Boolean Algebra Introduction Logical arguments are valid (or not) by virtue of their form, not...

Boolean Algebra

IntroductionLogical arguments are valid (or not) by virtue of their form, not content

ExampleAll men are mortal (premise 1)Harry is a man (premise 2)-------------------------------------------Harry is mortal (conclusion)

This regularity allows implementation of logic by computers, which have no capacity to understand meaning

Boolean Algebra

Logical variables

P, Q, YourMamma, Waverly

take values 0 or 1 only

Logical operators OR operator AND operator

NOT operator

P1 P2 P1 + P2

0011

0101

0111

P1 P2 P1 P2

0011

0101

0001

P P’

01

10

Boolean Algebra

Truth tablesArbitrary functions can be composed from these elementary operators acting on variables

Example

Properties of truth tables1) If there are n variables, there will be 2n rows2) Rows are listed in counting order3) Intermediate results may be shown in inserted columns

P1 P2 P3 (P1 + P2) (P1 + P2) P3 [(P1 + P2) P3]΄

00001111

00110011

01010101

00111111

00010101

11101010

Boolean Algebra

Other logical operatorsOR, AND, and NOT operators are sufficient to represent any logical function; however, other operators will prove useful in digital design

NAND and NOR

XOR

P Q P NAND Q

0 0 1 1

0 1 0 1

1 1 1 0

P Q P NOR Q

0 0 1 1

0 1 0 1

1 0 0 0

P1 P2 P1 XOR P2

0 0 1 1

0 1 0 1

0 1 1 0

Boolean Algebra

Tautology, Equivalence, and Logical LawsA tautology is a function that is always true

Example

Equivalence operator

P P ́ P + P ́0 1

1 0

1 1

P1 P2 P1 ≡ P2 0 0 1 1

0 1 0 1

1 0 0 1

Boolean Algebra

Tautology, Equivalence, and Logical Laws (cont.)If an equivalence is a tautology, then the two expressions are interchangeable. Why? Because a function is completely defined by its truth values.

Example

This is one form of DeMorgan’s Law

P1 P2 (P1 + P2) ́ (P1 ́ P2 )́ (P1 + P2) ́≡ (P1 ́ P2 )́ 0 0 1 1

0 1 0 1

1 0 0 0

1 0 0 0

1 1 1 1

Boolean Algebra

SimplificationGoal: Reduce size of a functionMethod 1: Successive application of equivalence theoremsSample theoremstheorem dual

(T1) X + 0 ≡ X (T1)d X ∙ 1 ≡ X

(T2) X + 1 ≡ 1 (T2)d X ∙ 0 ≡ 0

(T3) X + X ≡ X (T3)d XX ≡ X

(T4) X + X΄ ≡ 1 (T4)d XX’ ≡ 0

(T5) (X΄)΄ ≡ X

(T6) X + Y ≡Y + X (T6)d XY ≡ YX

(T7) (X + Y)΄ ≡ (X΄ ∙ Y΄) (T7)d (XY)΄ ≡ (X΄ + Y΄)

(T8) (X + Y) + Z ≡ X + (Y + Z) (T8)d (XY) Z X(YZ)

(T9) XY + XZ ≡ X(Y + Z) (T9)d (X + Y) (X + Z) ≡ X + YZ

(T10) X + XY≡ X (T10)d X(X + Y) ≡ X

Boolean Algebra

Simplification (method 1)Fact about the theorems

1) 0 and 1 are constants that always take these values2) AND can be implicit (as with multiplication in numerical algebra)3) All equivalences can be proven with a truth table4) Every theorem is accompanied by its dual obtained by switching

all 0’s and 1’s and all ∙’s and +’s. When these switches are made, a new valid theorem is obtained.5) Every symbol in every identity can be systematically replaced by arbitrary expressions.6) A number of theorems are identical to those of numerical algebra(e.g., (T8) and (T9), but not (T9)d)

Boolean Algebra

Simplification (method 1)Example 1step justification1. PQ + PQ΄ given2. P(Q + Q΄) (T9)3. P(1) (T4)4. P (T1)d

Example 2step justification

1. P1P2 + P1΄P3 + P2P3 given

2. P1P2 + P1΄P2 + P2P3(1) (T1)d

3. P1P2 + P1’P3 + P2P3(P1 + P1΄) (T4)4. P1P2 + P1΄P3 + P1P2P3 + P1΄P2P3 (T9), (T6)d

5. P1P2 + P1P2P3 + P1΄P3 + P1΄P2P3 (T6)6. P1P2(1 + P3) + P1’P3(1 + P2) (T9), two applications7. P1P2(1) + P1΄P3(1) (T2)8. P1P2 + P1΄P3 (T1)

