1. Circuit analysis summary
After finding the circuit inputs and outputs, you can come up with either an expression or a truth table to describe what the circuit does.
You can easily convert between expressions and truth tables.
Find the circuit’s
inputs and outputs
Find a Boolean
expression
for the circuit
Find a truth table
for the circuit
CS231 Boolean Algebra

2. Boolean operations summary
We can interpret high or low voltage as representing true or false.
A variable whose value can be either 1 or 0 is called a Boolean variable.
AND, OR, and NOT are the basic Boolean operations.
We can express Boolean functions with either an expression or a truth table.
Every Boolean expression can be converted to a circuit.
Nex, we’ll look at how Boolean algebra can help simplify expressions, which in turn will lead to simpler circuits.
CS231 Boolean Algebra

3. Boolean algebra
Last time we talked about Boolean functions, Boolean expressions, and truth tables.
Today we’ll learn how to how use Boolean algebra to simplify Booleans expressions.
Last time, we saw this expression and converted it to a circuit:
(x + y’)z + x’
Can we make this circuit “better”?
Cheaper: fewer gates
Faster: fewer delays from inputs to outputs
June 11, 2002
© Howard Huang

4. Expression simplification
Normal mathematical expressions can be simplified using the laws of algebra
For binary systems, we can use Boolean algebra, which is superficially similar to regular algebra
There are many differences, due to
having only two values (0 and 1) to work with
having a complement operation
the OR operation is not the same as addition
CS231 Boolean Algebra

5. Formal definition of Boolean algebra
A Boolean algebra requires
A set of elements B, which needs at least two elements (0 and 1)
Two binary (two-argument) operations OR and AND
A unary (one-argument) operation NOT
The axioms below must always be true (textbook, p. 33)
The magenta axioms deal with the complement operation
Blue axioms (especially 15) are different from regular algebra
CS231 Boolean Algebra

6. Comments on the axioms
The associative laws show that there is no ambiguity about a term such as x + y + z or xyz, so we can introduce multiple-input primitive gates:
The left and right columns of axioms are duals
exchange all ANDs with ORs, and 0s with 1s
The dual of any equation is always true
CS231 Boolean Algebra

7. Are these axioms for real?
We can show that these axioms are valid, given the definitions of AND, OR and NOT
The first 11 axioms are easy to see from these truth tables alone. For example, x + x’ = 1 because of the middle two lines below (where y = x’)
CS231 Boolean Algebra

8. Proving the rest of the axioms
We can make up truth tables to prove (both parts of) DeMorgan’s law
For (x + y)’ = x’y’, we can make truth tables for (x + y)’ and for x’y’
In each table, the columns on the left (x and y) are the inputs. The columns on the right are outputs.
In this case, we only care about the columns in blue. The other “outputs” are just to help us find the blue columns.
Since both of the columns in blue are the same, this shows that (x + y)’ and x’y’ are equivalent
CS231 Boolean Algebra

9. Simplification with axioms
We can now start doing some simplifications
x’y’ + xyz + x’y
= x’(y’ + y) + xyz [ Distributive; x’y’ + x’y = x’(y’ + y) ]
= x’1 + xyz [ Axiom 7; y’ + y = 1 ]
= x’ + xyz [ Axiom 2; x’1 = x’ ]
= (x’ + x)(x’ + yz) [ Distributive ]
= 1  (x’ + yz) [ Axiom 7; x’ + x = 1 ]
= x’ + yz [ Axiom 2 ]
CS231 Boolean Algebra

10. Let’s compare the resulting circuits
Here are two different but equivalent circuits.
In general the one with fewer gates is “better”:
It costs less to build
It requires less power
But we had to do some work to find the second form
CS231 Boolean Algebra

11. Some more laws
Here are some more useful laws (p. 37). Notice the duals again!
We can prove these laws by either
Making truth tables:
Using the axioms:
x + x’y = (x + x’)(x + y) [ Distributive ]
= 1  (x + y) [ x + x’ = 1 ]
= x + y [ Axiom 2 ]
CS231 Boolean Algebra

