimal sum:
f = x-z- + yz + wy-
Minimal products:
f = (w+y+z-)(x
-+y-+z)(w+x
-+y)
f = (w+y+z-)(x
-+y-+z)(w+x
-+z)
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4.8. (continued)
(f)
Minimal sum:
f = w-x-+ y-z + wz + x
-y
Minimal product:
f = (x-+z)(w+x
-+y-)(w-+y+z)
(g)
Minimal sum:
f = yz + x-z + xy
-z-
Minimal product:
f = (x+z)(y-+z)(x
-+y+z
-)
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4.8. (continued)
(h)
Minimal sums:
f = xy-+ xz + w
-x-y + wyz
-
f = xy-+ wx + w
-yz + x
-yz-
Minimal product:
f = (x+y)(w-+x+z-)(w+x-+y-+z)
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4.8. (continued)
(i)
Minimal sum:
f = wx- + x-z- + x-y- + w-y-z- + wy-z
Minimal product:
f = (x-+y-)(w+x
-+z-)(w+y
-+z-)(w-+x-+z)
(j)
Minimal sum:
f = wx- + x-y + wyz + w-xz-
Minimal products:
f = (w+x+y)(w+x-+z-)(w-+x-+z)(w
-+x-+y)
f = (w+x+y)(w+x-+z-)(w-+x-+z)(x
-+y+z
-)
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4.9. (a)
Minimal sums:
f = w-x-z- + xyz + x-yz- + wx-y-z + w-xy
f = w-x-z-+ xyz + x
-yz-+ wx
-y-z + w
-yz-
Minimal products:
f = (x-+y)(w
-+y+z)(x+y
-+z-)(w-+x-+z)(w+x+z
-)
f = (x-+y)(w
-+y+z)(x+y
-+z-)(w-+x-+z)(w+y+z
-)
(b)
Minimal sum:
f = x-y-+ w-x + w
-z-
Minimal product:
f = (w-+x-)(w-+y-)(x+y
-+z-)
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4.9. (continued)
(c)
Minimal sum:
f = x-y-z-+ xyz + w
-xy + w
-y-z
Minimal products:
f = (x+y-)(x-+y+z)(w
-+y+z
-)(w-+y-+z)
f = (x+y-)(x-+y+z)(w
-+y+z
-)(w-+x-+z)
(d)
Minimal sum:
f = y-z + w
-z + x
-z + w
-xy-
Minimal product:
f = (x+z)(w-+z)(y
-+z)(w
-+x-+y-)
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4.9. (continued)
(e)
Minimal sums:
f = xyz + xy-z-+ wy
-z-+ w-x-y-z + wxy
f = xyz + xy-z-+ wy
-z-+ w-x-y-z + wxz
-
Minimal products:
f = (x+y-)(x-+y+z
-)(w+x+z)(w+y
-+z)(w
-+x+z
-)
f = (x+y-)(x-+y+z
-)(w+x+z)(w+y
-+z)(w
-+y+z
-)
(f)
Minimal sums:
f = y-z-+ wz + w
-y
f = w-z-+ wy
-+ yz
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4.9. (continued)
Minimal product:
f = (w+y+z-)(w-+y-+z)
(g)
Minimal sum:
f = wx + y-z + wy
-+ xz
Minimal product:
f = (w+z)(x+y-)
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4.9. (continued)
(h)
Minimal sums:
f = x-z-+ xz + wy
-+ yz
-
f = x-z-+ xz + wy
-+ xy
Minimal product:
f = (w+x+z-)(x+y
-+z-)(w+x
-+y+z)
(i)
Minimal sum:
f = x-z + wz
-
Minimal product:
f = (x-+z-)(w+z)
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4.9. (continued)
(j)
Minimal sums:
f = x-y-+ w-y-
+ wxz + xyz-
f = x-y-+ y-z + w
-xz-+ wxy
Minimal product:
f = (x+y-)(w+y
-+z-)(w-+x-+y+z)
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4.10. (a)
The prime implicants are x-z-
__, w-y-, xz __, and wy. The
essential prime implicants are underlined.
The prime implicates are x-+z and x+z
-. Both prime
implicates are essential.
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4.10. (continued)
(b)
The prime implicants are x, w-z, and wz-. All three
prime implicants are essential.
The prime implicates are w+z___, w-+x+z
-_____, and w
-+y-+z-. The
essential prime implicates are underlined.
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4.10. (continued)
(c)
The prime implicants are wx__, yz-, w-y __, and xy. The
essential prime implicants are underlined.
The prime implicates are w+y___, x+y, w-+x ___, and x+z. The
essential prime implicates are underlined.
