1. Combinational Logic (mostly review!)
Logic functions, truth tables, and switches
NOT, AND, OR, NAND, NOR, XOR, . . .
Minimal set
Axioms and theorems of Boolean algebra
Proofs by re-writing
Proofs by perfect induction
Gate logic
Networks of Boolean functions
Time behavior
Canonical forms
Two-level
Incompletely specified functions
Two-level minimization
CS Spring – Lec #3: Combinational Logic - 1

2. Possible Logic Functions of Two Variables
16 possible functions of 2 input variables:
2**(2**n) functions of n inputs X F Y
X Y possible functions (F0–F15) 1
not X
X and Y
X or Y
not Y X Y
X xor Y
X = Y
X nor Y not (X or Y)
X nand Y not (X and Y)
CS Spring – Lec #3: Combinational Logic - 2

3. Cost of Different Logic Functions
Some are easier, others harder, to implement
Each has a cost associated with the number of switches needed
0 (F0) and 1 (F15): require 0 switches, directly connect output to low/high
X (F3) and Y (F5): require 0 switches, output is one of inputs
X' (F12) and Y' (F10): require 2 switches for "inverter" or NOT-gate
X nor Y (F4) and X nand Y (F14): require 4 switches
X or Y (F7) and X and Y (F1): require 6 switches
X = Y (F9) and X  Y (F6): require 16 switches
Because NOT, NOR, and NAND are the cheapest they are the functions we implement the most in practice
CS Spring – Lec #3: Combinational Logic - 3

4. Minimal Set of Functions
Implement any logic functions from NOT, NOR, and NAND?
For example, implementing X and Y is the same as implementing not (X nand Y)
Do it with only NOR or only NAND
NOT is just a NAND or a NOR with both inputs tied together
and NAND and NOR are "duals", i.e., easy to implement one using the other
Based on the mathematical foundations of logic: Boolean Algebra
X Y X nor Y
X Y X nand Y
X nand Y  not ( (not X) nor (not Y) ) X nor Y  not ( (not X) nand (not Y) )
CS Spring – Lec #3: Combinational Logic - 4

5. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 5
Algebraic Structure
Consists of
Set of elements B
Binary operations { + , • }
Unary operation { ' }
Following axioms hold:
1. set B contains at least two elements, a, b, such that a  b 2. closure: a + b is in B a • b is in B 3. commutativity: a + b = b + a a • b = b • a 4. associativity: a + (b + c) = (a + b) + c a • (b • c) = (a • b) • c 5. Identity: a + 0 = a a • 1 = a 6. distributivity: a + (b • c) = (a + b) • (a + c) a • (b + c) = (a • b) + (a • c) 7. complementarity: a + a' = a • a' = 0
CS Spring – Lec #3: Combinational Logic - 5

6. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 6
Boolean Algebra
Boolean algebra
B = {0, 1}
+ is logical OR, • is logical AND
' is logical NOT
All algebraic axioms hold
CS Spring – Lec #3: Combinational Logic - 6

7. Logic Functions and Boolean Algebra
Any logic function that can be expressed as a truth table can be written as an expression in Boolean algebra using the operators: ', +, and •
X Y X • Y
1 1 1
X Y X' X' • Y
X Y X' Y' X • Y X' • Y' ( X • Y ) + ( X' • Y' )
( X • Y ) + ( X' • Y' ) º X = Y
Boolean expression that is true when the variables X and Y have the same value
and false, otherwise
X, Y are Boolean algebra variables
CS Spring – Lec #3: Combinational Logic - 7

8. Axioms and Theorems of Boolean Algebra
Identity X + 0 = X 1D. X • 1 = X
Null X + 1 = 1 2D. X • 0 = 0
Idempotency: X + X = X 3D. X • X = X
Involution: (X')' = X
Complementarity: X + X' = 1 5D. X • X' = 0
Commutativity: X + Y = Y + X 6D. X • Y = Y • X
Associativity: (X + Y) + Z = X + (Y + Z) 7D. (X • Y) • Z = X • (Y • Z)
CS Spring – Lec #3: Combinational Logic - 8

9. Axioms and Theorems of Boolean Algebra (cont’d)
Distributivity: X • (Y + Z) = (X • Y) + (X • Z) 8D. X + (Y • Z) = (X + Y) • (X + Z)
Uniting: X • Y + X • Y' = X 9D. (X + Y) • (X + Y') = X
Absorption: 10. X + X • Y = X 10D. X • (X + Y) = X 11. (X + Y') • Y = X • Y 11D. (X • Y') + Y = X + Y
Factoring: 12. (X + Y) • (X' + Z) = 12D. X • Y + X' • Z = X • Z + X' • Y (X + Z) • (X' + Y)
Consensus: 13. (X • Y) + (Y • Z) + (X' • Z) = 13D. (X + Y) • (Y + Z) • (X' + Z) = X • Y + X' • Z (X + Y) • (X' + Z)
CS Spring – Lec #3: Combinational Logic - 9

10. Axioms and Theorems of Boolean Algebra (cont’d)
deMorgan's: 14. (X + Y + ...)' = X' • Y' • D. (X • Y • ...)' = X' + Y' + ...
Generalized de Morgan's: 15. f'(X1,X2,...,Xn,0,1,+,•) = f(X1',X2',...,Xn',1,0,•,+)
Establishes relationship between • and +
CS Spring – Lec #3: Combinational Logic - 10