12. The complement of a function
The complement of a function always outputs 0 where the original function outputted 1, and 1 where the original produced 0.
In a truth table, we can just exchange 0s and 1s in the output column(s)
f(x,y,z) = x(y’z’ + yz)
CS231 Boolean Algebra

13. Complementing a function algebraically
You can use DeMorgan’s law to keep “pushing” the complements inwards
You can also take the dual of the function, and then complement each literal
If f(x,y,z) = x(y’z’ + yz)…
…the dual of f is x + (y’ + z’)(y + z)…
…then complementing each literal gives x’ + (y + z)(y’ + z’)…
…so f’(x,y,z) = x’ + (y + z)(y’ + z’)
f(x,y,z) = x(y’z’ + yz)
f’(x,y,z) = ( x(y’z’ + yz) )’ [ complement both sides ]
= x’ + (y’z’ + yz)’ [ because (xy)’ = x’ + y’ ]
= x’ + (y’z’)’ (yz)’ [ because (x + y)’ = x’ y’ ]
= x’ + (y + z)(y’ + z’) [ because (xy)’ = x’ + y’, twice]
CS231 Boolean Algebra

14. Standard forms of expressions
We can write expressions in many ways, but some ways are more useful than others
A sum of products (SOP) expression contains:
Only OR (sum) operations at the “outermost” level
Each term that is summed must be a product of literals
The advantage is that any sum of products expression can be implemented using a two-level circuit
literals and their complements at the “0th” level
AND gates at the first level
a single OR gate at the second level
This diagram uses some shorthands…
NOT gates are implicit
literals are reused
this is not okay in LogicWorks!
f(x,y,z) = y’ + x’yz’ + xz
CS231 Boolean Algebra

15. Minterms
A minterm is a special product of literals, in which each input variable appears exactly once.
A function with n variables has 2n minterms (since each variable can appear complemented or not)
A three-variable function, such as f(x,y,z), has 23 = 8 minterms:
Each minterm is true for exactly one combination of inputs:
x’y’z’ x’y’z x’yz’ x’yz
xy’z’ xy’z xyz’ xyz
Minterm Is true when… Shorthand
x’y’z’ x=0, y=0, z=0 m0
x’y’z x=0, y=0, z=1 m1
x’yz’ x=0, y=1, z=0 m2
x’yz x=0, y=1, z=1 m3
xy’z’ x=1, y=0, z=0 m4
xy’z x=1, y=0, z=1 m5
xyz’ x=1, y=1, z=0 m6
xyz x=1, y=1, z=1 m7
CS231 Boolean Algebra

16. Sum of minterms form
Every function can be written as a sum of minterms, which is a special kind of sum of products form
The sum of minterms form for any function is unique
If you have a truth table for a function, you can write a sum of minterms expression just by picking out the rows of the table where the function output is 1.
f = x’y’z’ + x’y’z + x’yz’ + x’yz + xyz’
= m0 + m1 + m2 + m3 + m6
= m(0,1,2,3,6)
f’ = xy’z’ + xy’z + xyz
= m4 + m5 + m7
= m(4,5,7)
f’ contains all the minterms not in f
CS231 Boolean Algebra

17. The dual idea: products of sums
Just to keep you on your toes...
A product of sums (POS) expression contains:
Only AND (product) operations at the “outermost” level
Each term must be a sum of literals
Product of sums expressions can be implemented with two-level circuits
literals and their complements at the “0th” level
OR gates at the first level
a single AND gate at the second level
Compare this with sums of products
f(x,y,z) = y’ (x’ + y + z’) (x + z)
CS231 Boolean Algebra

