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EECS 465: Digital Systems. Lecture Notes # 2. Two-Level Minimization Using Karnaugh Maps. SHANTANU DUTT. Department of Electrical & Computer Engineering University of Illinois, Chicago Phone: (312) 355-1314: e-mail: dutt@ece.uic.edu URL: http://www.ece.uic.edu/~dutt. 2.

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eecs 465 digital systems
EECS 465: Digital Systems

Lecture Notes # 2

Two-Level Minimization Using Karnaugh Maps

SHANTANU DUTT

Department of Electrical & Computer Engineering

University of Illinois, Chicago

Phone: (312) 355-1314: e-mail: dutt@ece.uic.edu

URL: http://www.ece.uic.edu/~dutt

slide2

2

Concepts in 2-Level Minimization

  • Defn: An implicant g of a function is a product term (e.g., g=xyz) s.t. when g=1  f=1
  • Defn: A minterm is an implicant whose product term representation consists of all n variables of an n-var. function (each in either compl.or uncompl. form)
  • Defn: Let g, and h be 2 implicants of a function f. g is said to cover h if h=1  g=1 (e.g., g = xz, h = xyz)
  • Defn: Let G = {g1, .., gk} be a set of implicants of func f, and let h be another implicant of f. Then G is covers h if h=1  g1+ .. + gk =1
  • Defn: An SOP or POS expression is also called a 2-level expression
  • Defn: A 2-level AND-OR (OR-AND) gate realizationof an SOP
  • (POS) expression is one in which all product terms (OR
  • terms in POS) in the SOP (POS) expression are implemented
  • by multiple input AND (OR) gates & the ORing (ANDing)
  • of the product (OR) terms is realized by a multiple input OR
  • (AND) gate.
slide3

3

E.g.

f1 = AB + BC + ACD --- 2-level. (SOP)

f2 = ( B+C )( A+D )( C+D ) --- 2-level (POS)

f3 =AB + AC( B+D ) --- not SOP or POS

level 1 I/Ps --- not 2-level

--- 3 level ( can be realized

directly by 3 levels of gates)

A

B

A

f3

C

B

level 2

I/Ps

D

level 2

gates

level 3

gate

level 0

I/Ps

level 1

gates

slide4

4

A

AB + BC + ACD

B

f1

B

C

A

C

Level 2

D

Level 1

B

(B + C)(A + D)(C + D)

C

f2

A

D

C

D

slide5

5

  • Defn.

A gate in a logic circuit is a level-1 gate if all its inputs are

primary inputs (i.e., literals, A, A, B, B, etc.)

A gate is a level-i gate i > 1 if the highest-level gate whose

output feeds the (level-i) gate is a level-(i-1) gate.

  • NOTE:

If a given expr. is not 2-level, we can convert it to a

2-level one using the distributive law.

The goal of 2-level minimization is to minimize the number of

literals (a literal is a var. in complemented or uncompl. form)

in a 2-level logic expr..

This approx. reduces the total # of inputs over all gates in the

circuit in a 2-level gate realization.

slide6

6

E.g. f = ABC + ABC + ABC (non-minimized 9 literals)

A

B

(12 gate I/Ps)

= Circuit complexity

C

A

f

B

C

A

B

C

f = AB + AC (minimized 4 literals)

A

f

B

(6 gate I/Ps)

= Circuit complexity

A

C

Note: Minimizing # of literals + # of product terms  (approx) minimizing total # of gate inputs

slide7

7

Multilevel Minimization (Example):

f = x1x2x3 + x2x3x4 + x1x5 + x4x5 (2-level minimized)

Apply distributive law (factoring)

f = x2x3(x1 + x4) + x5(x1 + x4)

= (x2x3 + x5)(x1 + x4) -- Multilevel expr.

Gate Impl.:

x2

x3

f

x5

Multilevel

circuit

x1

x4

slide8

8

Defn.