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4.11. (a)
Minimal sum:
f = x-z- + xyz + w-y-z
Minimal product:
f = (x-+z)(w
-+y+z
-)(x+y
-+z-)
(b)
Minimal sum:
f = wy-+ x-yz-
Minimal products:
f = (w+y)(x-+y-)(w+z
-)
f = (w+y)(x-+y-)(y-+z-)
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4.11. (continued)
(c)
Minimal sum:
f = y-z + xz + wz-
Minimal product:
f = (w+z)(x+y-+z-)
(d)
Minimal sums:
f = wx + wz + yz
f = wx + wz + xz
f = wx + wz + xy
Minimal product:
f = (x+z)(w+y)
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4.11. (continued)
(e)
Minimal sums:
f = x-z- + yz-
f = x-z-+ w-z-
Minimal products:
f = z-(x-+y)
f = z-(w-+x-)
(f)
Minimal sum:
f = wx-+ x-z + w
-xz-+ wy
-
Minimal product:
f = (w+x+z)(w+x-+z-)(w-+x-+y-)
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4.11. (continued)
(g)
Minimal sum:
f = x-z- + wy- + w-xy
Minimal product:
f = (w+y)(w-+x-+y-)(x+y
-+z-)
(h)
Minimal sum:
f = w-y + xy
-+ y-z + xz
Minimal products:
f = (x+y+z)(w-+x+y
-)(x-+y-+z)
f = (x+y+z)(w-+x+y
-)(w-+y-+z)
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4.11. (continued)
(i)
Minimal sum:
f = w-y-+ y-z + wyz
-+ w-x
Minimal product:
f = (w-+y+z)(w+x+y
-)(w-+y-+z-)
(j)
Minimal sums:
f = w-z + wz
-+ w-xy-
f = w-z + wz
-+ xy
-z-
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4.11. (continued)
Minimal products:
f = (w-+z-)(y
-+z)(w+x+y)
f = (w-+z-)(y
-+z)(w+x+z)
4.12. (a)
Minimal sums:
f = w-x + wx
-y + wz
f = w-x + wx-y + y-z
Minimal products:
f = (y+z)(w+x)(w-+x-+y-)
f = (y+z)(w+x)(w-+x-+z)
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4.12. (continued)
(b)
Minimal sums:
f = wx-y- + w-y-z + wxy
f = wx-y-+ w
-y-z + xyz
-
Minimal products:
f = (x+y-)(w+z)(w
-+x-+y)(y
-+z-)
f = (x+y-)(w+z)(w
-+x-+y)(w+y
-)
(c)
Minimal sum:
f = x-z-+ w-y-
+ wxz
Minimal product:
f = (x+z-)(w-+x-+z)(w+x
-+y-)
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4.12. (continued)
(d)
Minimal sums:
f = w-x-z + w
-xz-+ wxy
-
f = w-x-z + w
-xz-+ xy
-z-
Minimal product:
f = (x+z)(x-+z-)(w-+y-)
(e)
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4.12. (continued)
Minimal sums:
f = w-y-+ y-z + w
-x-
f = w-y-+ y-z + x
-y
Minimal product:
f = (w-+z)(x
-+y-)
(f)
Minimal sum:
f = xz-+ w-y + x
-y-z
Minimal product:
f = (x-+z-)(x+y+z)(w
-+x+y
-)
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4.12. (continued)
(g)
Minimal sum:
f = wx + w-x-y + wy
-z-+ xz
-
Minimal products:
f = (w+y)(w+x-+z-)(w-+x+y
-)(w-+x+z
-)
f = (w+y)(w+x-+z-)(w-+x+y
-)(x+y+z
-)
(h)
Minimal sum:
f = w-y + wz
-+ w-xz
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4.12. (continued)
Minimal products:
f = (w-+z-)(w+y+z)(w+x+y)
f = (w-+z-)(w+y+z)(x+y+z
-)
(i)
Minimal sum:
f = xy + xz + wyz + wy-z-
Minimal products:
f = (w+x)(w+y+z)(x+y+z-)(w-+y-+z)
f = (w+x)(w+y+z)(x+y+z-)(x+y
-+z)
(j)
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4.12. (continued)
Minimal sum:
f = xy + wz-+ w-y-z
Minimal product:
f = (w+z)(x+y-)(w-+y+z-)
4.13.
To construct f2:
If g=1 and f1=1, then f
2=1.
If g=0 and f1=0, then f
2=-.
If g=0 and f1=1, then f
2=0.
(Note: g=1 and f1=0 can not occur if the problem is
solvable.)
Using the above rules, the map for f2is:
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4.13. (continued)
Minimal sum:
f2= xy
-+w-z
Minimal products:
f2 = (x+z)(y-+z)(w-+x) (Shown above)
Other possible minimal products:
f2= (x+z)(y
-+z)(w
-+y-)
f2= (x+z)(x
-+y-)(w-+x)
f2= (x+z)(x
-+y-)(w-+y-)
4.14. (a)
Minimal sums:
f = v-y-z + vwy + vxy
-+ v-wx-z
f = v-y-z + vwy + vxy
-+ wx
-yz
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4.14. (continued)
Minimal product:
f = (w+y-)(v+z)(v
-+x+y)(v+x
-+y-)
(b)
Minimal sum:
f = yz + v-wy-+ vw
-y-z-+ vwxy
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4.14. (continued)
Minimal products (shown above):
f = (v-+w-+y)(v+y
-+z)(w+y
-+z)(x+y
-+z)(v+w+y)(v
-+y+z
-)
f = (v-+w-+y)(v+y
-+z)(w+y
-+z)(x+y
-+z)(v+w+y)(w+y+z
-)
Another minimal product (not shown):
f = (v-+w-+y)(v+y
-+z)(w+y
-+z)(x+y
-+z)(w+y+z
-)(v+w+z)
(c)
Minimal sums (shown above):
f = vwz-+ w
-x-z + vy
-z-+ v-w-xy-+ v
-wxy + v
-x-yz
f = vwz-+ w-x-z + vy
-z-+ v-w-xy-+ v
-wxy + v
-wyz
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4.14. (continued)
Another minimal sum (not shown):
f = vwz-+ w
-x-z + vy
-z-+ v-w-xy-
+ v-wyz + wxyz
-
Minimal product:
f = (v+x+z)(v+w-+y)(v
-+w-+z-)(w+x
-+y-)(v-+x-+z-)(w+y
-+z)
(d)
Minimal sum:
f = w-z + y
-z + wx
-y + vwy
-
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4.14. (continued)
Minimal product:
f = (w+z)(w-+x-+y-)(v+y+z)
4.15.
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4.15. (continued)
Minimal sums:
f = v-wz-+ uvw
-+ v-w-x + uvz
-
f = v-wz-+ uvw
-+ v-w-x + uwz
-
Minimal product:
f = (u+v-)(w-+z-)(v+w+x)
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4.16.
x y z f
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
Minimal sum (cost=9): f = xy + xz + yz
Minimal product (cost=9): f = (x+y)(x+z)(y+z)
Realization using the minimal sum:
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4.17.
Minimal sum (cost=20):
p = wz + xy-z + xyz- + x-yz + w-x-y-z-
Minimal product (cost=20):
p = (w-+z)(x
-+y+z)(x
-+y-+z-)(x+y
-+z)(w+x+y+z
-)
Comparing Tables 3.10 and 3.12, it is seen that if the
don't-cares for wxyz=1010, 1100, and 1111 are assigned
values of logic-1, the two tables become identical.
Since Table 3.10 is describable by the expression
p=(w-rx)r(yrz), this same expression can be used for
Table 3.12.