18. Maxterms
A maxterm is a sum of literals, in which each input variable appears exactly once.
A function with n variables has 2n maxterms
The maxterms for a three-variable function f(x,y,z):
Each maxterm is false for exactly one combination of inputs:
x’ + y’ + z’ x’ + y’ + z x’ + y + z’ x’+ y + z
x + y’ + z’ x + y’ + z x + y + z’ x + y + z
Maxterm Is false when… Shorthand
x + y + z x=0, y=0, z=0 M0
x + y + z’ x=0, y=0, z=1 M1
x + y’ + z x=0, y=1, z=0 M2
x + y’ + z’ x=0, y=1, z=1 M3
x’ + y + z x=1, y=0, z=0 M4
x’ + y + z’ x=1, y=0, z=1 M5
x’ + y’ + z x=1, y=1, z=0 M6
x’ + y’ + z’ x=1, y=1, z=1 M7
CS231 Boolean Algebra

19. Product of maxterms form
Every function can be written as a unique product of maxterms
If you have a truth table for a function, you can write a product of maxterms expression by picking out the rows of the table where the function output is 0. (Be careful if you’re writing the actual literals!)
f = (x’ + y + z)(x’ + y + z’)(x’ + y’ + z’)
= M4 M5 M7
= M(4,5,7)
f’ = (x + y + z)(x + y + z’)(x + y’ + z)
(x + y’ + z’)(x’ + y’ + z)
= M0 M1 M2 M3 M6
= M(0,1,2,3,6)
f’ contains all the maxterms not in f
CS231 Boolean Algebra

20. Minterms and maxterms are related
Any minterm mi is the complement of the corresponding maxterm Mi
For example, m4’ = M4 because (xy’z’)’ = x’ + y + z
Minterm Shorthand
x’y’z’ m0
x’y’z m1
x’yz’ m2
x’yz m3
xy’z’ m4
xy’z m5
xyz’ m6
xyz m7
Maxterm Shorthand
x + y + z M0
x + y + z’ M1
x + y’ + z M2
x + y’ + z’ M3
x’ + y + z M4
x’ + y + z’ M5
x’ + y’ + z M6
x’ + y’ + z’ M7
CS231 Boolean Algebra

21. Converting between standard forms
We can convert a sum of minterms to a product of maxterms
In general, just replace the minterms with maxterms, using maxterm numbers that don’t appear in the sum of minterms:
The same thing works for converting from a product of maxterms to a sum of minterms
From before f = m(0,1,2,3,6)
and f’ = m(4,5,7)
= m4 + m5 + m7
complementing (f’)’ = (m4 + m5 + m7)’
so f = m4’ m5’ m7’ [ DeMorgan’s law ]
= M4 M5 M7 [ By the previous page ]
= M(4,5,7)
f = m(0,1,2,3,6)
= M(4,5,7)
CS231 Boolean Algebra

22. Summary so far
So far:
A bunch of Boolean algebra trickery for simplifying expressions and circuits
The algebra guarantees us that the simplified circuit is equivalent to the original one
Introducing some standard forms and terminology
Next:
An alternative simplification method
We’ll start using all this stuff to build and analyze bigger, more useful, circuits
CS231 Boolean Algebra

23. Karnaugh maps
So far, we saw applications of Boolean logic to circuit design.
The basic Boolean operations are AND, OR and NOT.
These operations can be combined to form complex expressions, which can also be directly translated into a hardware circuit.
Boolean algebra helps us simplify expressions and circuits.
Next, we’ll look at a graphical technique for simplifying an expression into a minimal sum of products (MSP) form:
There are a minimal number of product terms in the expression.
Each term has a minimal number of literals.
Circuit-wise, this leads to a minimal two-level implementation.
CS231 Boolean Algebra