Two implicants (or product terms) of a function f are said to

be adjacent (logically) if they have the same literals

except in one variable xi which occurs in uncomplemented form

(xi) in one implicant and in complemented form (xi) in the

other. The 2 implicants are said to differ in xi.

E.g.

ABC, ABC (adjacent implicants, differ in B)

ABD, ABC (not adjacent)

NOTE:

Adjacent implicants can be combined into one implicant

by the combining theorem (ABC + ABC = AC)

slide9

9

Karnaugh Maps

2-variable TT outputs

A

B

Z1

Z2

Binary

place-value

ordering

(Binary

ordering)

0 0 0 0

Physically adjacent

but not logically

adjacent.

0 1 1 1

1 0 0 1

1 1 1 0

logically adjacent but not physically adjacent

AB + AB

B

Another ordering of inputs ( Gray-code ordering)

A B Z1

0 0 0

0 1 1

1 1 1

1 0 0

Z1 = B

Physically as well as

logically adjacent.

slide10

10

n-variable Gray-code ordering

Defn.

Gi = i-bit Gray-code ordering

rev(Gi) = reverse order of Gi

Gn = 0 Gn-1

1 rev(Gn-1)

G1 = 0 Base.

1

G2 = 0 G1

1 rev(G1)

0 0

0 1

1 1

1 0

A B C

0 0 0

0 0 1

0 1 1

0 1 0

1 1 0

1 1 1

1 0 1

1 0 0

=

G3 = 0 G2

1 rev(G2)

=

Convert to a 2-dimensional

Gray-code ordered TT

x

110

AB

C

00 01 11 10

y

0

1

Throw away variable (A)

changing across the 2 squares.

We thus obtain BC.

1

1

001

slide11

11

Example ( 3-var. K-Map)

Consensus = BC

AB

C

Function f:

00 01 11 10

0

1

1 1 1

1 1

f = AB + AB + AC + BC

Four Prime Implicants AB, BC, AC, AB formed. Not all these PIs

are needed in the expression. Only a minimum set that covers all

minterms is needed.

Defn. A Prime Implicant (PI) of a function f is an implicant of f that

is not covered by any other implicant of f.

Defn. An Essential PI is a PI that covers/includes at least one minterm

that is not covered by any other PI.

slide12

12

In the above example, AB & AB are essential PIs. These will need

to be present in any SOP expr. of the function. To form a minimal

set, the PI BC can be used. Thus f = AB + AB + BC is a minimized

expression.

Larger than 2-minterm PIs can also exist:

AB

AB

E.g.,

C

C

00 01 11 10

00 01 11 10

0

1

1 1

1 1

0

1

1 1 1 1

AB

AB

B

C

B

2-minterm implicants (AB, AB in the 1st example above) can

be merged if they are logically adjacent to form a 4-minterm

implicant and so on.

slide13

13

When an implicant can not be “grown” any further, then it is a PI.

Defn. In a K-map, 2 implicants can be logically adjacent if they

are symmetric,

i.e., if : (1) They are disjoint (no common minterms).

(2) They cover the same # of minterms.

(3) Each minterm in one implicant is adjacent to a

unique minterm in the other implicant.

AB

More Examples:

Both PIs

(A, BC)

are

essential.

C

00 01 11 10

AB

Redrawn

0

1

00 01 11 10

1 1

1 1 1

C

0

1

1 1

1 1 1

Not symmetric

Thus, f = A + BC

is minimized.

Symmetric ( logically adjacent)

2-minterm implicants : Can be “merged” to form

a 4-minterm implicant, which will be a PI.

slide14

14

4-variable K-Map:

AB

CD

00 01 11 10

00

01

11

10

slide15

15

Example:

f = AC + D

Instead of starting with 2-minterm

implicants & growing them, you

can form larger implicants directly

by grouping power of 2 (2, 4, 8,

etc.) # of minterms so as to form

a rectangle or square ( a convex

region).