Note: This simplification involves expansion before contraction

Boolean Algebra

Simplification (method 2)Motivation

Simplification with equivalence theorems is unsystematic(witness the 2nd example)

AlternativeReduce function to sum of minterms (SOP) or product of maxterms

(POS), then simplify on a Karnaugh map

Boolean Algebra

Simplification (method 2)Minterms and Maxterms

minterm product of variables with variable negated if corresponding value is 0

maxterm sum of variables with variable negated if corresponding value is 1

Minterms and maxterms for 4 variable functions

# W X Y Z minterm maxterm

0123456789101112131415

0000000011111111

0000111100001111

0011001100110011

0101010101010101

W΄X΄Y΄Z΄

W΄X΄Y΄ZW΄X΄YZ΄

W΄X΄YZW΄XY΄Z΄

W΄XY΄ZW΄XYZ΄

W΄XYZWX΄Y΄Z΄

WX΄Y΄ZWX΄YZ΄

WX΄YZWXY΄Z΄

WXY΄ZWXYZ΄

WXYZ

W + X + Y + ZW + X + Y + Z΄

W + X + Y΄ + ZW + X + Y΄ + Z΄

W + X΄+ Y + ZW + X΄ + Y + Z΄

W + X΄+ Y΄ + ZW + X΄+ Y΄ + Z΄

W΄+ X + Y + ZW΄ + X + Y + Z΄

W΄ + X + Y΄ + ZW΄ + X + Y΄ + Z΄

W΄ + X΄ + Y + ZW΄ + X΄ + Y + Z΄

W΄+ X΄ + Y΄ + ZW΄ + X΄ + Y΄ + Z΄

Boolean Algebra Simplification (method 2)

TheoremEvery function can be rewritten as 1) Sum of the minterms where the function is 1 (SOP), or2) Product of the maxterms where the function is 0 (POS)

Example F = Y΄(X + W) + Z΄ SOP: F = W΄X΄Y΄Z΄ + W΄X΄YZ΄ + W΄XY΄Z΄ + W΄XY΄Z +

W΄XYZ΄ + WX΄Y΄Z΄ + WX΄Y΄Z΄’ + WX΄YZ΄ + WXY΄Z’ + WXY΄Z + WXYZ΄, or

F = WXYZ(0,2,4,5,6,8,9,10,12,13,14). POS: F = (W + X + Y + Z΄)(W + X + Y΄ + Z΄)(W + X΄+Y’+ Z΄)

(W΄ + X + Y΄ + Z΄) (W΄+ X΄ + Y΄ + Z΄), or

F = WXYZ(1,3,7,11,15).

# W X Y Z F = Y΄ (X + W) + Z΄

0123456789

101112131415

0000000011111111

0000111100001111

0011001100110011

0101010101010101

1010111011101110

Boolean Algebra Simplification (method 2, aside)

How many operators are necessary?AND, OR and NOT can be used to represent any SOPand therefore any function

But, an SOP with only positive literals can be rendered with only NAND’s by replacing all operators with NAND Example AB + CD ≡ (A NAND B) NAND (C NAND D)

And, can achieve NOT as follows: P’ ≡ P NAND P

Therefore, NAND suffices to represent any function.

Example AC’ + A’B + BC ≡ (A NAND (C NAND C)) NAND ((A NAND A) NAND B) NAND (B

NAND C)


Karnaugh maps

Procedure

1) Enter the minterms for the function on the map.

2) Determine all the prime implicants of the function.

3) Determine the smallest subset of these prime implicants that cover all the prime implicants.4) Name these implicants and construct the final expression.


Karnaugh maps

Step 1:Enter terms F = W΄XYZ + WXY΄ + WXY + X΄YZ + W΄X΄Y΄Z΄

= WXYZ(0,7,8,9,12,13,14,15)

Note: Gray code ordering on map

W XY Z

0 0 0 1 1 1 1 0

0 0

0 1

1 1

1 0

0

1

2

3

4

5

6

7

1 2

1 3

1 4

1 5

8

9

1 0

1 1

1

1

1

1

1

1

1

1


Karnaugh maps

Step 2: Determine prime implicants (PI’s)

What is a PI?i. Consists of 2n minterms, where n is an integer

ii. It forms a continuous rectangle over the 3D torus formed by joining the ends of the map

iii. It is not subsumed by a larger PI

0

1

4

5

8 1 2

9 1 3

2

3

6

7

1 0 1 4

11 1 5

Boolean Algebra

Simplification (method 2)Karnaugh maps

Step 2: Determine prime implicants (PI’s)