24. Review: Standard forms of expressions
We can write expressions in many ways, but some ways are more useful than others
A sum of products (SOP) expression contains:
Only OR (sum) operations at the “outermost” level
Each term that is summed must be a product of literals
The advantage is that any sum of products expression can be implemented using a two-level circuit
literals and their complements at the “0th” level
AND gates at the first level
a single OR gate at the second level
This diagram uses some shorthands…
NOT gates are implicit
literals are reused
this is not okay in LogicWorks!
f(x,y,z) = y’ + x’yz’ + xz
CS231 Boolean Algebra

25. Terminology: Minterms
A minterm is a special product of literals, in which each input variable appears exactly once.
A function with n variables has 2n minterms (since each variable can appear complemented or not)
A three-variable function, such as f(x,y,z), has 23 = 8 minterms:
Each minterm is true for exactly one combination of inputs:
x’y’z’ x’y’z x’yz’ x’yz
xy’z’ xy’z xyz’ xyz
Minterm Is true when… Shorthand
x’y’z’ x=0, y=0, z=0 m0
x’y’z x=0, y=0, z=1 m1
x’yz’ x=0, y=1, z=0 m2
x’yz x=0, y=1, z=1 m3
xy’z’ x=1, y=0, z=0 m4
xy’z x=1, y=0, z=1 m5
xyz’ x=1, y=1, z=0 m6
xyz x=1, y=1, z=1 m7
CS231 Boolean Algebra

26. Terminology: Sum of minterms form
Every function can be written as a sum of minterms, which is a special kind of sum of products form
The sum of minterms form for any function is unique
If you have a truth table for a function, you can write a sum of minterms expression just by picking out the rows of the table where the function output is 1.
f = x’y’z’ + x’y’z + x’yz’ + x’yz + xyz’
= m0 + m1 + m2 + m3 + m6
= m(0,1,2,3,6)
f’ = xy’z’ + xy’z + xyz
= m4 + m5 + m7
= m(4,5,7)
f’ contains all the minterms not in f
CS231 Boolean Algebra

27. Re-arranging the truth table
A two-variable function has four possible minterms. We can re-arrange these minterms into a Karnaugh map.
Now we can easily see which minterms contain common literals.
Minterms on the left and right sides contain y’ and y respectively.
Minterms in the top and bottom rows contain x’ and x respectively.
CS231 Boolean Algebra

28. Karnaugh map simplifications
Imagine a two-variable sum of minterms:
x’y’ + x’y
Both of these minterms appear in the top row of a Karnaugh map, which means that they both contain the literal x’.
What happens if you simplify this expression using Boolean algebra?
x’y’ + x’y = x’(y’ + y) [ Distributive ]
= x’  1 [ y + y’ = 1 ]
= x’ [ x  1 = x ]
CS231 Boolean Algebra

29. More two-variable examples
Another example expression is x’y + xy.
Both minterms appear in the right side, where y is uncomplemented.
Thus, we can reduce x’y + xy to just y.
How about x’y’ + x’y + xy?
We have x’y’ + x’y in the top row, corresponding to x’.
There’s also x’y + xy in the right side, corresponding to y.
This whole expression can be reduced to x’ + y.
CS231 Boolean Algebra

30. A three-variable Karnaugh map
For a three-variable expression with inputs x, y, z, the arrangement of minterms is more tricky:
Another way to label the K-map (use whichever you like):
CS231 Boolean Algebra

31. Why the funny ordering?
With this ordering, any group of 2, 4 or 8 adjacent squares on the map contains common literals that can be factored out.
“Adjacency” includes wrapping around the left and right sides:
We’ll use this property of adjacent squares to do our simplifications.
x’y’z + x’yz
= x’z(y’ + y)
= x’z  1
= x’z
x’y’z’ + xy’z’ + x’yz’ + xyz’
= z’(x’y’ + xy’ + x’y + xy)
= z’(y’(x’ + x) + y(x’ + x))
= z’(y’+y)
= z’
CS231 Boolean Algebra

