Boolean algebra, like any other axiomatic mathematical structure or algebraic system, can be characterized by specifying a number of fundamental things: 1. The domain of the algebra, that is, the set of elements over which the algebra is defined 2. A set of operations to be performed on the elements 3. A set of postulates, or axioms, accepted as premises without proof 4. A set of consequences called theorems, laws, or rules, which are deduced from the postulates
Trang 1www.uet.vnu.edu.vn/~tutx
Trang 2CÔNG
H Ọ C
NGH Ệ History
in his book (a treatise): “An Investigation of the Laws of Thought”
no application was made of Boolean Algebra until the late 1930s
Nakashima in Japan (1937) and Shannon at MIT (1938), each
independently applied the algebra of Boole to the analysis of networks of relays (in telephone systems)
Trang 3In 1904, Huntington found that
all of the results and implications of the algebra described by Boole could
be derived from only six basic postulates.
B is the set of elements or constants of the algebra
the symbols + and · are two binary operators(*)
the overbar ¯ is a unary operator(*)
is a Boolean algebra if the following hold true:
(*) The terms binary operator and unary operator refer to the number of arguments
involved in the operation: two or one, respectively.
Trang 4There exists a 0 element in B such that
for every element a in B, 0 + a = a + 0 = a
There exists a 1 element in B such that
for every element a in B, a = a · 1 = a
Trang 53. Commutativity (giao hoán)
Trang 6CÔNG
H Ọ C
NGH Ệ The Huntington postulates
5. ∀ a ∈ B, there exists an element a-bar in the set B such that
a a
Switching algebra is a Boolean algebra in which the number of
elements in the set B is precisely 2.
Trang 7is a Boolean algebra in which the number of elements in the set B is
precisely 2
The two binary operators, represented by the signs + and ·, are called
the OR and the AND, respectively.
The unary operator, represented by the overbar ¯ , is called the NOT or
the complement operator.
Trang 8a a
a
=
⋅
= +
a
a ⋅
a a
a a
a
a a a
a
a a
a
a a
a a a
a a
a a
a a
=
+
=
⋅ +
=
+
⋅ +
=
⋅ +
=
0
P-5(i)P-4(ii)P-5(ii)P-2(i)
Proof
(i)
(ii)
Trang 9Theorem 2
1 1
1
0 0
0
= +
= +
a a
) 0 (
0
0 0
⋅
=
⋅ +
=
a a
a a
a a
a
P-5(ii)P-4(i)P-2(i)P-5(ii)
P-2(ii)P-5(i)P-4(ii)P-2(ii)P-5(i)
Trang 10Assume that a has 2 distinct
complements (not equal), a-bar and b
Then by P-5, we must have that:
0
and
0 and
1
and
= +
a a b
a
a a
b
a
b a
b a
b a a
a
b a
a a
⋅
=
⋅ +
=
⋅ +
b a
a b
a b a
b
a a
b b
⋅
=
⋅ +
=
⋅ +
P-4(i)
P-2(i)
Trang 11From this point, we will restrict our attention to switching algebras only.
Switching algebra is basically a two-element Boolean algebra which,
obviously, has the two elements 0 and 1.
1 1
1
0 0
1
0 1
0
0 0
0
z = x · y y
x
Theorem 1(ii)Theorem 2(i)Postulate 3(ii)Theorem 1(ii)
Trang 121 0
1
1 1
0
0 0
0
z = x + y y
1 0
Trang 13NAND operation
0 1
1
1 0
1
1 1
0
1 0
0
z = x · y y
1
0 0
1
0 1
0
1 0
0
z = x + y y
Trang 141 0
1
1 1
0
0 0
0
z = x ⊕ y y
x y
x
- Construct XOR gate from NOT, AND, and OR ?
Trang 15NXOR operation
1 1
1
0 0
1
0 1
0
1 0
0
z = x ⊕ y y
x y
x
- Construct NXOR gate from NOT, AND, and OR ?
Trang 1601
0
x ( ) x
Since the left column is identical to the right column and we have listed all possibilities, we have proved the result
Trang 17Theorem 5
Let x and y be two switching variables Then
( x y ) x x
x y
x
x
= +
+ 1
1 1
1
1 0
+ 1
0 0
1
0 0
+ 0
0 1
0
0 0
+ 0
0 0
0
+
Trang 18x z
y
x
+ +
= +
y x
y x
x
⋅
= +
⋅
+
=
⋅ +
( x y ) ( x z ) ( y z ) ( x y ) ( x z )
z x y
x z
y z
x y
x
+
⋅ +
= +
⋅ +
⋅ +
⋅ +
⋅
=
⋅ +
⋅ +
⋅
(i) (ii)
(i) (ii)
(i) (ii)
Trang 19y x
Proof
This theorem may easily verified by complete enumeration
Exercise for Students
Trang 20CÔNG
H Ọ C
NGH Ệ Algebra Implications – De Morgan’s Theorem
x y
x
w z
y x
w z
y x
w z
y x
w z
y x
⋅
⋅ +
⋅
=
⋅ +
⋅
=
⋅ +
⋅
=
+ +
Trang 21Corollary of De Morgan’s Theorem
n n
x x
x x
x x
x x
x x
x x
+ +
+
+
L L
L L
21
21
21
21
Expand the following function:
( x ⋅ y + z ⋅ y + w ⋅ z ⋅ x + z )
Trang 22each assignment This list is called Truth Table.