AB

CD

00 01 11 10

00

1 1 1 1

1 1

01

AC

11

10

1 1 1 1

Convex

D

Concave

AB

CD

00 01 11 10

Invalid grouping

(region formed is concave)

f = AB + AC + BCD

00

01

11

10

1 1

1 1

Valid

groupings

1 1 1

1

slide16

16

Another Example:

AB

CD

BD

TT:

00 01 11 10

A B C D Z

0 0 0 0 1

0 0 0 1 0

0 0 1 0 1

0 0 1 1 1

0 1 0 0 0

0 1 0 1 1

0 1 1 0 1

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 1

1 1 0 0 0

1 1 0 1 0

1 1 1 0 1

1 1 1 1 1

00

01

1 1

ABD

1

C

11

10

1 1 1 1

Binary

order

1 1 1 1

f = C + BD + ABD

slide17

17

Don’t Cares in K-Maps

Many times certain input combinations are invalid for a function

(E.g. BCD to 7-segment display functions; see pp. 212-214 in

Katz text )

For such combinations, we do not care what the output is, and

we put an ‘x’ in the o/p column for the TT

AB

CD

00 01 11 10

A B C f

0 0 0 1

0 0 1 0

0 1 0 1

0 1 1 x

1 0 0 0

1 0 1 1

1 1 0 x

1 1 1 x

0

1

x x 1 0

0 1 1 0

f = BC + AB

by considering all

x’s as 0s

f = B

The x’s can, however, be profitably

used to make larger PIs & thus

further simplify the function.

by selectively

considering the 010

cell’s x as 1

slide18

18

In a K-map, the ‘x’s can be profitably used to form larger

implicants (which have fewer literals)

However, when we need to form a minimal PI cover, we need to

worry only about covering the minterms ( the 1s), and not the Xs.

• Thus as far as x’s are concerned, we can have the cake and eat it too!

Example:

f(A,B,C,D) =

  • m(1,3,5,7,9) +
  • d(2,6,12,13)

MSBs

AB

A B C D

0 0 0 0

0 0 0 1

0 0 1 0

0 0 1 1

CD

00 01 11 10

0 4 12 x 8

1 1 5 1 13 x 9 1

3 1 7 1 15 11

2 x 6 x 14 10

00

01

LSBs

CD

11

10

AD

f = CD + AD

or CD + AC

If input comb. 2,6 occur

o/p is 0.

AC

If inputs 2,6 occur o/p = 1.

Which to use for lower power?

slide19

19

STEPS IN K-MAP MINIMIZATION (sop EXPR.)

STEP1: Form all PIs by including the Xs with the 1s to form larger

PIs. (Do not form PIs covering only Xs !)

STEP2: Identify all essential PIs by visiting each minterm (1s) &

checking if it is covered by only one PI. If so, then that

PI is an essential PI.

STEP3: Identify all 1s not covered by essential PIs. Select a minimal- cost set of PIs S to cover them. This requires some trial-and error (in fact, this is a hard computational problem).

STEP4: Minimal SOP Expr. for function f:

f =

  • essential PIs +
  • Pi
  • S

Pi

PI Pi is in set S

slide20

20

Minimal POS Expr. from a K-Map :

METHOD 1: Obtain minimal SOP expression for the complement

function f, and complement this SOP expression to get a minimal

POS expr. for f (using De-Morgan’s Law)

Big M notation

Example:

F (A,B,C,D) =

  • M(0,2,4,8,10,11,14,15)
  • D(6,12,13)

Xs of F

0s of F

AB

F

F

AB

CD

CD

00 01 11 10

D

1 1 x 1

0 0 x 0

0 0 1 1

1 x 1 1

0 0 x 0

1 1 x 1

00

01

11

10

00

01

11

10

Complement

0 4 12 8

1 5 13 9

3 1 7 1 15 0 11 0

2 0 6 x 14 0 10 0

AC

F =D + AC F = D + AC = D(A +C)

slide21

21

METHOD2 : Direct Method :

(1) Obtain all prime implicates(PTs) (largest groups of 0s & Xs

of size 2 , i

  • 0, forming a convex region, that can not be “grown”

any further).

