5 Variable Karnaugh Map Example

Five Variable Maps

Printed; 11/02/04 Department of Electronics, Carleton UniversityModified; February 11, 2004 © John Knight Digital Circuits p. 91

Five Variable Maps

b

c

d

1 1

01

1a 1

1

b

c

d

1 1

11

1a 1

1Two methods:

1) 32 squares

2) Enter variables (letters) on the 16 square map

b

c

d

1 1

e

e1a 1

1

This map fore=1

Best for where “e” has simple relations:

Think of two maps on top of each other

1

0This map for

e=0

One can circle squares:on either level,or between levels.

b

c

d

1 1

01

1a 1

1

1b

c

d

1 1

01

1a 1

1

1

1

This is the same as putting a “1” on the e=0 map

Can circle

This is the same as putting a “1” on the e=1 map

e 1 e 1 1 1

e eCannot circle

“e” is in only a few squaresor “e” is in almost all the squares.

Slide 46


Five Variable Maps Five Variable Maps

Five Variable MapsWrap Around

To circle between layers, the layers must have a “1” in thesame position on both layers.These are the squares which differ by only one input bit.

Use of variable entered maps

For complicated functions using the double map is usuallyeasier.

For many functions one of the variables has a simplerelationship. Then the variable entered map is simpler.

58.• PROBLEM

Plot the variable entered map on the right on the 5-variablemap on its left.

b

c

d

1

1

11a 1

1

1

1

b

c

d

a

b

c

d

1

1a 1

11

b

c

d

a

1

e=1

e=0

e=1

e=0OK

b

c

d

a

b

c

d

a

e=1

e=0b

c

d

a

Variable entered map

11

1e1

e

e

Comment on Slide 46



Five Variable Maps

b

c

d

1 1

01

1a 1

1

b

c

d

1 1

11

a 1

1

Method 1: Dual 4-Variable maps

b

c

d

1 1

e

a 1

e=1 1

0e=0

b

0

ab

0

aMethod 2: Variable Entered Maps

1F = e(b·d)

+ acd + ab

+ e(cd)

1

1

11

1

11

1

1

e

F = e(b·d)

+ acd + ab

+ e(cd)

must also be in another circle“1s” in circles containing e

containing e or all “1s”.

must be in another circle“1s” in circles containing e

containing e or all “1s”.

Terms for circles

containing e are ANDed with e

1 e1

1 ee

1 e1

1 ee

or

or

Terms for circles

containing e are ANDed with e

e=0 map are ANDed with eTerms for circles only on the

Terms for circles only on thee=1 map are ANDed with e

Terms for circles on both mapsdon’t mention e.

CommentonSlide63Slide 47


Five Variable Maps Five-Variable Maps

Comment on Slide 47

Five-Variable MapsThe “0” is specifically entered on one map, just to remind you that it is a “1” on the other map.

Normally “0” are left blank to reduce the clutter.

59.• PROBLEM

F = (abc + abd + cb)e + (abd + ac + adb)e

= (abc + cb)e + (ac + adb)e +abd

Plot F on the 5 variable map on the right.

Plot F on the variable entered map on the right.

60.• PROBLEM (based on the last problem)

Circle the 5 variable map and reduce F to 12 letters.

Circle the variable entered map and reduce F to 4 terms of 3letters each.

b

c

d

a

b

c

d

a

e=1

e=0b

c

d

a

Variable entered map



Five Variable Maps

1

1

d

11

1

d

1

1

Method 1: Dual maps

e=1 0

e=0

Method 2: Variable entered

1

F = e(abd) + bd + e(abcd)

dd

b

c

d

1

1

a d

1

b

c

d

1

1

d

a 1

1

e=1 0

e=0

ba

ba

1dd

b

c

d

1

1

a d

1

b

c

d

1

1

d

a 1

1

e=1 0

e=0

ba

ba

1dd

F = e(abd) + bd + abcd

John’s solution Tom’s solution

1

e/d1d/e

e

e/d

d/0

1

e/d1d/e

e

e/d

d/0

F = e(abd) + bd + e(abcd)

ba

d

c

Tom says,“The interaction ofd and e is too complex.Use dual maps.”