Example

W X0 0 0 1 1 0

0 1

1 1

1 0

1 1

1 1 1

1

1 1

1

1 2

3

41

0 0

Y Z


Karnaugh maps

Step 3: Find minimal set of PI’si. Find all essential prime implicants (those that contain a minterm not covered by any other PI)ii. Cover rest of minterms with other PI’s

Example

All are necessary (all have distinguished minterms)

W X0 0 0 1 1 0

0 1

1 1

1 0

1 1

1 1 1

1

1 1

1

1 2

3

41

0 0

Y Z


Karnaugh maps

Step 4: Name the PI’s in the minimal set

Rule: The name of the PI will correspond to a product of the variables that don’t change, with the variable negated if it always 0, and not negated otherwise.

Example (PI 1)

WXY΄Z΄ + WXY΄Z + WXYZ΄ + WXY΄Z΄≡ WX(Y΄Z΄ + Y΄Z + YZ΄ + Y΄Z΄)≡ WX(Y΄(Z΄ + Z) + Y(Z΄ + Z)≡ WX(Y΄(1) + Y(1))≡ WX(Y΄(1) + Y(1))≡ WX(Y΄ + Y)≡ WX(1)≡ WX

W X0 0 0 1 1 0

0 1

1 1

1 0

1 1

1 1 1

1

1 1

1

1 2

3

41

0 0

Y Z


Karnaugh maps

Step 4: Name the PI’s in the minimal set.

Full set of PI’s for this example:

WX + WY΄ + X΄Y΄Z΄ + XYZ

Comparison with original expression

W΄XYZ + WXY΄ + WXY + X΄YZ + W΄X΄Y΄Z

(5 product terms reduced to 4, 17 literals to 10)

W X0 0 0 1 1 0

0 1

1 1

1 0

1 1

1 1 1

1

1 1

1

1 2

3

41

0 0

Y Z


Karnaugh maps

Another example WXYZ(0,1,2,3,4,5,6,14,15)

PI’sW΄X΄ comprising minterms 0,1,2,3W΄Y΄ comprising minterms 0,1,4,5W΄Z΄ comprising minterms 0,2,4,6XYZ΄ comprising minterms 6,14 not essentialWXY comprising minterms 14,15

Essential PI’s cover all minterms W΄X΄ + W΄Y΄ + W΄Z΄ + WXY

Y Z0 0 0 1 1 1 1 0

0 0

0 1

1 1

1 0

0

1

2

3

4

5

6

7

1 2

1 3

1 4

1 5

8

9

1 0

1 1

1

1

1 1 1

1

1

1

1

Y Z1 1 1 0

0 0

0 1

1 1

1 0

1

1

1 1 1

1

1

1

1

2

4

51

W X0 0 0 1

3


Karnaugh maps, POS minimization

WXYZ(0,1,2,3,4,5,6,14,15)

PI’sW΄ + X, comprising maxterms 8,9,10,11W΄ + Y, comprising maxterms 8,9,12,13W + X΄ + Y΄ + Z΄, comprising maxterm 7

Essential PI’s cover all minterms (W΄ + X)(W΄ + Y)(W + X΄ + Y΄ + Z΄) slightly better than SOP minimization

W XY Z

0 0 0 1 1 1 1 0

0 0

0 1

1 1

1 0

1

1

1 1 1

1

1

1

1

1

1

2

3


Karnaugh maps, a hard example

SOP and POS minimization does not work

But, if target representation is changed, a single term results! (W XOR X) XOR (Y XOR Z)

Furthermore, a small counting machine covers minterms regardless of number of variables

MORAL: Finding the optimal circuit is non-trivial. This makes Digital Design an art as well as an engineering task.

W XY Z

0 0 0 1 1 1 1 0

0 0

0 1

1 1

1 0

0

1

2

3

4

5

6

7

1 2

1 3

1 4

1 5

8

9

1 0

1 1

1

1

1

1 1

1

1

1

p

Boolean Algebra

Simplification (method 2)Karnaugh maps

Summary of minimization

o r ig in a lex p r es s io n

tr u th tab le

m in ter m s o nKar n au g h

m in im izedS O P

m in im izedP O S

m in im u m o fm in im u m s

m in im als e t o f P I 's

Boolean Algebra

Summary of topicsFunctionsTruth tablesTautologies and equivalenceSimplification with equivalenceSimplification with Karnaugh maps

minterms and maxtermsPrime ImplicantsSOP and POS minimization

Simplification depends on representation (the art of design)

Boolean Algebra Introduction Logical arguments are valid (or not) by virtue of their form, not...

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