11. Axioms and Theorems of Boolean Algebra (cont’d)
Duality
Dual of a Boolean expression is derived by replacing • by +, + by •, 0 by 1, and 1 by 0, and leaving variables unchanged
Any theorem that can be proven is thus also proven for its dual!
Meta-theorem (a theorem about theorems)
Duality: 16. X + Y  X • Y • ...
Generalized duality: 17. f (X1,X2,...,Xn,0,1,+,•)  f(X1,X2,...,Xn,1,0,•,+)
Different than deMorgan’s Law
This is a statement about theorems
This is not a way to manipulate (re-write) expressions
CS Spring – Lec #3: Combinational Logic - 11

12. Proving Theorems (Rewriting Method)
Using the axioms of Boolean algebra:
e.g., prove the theorem: X • Y + X • Y' = X
e.g., prove the theorem: X + X • Y = X
distributivity (8) X • Y + X • Y' = X • (Y + Y')
complementarity (5) X • (Y + Y') = X • (1)
identity (1D) X • (1) = X ü
identity (1D) X + X • Y = X • X • Y
distributivity (8) X • X • Y = X • (1 + Y)
identity (2) X • (1 + Y) = X • (1)
identity (1D) X • (1) = X ü
CS Spring – Lec #3: Combinational Logic - 12

13. Proving Theorems (Perfect Induction)
Using perfect induction (complete truth table):
e.g., de Morgan's:
X Y X' Y' (X + Y)' X' • Y'
(X + Y)' = X' • Y' NOR is equivalent to AND with inputs complemented 1 1
X Y X' Y' (X • Y)' X' + Y'
(X • Y)' = X' + Y' NAND is equivalent to OR with inputs complemented 1 1
CS Spring – Lec #3: Combinational Logic - 13

14. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 14
Simple Example
1-bit binary adder
Inputs: A, B, Carry-in
Outputs: Sum, Carry-out A B
Cin
Cout S
A B Cin S Cout
0 1 0
0 1 1
1 1 0
1 1 1 1
S = A' B' Cin + A' B Cin' + A B' Cin' + A B Cin
Cout = A' B Cin + A B' Cin + A B Cin' + A B Cin
CS Spring – Lec #3: Combinational Logic - 14

15. Apply the Theorems to Simplify Expressions
Theorems of Boolean algebra can simplify Boolean expressions
e.g., full adder's carry-out function (same rules apply to any function)
Cout = A' B Cin + A B' Cin + A B Cin' + A B Cin
= A' B Cin + A B' Cin + A B Cin' + A B Cin + A B Cin
= A' B Cin + A B Cin + A B' Cin + A B Cin' + A B Cin
= (A' + A) B Cin + A B' Cin + A B Cin' + A B Cin
= (1) B Cin + A B' Cin + A B Cin' + A B Cin
= B Cin + A B' Cin + A B Cin' + A B Cin + A B Cin
= B Cin + A B' Cin + A B Cin + A B Cin' + A B Cin
= B Cin + A (B' + B) Cin + A B Cin' + A B Cin
= B Cin + A (1) Cin + A B Cin' + A B Cin
= B Cin + A Cin + A B (Cin' + Cin)
= B Cin + A Cin + A B (1)
= B Cin + A Cin + A B
CS Spring – Lec #3: Combinational Logic - 15

16. From Boolean Expressions to Logic Gates
X Y
NOT X' X ~X
AND X • Y XY X  Y
OR X + Y X  Y X Y
X Y Z
1 1 1 X Y Z
X Y Z
1 1 1 X Z Y
CS Spring – Lec #3: Combinational Logic - 16

17. From Boolean Expressions to Logic Gates (cont’d)
X Y Z
1 1 0
NAND
NOR
XOR X Y
XNOR X = Y X Z Y
X Y Z
1 1 0 X Z Y
X Y Z
1 1 0 X
X xor Y = X Y' + X' Y X or Y but not both ("inequality", "difference") Z Y
X Y Z
1 1 1
X xnor Y = X Y + X' Y' X and Y are the same ("equality", "coincidence") X Z Y
CS Spring – Lec #3: Combinational Logic - 17

18. From Boolean Expressions to Logic Gates (cont’d)
More than one way to map expressions to gates
e.g., Z = A' • B' • (C + D) = (A' • (B' • (C + D))) T2 T1
use of 3-input gate A Z A B T1 B Z C C T2 D D
CS Spring – Lec #3: Combinational Logic - 18

19. Waveform View of Logic Functions
Just a sideways truth table
But note how edges don't line up exactly
It takes time for a gate to switch its output!
time
change in Y takes time to "propagate" through gates
CS Spring – Lec #3: Combinational Logic - 19

20. Choosing Different Realizations of a Function
A B C Z
two-level realization (we don't count NOT gates)
multi-level realization (gates with fewer inputs)
XOR gate (easier to draw but costlier to build)
CS Spring – Lec #3: Combinational Logic - 20

21. Which Realization is Best?
Reduce number of inputs
literal: input variable (complemented or not)
can approximate cost of logic gate as 2 transistors per literal
why not count inverters?
Fewer literals means less transistors
smaller circuits
Fewer inputs implies faster gates
gates are smaller and thus also faster
Fan-ins (# of gate inputs) are limited in some technologies
Reduce number of gates
Fewer gates (and the packages they come in) means smaller circuits
directly influences manufacturing costs
CS Spring – Lec #3: Combinational Logic - 21