4.18. Truth table:
xi
yi
ci
ci+1
si
0 0 0 0 0
0 0 1 0 1
0 1 0 0 1
0 1 1 1 0
1 0 0 0 1
1 0 1 1 0
1 1 0 1 0
1 1 1 1 1
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4.18. (continued)
For the ci+1
output:
Minimal sum (cost=9):
ci+1
= xiyi+ x
ici+ y
ici
Minimal product (cost=9)
ci+1
= (xi+yi)(x
i+ci)(y
i+ci)
For the sioutput:
Minimal sum (cost=16):
si= x-iy-ici
+ x-iyic-i+ x
iy-ic-i+ x
iyici
Minimal product (cost=16)
si= (x
i+yi+c
i)(x
i+y-i+c-i)(x-i+y
i+c-i)(x-i+y-i+ c
i)
However,
si= x-
iy-ici+ x-
iyic-i+ x
iy-ic-i
+ xiyici
= x-i(y-ici+y
ic-i) + x
i(y-ic-i+ y
ici)
= x-i(yirc
i) + x
i(yirc
i)
_______
= xir (y
irc
i)
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4.18. (continued)
Realization:
4.19. Truth table:
x
i
y
i
b
i
b
i+1
d
i
0 0 0 0 0
0 0 1 1 1
0 1 0 1 1
0 1 1 1 0
1 0 0 0 1
1 0 1 0 0
1 1 0 0 0
1 1 1 1 1
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4.19. (continued)
For the bi+1
output:
Minimal sum (cost=9):
bi+1
= x-iyi+ x-ibi+ y
ibi
Minimal product (cost=9):
bi+1
= (x-i+yi)(x-
i+bi)(y
i+bi)
For the dioutput:
Minimal sum (cost=16):
di= x-iy-ibi
+ x-iyib-i+ x
iy-ib-i+ x
iyibi
Minimal product (cost=16):
di= (x
i+yi+b
i)(x
i+y-i+b-i)(x-i+y
i+b-i)(x-i+y-i+bi)
However,
di= x-
iy-ibi+ x-
iyib-i+ x
iy-ib-i
+ xiyibi
= x-(y-ibi+yib-i) + x
i(y-ib-i+yibi)
= x-i(yirb
i) + x
i(yirb
i)
_______
= xir (y
irb
i)
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4.19. (continued)
Realization:
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4.20.
8 4 2 1
w x y z f
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0
0 1 0 1 0
0 1 1 0 0
0 1 1 1 0
1 0 0 0 0
1 0 0 1 0
1 0 1 0 1
1 0 1 1 1
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1
Minimal sum (cost=6):
f = wx + wy
Minimal product (cost=4):
f = w(x+y)
Realization:
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4.21. Let:
F=1 when the pressure in the fuel tank is equal to or above
a required minimum.
F=0 when the pressure in the fuel tank is below a required
minimum.
X=1 when the oxidizer tank pressure is equal to or above a
required minimum.
X=0 when the oxidizer tank pressure is below a required
minimum.
T=1 when there are 10 min. or less to lift off.
T=0 when there are more than 10 min. to lift off.
P=1 when the panel light is on.
P=0 when the panel light is off.
From the problem statement we can write the equation for
the panel light to be on:
P = FXT + F-XT-+ X-T-
Minimal sum (cost=10):
P = X-T-+ F-T-+ FXT
Minimal product (cost=10):
P = (F+T-)(X+T-)(F-+X-+T)
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4.21. (continued)
Realization using the minimal sum:
4.22. Let w, x, y, and z denote the bits of the 2421 code groups
and A, B, C, and D denote the bits of the 8421 code groups
where w and A are the most significant bits of their
respective code groups.
w x y z A B C D
0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 1
0 0 1 0 0 0 1 0
0 0 1 1 0 0 1 1
0 1 0 0 0 1 0 0
0 1 0 1 - - - -
0 1 1 0 - - - -
0 1 1 1 - - - -
1 0 0 0 - - - -
1 0 0 1 - - - -
1 0 1 0 - - - -
1 0 1 1 0 1 0 1
1 1 0 0 0 1 1 0
1 1 0 1 0 1 1 1
1 1 1 0 1 0 0 0
1 1 1 1 1 0 0 1
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4.22. (continued)
A = xy B = xy-+ wx
-(cost=6)
B = (w+x)(x-+y-) (cost=6)
C = wy-+ w-y (cost=6) D = z
C = (w+y)(w-+y-) (cost=6)
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4.23.
7 5 3 6-
w x y z f
0 0 0 0 1
0 0 0 1 -
0 0 1 0 1
0 0 1 1 -
0 1 0 0 0
0 1 0 1 -
0 1 1 0 0
0 1 1 1 1
1 0 0 0 0
1 0 0 1 1
1 0 1 0 -
1 0 1 1 0
1 1 0 0 -
1 1 0 1 0
1 1 1 0 -
1 1 1 1 0
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4.23. (continued)
Minimal sum (cost=10):
f = w-x-+ w-z + x
-y-z
Minimal products (cost=12):
f = (x-+z)(w-+z)(w-+y-)(w-+x-)
f = (x-+z)(w
-+z)(w
-+y-)(x-+y)
Realization:
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4.24.
