Two Level Networks. Two-Level Networks Slide 2 SOPs A function has, in general many SOPs Functions...

Two Level Networks

Two-Level Networks Slide 2

SOPs

• A function has, in general many SOPs• Functions can be simplified using Boolean

algebra

Compare theimplementationof the original to the finalrepresentation


Don’t Care Conditions

• In practice there are combinations that will never occur

• we may pick the most convenient assignment• A BCD input is greater than 5

ABCD F

0000 0

0001 0

0010 0

0011 0

0100 0

0101 0

0110 1

0111 1

ABCD F

1000 1

1001 1

1010 -

1011 -

1100 -

1101 -

1110 -

1111 -

F = m(6,7,8,9)+ d(10,11,12,13,14,15)

don’t cares = 0 F= ABC + ABC

don’t cares = 1 F= BC + A


Karnaugh MapBC

00 01 11 10

A 0 m0 m1 m3 m2

1 m4 m5 m7 m6ABC F

000 0

001 1

010 0

011 1

100 0

101 1

110 1

111 1

BC

00 01 11 10

A 0 0 1 1 0

1 0 1 1 1

F = C + AB

Adjacent cells are at Hamming distance 1


Examples

• F = m(1,4,5)+ d(0)• F = m(0,3,5,6)• F = m(0,2,3,4,5,7)


4-variable Karnaugh MapsCD

00 01 11 10

AB

00 m0 m1 m3 m2

01 m4 m5 m7 m6

11 m12 m13 m15 m14

10 m8 m9 m11 m10

•F = m(1,4,5,6,14)+ d(3,7)•F = m(0,4,5,6,9,12,13)•F = m(0,2,5,6,7,8,10,13)


Prime Implicants• Given two logic functions f and g, if g(x) = 1 for all x

such that f(x) = 1, then g contains f, denoted f ≤ g.• If a logic function f contains a product c, then c is an

implicant of f. Furthermore, if c is a minterm, then c is a minterm of f.

• A product P is a sub-product of Q, if all the literals in P also appear in Q.

• Let P be an implicant of a logic function f. If no other implicant Q of f is a sub-product of P, then P is a prime implicant (PI) of f.


Prime Implicants in K maps

• largest blocks (loops) of 2i 1-cells (may include don’t care cells)

• Show examples


Minimum SOP• Consider an SOP that consists of PIs. If any SOP that is

obtained by removing any of the PIs does not represent the original function f, then the original SOP is an irredundant sum-of-product expression (ISOP) of f.

• Among the ISOPs for f, one with the minimum number of products is a minimum sum-of-product expression (MSOP). Among the MSOPs for f, one with the minimum number of literals is a exact minimum sum-of-product expression (exact MSOP) of f.

• Let m be a minterm of a logic function f. If only one PI of f covers m, the m is a distinguished minterm. Among all the PIs of f, if there is only one that covers m, then it is an essential prime implicant.


Example

• F = m(2,5,6,7,10,11)

F = ABD + ABC + ACD + BCD + ABC (SOP)

F = ABD + ABC + BCD + ABC (ISOP)

F = ABC + ACD + ABC (MSOP)


Quine-McCluskey (Tabular) Minimization

• Two step process utilizing tabular listings to: – Identify prime implicants (implicant tables)– Identify minimal cover (cover tables)

• All work is done in tabular form– Number of variables is not a limitation

• It does affect the size of the table– Basis for many computer implementations

• Proper organization and term identification are key factors for correct results


Quine-McCluskey Minimization

• Minterms are initially listed one per line in groups– Each group contains terms with the same number of

true and complemented variables– Terms are listed in numerical order within group

• Terms and implicants are identified using one of three common notations– full variable form– cellular form– 1,0,- form


Notation Forms• Full variable form - variables and complements

in algebraic form– hard to identify when adjacency applies– very easy to make mistakes

• Cellular form - terms are identified by their decimal index value– Easy to tell when adjacency applies; indexes must

differ by power of two (one bit)– Implicants identified by term nos. separated by

comma; differing bit pos. in () following terms


Notation Forms (cont.)

• 1,0,- form - terms are identified by their binary index value– Easier to translate to/from full variable form– Easy to identify when adjacency applies, one bit is

different– - shows variable(s) dropped when adjacency is

used

• Different forms may be mixed during the minimization


Example of Different NotationsF(A, B, C, D) = m(4,5,6,8,10,13)

1 ABCD 4 0100ABCD 8 1000

2 ABCD 5 0101 ABCD 6 0110 ABCD 10 10103 ABCD 13 1101

Full variable Cellular 1,0,-


Implication Table

• Quine-McCluskey Method– Tabular method to

systematically find all prime implicants

– ƒ(A,B,C,D) = Σ m(4,5,6,8,9, 10,13) + Σ d(0,7,15)

– Part 1: Find all prime implicants– Step 1: Fill Column 1 with

active-set and DC-set minterm indices. Group by number of true variables.

Implication Table

Column I

0000 0100 1000 0101 0110 1001 1010 0111 1101

1111

NOTE: DCs are included in this step!


Minimization - First Pass• Quine-McCluskey Method

– Tabular method to systematically find all prime implicants

– ƒ(A,B,C,D) = Σ m(4,5,6,8,9,10,13) + Σ d(0,7,15)

– Part 1: Find all prime implicants– Step 2: Apply Adjacency - Compare

elements of group with N 1's against those with N+1 1's. One bit difference implies adjacent. Eliminate variable and place in next column.