22. Which is the Best Realization? (cont’d)
Reduce number of levels of gates
Fewer level of gates implies reduced signal propagation delays
Minimum delay configuration typically requires more gates
wider, less deep circuits
How do we explore tradeoffs between increased circuit delay and size?
Automated tools to generate different solutions
Logic minimization: reduce number of gates and complexity
Logic optimization: reduction while trading off against delay
CS Spring – Lec #3: Combinational Logic - 22

23. Are All Realizations Equivalent?
Under the same inputs, the alternative implementations have almost the same waveform behavior
Delays are different
Glitches (hazards) may arise
Variations due to differences in number of gate levels and structure
Three implementations are functionally equivalent
CS Spring – Lec #3: Combinational Logic - 23

24. Implementing Boolean Functions
Technology independent
Canonical forms
Two-level forms
Multi-level forms
Technology choices
Packages of a few gates
Regular logic
Two-level programmable logic
Multi-level programmable logic
CS Spring – Lec #3: Combinational Logic - 24

25. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 25
Canonical Forms
Truth table is the unique signature of a Boolean function
Many alternative gate realizations may have the same truth table
Canonical forms
Standard forms for a Boolean expression
Provides a unique algebraic signature
CS Spring – Lec #3: Combinational Logic - 25

26. Sum-of-Products Canonical Forms
Also known as disjunctive normal form
Also known as minterm expansion
F =
+ ABC'
+ ABC
A'B'C
+ AB'C
+ A'BC
F =
A B C F F'
F' = A'B'C' + A'BC' + AB'C'
CS Spring – Lec #3: Combinational Logic - 26

27. Sum-of-Products Canonical Form (cont’d)
Product term (or minterm)
ANDed product of literals – input combination for which output is true
Each variable appears exactly once, in true or inverted form (but not both)
F in canonical form: F(A, B, C) = m(1,3,5,6,7) = m1 + m3 + m5 + m6 + m7
= A'B'C + A'BC + AB'C + ABC' + ABC
canonical form  minimal form F(A, B, C) = A'B'C + A'BC + AB'C + ABC + ABC'
= (A'B' + A'B + AB' + AB)C + ABC'
= ((A' + A)(B' + B))C + ABC' = C + ABC' = ABC' + C
= AB + C
A B C minterms A'B'C' m0
0 0 1 A'B'C m1
0 1 0 A'BC' m2
0 1 1 A'BC m3
1 0 0 AB'C' m4
1 0 1 AB'C m5
1 1 0 ABC' m6
1 1 1 ABC m7
short-hand notation for minterms of 3 variables
CS Spring – Lec #3: Combinational Logic - 27

28. Product-of-Sums Canonical Form
Also known as conjunctive normal form
Also known as maxterm expansion
F = F =
(A + B' + C)
(A' + B + C)
(A + B + C)
A B C F F'
F' = (A + B + C') (A + B' + C') (A' + B + C') (A' + B' + C) (A' + B' + C')
CS Spring – Lec #3: Combinational Logic - 28

29. Product-of-Sums Canonical Form (cont’d)
Sum term (or maxterm)
ORed sum of literals – input combination for which output is false
Each variable appears exactly once, in true or inverted form (but not both)
F in canonical form: F(A, B, C) = M(0,2,4) = M0 • M2 • M4
= (A + B + C) (A + B' + C) (A' + B + C)
canonical form  minimal form F(A, B, C) = (A + B + C) (A + B' + C) (A' + B + C)
= (A + B + C) (A + B' + C)
(A + B + C) (A' + B + C)
= (A + C) (B + C)
A B C maxterms A+B+C M0
0 0 1 A+B+C' M1
0 1 0 A+B'+C M2
0 1 1 A+B'+C' M3
1 0 0 A'+B+C M4
1 0 1 A'+B+C' M5
1 1 0 A'+B'+C M6
1 1 1 A'+B'+C' M7
short-hand notation for maxterms of 3 variables
CS Spring – Lec #3: Combinational Logic - 29

30. S-o-P, P-o-S, and deMorgan’s Theorem
Sum-of-products
F' = A'B'C' + A'BC' + AB'C'
Apply de Morgan's
(F')' = (A'B'C' + A'BC' + AB'C')'
F = (A + B + C) (A + B' + C) (A' + B + C)
Product-of-sums
F' = (A + B + C') (A + B' + C') (A' + B + C') (A' + B' + C) (A' + B' + C')
(F')' = ( (A + B + C')(A + B' + C')(A' + B + C')(A' + B' + C)(A' + B' + C') )'
F = A'B'C + A'BC + AB'C + ABC' + ABC
CS Spring – Lec #3: Combinational Logic - 30

31. Four Alternative Two-level Implementations of F = AB + C
canonical sum-of-products minimized sum-of-products canonical product-of-sums minimized product-of-sums B F1 C F2 F3 F4
CS Spring – Lec #3: Combinational Logic - 31

32. Waveforms for the Four Alternatives
Waveforms are essentially identical
Except for timing hazards (glitches)
Delays almost identical (modeled as a delay per level, not type of gate or number of inputs to gate)
CS Spring – Lec #3: Combinational Logic - 32