Inputs Outputs
8 4 2 1
w x y z a b c d e f g
0 0 0 0 1 1 1 1 1 1 0
0 0 0 1 0 1 1 0 0 0 0
0 0 1 0 1 1 0 1 1 0 1
0 0 1 1 1 1 1 1 0 0 1
0 1 0 0 0 1 1 0 0 1 1
0 1 0 1 1 0 1 1 0 1 1
0 1 1 0 1 0 1 1 1 1 1
0 1 1 1 1 1 1 0 0 0 0
1 0 0 0 1 1 1 1 1 1 1
1 0 0 1 1 1 1 1 0 1 1
1 0 1 0 - - - - - - -
1 0 1 1 - - - - - - -
1 1 0 0 - - - - - - -
1 1 0 1 - - - - - - -
1 1 1 0 - - - - - - -
1 1 1 1 - - - - - - -
a = w + y + xz + x-z-
b = x-+ y-z-+ yz
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4.24. (continued)
c = x + y-+ z d = w + yz
-+ x-y + x
-z-+ xy
-z
e = x-z-+ yz
-f = w + y
-z-+ xy
-+ xz
-
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4.24. (continued)
g = w + xy-+ x
-y + yz
-or g = w + xy
-+ x-y + xz
-
4.25. (a)
wxyz wxyz wxyz
0 0000 T 0,2 00-0 T 0,2,8,10 -0-0 A
2 0010 T 0,4 0-00 T 0,4,8,12 --00 B
4 0100 T 0,8 -000 T 8,10,12,14 1--0 C
8 1000 T 2,3 001- D
3 0011 T 2,10 -010 T
10 1010 T 4,12 -100 T
12 1100 T 8,10 10-0 T
13 1101 T 8,12 1-00 T
14 1110 T 10,14 1-10 T
12,13 110- E
12,14 11-0 T
The prime implicants are:
A: x-z-
B: y-z-
C: wz-
D: w-x-y
E: wxy-
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4.25. (continued)
(b)
vwxyz vwxyz vwxyz
4 00100 T 4,5 0010- T 4,5,6,7 001-- A
5 00101 T 4,6 001-0 T 10,14,26,30 -1-10 B
6 00110 T 5,7 001-1 T
9 01001 E 6,7 0011- T
10 01010 T 6,14 0-110 C
7 00111 T 10,14 01-10 T
14 01110 T 10,26 -1010 T
19 10011 F 14,30 -1110 T
26 11010 T 26,30 11-10 T
30 11110 T 30,31 1111- D
31 11111 T
The prime implicants are:
A: v-w-x
B: wyz-
C: v-xyz-
D: vwxy
E: v-wx-y-z
F: vw-x-yz
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4.25. (continued)
(c)
wxyz wxyz wxyz
4 0100 T 4,12 -100 C 9,11,13,15 1--1 A
9 1001 T 9,11 10-1 T 12,13,14,15 11-- B
12 1100 T 9,13 1-01 T
7 0111 T 12,13 110- T
11 1011 T 12,14 11-0 T
13 1101 T 7,15 -111 D
14 1110 T 11,15 1-11 T
15 1111 T 13,15 11-1 T
14,15 111- T
The prime implicants are:
A: wz
B: wx
C: xy-z-
D: xyz
(d)
wxyz wxyz wxyz
0 0000 T 0,1 000- T 0,1,2,3 00-- A
1 0001 T 0,2 00-0 T 1,3,5,7 0--1 B
2 0010 T 1,3 00-1 T 2,3,6,7 0-1- C
3 0011 T 1,5 0-01 T
5 0101 T 1,9 -001 D
6 0110 T 2,3 001- T
9 1001 T 2,6 0-10 T
10 1010 T 2,10 -010 E
12 1100 F 3,7 0-11 T
7 0111 T 5,7 01-1 T
6,7 011- T
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4.25. (continued)
The prime implicants are:
A: w-x-
B: w-z
C: w-y
D: x-y-z
E: x-yz-
F: wxy-z-
4.26. (a)
vwxyz vwxyz vwxyz
1 00001 T 1,3 000-1 T 1,3,17,19 -00-1 A
3 00011 T 1,17 -0001 T 6,14,22,30 --110 B
6 00110 T 3,11 0-011 D 10,11,14,15 01-1- C
10 01010 T 3,19 -0011 T
12 01100 T 6,14 0-110 T
17 10001 T 6,22 -0110 T
20 10100 T 10,11 0101- T
24 11000 G 10,14 01-10 T
11 01011 T 12,14 011-0 E
14 01110 T 17,19 100-1 T
19 10011 T 20,22 101-0 F
22 10110 T 11,15 01-11 T
15 01111 T 14,15 0111- T
29 11101 H 14,30 -1110 T
30 11110 T 22,30 1-110 T
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4.26. (continued)
The prime implicates are:
A: w+x+z-
B: x-+y-+z
C: v+w-+y-
D: v+x+y-+z-
E: v+w-+x-+z
F: v-+w+x
-+z
G: v-+w-+x+y+z
H: v-+w-+x-+y+z
-
(b)
wxyz wxyz wxyz
0 0000 T 0,2 00-0 T 0,2,8,10 -0-0 A
2 0010 T 0,4 0-00 T 0,4,8,12 --00 B
4 0100 T 0,8 -000 T 4,5,12,13 -10- C
8 1000 T 2,3 001- D
3 0011 T 2,10 -010 T
5 0101 T 4,5 010- T
10 1010 T 4,12 -100 T
12 1100 T 8,10 10-0 T
13 1101 T 8,12 1-00 T
5,13 -101 T
12,13 110- T
The prime implicates are:
A: x+z
B: y+z
C: x-+y
D: w+x+y-
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4.27. (a)
wxyz wxyz
4 0100 T 4,5 010- A
5 0101 T 4,12 -100 B
12 1100 T 5,7 01-1 C
7 0111 T 12,14 11-0 D
14 1110 T 7,15 -111 E
15 1111 T 14,15 111- F
m4
m5
m7
m12
m14
m15
A: w-xy-
B: xy-z-
C: w-xz
D: wxz-
E: xyz
F: wxy
p = (A+B)(A+C)(C+E)(B+D)(D+F)(E+F)
= (B+AD)(C+AE)(F+DE)
= (BC+ABE+ACD+ADE)(F+DE)
= BCF + BCDE + ABEF + ABDE + ACDF + ACDE + ADEF + ADE
= BCF + ADE + BCDE + ABEF + ACDF
There are five irredundant disjunctive normal formulas:
f1= B + C + F = xy
-z-+ w-xz + wxy
f2= A + D + E = w
-xy-+ wxz
-+ xyz
f3 = B + C + D + E = xy-z- + w-xz + wxz- + xyz
f4= A + B + E + F = w
-xy-+ xy
-z-
+ xyz + wxy
f5= A + C + D + F = w
-xy-+ w-xz + wxz
-+ wxy
The costs of f1and f
2are 12; while the costs of f
3,
f4, and f
5are 16. Therefore, f
1and f
2are minimal
sums.