Slide 48


Five Variable Maps Five Variable Maps With Don’t Cares

Comment on Slide 48

Five Variable Maps With Don’t CaresMethod 1: Dual 4-variable maps

The extension to 5 variables is straight forward. As before, “d” can be optionally circled like “1”.

Method 2: Variable entered mapsThis is much harder.Some new notation must be devised.

A “d” by itself in a square means a “d” in both the upper and lower levels.

If a “d” is on only one level, the value on the other level must be specified.

“d / e” means “d” on the top level and “1” on the lower (e=1) level.

“e / d” means “1” on the upper level, the e level, and “d” on the lower level.

One must also have “0 / d” and “d /0”

The result is very difficult to circle properly.

However it may be useful if e appears in a very simple way.

Method 3: Split-square mapsA third method for 5 variables would be to use a “/” if thesymbols differed between the levels.

Thus 1/0, 0/1, d/0, d/1, 0/d and 1/d would be introduced.

You can judge if this is an improvement over Method 2. 1

1/d1d/1

1/0

1/d

d/0

ba

cMethod 3 map

Multiple Output Maps Five Variable Maps


Multiple Output Maps

b

c

d

11

1

1a1

111 b

c

d

11

1

1a1

111

b

c

d

11

1

1a1

111

b

c

d

11

1

1a1

111

Two or more outputsSame inputs

Need two mapsNeed two circuits

Often one can share some gates

ad b

ad

b

c

a

dc

a

dc

ad

c

b

b

b

ad b

ad

c

dc

acb

Map of F Map of G

GF

Find the circuits for F and G

We optimized maps individuallygot one common gate abd.

11 gates37 gate inputs29 letters (literals)

F = a·c·d+bcd

+u+abd+abc

G=b·c·d+acd

+u+abc+acd

u=abd

size measures

13 letters 13 letters

3 letters

Slide 49


Multiple Output Maps Five Variable Maps With Don’t Cares

Comment on Slide 49

Circuit With Two Outputs

Multiple OutputsThe change between multiple outputs and single outputs

With multiple outputs, one can often find common gates that can be used for both outputs.Often these common gates are not optimum for either individual circuit, but are optimum for the whole circuit.

In the example• In this slide the circuits were optimized individually with a half-hearted effort to find common terms.

• In the next slide, common terms were agressively sought out.

Circuit complexity

There are several ways to estimate the size of the circuit. The same measures also estimate power dissipationwhich is now likely to be more important than size.

• Inverters are not usually counted in the gate count. This is because most will be absorbed when onedoes a AND/OR to NAND/NOR conversion.

• The number of gates.

• The number of gate inputs. This admits that multi-input gates are larger.

• The number of letters on the right hand side of the expressions. This is easy to do, and is usuallyconsidered the best estimate.

Note these are relative estimates; used for estimating if one circuit bigger than another. An exact estimate isusually not needed since modern logic is quite inexpensive. It is like estimating the cost of something as $10.00and quibbiling whether it was really $9.98.

Multiple Output Maps Multiple Outputs, Finding Common Terms


Multiple Outputs, Finding Common Terms

ad

c

acbc

a

b

c

d

11

1

1a1

111 b

c

d

11

1

1a1

111

b

c

d

11

1

1a1

11

b

c

d

11

1

1a1

111

• Identify common squares on both maps

• Check, sometimes this doesn’t help.

ad b

a

d

b

ca

c

db

b

b

a

c

1

Try to share terms

• Circle common termseven if the individual mapsallow larger circles

• Here changed 3-input AND to 4-inputand removed another 3-input.

Map of F Map of G

F =v+w+u

+abc+abc

G=v+w+u

+acd+ab·c

u=abd

v=a·b·c·d

w=abcd

F

G9 gates33 gate inputs29 letters (literals)

size measuresPrev Thisslide slide293711

acdabc11 gate inputs 11 gate inputs

11 gate inputs

Slide 50


Multiple Output Maps Multiple Outputs

Comment on Slide 50

Multiple OutputsCollecting the u+v+w terms would reduce the number of letters and gate inputs, but will increase the number ofgates. However the total logic is clearly reduced.

