Two level simplification
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Two-Level Simplification. All Boolean expressions can be represented in two-level forms Sum-of-products Product-of-sums. Canonical S.O.P. form. Canonical forms are very easy to produce Just read them off of a truth table But, they’re not the most efficient representation.

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Two-Level Simplification

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Two level simplification

Two-Level Simplification

  • All Boolean expressions can be represented in two-level forms

    • Sum-of-products

    • Product-of-sums

Canonical S.O.P. form

  • Canonical forms are very easy to produce

    • Just read them off of a truth table

    • But, they’re not the most efficient representation

Reduced S.O.P. form

  • Reduced two-level forms are more efficient


Venn diagrams

B

A

B=1

B=0

A=0

A’B’

A’B

AB

AB’

A=1

Venn Diagrams

Consider a Venn Diagram for 2 sets, A and B

A’B’

A’B

AB

AB’


Karnaugh maps

A

B

F

0

0

0

0

1

1

F(A,B)

1

0

1

B

0

1

1

1

0

A

0

00

01

1

10

11

Karnaugh maps

2-variable K-map

1

0

Space for A’B’

1

0

Space for A’B

Space for AB’

Space for AB


Karnaugh maps1

C

CD

C

0

1

00

01

11

10

AB

AB

000

001

0000

0001

0011

0010

00

00

m1

m1

m2

m0

m0

m3

010

011

0100

0101

0111

0110

01

01

m3

m5

m6

m2

m4

m7

B

B

C

110

111

1100

1101

1111

1110

11

11

m7

m13

m14

m6

m12

m15

A

A

100

101

1000

1001

1011

1010

10

10

m5

m9

m10

m4

m8

m11

D

Karnaugh maps

K-maps can represent up to four variables easily

f(A,B,C)

f(A,B,C,D)

3-variable

K-map

4-variable

K-map

Numbering Scheme: 00, 01, 11, 10

Gray Code — only a single bit changes from one number to the next


Filling in a k map

C

CD

C

0

1

00

01

11

10

AB

AB

00

00

f(A,B,C)

01

01

B

B

C

11

11

A

A

10

10

D

0

0

0

0

0

0

0

0

0

0

0

Filling in a K-map

F(A,B,C,D) = ABC’D’ + AB’CD’ + ABC’D + AB’CD + A’BCD

F (A,B,C) = A’B’C’ + ABC’ + A’B’C + AB’C

1

1

1

1

1

0

0

1

1

1

0

0

1

4-variable

K-map

3-variable

K-map


Finding combinations with k maps

B

0

1

A

0

0

1

1

0

1

B

0

1

A

0

1

1

1

0

0

Finding Combinations with K-maps

We can combine A’B and AB

F = A’B + AB

= B

We can combine A’B’ and A’B

G = A’B’ + A’B

= A’

With Karnaugh maps, adjacent 1’s mean we can combine them


Adjacencies in the k map

C

C

0

1

AB

00

01

B

11

A

10

Adjacencies in the K-map

Neighbors

Wrap from top to bottom

Wrap from left to right


3 variable k map examples

F

F

C

C

1

1

1

1

C

C

0

0

1

1

AB

AB

1

1

0

0

00

00

0

0

0

0

01

01

B

B

0

0

1

1

11

11

A

A

10

10

3-variable K-map examples

F(C,B,A) = A’BC’ + AB’C+ A’B’

In the K-map, adjacency wraps from left to right

and from top to bottom

F(C,B,A) = A’C’+ B’C

Same function, alternative “circling”

Note: Larger circles are better


3 variable k map examples1

G

C

0

0

C

0

1

AB

1

1

00

1

1

01

B

0

0

11

A

10

3-variable K-map examples

We can use the combining theorem on larger units as well.

G(A,B,C) = A’BC’ + A’BC + ABC’ + ABC

= A’B(C’ + C) + AB(C’ + C)

= A’B + AB

= B(A’ + A)

= B

  • What can we circle?