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4.27. (continued)
(b)
wxyz wxyz wxyz
0 0000 T 0,4 0-00 T 0,4,8,12 --00 A
4 0100 T 0,8 -000 T 4,5,12,13 -10- B
8 1000 T 4,5 010- T 8,9,10,11 10-- C
3 0011 T 4,12 -100 T 8,9,12,13 1-0- D
5 0101 T 8,9 100- T
9 1001 T 8,10 10-0 T
10 1010 T 8,12 1-00 T
12 1100 T 3,7 0-11 E
7 0111 T 3,11 -011 F
11 1011 T 5,7 01-1 G
13 1101 T 5,13 -101 T
9,11 10-1 T
9,13 1-01 T
10,11 101- T
12,13 110- T
m4
m5
m8
m9
m12
m13
A: y-z-
B: xy-
C: wx-
D: wy-
G: w-xz
(Note: Prime implicants E and F do not appear in the
prime-implicant table since they do not cover any of
the minterms having functional values of 1.)
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4.27. (continued)
p = (A+B)(B+G)(A+C+D)(C+D)(A+B+D)(B+D)
= (A+B)(B+G)(C+D)(B+D)
= (B+AG)(D+BC)
= BD + BC + ADG + ABCG
= BD + BC + ADG
There are three irredundant disjunctive normal
formulas:
f1= B + D = xy
-+ wy
-
f2= B + C = xy
-+ wx
-
f3= A + D + G = y
-z-+ wy
-+ w-xz
The costs of f1
and f2are 6 and the cost of f
3is 10.
Therefore, f1and f
2are minimal sums.
4.28. (a)
wxyz wxyz wxyz
4 0100 T 4,5 010- B 9,11,13,15 1--1 A
8 1000 T 8,9 100- C
3 0011 T 3,11 -011 D
5 0101 T 5,13 -101 E
9 1001 T 9,11 10-1 T
11 1011 T 9,13 1-01 T
13 1101 T 11,15 1-11 T
14 1110 T 13,15 11-1 T
15 1111 T 14,15 111- F
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4.28. (continued)
M3
M4
M5
M8
M9
M11
M13
M14
M15
A: w-+z-
B: w+x-+y
C: w-+x+y
D: x+y-+z-
E: x-+y+z
-
F: w-+x-+y-
p = DB(B+E)C(A+C)(A+D)(A+E)F(A+F)
= BCDF(A+E)= ABCDF + BCDEF
There are two irredendant conjunctive normal formulas:
f1= ABCDF = (w
-+z-)(w+x
-+y)(w
-+x+y)(x+y
-+z-)(w-+x-+y-)
f2= BCDEF = (w+x
-+y)(w
-+x+y)(x+y
-+z-)(x-+y+z
-)(w-+x-+y-)
The cost of f1is 19 and the cost of f
2is 20.
Therefore, f1 is the minimal product.
(b)
wxyz wxyz wxyz
0 0000 T 0,8 -000 B 5,7,13,15 -1-1 A
8 1000 T 8,9 100- C
5 0101 T 5,7 01-1 T
6 0110T
5,13 -101T
9 1001 T 6,7 011- D
7 0111 T 9,13 1-01 E
13 1101 T 7,15 -111 T
15 1111 T 13,15 11-1 T
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4.28. (continued)
M0
M6
M7
M8
M9
M13
A: x-+z-
B: x+y+z
C: w-+x+y
D: w+x-+y-
E: w-+y+z
-
p = BD(A+D)(B+C)(C+E)(A+E)
= BD(C+E)(A+E)
= BD(E+AC)
= BDE + ABCD
There are two irredundant conjunctive normal formulas:
f1= BDE = (x+y+z)(w+x
-+y-)(w-+y+z
-)
f2= ABCD = (x
-+z-)(x+y+z)(w
-+x+y)(w+x
-+y-)
The cost of f1is 12 and the cost of f
2is 15.
Therefore, f1is the minimal product.
4.29. (a) Referring to Table P4.29a, c2dominates c
1, c
6
dominates c1, c6 dominates c2, c3 dominates c7, and c5dominates c
7. Deleting the dominating columns c
2, c
3,
c5, and c
6, the table reduces to
c1
c4
c7
cost
r1
3
r2
4
r4
4
r5 4
r6
6
r7
6
(Note: r3no longer appears since it has no crosses in
the remaining columns.)
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4.29. (continued)
r1equals r
5, r
2equals r
7, and r
4equals r
6. Deleting
those rows with the higher costs, i.e., r5, r
6, and r
7,
the table becomes
c1
c4
c7
cost
**r1
3
**r2
4
**r4
4
Since only single crosses exist in each column, the
minimal cover consists of rows r1, r
2, and r
4.
(b) Referring to Table P4.29b, r3is an essential prime
implicant since c11
has a single cross. Therefore, r3
must be selected and r3and c
11can be deleted from the
table. Also, c1dominates c
8, c
2dominates c
5, c
9
dominates c3, and c
10dominates c
6. Thus, columns c
1,
c2, c
9, and c
10are deleted. The table now reduces to
c3
c4
c5
c6
c7
c8
cost
r1
3
r2
3
r4
4
r5
4
r6
5
r7 6
r8
7
r9
7
Minimal cover: {r3,...}
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4.29. (continued)
r1now dominates r
5, r
1dominates r
6, r
8dominates r
2,
r7dominates r
4, and r
5dominates r
6. Rows r
5and r
6
are deleted since they have a higher cost than their
dominating rows. However, r2 and r4 cannot be deleted
since they have a lower cost than their dominating
rows. The resulting table is
c3
c4
c5
c6
c7
c8
cost
**r1
3
r2
3
r4
4
r7
6
r8
7
r9
7
Minimal cover: {r3,...}
Since column c4 has a single cross, row r1 is selected
to appear in the minimal cover. After deleting row r1
and columns c4, c
5, and c
7, the table reduces to
c3
c6
c8
cost
r2
3
r4
4
r7 6
r8
7
r9
7
Minimal cover: {r1,r3,...}
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4.29. (continued)
r4equals r
7and the cost of r
4is less than the cost
of r7. Therefore, delete r
7. Also, r
8dominates r
9.
r9is deleted since it has equal cost. The table
becomes
c3
c6
c8
cost
r2
3
**r4
4
**r8
7
Minimal cover: {r1,r3,...}
Finally, since columns c3and c
8have single crosses,
rows r4and r
8must appear in the minimal cover. This
completes the covering of the prime-implicant table.
