Boolean Algebra (1)

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Functions

• Computers take inputs and produce outputs, just like functions in math!

• Mathematical functions can be expressed in two ways:

• We can represent logical functions in two analogous ways too:

– A finite, but non-unique Boolean expression.

– A truth table, which will turn out to be unique and finite.

x y f(x,y)

0 0 0

… … …

2 2 6

… … …

23 41 87

… … …

f(x,y) = 2x + y= x + x + y= 2(x + y/2)= ...

An expression isfinite but not unique

A function table isunique but infinite

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Basic Boolean operations

• There are three basic operations for logical values.

x y xy

0 0 0

0 1 0

1 0 0

1 1 1

x y x+y

0 0 0

0 1 1

1 0 1

1 1 1

x x

0 1

1 0

AND (product)of two inputs

OR (sum) of two inputs

NOT(complement)on one input

xy, or xy x + y

Operation:

Expression:

Truth table:

x

4

Note Simbols

'

'

'

zz

yy

xx

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Boolean expressions

• We can use these basic operations to form more complex expressions:

f(x,y,z) = (x + y’)z + x’

• Some terminology and notation:

– f is the name of the function.

– (x,y,z) are the input variables, each representing 1 or 0. Listing the inputs is optional, but sometimes helpful.

– A literal is any occurrence of an input variable or its complement. The function above has four literals: x, y’, z, and x’.

• Precedences are important, but not too difficult.

– NOT has the highest precedence, followed by AND, and then OR.

– Fully parenthesized, the function above would be kind of messy:

f(x,y,z) = (((x +(y’))z) + x’)

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Truth tables

• A truth table shows all possible inputs and outputs of a function.

• Remember that each input variable represents either 1 or 0.

– Because there are only a finite number of values (1 and 0), truth tables themselves are finite.

– A function with n variables has 2n possible combinations of inputs.

• Inputs are listed in binary order—in this example, from 000 to 111.

x y z f(x,y,z)

0 0 0 1

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 0

1 1 1 1

f(0,0,0) = (0 + 1)0 + 1 = 1f(0,0,1) = (0 + 1)1 + 1 = 1f(0,1,0) = (0 + 0)0 + 1 = 1f(0,1,1) = (0 + 0)1 + 1 = 1f(1,0,0) = (1 + 1)0 + 0 = 0f(1,0,1) = (1 + 1)1 + 0 = 1f(1,1,0) = (1 + 0)0 + 0 = 0f(1,1,1) = (1 + 0)1 + 0 = 1

f(x,y,z) = (x + y’)z + x’

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Primitive logic gates

• Each of our basic operations can be implemented in hardware using a primitive logic gate.

– Symbols for each of the logic gates are shown below.

– These gates output the product, sum or complement of their inputs.

Logic gate:

AND (product)of two inputs

OR (sum) of two inputs

NOT(complement)on one input

xy, or xy x + y x’

Operation:

Expression:

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Expressions and circuits

• Any Boolean expression can be converted into a circuit by combining basic gates in a relatively straightforward way.

• The diagram below shows the inputs and outputs of each gate.

• The precedences are explicit in a circuit. Clearly, we have to make sure that the hardware does operations in the right order!

(x + y’)z + x’

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Circuit analysis

• Circuit analysis involves figuring out what some circuit does.

– Every circuit computes some function, which can be described with Boolean expressions or truth tables.

– So, the goal is to find an expression or truth table for the circuit.

• The first thing to do is figure out what the inputs and outputs of the overall circuit are.

– This step is often overlooked!

– The example circuit here has three inputs x, y, z and one output f.

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Write algebraic expressions...

• Next, write expressions for the outputs of each individual gate, based on that gate’s inputs.

– Start from the inputs and work towards the outputs.

– It might help to do some algebraic simplification along the way.

• Here is the example again.

– We did a little simplification for the top AND gate.

– You can see the circuit computes f(x,y,z) = xz + y’z + x’yz’

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...or make a truth table

• It’s also possible to find a truth table directly from the circuit.

• Once you know the number of inputs and outputs, list all the possible input combinations in your truth table.

– A circuit with n inputs should have a truth table with 2n rows.

– Our example has three inputs, so the truth table will have 23 = 8 rows. All the possible input combinations are shown.

x y z f

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

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Simulating the circuit

• Then you can simulate the circuit, either by hand or with a program like LogicWorks, to find the output for each possible combination of inputs.

• For example, when xyz = 101, the gate outputs would be as shown below.

– Use truth tables for AND, OR and NOT to find the gate outputs.

– For the final output, we find that f(1,0,1) = 1.

x y z f

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1 1

1 1 0

1 1 1

1

0

1

1

0

0

1

1

0

1

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Finishing the truth table

• Doing the same thing for all the other input combinations yields the complete truth table.

• This is simple, but tedious.

x y z f

0 0 0 0

0 0 1 1

0 1 0 1

0 1 1 0

1 0 0 0

1 0 1 1

1 1 0 0

1 1 1 1

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Expressions and truth tables

• Remember that if you already have a Boolean expression, you can use that to easily make a truth table.

• For example, since we already found that the circuit computes the function f(x,y,z) = xz + y’z + x’yz’, we can use that to fill in a table:

– We show intermediate columns for the terms xz, y’z and x’yz’.