F = abc + abc + x (7 letters, 9 inputs, 3 gates)

G = acd + ab·c + x (7 letters, 9 inputs, 3 gates)

x = a·b·c·d + abcd + abd (11 letters, 14 inputs, 4 gates))

Total: 25 letters, 10 gates, 32 gate inputs

61.• PROBLEM

Find the Σ of Π expressions with minimallogic for the two-output circuit E,F.

Soln has 5 gates.

If it does not have to be pure Σ of Π, it can bedone in 5 two-input gates, or, with factoring,4 gates.

62.• PROBLEM

Find the minimum circuit with the three outputs defined by the maps below. This is a hard problem. You shouldread over the example for the 7-segment display drivers before attempting it. .

X

W

Z

Y

00

01

11

10

00 01 11 10WXYZ

1d1 11

d

d

1

X

W

Z

Y

1 d

00

01

11

10

00 01 11 10WXYZ

1d

d

d1

1

1

E

W

YX

Z

E

F

F

X

W

Z

Y

1

1

X

W

Z

Y

W

Z

Y

11

11 1

11

d1

1

00

01

11

10

00 01 11 10WXYZ

00

01

11

10

00 01 11 10WXYZ

00

01

11

10

00 01 11 10WXYZ

1

d11

1

1 1 11

11

1 1

1

1

d

E= F= G=

Multiple Output Maps 7-Segment Display Driver,


7-Segment Display Driver,

a

b

c

d

e

fg

Digits with “a” lit Digits with “b” lit Digits with “c” lit

Digits with “d” lit Digits with “g” litDigits with “e” lit

ZYXW

WX

Decimal digits displayedDesign this

BCD Digits in binary

00011110

YZWX 00 01 11 10

0000010011001000 1001

110101010001 0011

011111111011 1010

1110011000100000

0100

1000 1001

010100110111 0110

0010

LO

GIC

bafgedc

logic

Digits with “f” lit

Design Driver Logic

YZ

00

01

11

10

00 01 11 10

4 inputs, 7 outputs7 maps, each with 6 don’t cares

Generate MapsChoose segment “a”find all the squareswhere “a” is lit.

Repeat for “b”, “c”, . . .

Slide 51


Multiple Output Maps Seven-Segment Display Driver

Comment on Slide 51

Seven-Segment Display DriverDesign Example

The slide above, shows the bars in a seven segment display such as is used in many automotive dashboarddisplays, or other bright displays1. All the digits from 0 through nine can be shown by lighting the proper bars.

Design a circuit which takes a binary-coded-decimal (BCD) digit in on leads W,X,Y and Z and sends out theproper signals to light the 7-segment display on leads a, b, c, d, e, f, and g. Binary-coded decimal (BCD) digitsonly go from 0 to 9. The other numbers, 10 through 15 will never be received as inputs. Utilize this fact in yoursolution.

63.• PROBLEM

The digits on the right have a revised form for 1, 7, 6 and 9.

Derive the equations for the display drivers. Keep the same notation for all items thatdo not change.

Minimize the equations for multiple outputs, as is done in the next few pages. If youhave new terms, the letters H, Q, S, T, U and V have not been used.

1. The bright light-emitting diode displays use 7-segments as shown. The dimmer watch and control panel displays are usuallyliquid crystal and have more complex driver logic.

WX\YZ

Revised display

Multiple Output Maps 7-Segment Display Driver,


Maps for 7-Segment Display Driver

1 1 1 1 1 1 1 1 1 1

1 1 111

11111111

11 1 1 1 1

1

11111

1 1 1 1 1

11 11111

b c

d gfe

a

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

1 1 111

1

11111

1 1 1 1 1

11 11111

d gfe

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

1

1

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

Minimization

Digits with “a” lit

Transfer lit segment mapsto Karnaugh maps

Look for isolated “1”s with no neighbors.Look for isolated pairs of “1”s with no neighbors.These will always have to be circled individually.

Expand circles to include “d”

Slide 52


Multiple Output Maps BCD Display

Comment on Slide 52

BCD Display

Solution to Prob 62.• See also prob 74.• .