33. Mapping Between Canonical Forms
Minterm to maxterm conversion
Use maxterms whose indices do not appear in minterm expansion
e.g., F(A,B,C) = m(1,3,5,6,7) = M(0,2,4)
Maxterm to minterm conversion
Use minterms whose indices do not appear in maxterm expansion
e.g., F(A,B,C) = M(0,2,4) = m(1,3,5,6,7)
Minterm expansion of F to minterm expansion of F'
Use minterms whose indices do not appear
e.g., F(A,B,C) = m(1,3,5,6,7) F'(A,B,C) = m(0,2,4)
Maxterm expansion of F to maxterm expansion of F'
Use maxterms whose indices do not appear
e.g., F(A,B,C) = M(0,2,4) F'(A,B,C) = M(1,3,5,6,7)
CS Spring – Lec #3: Combinational Logic - 33

34. Incompletely Specified Functions
Example: binary coded decimal increment by 1
BCD digits encode decimal digits 0 – 9 in bit patterns 0000 – 1001
A B C D W X Y Z
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
off-set of W
on-set of W
don't care (DC) set of W
these inputs patterns should never be encountered in practice – "don't care" about associated output values, can be exploited in minimization
CS Spring – Lec #3: Combinational Logic - 34

35. Notation for Incompletely Specified Functions
Don't cares and canonical forms
So far, only represented on-set
Also represent don't-care-set
Need two of the three sets (on-set, off-set, dc-set)
Canonical representations of the BCD increment by 1 function:
Z = m0 + m2 + m4 + m6 + m8 + d10 + d11 + d12 + d13 + d14 + d15
Z =  [ m(0,2,4,6,8) + d(10,11,12,13,14,15) ]
Z = M1 • M3 • M5 • M7 • M9 • D10 • D11 • D12 • D13 • D14 • D15
Z =  [ M(1,3,5,7,9) • D(10,11,12,13,14,15) ]
CS Spring – Lec #3: Combinational Logic - 35

36. Simplification of Two-level Combinational Logic
Finding a minimal sum of products or product of sums realization
Exploit don't care information in the process
Algebraic simplification
Not an algorithmic/systematic procedure
How do you know when the minimum realization has been found?
Computer-aided design tools
Precise solutions require very long computation times, especially for functions with many inputs (> 10)
Heuristic methods employed – "educated guesses" to reduce amount of computation and yield good if not best solutions
Hand methods still relevant
Understand automatic tools and their strengths and weaknesses
Ability to check results (on small examples)
CS Spring – Lec #3: Combinational Logic - 36

37. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 37
The Uniting Theorem
Key tool for simplification: A (B' + B) = A
Essence of simplification:
Find two element subsets of the ON-set where only one variable changes its value – this single varying variable can be eliminated and a single product term used to represent both elements
F = A'B'+AB' = (A'+A)B' = B'
A B F
0 0 1
0 1 0
1 0 1
1 1 0
B has the same value in both on-set rows – B remains
A has a different value in the two rows – A is eliminated
CS Spring – Lec #3: Combinational Logic - 37

38. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 38
Boolean Cubes
Visual technique for indentifying when the uniting theorem can be applied
n input variables = n-dimensional "cube"
2-cube X Y 11 00 01 10
1-cube X 1
3-cube X Y Z
000
111
101
4-cube W X Y Z
0000
1111
1000
0111
CS Spring – Lec #3: Combinational Logic - 38

39. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 39
Boolean Cubes
Visual technique for indentifying when the uniting theorem can be applied
n input variables = n-dimensional "cube"
2-cube X Y 11 00 01 10
1-cube X 1
1111
3-cube X Y Z
000
111
101
CS Spring – Lec #3: Combinational Logic - 39
0000

40. Mapping Truth Tables onto Boolean Cubes
Uniting theorem combines two "faces" of a cube into a larger "face"
Example:
two faces of size 0 (nodes) combine into a face of size 1(line)
A varies within face, B does not this face represents the literal B' F
A B F
0 0 1
0 1 0
1 0 1
1 1 0 A B 11 00 01 10
ON-set = solid nodes OFF-set = empty nodes DC-set = 'd nodes
CS Spring – Lec #3: Combinational Logic - 40

41. Three Variable Example
Binary full-adder carry-out logic
(A'+A)BCin
A B Cin Cout
AB(Cin'+Cin) A B C
000
111
101
A(B+B')Cin
the on-set is completely covered by the combination (OR) of the subcubes of lower dimensionality - note that “111” is covered three times
Cout = BCin+AB+ACin
CS Spring – Lec #3: Combinational Logic - 41

42. Higher Dimensional Cubes
Sub-cubes of higher dimension than 2
F(A,B,C) = m(4,5,6,7)
on-set forms a square i.e., a cube of dimension 2
represents an expression in one variable i.e., 3 dimensions – 2 dimensions A B C
000
111
101
100
001
010
011
110
A is asserted (true) and unchanged
B and C vary
This subcube represents the
literal A
CS Spring – Lec #3: Combinational Logic - 42

43. m-Dimensional Cubes in an n-Dimensional Boolean Space
In a 3-cube (three variables):
0-cube, i.e., a single node, yields a term in 3 literals
1-cube, i.e., a line of two nodes, yields a term in 2 literals
2-cube, i.e., a plane of four nodes, yields a term in 1 literal
3-cube, i.e., a cube of eight nodes, yields a constant term "1"
In general,
m-subcube within an n-cube (m < n) yields a term with n – m literals
CS Spring – Lec #3: Combinational Logic - 43

44. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 44
Karnaugh Maps
Flat map of Boolean cube
Wrap–around at edges
Hard to draw and visualize for more than 4 dimensions
Virtually impossible for more than 6 dimensions
Alternative to truth-tables to help visualize adjacencies
Guide to applying the uniting theorem
On-set elements with only one variable changing value are adjacent unlike the situation in a linear truth-table
A B F
0 0 1
0 1 0
1 0 1
1 1 0
0 2
1 3 1 A B
CS Spring – Lec #3: Combinational Logic - 44

45. Karnaugh Maps (cont’d)
Numbering scheme based on Gray–code
e.g., 00, 01, 11, 10
Only a single bit changes in code for adjacent map cells
0 2
1 3 00 01 AB C 1
6 4
7 5 11 10 B A
0 4
1 5
12 8
13 9 D A
3 7
2 6
15 11
14 10 C B
0 2
1 3
6 4
7 5 C B A
13 = 1101= ABC’D
CS Spring – Lec #3: Combinational Logic - 45

46. Adjacencies in Karnaugh Maps
Wrap from first to last column
Wrap top row to bottom row
011
111 C B A
110
010
001 B C
101
000
100 A
CS Spring – Lec #3: Combinational Logic - 46

47. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 47
Karnaugh Map Examples
F =
Cout =
f(A,B,C) = m(0,4,6,7)
1 1
0 0 B A B’
0 0
0 1
1 0
1 1
Cin B A
+ BCin
+ ACin AB
1 0
0 0
0 1
1 1 C B A
+ B’C’
obtain the complement of the function by covering 0s with subcubes
+ AB’ AC
CS Spring – Lec #3: Combinational Logic - 47

48. More Karnaugh Map Examples
0 0
1 1 C B A A
G(A,B,C) =
1 0
0 0
0 1
1 1 C B A
= AC + B’C’
F(A,B,C) = m(0,4,5,7)
0 1
1 1
1 0
0 0 C B A
F' simply replace 1's with 0's and vice versa
= BC’ + A’C
F'(A,B,C) =  m(1,2,3,6)
CS Spring – Lec #3: Combinational Logic - 48

49. Karnaugh Map: 4-Variable Example
F(A,B,C,D) = m(0,2,3,5,6,7,8,10,11,14,15) F =
+ B’D’ C
+ A’BD A A B C D
0000
1111
1000
0111
1 0
0 1
0 0
1 1 C D B
find the smallest number of the largest possible subcubes to cover the ON-set
(fewer terms with fewer inputs per term)
CS Spring – Lec #3: Combinational Logic - 49

50. Karnaugh Maps: Don’t Cares
f(A,B,C,D) = m(1,3,5,7,9) + d(6,12,13)
without don't cares
f =
+ B’C’D
A’D
0 0
1 1
X 0
X 1 D A
0 X B C
CS Spring – Lec #3: Combinational Logic - 50

51. Karnaugh Maps: Don’t Cares (cont’d)
f(A,B,C,D) = m(1,3,5,7,9) + d(6,12,13)
f = A'D + B'C'D without don't cares
f = with don't cares
A'D
by using don't care as a "1" a 2-cube can be formed rather than a 1-cube to cover this node
+ C'D
0 0
1 1
X 0
X 1 D A
0 X B C
don't cares can be treated as 1s or 0s depending on which is more advantageous
CS Spring – Lec #3: Combinational Logic - 51

52. Design Example: Two-bit Comparator
A B C D LT EQ GT
LT EQ GT
A B < C D A B = C D A B > C D A B C D N1 N2
block diagram
and
truth table
we'll need a 4-variable Karnaugh map for each of the 3 output functions
CS Spring – Lec #3: Combinational Logic - 52

53. Design Example: Two-bit Comparator (cont’d)
B C' D' + A C' + A B D'
0 0
1 0 D A
1 1
0 1 B C
1 0
0 1
0 0 D A B C
0 1
0 0
1 1 D A
1 0 B C
A'B'C'D' + A'BC'D + ABCD + AB'CD’
A' B' D + A' C + B' C D
K-map for LT
K-map for EQ
K-map for GT
LT =
EQ =
GT =
= (A xnor C) • (B xnor D)
LT and GT are similar (flip A/C and B/D)
CS Spring – Lec #3: Combinational Logic - 53

54. Design Example: Two-bit Comparator (cont’d)
A B C D EQ
two alternative
implementations of EQ
with and without XOR
XNOR is implemented with at least 3 simple gates
CS Spring – Lec #3: Combinational Logic - 54

55. Design Example: 2x2-bit Multiplier
A2 A1 B2 B1 P8 P4 P2 P1 P1
P2 P4 P8 A1
A2 B1 B2
block diagram
and
truth table
4-variable K-map
for each of the 4
output functions
CS Spring – Lec #3: Combinational Logic - 55

56. Design Example: 2x2-bit Multiplier (cont’d)
0 0 B1 A2
1 0 A1 B2
0 0 B1 A2
0 1
1 1 A1 B2
K-map for P8
K-map for P4
P4 = A2B2B1' + A2A1'B2
P8 = A2A1B2B1
0 0
1 1 B1 A2
0 1
1 0 A1 B2
0 0
0 1
1 0 B1 A2 A1 B2
K-map for P2
K-map for P1
P1 = A1B1
P2 = A2'A1B2 + A1B2B1' + A2B2'B1 + A2A1'B1
CS Spring – Lec #3: Combinational Logic - 56

57. Design Example: BCD Increment by 1
I8 I4 I2 I1 O8 O4 O2 O
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X O1
O2 O4 O8 I1
I2 I4 I8
block diagram
and
truth table
4-variable K-map for each of the 4 output functions
CS Spring – Lec #3: Combinational Logic - 57