The minimal cover consists of rows r1, r
3, r
4, and r
8.
(c) Referring to Table P4.29c, c
5
dominates c
1
, c
2
equals
c6, and c
3dominates c
7. Deleting c
3, c
5, and c
6, the
table reduces to
c1
c2
c4
c7
cost
r1
3
r2
4
r3
4
r4
5
r5
5
r6
6
r7
7
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4.29. (continued)
r6dominates r
1, but cost prevents deleting r
1.
However, since r2dominates r
5, r
2dominates r
7, r
3
dominates r7, and r
4dominates r
5, and since the
dominated rows have a greater or equal cost than the
dominating rows, delete r5and r
7. The table now
becomes
c1
c2
c4
c7
cost
r1
3
r2
4
r3
4
r4
5
r6
6
The resulting table cannot be reduced any further.
Petrick's method should be applied.
p = (r1+r6)(r
3+r4+r6)(r
2+r3)(r
2+ r
4)
= (r6+r1r3+r1r4)(r2+r3r4)= r
2r6+ r
3r4r6+ r
1r2r3+ r
1r3r4+ r
1r2r4+ r
1r3r4
= r2r6+ r
3r4r6+ r
1r2r3+ r
1r3r4+ r
1r2r4
Irredundant
cover
cost
r2, r
610
r3, r
4, r
615
r1, r
2, r
311
r1, r
3, r
412
r1, r
2, r
412
The minimal cost cover consists of rows r2and r
6.
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4.30.
wxyz wxyz
2 0010 T 2,3 001- A
4 0100 T 2,10 -010 B
3 0011 T 4,5 010- C
5 0101 T 4,12 -100 D
10 1010 T 3,7 0-11 E
12 1100 T 5,7 01-1 F
7 0111 T 10,14 1-10 G
14 1110 T 12,14 11-0 H
15 1111 T 7,15 -111 I
14,15 111- J
m3
m4
m5
m7
m10
m12
m14
m15
cost
A 4
B 4
C 4
D 4
E 4
F 4
G 4
H 4
I 4
J 4
Row E dominates row A and row G dominates row B. Since
all rows have equal cost, delete rows A and B. The
reduced table is
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4.30. (continued)
m3
m4
m5
m7
m10
m12
m14
m15
cost
C 4
D 4
**E 4
F 4
**G 4
H 4
I 4
J 4
Select prime implicants E and G for a minimal sum since
columns m3and m
10each have a single cross. Delete
rows E and G as well as the columns having crosses in
these rows. The reduced table becomes
m4
m5
m12
m15
cost
C 4
D 4
F 4
H 4
I 4
J 4
Minimal sum: f = E + G +...
Row C dominates row F, row D dominates row H, and rows
I and J are equal. Since all rows have the same cost,
delete rows F, H, and J. The resulting table is
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4.30. (continued)
m4
m5
m12
m15
cost
**C 4
**D 4
**I 4
Minimal sum: f = E + G + ...
Since columns m5, m
12, and m
15each have a single
cross, rows C, D, and I are selected for a minimal sum.
Furthermore, by selecting these rows, all the columns
of the above table are covered. The minimal sum is
f = C + D + E + G + I
= w-xy-+ xy
-z-+ w-yz + wyz
-+ xyz
(Note: If row I was deleted since it was equal to row
J, then the minimal cover would consist of rows C,
D, E, G, and J. Applying Petrick's method to the
original prime-implicant table would result in 16
minimal sums of which only 2 are readily found when
using table reduction procedures.)
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4.31.
1 T 1,3 (2) C 8,9,12,13 (1,4) A
8 T 1,9 (8) D 8,10,12,14 (2,4) B
3 T 8,9 (1) T
6 T 8,10 (2) T
9 T 8,12 (4) T
10 T 3,7 (4) E
12 T 6,7 (1) F
7 T 6,14 (8) G
13 T 9,13 (4) T
14 T 10,14 (4) T
12,13 (1) T
12,14 (2) T
Prime implicants:
wxyz
A: 1-0- 6 wy-
B: 1--0 6 wz-
C: 00-1 6 w-x-z
D: -001 6 x-y-z
E: 0-11 6 w-yz
F: 011- 6 w-xy
G: -110 6 xyz-
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4.31. (continued)
m1
m3
m6
m8
m9
m10
m12
m14
cost
A 3
*B 3
C 4
D 4
E 4
F 4
G 4
B is an essential prime implicant. Selecting row B and
deleting the columns having crosses in row B, the table
reduces to
m1
m3
m6
m9
cost
A 3
C 4
D 4
E 4
F 4
G 4
Minimal sum: f = B + ...
Even though row D dominates row A, row A cannot be deleted
because of cost. Row C dominates row E and rows F and G
are equal. Since these rows have the same cost, delete
rows E and G. The table becomes
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4.31. (continued)
m1
m3
m6
m9
cost
A 3
**C 4
D 4
**F 4
Minimal sum: f = B + ...
Since columns m3and m
6have single crosses, rows C and F
must be selected. After deleting rows C and F and the
columns having crosses in these rows, the reduced table is
m9
cost
A 3
D 4
Minimal sum: f = B + C + F + ...
Row D can be eliminated since it is equal to row A and has
a higher cost. This results in the table
m9
cost
**A 3
Minimal sum: f = B + C + F + ...