For example,
z y x
z y x
F ( , , ) = +
Trang 230 1
1 1
1 0
1 1
0 1
0 1
0 0
0 1
1 1
1 0
1 0
1 0
1 1
0 0
1 0
0 0
F (x, y, z) z
y x
z y x
z y x
- Write the Truth Table for the following switching function: F (x, y, z) = x ⋅ y ⋅ z + x ⋅ y
2n assignment
possibilities
Trang 240 0
1
0 1
0
0 0
0
F (x, y) y
x
F (x, y) ???
Trang 25Alternatives to the tabular representation
To list only the assignments for which a function is 1 or,
To list those for which the function is 0
F = 1 whenever x = 1 and y = 1
1 1
1
0 0
1
0 1
0
0 0
0
F (x, y) y
x
xy y
x
F ( , ) =
Trang 26CÔNG
H Ọ C
NGH Ệ Canonical forms: Minterm - Maxterm
0 1
1 1
0 0
1 1
1 1
0 1
0 0
0 1
0 1
1 0
1 0
1 0
1 1
0 0
0 0
0 0
F (x, y, z) z
y x
+ +
=
If a product term involves all the variables of a function, it
is referred to as a Minterm
A SOP is made of minterms only
is called a canonical minterm expression or expansion of the
function F(x, y, z)
Trang 27Sum of Products (SOP) expression
0 1
1 1
1 0
1 1
0 1
0 1
0 0
0 1
1 1
1 0
1 0
1 0
1 1
0 0
1 0
0 0
F (x, y, z) z
y x
z y x
z y yz
x y
x
x x
z y yz
x z
z y x
z xy yz
x z
y x z
y x z
y x
z y x F
+
=
+ +
=
+ +
+ +
=
+ +
+ +
=
=
) (
) (
) , , (
Write the SOP expression and simplify the function !
Trang 28CÔNG
H Ọ C
NGH Ệ Canonical forms: Minterm - Maxterm
0 1
1 1
1 0
1 1
0 1
0 1
0 0
0 1
1 1
1 0
1 0
1 0
1 1
0 0
1 0
0 0
F (x, y, z) z
y x
xyz z
y x z
y x z
y x
z y
x z
y x
z y
x
xyz z
y x z y x
xyz z
y x z
y x
z y x F z
y x F
+ +
+ +
+ +
=
=
=
) )(
)(
(
) , , ( )
, , (
Trang 29Product of Sums (POS) expression
0 1
1 1
1 0
1 1
1 1
0 1
0 0
0 1
0 1
1 0
0 0
1 0
1 1
0 0
0 0
0 0
F (x, y, z) z
y x
) )(
(
) )(
)(
(
) , , ( )
, , (
z y
x z y
x
z y
x z y
x z y
x
xyz z
y x yz
x z
y x z
y x
z y x F z
y x F
+ +
+ +
+ +
+ +
+ +
=
+ +
+ +
Trang 30A C
B A
F ( , , ) = +
C B A ABC
C B A BC
A
C B B
A C
C B
A
C A
B
A
+ +
+
=
+ +
+
=
⋅
⋅ +
⋅
=
) (
) (
1 1
] [concensus
) )(
(
) )(
)(
(
) )(
)(
( 1
) )(
)(
)(
(
) )(
(
C A
B A
C B
C A
B A
C B
C A
B A
B C
A C
B A
A A
C B
A A B
A
AC B
A
+ +
=
+ +
+
=
+ +
+
⋅
=
+ +
+ +
=
+ +
=
+
=
4(ii)-P ))(
C C B A
B A
B
A
+++
+
=
++
=
++
=
+
+ +
+ +
+ +
+ +
=
Trang 3110/8/2010 Xuan-Tu Tran 31
A + B + C = M 0
A B C =m 7 C
B A
A + B + C = M 1
A B C =m 6 C
B A
A + B + C = M 2
A B C =m 5 C
B A
A + B + C = M 3
A B C =m 4 C
B A
A + B + C = M 4
A B C =m 3 C
B A
A + B + C = M 5
A B C =m 2 C
B A
A + B + C = M 6
A B C =m 1 C
B A
A + B + C = M 7
A B C =m 0 C
B A
C B
A
Maxterm (M) Minterm (m)
Variable
M = m
Trang 32A C
B A
C B A ABC
C B A BC
A
C B B
A C
C B
A
C A
B
A
+ +
+
=
+ +
+
=
⋅
⋅ +
⋅
=
) (
) (
1 1
) )(
)(
)(
( ) , ,
) 5 7 2 3 (
M m
Trang 33Construct a XOR gate from NOR and NOT
gates ?