    • Any rectangle that contains all ones

    • As long as its size is a power of two

      • 1, 2, 4, 8, 16, ...

      • No rectangles of 3, 5, 6, ...

Find the smallest number of the largest possible rectangles that cover all the 1’s at least once (overlapping circles are allowed)


4 variable k map example

CD

00

01

11

10

AB

F

00

1

1

1

1

m1

m2

m0

m3

01

m5

m6

m4

m7

1

1

1

1

B

C

11

m13

m14

m12

m15

A

0

1

0

1

10

m9

m10

m8

m11

1

0

1

1

D

4-variable K-map example

F(A,B,C,D) = åm(0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 13, 14)

Find the smallest number of the largest possible rectangles that cover all the 1’s

  • Start at upper left corner and search for 1’s:

    • Circled? – Go to next ‘1’

    • Not circled? – Circle largest term thatcontains this ‘1’ and go to next ‘1’

      • Tie? – Skip this square for now and come back to it later...

F(A,B,C,D) =

A’

+ BC’D

+ CD’

+ B’D’

+ B’C


K maps for xors and xnors

B

G

0

1

A

C

0

0

0

0

0

1

1

1

1

1

0

0

1

CD

CD

C

0

1

00

00

01

01

11

11

10

10

AB

AB

AB

1

1

1

1

1

0

0

0

0

0

1

0

1

00

00

00

0

0

0

0

0

1

1

1

1

1

01

01

01

B

B

B

1

1

1

1

1

0

0

0

0

0

C

C

11

11

11

Q

A

A

A

10

10

10

P

D

D

P= A Å B Å C Å D

K-maps for XORs and XNORs

F = A’B + AB’

= A Å B

G = A’B’C + A’BC’ + ABC’ + AB’C

= A Å B Å C

Q= A Å B Å C Å D


Product of sums

CD

00

01

11

10

AB

F

00

m1

m2

m0

m3

1

1

0

0

01

m5

m6

m4

m7

0

0

0

1

B

F = A’C + A’BD’ + AD

C

11

m13

m14

m12

m15

A

0

0

1

1

10

m9

m10

m8

m11

1

0

1

0

D

Product-of-Sums

We can circle 0’s to find a sum-of-products for the complement

F(A,B,C,D) = Sm(0,1,5,8,10,12,14)

F’ =

A’C

+ A’BD’

+ AD

DeMorgan’s Law

F = (A+C’)(A+B’+D)(A’+D’)

Product-of-Sums!

  • Circling 1’s gives S.O.P. for F

    • Complementing S.O.P. of F gives P.O.S. for F’

  • Circling 0’s gives S.O.P. for F’

    • Complementing S.O.P. for F’ gives P.O.S. for F


K maps and don t cares

CD

CD

00

00

01

01

11

11

10

10

AB

AB

F

F

00

00

0

0

1

1

1

1

0

0

m1

m1

m2

m2

m0

m0

m3

m3

01

01

0

0

1

1

1

1

x

x

m5

m5

m6

m6

m4

m4

m7

m7

B

B

C

C

11

11

m13

m13

m14

m14

m12

m12

m15

m15

A

A

x

x

x

x

0

0

0

0

10

10

m9

m9

m10

m10

m8

m8

m11

m11

0

0

1

1

0

0

0

0

D

D

K-maps and Don’t Cares

Invalid Inputs (Don’t Cares) can be treated as 1's or 0's if it is advantageous to do so

F(A,B,C,D) = Sm(1,3,5,7,9) + Sd(6,12,13)