E.g., 0000 vs. 0100 yields 0-00

0000 vs. 1000 yields -000

When used in a combination, mark with a check. If cannot be combined, mark with a star. These are the prime implicants.

Repeat until nothing left.

Implication Table

Column I

0000 0100 1000 0101 0110 1001 1010 0111 1101

1111

Column II

0-00 -000 010- 01-0 100- 10-0 01-1 -101 011- 1-01

-111 11-1


Minimization - Second Pass

Implication Table

Column I

0000 0100 1000 0101 0110 1001 1010 0111 1101

1111

Column II

0-00 * -000 * 010- 01-0 100- * 10-0 * 01-1 -101 011- 1-01 *

-111 11-1

Column III 01-- * -1-1 *

• Quine-McCluskey Method– Step 2 cont.: Apply Adjacency - Compare

elements of group with N 1's against those with N+1 1's. One bit difference implies adjacent. Eliminate variable and place in next column.

E.g., 0000 vs. 0100 yields 0-00

0000 vs. 1000 yields -000

When used in a combination, mark with a check. If cannot be combined, mark with a star. These are the prime implicants.

Repeat until nothing left.


Prime ImplicantsPrime Implicants:

D C A 00-0 =

C BA -100 =

D C B 000- =

D BA 0-10 =

D CA 01-1 = D B 1-1- =

B A --01 =

X 1 0 1

0 1 1 1

0 X X 0

0 1 0 1

AB 00 01 11

01

11

10 C

CD

A

D

B

00

10


Stage 2: find smallest set of prime implicants that cover the active-set

recall that essential prime implicants must be in final expression

Prime Implicants (cont.)Prime Implicants:

D C A 00-0 =

C BA -100 =

D C B 000- =

D BA 0-10 =

D CA 01-1 = D B 1-1- =

B A --01 =

X 1 0 1

0 1 1 1

0 X X 0

0 1 0 1

AB 00 01 11

01

11

10 C

CD

A

D

B

00

10


rows = prime implicantscolumns = ON-set elementsplace an "X" if ON-set element is covered by the prime implicant

Coverage Chart

0,4(0-00)

0,8(-000)

8,9(100-)

8,10(10-0)

9,13(1-01)

4,5,6,7(01--)

5,7,13,15(-1-1)

4X

X

5

X

X

6

X

8

X

X

X

9

X

X

10

X

13

X

X

Coverage Table

Note: Don’t include DCs in coverage table; they don’t have to be covered by the final logic expression!


Coverage Table (cont.)

rows = prime implicantscolumns = ON-set elementsplace an "X" if ON-set element is covered by the prime implicant

If column has a single X, than theimplicant associated with the rowis essential. It must appear inminimum cover

0,4(0-00)

0,8(-000)

8,9(100-)

8,10(10-0)

9,13(1-01)

4,5,6,7(01--)

5,7,13,15(-1-1)

4X

X

5

X

X

6

X

8

X

X

X

9

X

X

10

X

13

X

X

0,4(0-00)

0,8(-000)

8,9(100-)

8,10(10-0)

9,13(1-01)

4,5,6,7(01--)

5,7,13,15(-1-1)

4X

X

5

X

X

6

X

8

X

X

X

9

X

X

10

X

13

X

X

Coverage Chart



Eliminate all columns covered byessential primes

0,4(0-00)

0,8(-000)

8,9(100-)

8,10(10-0)

9,13(1-01)

4,5,6,7(01--)

5,7,13,15(-1-1)

4X

X

5

X

X

6

X

8

X

X

X

9

X

X

10

X

13

X

X



Eliminate all columns covered byessential primes

Find minimum set of rows thatcover the remaining columns

0,4(0-00)

0,8(\000)

8,9(100-)

8,10(10-0)

9,13(1-01)

4,5,6,7(01--)

5,7,13,15(-1-1)

4X

X

5

X

X

6

X

8

X

X

X

9

X

X

10

X

13

X

X

0,4(0-00)

0,8(-000)

8,9(100-)

8,10(10-0)

9,13(1-01)

4,5,6,7(01--)

5,7,13,15(-1-1)

4X

X

5

X

X

6

X

8

X

X

X

9

X

X

10

X

13

X

X

BA DCA DBA F ++= BA DCA DBA F ++=


Cover Table Reduction

1. Detection of essential rows

2. Elimination of dominated rows

3. Elimination of dominated columns

4. Detection of Pseudo-essential rows

5. repeat step 2 to 4 while reductions are possible


Solution Methods for Cyclic Tables

• Example F = m(0,2,3,4,5,7)• naïve algorithm

1. Assume that a row is in the final solution, reduce the covering, and obtain a minimum cover

2. Assume that a row is not in the final solution, reduce the covering, and obtain a minimum cover

3. select the lower from (1) and (2)


Branch and Bound

1. Brancha) Choose one row from the covering table. Let it be A.b) Assume A is in the final solution---find minimal coverc) Assume A is not in the final solution---find minimal

coverd) Select smaller of (b) and (c)

2. Bounda) if the cost of the solution is equal to the lower bound of

the cost of the minimum cover, then stop the search.b) if the solution of the current problems cannot be better

than the best solution found, then stop the search.


Algebraic Method for Cyclic Tables

ma mb mc md me

P1 X X X

P2 X X

P3 X X

P4 X X

P5 X X

P6 X X

Let gi = 1 iff Pi is present in the SOP


Algebraic Method for Cyclic Tables

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P1 + P5 is the MSOP

Two Level Networks. Two-Level Networks Slide 2 SOPs A function has, in general many SOPs Functions...

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