Selecting row A, the minimal sum is
f = A + B + C + F = wy-+ wz
-+ w-x-z + w
-xy
(Note: At an earlier point, row F could have been deleted
since it was equal to row G. In this case, a minimal
sum is
f = A + B + C + G = wy-+ wz
-+ w
-x-z + xyz
-)
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4.32.
xyz xyz
0 000 f1_
f3
F 0,2 (2) 0-0_ _
f3A
2 010_
f2f3T 2,3 (1) 01-
_ _f3B
3 011 f1_
f3
G 2,6 (4) -10_
f2f3C
5 101 f1f2_
H 3,7 (4) -11 f1_ _
D
6 110_
f2f3T 5,7 (2) 1-1 f
1_ _
E
7 111 f1_ _
T
f1
f2
f3
m0 m3 m5 m7 m2 m5 m6 m0 m2 m3 m6
A: x-z-
B: x-y
C: yz-
D: yz
E: xz
F: x-y-z-
G: x
-
yz
H: xy-z
p = F1(D1+G1)(E
1+H1)(D
1+E1)C2H2C2(A3+F
3)(A
3+B3+C3)(B
3+G3)C3
= F1C2H2C3(D1+G1)(E
1+H1)(D
1+E1)(A
3+F3)(B
3+G3)
= F1C2H2C3(D1+G1)(E
1+D
1H1)(A
3+F3)(B
3+G3)
= F1C2H2C3(D1E1+D1H1+E1G1+D1G1H1)(A
3B3+A3G3+B3F3+F3G3)
= F1C2H2C3(D1E1A3B3+D1E1A3G3+D1E1B3F3+D1E1F3G3+D1H1A3B3
+D1H1A3G3+D1H1B3F3+D1H1F3G3+E1G1A3B3+E1G1A3G3
+E1G1B3F3+E1G1F3G3)
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4.32. (continued)
= A3B3C2C3D1E1F1H2+ A
3C2C3D1E1F1G3H2+ B
3C2C3D1E1F1F3H2
+ C2C3D1E1F1F3G3H2+ A
3B3C2C3D1F1H1H2
+ A3C2C3D1F1G3H1H2+ B
3C2C3D1F1F3H1H2
+ C2C3D1F1F3G3H1H2 + A3B3C2C3E1F1G1H2
+ A3C2C3E1F1G1G3H2+ B
3C2C3E1F1F3G1H2
+ C2C3E1F1F3G1G3H2
Product terms Distinct terms Input terminals (E"i+G$
j)
A3B3C2C3D1E1F1H2
7 8 + 16 = 24
A3C2C3D1E1F1G3H2
7 8 + 17 = 25
B3C2C3D1E1F1F3H2 6 8 + 14 = 22
C2C3D1E1F1F3G3H2
6 8 + 15 = 23
A3B3C2C3D1F1H1H2
6 8 + 14 = 22
A3C2C3D1F1G3H1H2
6 8 + 15 = 23
B3C2C3D1F1F3H1H2
5 * 8 + 12 = 20 *
C2C3D1F1F3G3H1H2
5 * 8 + 13 = 21
A3B3C2C3E1F1G1H2
7 8 + 17 = 25
A3C2C3E1F1G1G3H2 6 8 + 15 = 23
B3C2C3E1F1F3G1H2
6 8 + 15 = 23
C2C3E1F1F3G1G3H2
5 * 8 + 13 = 21
There are three multiple-output minimal sums when the cost
is based on number of distinct terms:
f1= yz + x
-y-z-+ xy
-z f
1= yz + x
-y-z-+ xy
-z
f2= yz
-+ xy
-z f
2= yz
-+ xy
-z
f3= x-y + yz- + x-y-z- f
3= yz- + x-y-z- + x-yz
f1= xz + x
-y-z-+ x-yz
f2= yz
-+ xy
-z
f3= yz
-+ x-y-z-
+ x-yz
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4.32. (continued)
There is only one multiple-output minimal sum when the cost
is based on number of input terminals in the realization:
f1= yz + x
-y-z-+ xy
-z
f2 = yz- + xy-z
f3= x-y + yz
-+ x-y-z-
4.33. (a)
xyz xyz xyz
0 000 f1f2_
T 0,2 (2) 0-0 f1f2_
B 0,2,4,6 (2,4) --0 f1_ _
A
2 010 f1f2_
T 0,4 (4) -00 f1_ _
T
4 100 f1_
f3T 2,3 (1) 01- f1_ _
C
3 011 f1_
f3F 2,6 (4) -10 f
1_ _
T
5 101_
f2f3G 4,5 (1) 10-
_ _f3D
6 110 f1_
f3T 4,6 (2) 1-0 f
1_
f3E
f1
f2
f3
m0
m2
m3
m4
m6
m0
m2
m5
m3
m4
m5
m6
cost
A: z- 1
*2 B: x-z-
3,4
C: x-y 3
D: xy-
3
*3 E: xz-
3,4
*3 F: x-yz 4,5
*2 G: xy-z 4,5
f1= ... f
2= x-z-+ xy
-z f
3= xz
-+ x-yz + ...
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4.33. (continued)
Reducing the table:
f
1
f
3m0
m2
m3
m4
m6
m5
cost
A: z-
1
B: x-z-
1
C: x-y 3
D: xy-
3
E: xz-
1
F: x-yz 1
G: xy-z 1
f1= ... f
2= x-z-+ xy
-z f
3= xz
-+ x-yz + ...
After deleting dominated and equal rows B, E, and D,
the table becomes
f1 f3
m0
m2
m3
m4
m6
m5
cost
*1 A: z-
1
C: x-y 3
F: x-yz 1
*3 G: xy-z 1
f1 = z-+ ... f2 = x
-z-+ xy
-z f3 = xz
-+ x-yz + xy
-z
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4.33. (continued)
Reducing the table:
f
1m3
cost
C: x-y 3
F: x-yz 1
f1= z-+ ... f
2= x-z-+ xy
-z f
3= xz
-+ x-yz + xy
-z
To cover the remain column, select F. The multiple-
output minimal sum is
f1= z-+ x-yz
f2= x-z-+ xy
-z
f3= xz
-+ x-yz + xy
-z
(b)
xyz xyz xyz
0 000_ _
f
3
T 0,1 (1) 00-_ _
f
3
C 1,3,5,7 (2,4) --1_
f
2
_A
1 001 f1f2f3J 0,4 (4) -00
_ _f3D 4,5,6,7 (1,2) 1--
_f2_
B
2 010 f1_ _
T 1,3 (2) 0-1_
f2_
T
4 100_
f2f3T 1,5 (4) -01 f
1f2_
E
3 011_
f2_
T 2,6 (4) -10 f1_ _
F
5 101 f1f2_
T 4,5 (1) 10-_
f2_
T
6 110 f1f2f3K 4,6 (2) 1-0
_f2f3G
7 111 f
1
f
2
_T 3,7 (4) -11
_f
2
_T
5,7 (2) 1-1 f1f2_
H
6,7 (1) 11- f1f2_
I
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4.33. (continued)
f1
f2
f3
m1
m2
m5
m3
m5
m6
m7
m1
m4
m6
cost
*2 A: z 1
B: x 1
C: x-y-
3
D: y-z-
3
E: y-z 3,4
*1 F: yz-
3
G: xz-
3,4
H: xz 3,4
I: xy 3
J: x-y-z 4,5
K: xyz-
4,5
f1= yz
-+ ... f
2= z + ... f
3= ...