(2) Identify all essential PTs (those that cover at least one 0 not

covered by any other PT).

(3) Choose a minimal set T of PTs to cover the 0s not covered by

the set of essential PTs.

(4) The expression for a PT is an OR term obtained by discarding

all changing variables, & keeping variables complemented if they

are constant at 1 or uncomplemented if they are constant at 0 &

taking the OR (sum) of these literals.

(5) Minimal POS Expr.:

f =

  • (essential PTs)
  • qi

qi

T

slide22

22

Example: (Direct Method)

AB

(OR terms)

CD

00 01 11 10

0 0 x 0

1 1 x 1

1 1 0 0

0 x 0 0

00

01

11

10

Prime Implicate

( Groups of 0s that cannot

be grown any further)

Both D and A + C are

Essential Prime Implicates.

D

(A + C)

F = D( A + C )

( Product of Prime Implicates )

slide23

23

5-variable K-map: f(A,B,C,D,E)---Juxtapose two 4-var. K-submaps

one for the MSB A=0 and the other for A=1:

BC

BC

DE

00 01 11 10

00 01 11 10

DE

00

01

11

10

00

01

11

10

0 4 12 8

1 5 13 9

3 7 15 11

2 6 14 10

16 20 28 24

17 21 29 25

19 23 31 27

18 22 30 26

A=0

A=1

4-var. K-submap

4-var. K-submap

Adjacencies of a square: Adjacent squares of the 4-var. K-submap

plus the “corresponding” or “mirror” square in the other 4-var.

K-submap.

Example:

BC

BC

00 01 11 10

DE

DE

00 01 11 10

00

01

11

10

16 20 28 x 24

17 21 29 1 25

19 x 23 1 31 x 27

18 22 1 30 1 26

00

01

11

10

0 1 4 12 1 8

1 5 13 1 9

3 1 7 1 15 x 11

2 6 14 x 10

A=0

A=1

slide24

24

6-variable K-map: f(A,B,C,D,E,F)--Juxtapose two 5-var. K-submaps

one for the MSB A=0 and other for A=1

CD

CD

EF

00 01 11 10

00 01 11 10

EF

00

01

11

10

00

01

11

10

0 4 12 8

1 5 13 9

3 7 15 11

2 6 14 10

16 20 28 24

17 21 29 25

19 23 31 27

18 22 30 26

5-var.

K-submap

A=0

B=0

B=1

CD

CD

00 01 11 10

EF

EF

00 01 11 10

00

01

11

10

00

01

11

10

48 52 60 56

49 53 61 57

51 55 63 59

50 54 62 58

32 36 44 40

33 37 45 41

35 39 47 43

34 38 46 42

5-var.

K-submap

A=1

B=1

B=0

Adjacencies of a square: Adjacent squares of the 5-var.

K-submap plus the “corresponding” or “mirror” square

in the other 5-var. K-submap.

slide25

25

6-variable K-map: f(A,B,C,D,E,F,)---Example:

CD

CD

EF

00 01 11 10

00 01 11 10

EF

00

01

11

10

00

01

11

10

0 4 12 8

1 5 13 9

3 7 15 11

2 6 14 10

16 20 28 24

17 21 29 25

19 23 31 27

18 22 30 26

A=0

B=0

B=1

CD

CD

00 01 11 10

EF

EF

00 01 11 10

00

01

11

10

00

01

11

10

48 52 60 56

49 53 61 57

51 55 63 59

50 54 62 58

32 36 44 40

33 37 45 41

35 39 47 43

34 38 46 42

A=1

B=1

B=0