1. Representing Boolean functions
Section 10.2
Representing Boolean functions

2. Boolean function representation: 2 important concepts
Any Boolean function may be represented by a Boolean sum of Boolean products of the variables and their complements
All Boolean functions can be represented using only 1 operator
This section concerns itself with illustrating these concepts

3. Example 1
Construct Boolean expressions representing functions with the given values:
x y z F G
F has value 1 only when x=1, y=1, and z=0; one
such expression is the Boolean product xyz
G has value 1 when either x and z are 1, y is 0 or
when x and z are 0, y is 1
The first condition is met by xyz
The second is met by xyz
So G can be represented by the Boolean sum of
the 2 products:
x y z + x y z

4. Literals & minterms Literal: Boolean variable or its complement
For Boolean variables x1, x2, … xn the minterm is a Boolean product y1y2…yn where yi = xi or yi = xi
A minterm is a product of literals, with one literal for each variable

5. Minterms
A minterm has value 1 for exactly one combination of values of its variables:
Minterm y1y2…yn is 1 if and only if each yi=1 and this occurs if and only if
xi = 1 when yi = xi and xi = 0 when yi = xi
Example: find a minterm that equals 1 when x1 and x3 are 0 and x2, x4 and x5 are 1, and 0 otherwise
Solution: x1x2x3x4x5

6. Sum of products expansion
By taking Boolean sums of distinct minterms, we can build a Boolean expression with a specified set of values
Boolean sum of minterms has value 1 when exactly one of the minterms in the sum has value 1, and 0 for all other combinations of values of the variables
Given a Boolean function, a Boolean sum of minterms can be found with value 1 when the function is 1 and 0 when the function is 0; this is called the sum-of-products expansion or disjunctive normal form of the Boolean function

7. Example 2 Find the sum-of-products expansion for F(x,y,z) = (x+y)z
is the Boolean sum of 3 minterms
corresponding to the rows that
show 1 for the values of F:
xyz + xyz + xyz
x y z x+y z (x+y)z

8. Conjunctive normal form
Can find a Boolean expression that represents a Boolean function by taking a Boolean product of Boolean sums - this is the conjunctive normal form or product-of-sums expansion
We can find the product-of-sums from sum-of-products by taking duals; for function on previous slide, product-of-sums is:
(x + y + z)(x + y + z)(x + y + z)

9. Functional completeness
Every Boolean function can be expressed as a Boolean sum of minterms
Each minterm is the Boolean product of Boolean variables or their complements
So every Boolean function can be represented using Boolean operators ., +, and _
We say that the set {., +, _} is functionally complete

10. Functional Completeness
If one of the operators from the set {., +, _} can be expressed in terms of the other two, we can find a smaller set of functionally complete operators
We can do this using DeMorgan’s law:
x + y = x y
(We obtain this by taking complements of both sides of DeMorgan’s second law, then applying double complementation)

11. Functional completeness
Using DeMorgan’s law, as stated on the previous slide, we can eliminate sums; this means the set { ., _ } is functionally complete
Using DeMorgan’s first law we can similarly show that the set { +, _ } is functionally complete

12. Functionally complete sets containing one operator
12/8/2018
Functionally complete sets containing one operator
NAND: {|} (not and)
1|1 = 0, 1|0 = 1, 0|1 = 1, 0|0 = 1
NOR: {} (not or)
11 = 0, 10 = 0, 01 = 0, 00 = 1
NAND is functionally complete, since:
{ ., _ } is functionally complete and
x = x|x and
xy = (x|y)|(x|y)
We can similarly show that NOR is functionally complete
ch10.2