1. 3-7 Other Two-level Implementations

2. NAND and NOR logic implementation
the most important
from a practical point of view
Wired logic
1) the open-collector TTL NAND gates
a wired-AND logic Chap.10, ? Fig.10-11
? Fig.3-22(a)
F = (AB)’(CD)’= (AB + CD)’
AND-OR-INVERT function

3. 2) the NOR output of ECL gates
a wired-OR logic
? Fig.3-22(b)
F = (A+B)’+ (C+D)’= ((A+B)(C+D))’
OR-AND-INVERT function
Nondegenerate Forms
Two-level combinations of gates
AND, OR, NAND, NOR
16 possible combinations of two-level forms

4. 8 degenerate forms
degenerate to a single operation
Ex) AND-AND ==> AND of all input variables
OR-OR ==> OR
AND-NAND ==> NAND
OR-NOR ==> NOR
NAND-OR ==> AND-NAND ==> NAND
NAND-NOR ==> AND-AND ==> AND
NOR-AND ==> OR-NOR ==> NOR
NOR-NAND ==> OR-OR ==> OR

5. 8 nondegerate forms
an implementation in sum of products or
product of sums
dual
AND-OR OR-AND basic forms
NAND-NAND NOR-NOR the most practical
forms
NOR-OR NAND-AND
OR-NAND AND-NOR
AND-OR-INVERT implementation
NAND-AND and AND-NOR
equivalent forms of the AND-OR-INVERT

6. function ?
Fig.3-23 F = (AB+CD+E)’
(a) AND-NOR sum of product + invert
(b) AND-NOR
(c) NAND-AND
OR-AND-INVERT implementation
OR-NAND and NOR-OR
equivalent forms of the OR-AND-INVERT

7. function ?
Fig.3-24 F = ((A+B)(C+D)E)’
(a) OR-NAND product of sums + invert
(b) OR-NAND
(c) NOR-OR
Tabular Summary and Example
Table 3-4 implementation
? with other Two-level forms
AND-NOR and NAND-AND
OR-NAND and NOR-OR
AND-OR-INVERT form
F’ F
F’= sum of products form
for 0’s in the map

8. ∑ OR-NAND and NOR-OR OR-AND-INVERT form ---------- -------- F’ F
F’= product of sums forms
F = sum of products form
for 1’s in the map
Ex 3-11 ?
Implement the function of Fig.3-19(a)
( F(x,y,z) = ∑
(0,6) )

9. with the 4 two-level forms in Table 3-4 ?
F’= x’y + xy’+ z for 0’s in the map
F = (x’y + xy’+ z)’
in AND-OR-INVERT form
AND-NOR and NAND-AND implementation
? Fig.3-25(a)
F = x’y’z’+ xyz’for 1’s in the map
F’= (x + y + z)(x’+ y’+ z)
F = ((x + y + z)(x’+y’+z))’
in OR-AND-INVERT form

10. ∑ OR-NAND and NOR-OR implementation ? Fig.3-25(b)
3-8 Don’t-Care Conditions
don’t care condition
the unspecified minterms whose outputs
can be either 0 or 1, marked with X
==> used for further simplification
Ex 3-12 Simplify the Boolean Function
F(w,x,y,z) = ∑

11. ∑ (1,3,7,11,15) d(w,x,y,z) = (0,2,5) ? Fig.3-26 (a) F = yz + w’x’
(b) F = yz + w’z
the smallest literals and smallest terms
The choice
between 0 or 1 of don’t care term X
depends on the simplicity
Ex 3-12 F(w,x,y,z) = yz + w’x’ =

12. ∑ (1,3,5,7,11,15) Fig.3-26(b) (0,1,2,3,7,11,15) F(w,x,y,z) = yz + w’z
Both expressions
include minterms 1,3,7,11 and 15 of
minterm ‘1’
? Fig.3-26
F’= z’+ wy’ for 0’s
F(w,x,y,z) = (z’+ wy’)’
= z(w’+ y) ?
(1,3,5,7,11,15) Fig.3-26(b)

13. 3-9 The Tabulation Method
The map method of simplification
convenient for 3,4,5 numbers of variable
for reasonable selection of adjacent squares
disadvantage
rely on the recognition ability
of certain patterns
through trial-and-error procedure

14. The tabulation mathod
Quine-Macluskey method
a guaranteed specific step-by-step procedure
for a simplified standard-form expression
suitable for machine computation
for problems with many variables
tedious for human use
prone to mistakes because of routine and
monotonous process

15. The tabula method
1) find by an exhaustive search
all the terms(prime implicants )
that are candidates for inclusion
in the simplified function
2) choose among the prime implicants
for an expression
with the least number of literals
3-10 Determination of prime implicants

16. Find the prime implicants
by using a matching process
1) compare each minterm
with every other minterm
2) If two minterms differ in only one variable,
that variable is removed
(a term with one less literal is found)
3) repeat until no further elimination
of literals