Reducing the table:
f
1
f
2
f
3m1
m5
m6
m1
m4
m6
cost
B: x 1
C: x-y-
3
D: y-z-
3
E: y-z 3
G: xz-
3,4
H: xz 3
I: xy 3
J: x-y-z 4,5
K: xyz-
4,5
f1= yz
-+ ... f
2= z + ... f
3= ...
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4.33. (continued)
After deleting equal row I and dominated rows D, H, and
K, the table becomes
f1
f2
f3
m1
m5
m6
m1
m4
m6
cost
B: x 1
C: x-y-
3
*1 E: y-z 3
*3 G: xz-
3,4
J: x-y-z 4,5
f1= yz
-+ y
-z f
2= z + ... f
3= xz
-+ ...
Reducing the table:
f2
f3
m6
m1
cost
B: x 1
C: x
-
y
-
3
G: xz-
1
J: x-y-z 4,5
f1= yz
-+ y-z f
2= z + ... f
3= xz
-+ ...
After deleting equal rows G and J, the table becomes
f2
f3
m6
m1
cost
*2 B: x 1
*3 C: x-y-
3
f1= yz
-+ y
-z f
2= z + x f
3= xz
-+ x-y-
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4.34.
w x y z (a) (b) (c) (d) (e) (f)
0 0 0 0 0 1 0 0 0 1
0 0 0 1 0 0 1 0 0 0
0 0 1 0 1 0 0 0 0 1
0 0 1 1 1 1 1 0 1 0
0 1 0 0 1 1 0 1 1 0
0 1 0 1 1 1 0 0 1 1
0 1 1 0 0 0 0 0 0 0
0 1 1 1 0 0 0 1 1 0
1 0 0 0 0 1 1 1 1 1
1 0 0 1 0 1 1 0 0 1
1 0 1 0 1 0 1 0 0 1
1 0 1 1 0 1 1 0 1 1
1 1 0 0 1 1 1 1 1 0
1 1 0 1 1 1 0 1 1 1
1 1 1 0 0 0 1 0 0 0
1 1 1 1 0 0 1 1 1 1
(a)
f = x-yz-+ xy
-+ w
-x-y
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4.34. (continued)
f = (w-+x+z
-)(x+y)(x
-+y-) f = (w
-+y-+z-)(x+y)(x
-+y-)
(b)
f = x-yz + y
-z-+ xy
-+ wy
-
f = (y-+z)(w+x+y+z
-)(x-+y-)
(c)
f = x-z + wz
-+ wy
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4.34. (continued)
f = (w+z)(x-+y+z
-)(w+x
-)
(d)
f = xyz + xy-z-+ wy
-z-+ wxy
-f = xyz + xy
-z-+ wy
-z-
+ wxz
f = (y-+z)(x+z
-)(w+y+z
-)(w+x)
(e)
f = yz + wy-z-+ xy
-
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4.34. (continued)
f = (y-+z)(x+y+z
-)(w+x+z) f = (y
-+z)(x+y+z
-)(w+x+y)
(f)
f = wz + xy-z + x
-z-
f = (x-+z)(w+x+z-)(w+x-+y-) f = (x-+z)(w+x+z-)(w+y-+z-)
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4.35.
w x y z (a) (b) (c) (d) (e)
0 0 0 0 - - 0 0 0
0 0 0 1 0 1 1 0 0
0 0 1 0 1 0 - 0 1
0 0 1 1 1 - - - 1
0 1 0 0 - - 0 0 1
0 1 0 1 1 1 1 1 0
0 1 1 0 0 1 - 1 0
0 1 1 1 0 1 1 1 -
1 0 0 0 - 0 0 - 0
1 0 0 1 0 1 0 - -
1 0 1 0 - 0 1 0 1
1 0 1 1 - 1 1 0 -
1 1 0 0 1 1 0 1 0
1 1 0 1 0 1 - 1 1
1 1 1 0 1 0 0 1 1
1 1 1 1 0 0 0 0 1
(a)
f = wz-+ x-y + w
-xy-
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4.35. (continued)
f = (w-+z-)(x+y)(w+x
-+y-)
(b)
f = x-z + w-x + xy-
f = (x+z)(w-+x-+y-)
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4.35. (continued)
(c)
f = w-z + x
-y
f = (w+z)(w-+x-)(w-+y)
f = (y+z)(w-+x-)(w-+y)
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4.35. (continued)
(d)
f = w-xz + xyz
-+ wy
-
f = x(w+y+z)(w-+y-+z-)
(e)
f = wz + w-xy-z-+ wy + x
-y
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4.35. (continued)
f = (w+x-+z-)(w-+y+z)(x+y)(w+x
-+y-)
f = (w+y+z-)(w-+y+z)(x+y)(w+x
-+y-)
4.36. (a)
f = Ax-z + Bxz
-+ x-yz
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4.36. (continued)
(b)
f = Ay-+ B-x-z + Bx
-y + xy
-z-
(c)
f = ABz + Ayz + A-x-y
4.37. (a)
f = yz-+ w-y + xz
-+ wx
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4.37. (continued)
(b)
f = v-wy-z-+ vxyz
-+ w-xz + v
-xy-
+ v-wx
(c)
f = x-yz + vz
-+ wy
-z