Construct a XOR gate from NAND and
NOT gates ?
y x y
x y
x
Trang 34CÔNG
H Ọ C
NGH Ệ Simplification of Switching Functions
Related to the number of terms and the number of literals in each term of the expression
The complexity of an equation should be reduced
expressions
Use postulates and other results to reduce the form of an expression
algebraically (require a good deal of experience)
Use Karnaugh map (group of minterms are easily identified, limited
variables)
Trang 35Algebraic Manipulation
Simplified expression: contains a minimal number of literals and terms
Minimal SOP, Minimal POS
All theorems and postulates are used to find the minimal form
Depend on your experience !!!
3 basic results are usually used for minimizing SOP expressions:
xy z
x yz
xy z
x
y x
y x x
y y
x xy
+
= +
+
+
= +
= +
: 3 Result
: 2 Result
: 1 Result (Verified using Distributivity)
Theorem 6, part 2(i)
Theorem 6, part 3(i) (consensus)
Trang 36CÔNG
H Ọ C
NGH Ệ Simplification of Switching Functions
Example:
wxy xyz
xz w z
x w z
y x w
wxy z
w
xy z
xy w
z w
wxy xyz
z w x z
w x
+
=
+ +
=
+ +
+
=
) (
) (
) (
) (
xy z
x yz xy
z x
y x y x x
y y x xy
+
= +
+
+
= +
= +
: 3 Result
: 2 Result
: 1 Result
Trang 37Algebraic Manipulation (cont’d)
Example (for students):
D C B A CD
A BCD
ABC C
AB D
C B A
D C A CD
A AB
D C A CD
B CD
A B
A
CD A
BCD D
C A AB
CD A
BCD D
C B
A
CD A
BCD D
C B B
A
D C B A CD
A BCD
AB
D C B A CD
A BCD
C AB C
AB
+ +
=
+ +
+
=
+ +
+
=
+ +
+
=
+ +
+
=
+ +
+
=
+ +
+ +
=
)]
( )
( )
( [
] [
)]
( [
] ) (
) [(
xy z
x yz xy
z x
y x y x x
y y x xy
+
= +
+
+
= +
= +
:
3
Result
:
Trang 38CÔNG
H Ọ C
NGH Ệ Simplification of Switching Functions
Algebraic Manipulation (cont’d)
Example:
z z
w y
z x
z z
w y y
z x x
z x z
w y z
y
x
+ +
+ +
+
=
+ +
+ +
+
= +
+ +
+
+
)]
( [
) (
) (
None of the listed results can be applied for the above expression
It’s OK or NOT yet ?
xy z
x yz xy
z x
y x y x x
y y x xy
+
=+
+
+
=+
=+
Trang 39Prime implicants
An implicant is a "covering" (sum term or product term) of one or more
minterms in a sum of products (or maxterms in a product of sums) of a boolean function
In an SOP expression, each of the product terms an implicant of the function if it implies the function (e.g., if the product term is 1 then the
function is also 1).
Trang 40xyz yz
x w xyz
yz x
w
+
=
+ +
+
= ( ) ( )
wxyz yz
x w xyz
w yz
x w z
y x w
Each of the minterm is an implicant
Each of these products is also an implicant
yz
= is also an implicant.
Smaller in term of the number of literals
A prime implicant of a function is an implicant that cannot be covered by a
Trang 41Minimal expression
Any other expression having fewer terms and literals will not represent the original function
Trang 42M Karnaugh published an article describing a geometrical method for
finding a minimal closed cover (1953) Karnaugh map method
Based on mapping minterms onto a surface in such a way that minterms that differ in one literal are adjacent to each other on the surface
Reason: when 2 minterms differ in one literal, they can be combined to form
a product term which has this literal missing
For example, Ā BC and ABC differ in only one literal ( Ā and A) the sum reduces to A BC + ABC = BC
Trang 43Karnaugh map
All minterms in the column labeled B contain the
literal B, all those in the other column contain
B-bar Similar for the rows
1 in each square corresponds to a minterm of the
function f
Prime implicants can be found by grouping the 1
cells into as large a block of adjacent cells as possible.
For example, the pair of cells and
group together to give
B A
B
) , ( A B f
B A
B A
A A
B
B A B
Trang 441-cube: a single square
2-cube: two adjacent squares are taken together
4-cube: two adjacent 2-cubes or two 2-cubes
have a long edge in common.
The largest cube of 1s represent the prime
B A
B A
B A
f ( , ) = +
A
Trang 45Karnaugh map for 3 variables
Two 2-variable Karnaugh map are placed side by side
after reflecting one of the two 2-variable maps get a 3-variable map
z y x
z xy
xyz z
y x
z y x
Trang 46CÔNG
H Ọ C
NGH Ệ Simplification of Switching Functions
Karnaugh map for 4 variables
Two 3-variable Karnaugh map are
placed side by side after reflecting one of the two 3-variable maps
get a 4-variable map
How to find prime implicants for the
w
y