F = assuming x’s are zero

A’D

+ B’C’D

Tie! - Skip and come back

F = using don’t cares

A’D

+C’D

By treating this X as a "1", a largerrectangle can be formed


Example 2 bit comparator

C

D

F

F

F

A

B

1

2

3

AB = CD

A

0

0

0

0

0

F1

N

B

1

1

AB < CD

0

1

F2

1

1

0

AB > CD

C

1

F3

1

1

N

2

D

0

0

0

0

1

=, >, <

0

0

1

1

1

0

1

0

1

0

0

1

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

1

1

0

1

1

0

1

1

0

0

0

0

0

1

1

0

0

0

0

1

0

0

1

1

0

0

0

1

1

1

1

0

0

0

0

1

0

1

0

1

1

0

Example: 2-bit Comparator

Will need a 4-variable K-map for each of the 3 output functions

DC and BA are two-bitbinary numbers


K maps for 2 bit comparator

F2

F1

F3

CD

CD

CD

00

00

00

01

01

01

11

11

11

10

10

10

AB

AB

AB

0

0

0

1

0

0

0

0

0

1

1

1

00

00

00

m1

m1

m1

m2

m2

m2

m0

m0

m0

m3

m3

m3

0

0

0

1

1

0

0

0

0

0

1

1

01

01

01

m5

m5

m5

m6

m6

m6

m4

m4

m4

m7

m7

m7

B

B

B

0

0

1

0

1

1

0

1

0

0

0

0

11

11

11

A

A

A

m13

m13

m13

m14

m14

m14

m12

m12

m12

m15

m15

m15

0

0

0

1

1

1

0

0

0

0

1

0

C

C

C

10

10

10

m9

m9

m9

m10

m10

m10

m8

m8

m8

m11

m11

m11

D

D

D

K-maps for 2-bit comparator

F1 = AB==CD

F2 = AB<CD

F3 = AB>CD

F1 =

F2 =

A’B’C’D’

+ A’BC’D

+ ABCD

+ AB’CD’

A’B’D

+ A’C

+B’CD

-OR-

F3 =

BC’D

+AC’

+ABD’

F1’ =

B’D

+A’C

+BD’

+AC’

F1 = (B+D’)(A+C’)(B’+D)(A’+C)


Bcd decrement by 1

N2

N1

ABCDWXYZ

00001001

00010000

00100001

00110010

01000011

01010100

01100101

01110110

10000111

10011000

1010XXXX

1011XXXX

1100XXXX

1101XXXX

1110XXXX

1111XXXX

BCD Decrement by 1

  • BCD – Binary Coded Decimal

    • Represents ‘0’ through ‘9’ in four bits

  • Binary patterns for 10-15 are invalid inputs

  • Decrement by 1 function

    • N2 = N1 – 1

    • 0 – 1 = 9 (rolls over)


Bcd decrement by one

Y

W

Z

X

CD

CD

CD

CD

00

00

00

00

01

01

01

01

11

11

11

11

10

10

10

10

AB

AB

AB

AB

0

0

0

0

0

1

0

1

0

0

0

0

00

00

00

00

m1

m1

m1

m1

m2

m2

m2

m2

m0

m0

m0

m0

m3

m3

m3

m3

0

1

1

1

1

1

0

0

0

0

0

0

01

01

01

01

m5

m5

m5

m5

m6

m6

m6

m6

m4

m4

m4

m4

m7

m7

m7

m7

B

B

B

B

x

x

x

x

x

x

x

x

x

x

x

x

11

11

11

11

A

A

A

A

m13

m13

m13

m13

m14

m14

m14

m14

m12

m12

m12

m12

m15

m15

m15

m15

x

x

x

x

x

x

0

1

1

1

0

C

C

C

C

0

10

10

10

10

m9

m9

m9

m9

m10

m10

m10

m10

m8

m8

m8

m8

m11

m11

m11

m11

D

D

D

D

0

0

1

1

1

0

1

0

x

x

x

x

x

x

1

0

BCD Decrement by One

W =

A’B’C’D’

+AD

X =

BD

+BC

+AD’

Y =

CD

+BC’D’

+AD’

Z =

D’


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