32. Example K-map simplification
Let’s consider simplifying f(x,y,z) = xy + y’z + xz.
First, you should convert the expression into a sum of minterms form, if it’s not already.
The easiest way to do this is to make a truth table for the function, and then read off the minterms.
You can either write out the literals or use the minterm shorthand.
Here is the truth table and sum of minterms for our example:
f(x,y,z) = x’y’z + xy’z + xyz’ + xyz
= m1 + m5 + m6 + m7
CS231 Boolean Algebra

33. Unsimplifying expressions
You can also convert the expression to a sum of minterms with Boolean algebra.
Apply the distributive law in reverse to add in missing variables.
Very few people actually do this, but it’s occasionally useful.
In both cases, we’re actually “unsimplifying” our example expression.
The resulting expression is larger than the original one!
But having all the individual minterms makes it easy to combine them together with the K-map.
xy + y’z + xz = (xy  1) + (y’z  1) + (xz  1)
= (xy  (z’ + z)) + (y’z  (x’ + x)) + (xz  (y’ + y))
= (xyz’ + xyz) + (x’y’z + xy’z) + (xy’z + xyz)
= xyz’ + xyz + x’y’z + xy’z
CS231 Boolean Algebra

34. Making the example K-map
Next up is drawing and filling in the K-map.
Put 1s in the map for each minterm, and 0s in the other squares.
You can use either the minterm products or the shorthand to show you where the 1s and 0s belong.
In our example, we can write f(x,y,z) in two equivalent ways.
In either case, the resulting K-map is shown below.
f(x,y,z) = x’y’z + xy’z + xyz’ + xyz
f(x,y,z) = m1 + m5 + m6 + m7
CS231 Boolean Algebra

35. K-maps from truth tables
You can also fill in the K-map directly from a truth table.
The output in row i of the table goes into square mi of the K-map.
Remember that the rightmost columns of the K-map are “switched.”
CS231 Boolean Algebra

36. Grouping the minterms together
The most difficult step is grouping together all the 1s in the K-map.
Make rectangles around groups of one, two, four or eight 1s.
All of the 1s in the map should be included in at least one rectangle.
Do not include any of the 0s.
Each group corresponds to one product term. For the simplest result:
Make as few rectangles as possible, to minimize the number of products in the final expression.
Make each rectangle as large as possible, to minimize the number of literals in each term.
It’s all right for rectangles to overlap, if that makes them larger.
CS231 Boolean Algebra

37. Reading the MSP from the K-map
Finally, you can find the MSP.
Each rectangle corresponds to one product term.
The product is determined by finding the common literals in that rectangle.
For our example, we find that xy + y’z + xz = y’z + xy. (This is one of the additional algebraic laws from last time.)
CS231 Boolean Algebra

38. Practice K-map 1 Simplify the sum of minterms m1 + m3 + m5 + m6.
CS231 Boolean Algebra

39. Solutions for practice K-map 1
Here is the filled in K-map, with all groups shown.
The magenta and green groups overlap, which makes each of them as large as possible.
Minterm m6 is in a group all by its lonesome.
The final MSP here is x’z + y’z + xyz’.
CS231 Boolean Algebra

40. Four-variable K-maps We can do four-variable expressions too!
The minterms in the third and fourth columns, and in the third and fourth rows, are switched around.
Again, this ensures that adjacent squares have common literals.
Grouping minterms is similar to the three-variable case, but:
You can have rectangular groups of 1, 2, 4, 8 or 16 minterms.
You can wrap around all four sides.
CS231 Boolean Algebra

41. Example: Simplify m0+m2+m5+m8+m10+m13
The expression is already a sum of minterms, so here’s the K-map:
We can make the following groups, resulting in the MSP x’z’ + xy’z.
CS231 Boolean Algebra

42. K-maps can be tricky!
There may not necessarily be a unique MSP. The K-map below yields two valid and equivalent MSPs, because there are two possible ways to include minterm m7.
Remember that overlapping groups is possible, as shown above.
y’z + yz’ + xy
y’z + yz’ + xz
CS231 Boolean Algebra