58. Design Example: BCD Increment by 1 (cont’d)
0 0
X 1
X 0 I1 I8
0 1
X X I4 I2
O8 = I4 I2 I1 + I8 I1'
O4 = I4 I2' + I4 I1' + I4’ I2 I1 O2 = I8’ I2’ I1 + I2 I1'
O1 = I1'
0 1
X 0 I1 I8
1 0
X X I4 I2 O8 O4
0 0
1 1
X 0 I1 I8
X X I4 I2
1 1
0 0
X 1
X 0 I1 I8
X X I4 I2 O2 O1
CS Spring – Lec #3: Combinational Logic - 58

59. Definition of Terms for Two-level Simplification
Implicant
Single element of ON-set or DC-set or any group of these elements that can be combined to form a subcube
Prime implicant
Implicant that can't be combined with another to form a larger subcube
Essential prime implicant
Prime implicant is essential if it alone covers an element of ON-set
Will participate in ALL possible covers of the ON-set
DC-set used to form prime implicants but not to make implicant essential
Objective:
Grow implicant into prime implicants (minimize literals per term)
Cover the ON-set with as few prime implicants as possible (minimize number of product terms)
CS Spring – Lec #3: Combinational Logic - 59

60. Examples to Illustrate Terms
1 1
1 0 D A
0 0 B C
6 prime implicants: A'B'D, BC', AC, A'C'D, AB, B'CD
essential
minimum cover: AC + BC' + A'B'D
0 0
1 1
1 0 D A
0 1 B C
5 prime implicants: BD, ABC', ACD, A'BC, A'C'D
essential
minimum cover: 4 essential implicants
CS Spring – Lec #3: Combinational Logic - 60

61. Basic Two-Level Logic Simplification Algorithm
Enumerate all the minterms of the function
Enumerate all the primes of the function
Form the “Covering Table” of Primes vs Minterms
Choose the smallest set of primes that cover all the minterms
Every minterm is contained in at least one prime
CS Spring – Lec #3: Combinational Logic - 61

62. Ex: F=a’b’d’ + abd’ +a’b
ON = (0,2,4,5,6,7,12,14)
DC=15
OFF=(1,3,8,9,10,11,13)
CS Spring – Lec #3: Combinational Logic - 62

63. Ex: F=a’b’d’ + abd’ +a’b
Primes = a’d’,bd’,a’b,bc
CS Spring – Lec #3: Combinational Logic - 63

64. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 64
Covering Table
a’d’
0XX0
bd’
X1X0
a’b
01XX bc
X11X
0 (0000) 1
2 (0010)
4 (0100)
5 (0101)
6 (0110)
7 (0111)
12 (1100)
14 (1110)
Rows are minterms
Columns are primes
“1” in Square x/y if prime y covers minterm x
E.g., “1” in Square 6/01xx because prime a’b contains a’bcd’ (minterm 6)
CS Spring – Lec #3: Combinational Logic - 64

65. Solving the Covering Table
a’d’
0XX0
bd’
X1X0
a’b
01XX bc
X11X
0 (0000) 1
2 (0010)
4 (0100)
5 (0101)
6 (0110)
7 (0111)
12 (1100)
14 (1110)
Choose minimum set of columns so that every row is “covered” by a column
“Covered” == has a ‘1’ in some chosen column
CS Spring – Lec #3: Combinational Logic - 65

66. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 66
“Dominated” Rows
Row x dominates row y if covering x automatically covers y
Primes covering x are a subset of primes covering y
Prime 1
Prime 2
Prime 3
Minterm x 1
Minterm y
Must choose Prime 1 or Prime 2 to cover x
But then y automatically covered
=> x dominates y
CS Spring – Lec #3: Combinational Logic - 66

67. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 67
“Dominated” Columns
Column 1 dominates Column 2 if Column covers every row covered by Column 2
Minterms covered by 1 are a superset of minterms covered by 2
Prime 1
Prime 2
Minterm x 1
Minterm y
Minterm z
Choose prime 1 = > cover x or y or z
Choose prime 2 => cover x or z
Always better to choose 1 over 2 => 1 dominates 2
CS Spring – Lec #3: Combinational Logic - 67

68. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 68
Algorithm (again)
While there are dominated rows or columns
Eliminate dominated rows
Eliminate dominated columns ….
K-ary Essential Primes
Cyclic core 1 1 1 1 1 1
Solve Cyclic Core by branch-and-bound
Solution is k-ary essential primes + Cyclic Core Solution
CS Spring – Lec #3: Combinational Logic - 68

69. Solving the Covering Table
a’d’
0XX0
bd’
X1X0
a’b
01XX bc
X11X
0 (0000) 1
2 (0010)
4 (0100)
5 (0101)
6 (0110)
7 (0111)
12 (1100)
14 (1110)
Step 1: Eliminate “dominated” rows
Rows automatically “covered” by another
Ex: row 4 dominated by rows 0 and 2
Cover 0 => automatically cover 4
CS Spring – Lec #3: Combinational Logic - 69

70. Solving the Covering Table
a’d’
0XX0
bd’
X1X0
a’b
01XX bc
X11X
0 (0000) 1
2 (0010)
5 (0101)
12 (1100)
Done! (Not usually so lucky)
Minimum cover is a’d’+bd’+a’b
CS Spring – Lec #3: Combinational Logic - 70

71. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 71
What can go wrong?
Too Many primes!
Can have as many as 3n/n (McMullen)
Too Many Minterms!
In general, have about 2n/k
All subproblems are NP-Hard
Is cube c a prime of F?
Find the minimum cover of a covering table
NP-Hard == Computer Scientist-speak for “no fast algorithm exists (we believe)”
Best known algorithm = “try all solutions”
2**(3**n/n) possible solutions!
n = 4 implies 2**20 (approx 1,000,000) solutions
n = 5 implies 2**49 (approx 10,000,000,000,000,000,000 solutions)
Easy to generate problems with 100 or so variables
CS Spring – Lec #3: Combinational Logic - 71

72. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 72
What do we do about it?
In practice, things are not so bad!
Circuits people actually build usually have “reasonable” number of primes
Covering tables for circuits people actually build are usually easy to solve
=> Use algorithms that are in practice efficient for generating primes, covering table, solve covering table
Use heuristics
Don’t necessarily need minimum form
Given one sum-of-products, generate a better form
Iterate until we run out of time/get answer that is good enough
CS Spring – Lec #3: Combinational Logic - 72

73. What do we mean by “usually”?
Circuits People
Actually build
Circuits we can
prove we do
well on
Circuits we in practice do well on
CS Spring – Lec #3: Combinational Logic - 73

74. What do we mean by “usually”?
All 2^2^n n-input circuits
Circuits we see…
CS Spring – Lec #3: Combinational Logic - 74

75. We See only a tiny fraction of circuits
Complexity results mostly refer to circuits no one ever builds!
Humans only build a tiny fraction of circuits (say, less than 1,000,000 of 2^2^ input circuits)
Big discovery: mathematically characterize non-trivial circuits that (a) we build; and (b) our algorithms work well on
Nobody knows how to do this!
Until then…use our best algorithms on stuff we know how to build
CS Spring – Lec #3: Combinational Logic - 75

76. Example: Prime Generation
Problem: “in practice” efficiently generate all primes for a function
Solution: Use recursive-evaluation method on “cofactors”
CS Spring – Lec #3: Combinational Logic - 76

77. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 77
Cofactors
Partial evaluation of function
Fa = F evaluated at a=1
Fa’ = F evaluated at a=0
Example: F=a’b’d’ + abd’ +a’b
Fa = bd’
Fa’ = b’d’ + b
For any function F, variable a:
F = a Fa + a’ Fa’ (“Shannon Expansion”)
Example: F = a bd’ + a’(b’d’ + b) [Check it!]
CS Spring – Lec #3: Combinational Logic - 77

78. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 78
Cofactors and Primes
Theorem: Any Function F, variable a, Primes are maximal implicants of set:
{a Primes(Fa), a’ Primes(Fa’), Primes(Fa) Primes(Fa’)}
In other words
Find the primes of Fa, multiply each by a
Find the primes of Fa’, multiply each by a’
Multiply each prime of Fa by each prime of Fa’
Maximal members of these sets are primes of F
CS Spring – Lec #3: Combinational Logic - 78

79. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 79
Example
0 0
1 1
0 0
1 1
X 0
X 1 D A
0 X B C
Primes(Fa’) = d, bc
Fa’ =
F =
1 1
0 X
Primes(F) = a’d, a’bc, abc’,ac’d’, bc’d A
X 0
X 1
Primes(Fa) = bc’,c’d’
Fa =
0 0
CS Spring – Lec #3: Combinational Logic - 79

80. Similar Fast Techniques to Generate Rows
Recurse down cofactor tree, throwing out primes when they disagree
Stop when:
Cofactor of on-set is empty (no rows)
All the remaining primes cover the whole space (1 row, with those primes)
Still one very fast technique, “Espresso-Signature” (to follow approximate algorithm)
CS Spring – Lec #3: Combinational Logic - 80

81. Approximate Algorithm
Don’t try to generate minimum cover
Given a cover (sum-of-products for on-set and don’t-care set)
Generate a prime cover smaller than given cover
Keep generating smaller prime covers until you can’t anymore
Program – Espresso-II
CS Spring – Lec #3: Combinational Logic - 81

82. Espresso-II algorithm (approx)
Given Cover C = c1 + c2 +…+cn
While (we keep making improvement)
PrimeCover = 0
Foreach cube c_i of the cover, C
Find a prime p_i containing c_i that contains as many c_j’s as possible
PrimeCover = PrimeCover + p_i
We now have a prime cover p1+…+pn, Make the cover “irredundant” (example next slide). Result is a smaller prime cover
“Reduce” each cube away from a prime while still retaining a cover (this permits a different expansion in next iteration)
CS Spring – Lec #3: Combinational Logic - 82

83. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 83
Espresso-II example
F=a’b’d’ + abc’d’ + abc+ a’b, DC = abcd
Expand:
a’b’d’ => a’d’
abc’d’ => bd’
abc => bc
a’b => a’b
Prime Cover a’d’ + bd’ + bc + a’b
Prime bc is redundant: covered by bd’ + a’b + don’t-care
Irredundant cover is:
a’d’ + bd’ + a’b
Reduce to a’b’d’ + bd’ + a’bd for next iteration
CS Spring – Lec #3: Combinational Logic - 83