17. ∑ The prime implicants The remaining terms and all the terms
that did not match during the process
Ex 3-13 Simplify the following Boolean function
by using the tabulation method
F = ∑
(0,1,2,8,10,11,14,15) ?
step 1 : Table 3-5(a) minterms of 1’s
group into 5 sections

18. The 1st section, the number with no 1’s
The 2nd section, the number with one 1’s
The 3rd section, the number with two 1’s
The 4th section, the number with three 1’s
The 5th section, the number with four 1’s ?
step 2 : Table 3-5(b)
combine two minterms
that differ by only one variable
(compare the next section down only)

19. In the same section
more than one variable differs
두 숫자가 같지 않으면, minterm1이 1일 때,
minterm2가 0이고 minterm1이 0일 때,
minterm2가 1인 두쌍이 반드시 하나 이상있음
예) minterm1 :
minterm2 :
remove the unmatched variable of (a)
place a check mark to show
that they have been used
step 3 : The terms of column (b)
has only 3 variables

20. ? Table 3-5(c) with only 2 variables no check mark of 000-
because 000- does not match
with any other term
step 4 : the prime implicants
the unchecked terms in the table
because each checked terms
are changed into a simpler from
in the subsequent column
w’x’y’(000-) in column(b)
x’z’(-0-0) and wy(1-1-) in column(c)

21. A simplified expression
the sum of the prime implicants
F = w’x’y’+ x’z’+ wy
The map method, ? Fig.3-27
The sum of prime implicants
≠ the expression
with minimum number of terms
Ex 3-14
Comparing with decimal number
( in stead of binary )

22. When two minterms differ
by only one position,
the difference is a power of 2
Ex) (25) (17)
(8 = 23)
Table 3-6 ?
Example 3-13 with Decimal Notation
a) Group the minterms
b) Write two numbers whose difference
is a power of 2(1,2,4,8,16...)

23. Mark two numbers at right of (a)
The number in the ( )
the position of -
c) Compare only those terms
with the same number in the ( ) 0,2(2) 쭽
8 쭾 =====> 0,2,8,10(2,8)
8,10(2)
0,8(8)
2 쭾 =====> 0,2,8,10(2,8)
2,10(8)
10,11(1)
4 쭾 =====> 10,11,14,15(1,4)
14,15(1)
10,14(4)
1 쭾 =====> 10,11,14,15(1,4)
11,15(4)

24. The prime implicants
terms not checked in the table
Conversion to binary
0,1(1) ==> 0000,
0,2,8,10(2,8)
==> 0000
1000
Ex 3-14

25. ∑ Determine the prime implicants of the function F(w,x,y,z) =
(1,4,6,7,8,9,11,15)
Table 3-7 ?
The prime implicants
x’y’z, w’xz’, w’xy, xyz, wyz, wx’
F = x’y’z + w’xz’+ w’xy
+ xyz + wyz + wx’
not necessarily minimum number of terms
implicants marked in Table 3-8 ?

26. The map method
F = x’y’z + w’xz’+ xyz + wx’
3-11 Selection of Prime Implicants
Place X’s in each row
to show the composition of minterms
that make the prime implicants Choose a minimum set of prime implicants
that cover all the minterms
in the function
Ex 3-15
Minimize the function of Example 3-14 ?

27. Table 3-8 Prime Implicant Table
x’y’z is from 1,9 marked with X.
Inspect for column containing
only a single X
1, 4, 8, 10
Essential prime implicants
x’y’z, w’xz’, wx’
prime implicants with a single X
should be included

28. Choose the prime implicants ( xyz ) which cover the other minterms
than those covered
by essential prime implicants
marked in Table 3-8 ?
F = x’y’z + w’xz’+ xyz + wx’
The tabulation process with 1’s
the sum of products form
The tabulation process with 0’s
the product of sums form
A function with don’t-care conditions
the don’t-care terms included in the list of minterms for the determination of prime implicants

29. allows the derivation of prime implicants
with the least number of literals
not included in the prime implicant table
( because the don’t-care terms
do not have to be covered )
3-12 Concluding Remarks
Boolean-function simplification
minimization of the number of literals
of Boolean function in standard forms
( restriction )
sum of products form
product of sums form
map method
tabulation method

30. From Fig.3-15 ?
standard forms ==> no more than
two levels of gates
nonstandard forms ==> more than
the Gray-code sequence for the maps
a change in only one bit
between adjacent squares ?
Variations of the three-variable map Fig.3-29
Variations of the four-variable map Fig.3-30
(a) very popular ?
(b) original Veitch diagram
modified to (a) by Karnaugh

31. The tabulation method
error-prone in comparing numbers
over long lists
step-by-step procedure
suitable for computer mechanization
The map method
for more than 5 variables,
not the best simplified expression
For the functions with multiple outputs,
further simplification is possible
because of common gates
an extension of the tabulation method
for multiple-output circuits
practically important
only if a computer program of
the method is available
(because of tediousness
for human manipulation)