– Then, f is obtained by just OR’ing the intermediate columns.

x y z xz y’z x’yz’ f

0 0 0 0 0 0 0

0 0 1 0 1 0 1

0 1 0 0 0 1 1

0 1 1 0 0 0 0

1 0 0 0 0 0 0

1 0 1 1 1 0 1

1 1 0 0 0 0 0

1 1 1 1 0 0 1

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Truth tables and expressions

• The opposite is also true: it’s easy to come up with an expression if you already have a truth table.

• We saw that you can quickly convert a truth table into a sum of minterms expression. The minterms correspond to the truth table rows where the output is 1.

• You can then simplify this sum of minterms if desired—using a K-map, for example.

x y z f

0 0 0 0

0 0 1 1

0 1 0 1

0 1 1 0

1 0 0 0

1 0 1 1

1 1 0 0

1 1 1 1

f(x,y,z) = x’y’z + x’yz’ + xy’z + xyz

= m1 + m2 + m5 + m7

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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’sinputs and outputs

Find a Booleanexpression

for the circuit

Find a truth tablefor the circuit

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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.

• Next time, we’ll look at how Boolean algebra can help simplify expressions, which in turn will lead to simpler circuits.

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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

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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

1. x + 0 = x 2. x 1 = x

3. x + 1 = 1 4. x 0 = 0

5. x + x = x 6. x x = x

7. x + x’ = 1 8. x x’ = 0

9. (x’)’ = x

10. x + y = y + x 11. xy = yx Commutative

12. x + (y + z) = (x + y) + z 13. x(yz) = (xy)z Associative

14. x(y + z) = xy + xz 15. x + yz = (x + y)(x + z) Distributive

16. (x + y)’ = x’y’ 17. (xy)’ = x’ + y’ DeMorgan’s

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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

1. x + 0 = x 2. x 1 = x

3. x + 1 = 1 4. x 0 = 0

5. x + x = x 6. x x = x

7. x + x’ = 1 8. x x’ = 0

9. (x’)’ = x

10. x + y = y + x 11. xy = yx Commutative

12. x + (y + z) = (x + y) + z 13. x(yz) = (xy)z Associative

14. x(y + z) = xy + xz 15. x + yz = (x + y)(x + z) Distributive

16. (x + y)’ = x’y’ 17. (xy)’ = x’ + y’ DeMorgan’s

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Are these axioms for real?

• We can show that these axioms are true, 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’)

x y xy

0 0 0

0 1 0

1 0 0

1 1 1

x y x+y

0 0 0

0 1 1

1 0 1

1 1 1

x x’

0 1

1 0

x y x+y

0 0 0

0 1 1

1 0 1

1 1 1

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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

x y x + y (x + y)’ x y x’ y’ x’y’

0 0 0 1 0 0 1 1 1

0 1 1 0 0 1 1 0 0

1 0 1 0 1 0 0 1 0

1 1 1 0 1 1 0 0 0

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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 ]

1. x + 0 = x 2. x 1 = x

3. x + 1 = 1 4. x 0 = 0

5. x + x = x 6. x x = x

7. x + x’ = 1 8. x x’ = 0

9. (x’)’ = x

10. x + y = y + x 11. xy = yx Commutative

12. x + (y + z) = (x + y) + z 13. x(yz) = (xy)z Associative

14. x(y + z) = xy + xz 15. x + yz = (x + y)(x + z) Distributive

16. (x + y)’ = x’y’ 17. (xy)’ = x’ + y’ DeMorgan’s

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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

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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:

1. x + xy = x 4. x(x + y) = x

2. xy + xy’ = x 5. (x + y)(x + y’) = x

3. x + x’y = x + y 6. x(x’ + y) = xy

xy + x’z + yz = xy + x’z (x + y)(x’ + z)(y + z) = (x + y)(x’ + z)

x y x’ x’y x + x’y x y x + y

0 0 0 0 0

0 1 0 1 1

1 0 1 0 1

1 1 1 1 1

x + x’y = (x + x’)(x + y) [ Distributive ]= 1 (x + y) [ x + x’ = 1 ]= x + y [ Axiom 2 ]

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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)

x y z f(x,y,z)

0 0 0 1

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 0

1 0 1 0

1 1 0 1

1 1 1 0

x y z f’(x,y,z)

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 0

1 1 1 1

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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]

Boolean Algebra (2)

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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

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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’yzxy’z’ xy’z xyz’ xyz

Minterm Is true when… Shorthandx’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

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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.

x y z f(x,y,z) f’(x,y,z)

0 0 0 1 0

0 0 1 1 0

0 1 0 1 0

0 1 1 1 0

1 0 0 0 1

1 0 1 0 1

1 1 0 1 0

1 1 1 0 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

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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)

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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 + zx + y’ + z’ x + y’ + z x + y + z’ x + y + z

Maxterm Is false when… Shorthandx + 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

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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!)

x y z f(x,y,z) f’(x,y,z)

0 0 0 1 0

0 0 1 1 0

0 1 0 1 0

0 1 1 1 0

1 0 0 0 1

1 0 1 0 1

1 1 0 1 0

1 1 1 0 1

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

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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

Maxterm Shorthandx + 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

Minterm Shorthandx’y’z’ m0

x’y’z m1

x’yz’ m2

x’yz m3

xy’z’ m4

xy’z m5

xyz’ m6

xyz m7

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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)

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Summary

• So far:

– A bunch of Boolean algebra trickery for simplifying expressions and circuits

– The algebra guarantees us that the simplified circuit is equivalentto 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