84. Espresso-II presented here very simplified
See text for fuller description
What I left out
Details of EXPAND (how to pick the primes)
Details of IRREDUNDANT
Details of REDUCE
ESSENTIAL_PRIMES (identified early and tossed into don’t-care set for efficiency)
LAST_GASP (radical reduction to get new starting point)
MAKESPARSE (final pruning of literals in the output plane)
CS Spring – Lec #3: Combinational Logic - 84

85. Return to Exact Methods
Realization (mid-nineties): The only purpose of generating primes was to get to covering table!
Specifically: to get to “non-dominated” rows of covering table
Realization (beyond the scope of this class)
Can identify non-dominated rows without generating primes or minterms!
“Espresso Signature”: first radical revision of exact two-level synthesis algorithm
a’d’
0XX0
bd’
X1X0
a’b
01XX bc
X11X
0 (0000) 1
2 (0010)
5 (0101)
12 (1100)
CS Spring – Lec #3: Combinational Logic - 85

86. CS 150 - Spring 2008 – Lec #3: Combinational Logic - 86
Espresso-Signature
Espresso-Signature Algorithm
Generate non-dominated rows of covering table
This forms a cover, the “canonical cover” of a logic function
Use modified Espresso EXPAND to generate non-dominated columns of covering table
Solve the covering problem
Much more efficient than standard algorithm
CS Spring – Lec #3: Combinational Logic - 86

87. Two-Level Synthesis Summary
Two exact methods: Quine-McCluskey and Espresso-Signature
In practice, both highly efficient
Theoretically: exponential complexity
Algorithms work well because they fit circuits people actually design (though this set is not characterized well)
One approximate method shown, Espresso
Others (McBoole, e.g.) similar
Also theoretically exponential complexity!
In practice, fast, gives results within a few percent of optimum.
CS Spring – Lec #3: Combinational Logic - 87

88. Implementations of Two-level Logic
Sum-of-products
AND gates to form product terms (minterms)
OR gate to form sum
Product-of-sums
OR gates to form sum terms (maxterms)
AND gates to form product
CS Spring – Lec #3: Combinational Logic - 88

89. Two-level Logic Using NAND Gates
Replace minterm AND gates with NAND gates
Place compensating inversion at inputs of OR gate
CS Spring – Lec #3: Combinational Logic - 89

90. Two-level Logic Using NAND Gates (cont’d)
OR gate with inverted inputs is a NAND gate
de Morgan's: A' + B' = (A • B)'
Two-level NAND-NAND network
Inverted inputs are not counted
In a typical circuit, inversion is done once and signal distributed
CS Spring – Lec #3: Combinational Logic - 90

91. Two-level Logic Using NOR Gates
Replace maxterm OR gates with NOR gates
Place compensating inversion at inputs of AND gate
CS Spring – Lec #3: Combinational Logic - 91

92. Two-level Logic Using NOR Gates (cont’d)
AND gate with inverted inputs is a NOR gate
de Morgan's: A' • B' = (A + B)'
Two-level NOR-NOR network
Inverted inputs are not counted
In a typical circuit, inversion is done once and signal distributed
CS Spring – Lec #3: Combinational Logic - 92

93. Two-level Logic Using NAND and NOR Gates
NAND-NAND and NOR-NOR networks
de Morgan's law: (A + B)' = A' • B' (A • B)' = A' + B'
written differently: A + B = (A' • B')’ (A • B) = (A' + B')'
In other words ––
OR is the same as NAND with complemented inputs
AND is the same as NOR with complemented inputs
NAND is the same as OR with complemented inputs
NOR is the same as AND with complemented inputs OR OR
AND
AND
NAND
NAND
NOR
NOR
CS Spring – Lec #3: Combinational Logic - 93

94. Conversion Between Forms
Convert from networks of ANDs and ORs to networks of NANDs and NORs
Introduce appropriate inversions ("bubbles")
Each introduced "bubble" must be matched by a corresponding "bubble"
Conservation of inversions
Do not alter logic function
Example: AND/OR to NAND/NAND A B C D Z A B C D
NAND
CS Spring – Lec #3: Combinational Logic - 94

95. Conversion Between Forms (cont’d)
Example: verify equivalence of two forms A B C D Z A B C D Z
NAND
Z = [ (A • B)' • (C • D)' ]'
= [ (A' + B') • (C' + D') ]'
= [ (A' + B')' + (C' + D')' ]
= (A • B) + (C • D) ü
CS Spring – Lec #3: Combinational Logic - 95

96. Conversion Between Forms (cont’d)
Example: map AND/OR network to NOR/NOR network A B C D Z
NOR A B C D Z
NOR \A \B \C \D Z
Step 1
Step 2
conserve
"bubbles"
conserve
"bubbles"
CS Spring – Lec #3: Combinational Logic - 96

97. Conversion Between Forms (cont’d)
Example: verify equivalence of two forms
NOR \A \B \C \D Z A B C D Z
Z = { [ (A' + B')' + (C' + D')' ]' }'
= { (A' + B') • (C' + D') }'
= (A' + B')' + (C' + D')'
= (A • B) + (C • D) ü
CS Spring – Lec #3: Combinational Logic - 97

98. Combinational Logic Summary
Logic functions, truth tables, and switches
NOT, AND, OR, NAND, NOR, XOR, . . ., minimal set
Axioms and theorems of Boolean algebra
Proofs by re-writing and perfect induction
Gate logic
Networks of Boolean functions and their time behavior
Canonical forms
Two-level and incompletely specified functions
Simplification
Two-level simplification
CS Spring – Lec #3: Combinational Logic - 98