1 1 111

1

11111

1 1 1 1 1

11 111111

d gfe

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

1. Locate isolated “1”s.

2. Circle these isolated “1”sIsolated

dd d

ddd

d dddd d

Typical Minimization ProcedureThis should work fairly well as as a general procedure, but a clever person may find more efficient

3. Locate isolated pairs of “1”s in which

procedures for certain problems.

Isolatedpair

These are “1”s in a squarethat cannot be grouped withany other square exceptpossibly a “d” squares.

and expand the circle to include any “d”s.

neither “1” can be paired with another any other square.

4. Circle this pair and expand the circle to include any “d”s.

X

W

Z

Y

1

1

X

W

Z

Y

X

W

Z

Y

11

11 1

11

d1

1

00

01

11

10

00 01 11 10WXYZ

00

01

11

10

00 01 11 10WXYZ

00

01

11

10

00 01 11 10WXYZ

1

d11

1

E=W·Y+L G=WX+Z+XYF=WYZ+L

L=YZ+WX·Y+WXZ

1 1 11

11

1 1

1

1 20 letters, 27 gate inputs, 11 gates

d

Multiple Output Maps Maps for 7-Segment Display Driver ,


Maps for 7-Segment Display Driver ,

1 1 1 1 1 1 1 1 1 1

1 1 111

11111111

11 1 1 1 1

1

11111

1 1 1 1 1

11 11111

b c

d gfe

a

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

1

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

MinimizationLook for half-map circles (one letter terms)

These do not require an AND gate.Hence they can always be circled

ZY

fWX

Example: f needs only one AND gatefor three circles

without loss of potential gate sharing.

The green (light) circles are repeats of half-map circles.

On maps “b” and “c”,circle W is redundant

A common error is to add the dashed circle W to the “b” mapand on the “c” map.They are not needed.

Slide 53



Comment on Slide 53

Maps for 7-Segment Display Driver

5. Locate all circles which, with “d”s if needed, cover half of a map.

Since they only contain a single letter, they do not need an AND gate. The input can feed directly intothe OR gate.There is no advantage to sharing these terms between maps because there is no hardware to share.

1 1 111

1

11111

1 1 1 1 1

11 111111

d gfe

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

1 1 1 1 1 1 1 1 1 111111111

11 1 1 1 1b ca

XW

Z

Y

XW

Z

Y

XW

Z

Yd d

d d

d ddd

ddd d

d dddd d

d dddd d

d dddd d

d dddd d

d dddd d

There are some ten of them in this example.

6. It is easy to overdo this step Two of these circles cover no “1”s that are not covered

Remove such circles. The “b” and “c” maps have such useless circles.by other circles. The only new squares they cover contain “d’s and hence are useless.

Minimization (continued)

Circles that cover half the mapThese are representend by a single letter and are particularly good.

Multiple Output Maps Maps for 7-Segment Display Driver,


Maps for 7-Segment Display Driver,

1 1 111

1

11111

1 1 1 1 1

11 11111

d gfe

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

1 1 1 1 1 1 1 1 1 111111111

11 1 1 1 1b ca

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

1

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

Squares with only one partner: (See arrows )

• Square is isolated on another map.• Square has a different single partner

1

1 111

1

G Hboth partners join other circles.

Single circle if:

Else circle both partners.

on another map, and

They are unlikely to be shareable.

1

1 111

1

G H3 AND gates 4 AND gates

Expand circles to include “d”s; lighter (blue) circles.

“d”s don't count as partners.

One partner squares.

There is no example inthe BCD display maps.

Slide 54



Display Driver, One Partner Squares


Squares that have one partner

11

11

1

1 1

11 1

1

g

e

X

W

Z

Y

X

W

Z

Y

d dddd d

d dddd d

These squares can be encircled with one and only one other square.

There are two cases, depending on what is at the same position on the other maps“d”s don’t count as a partner.

1 1 111

11

a

XW

Z

Yd dd ddd

1 1 111

1

d

X

W

Z

Yd dddd d

Map “a” has such a square. It can be given a single circleor a double square circle.Looking at map “d” one sees it has to have a singlecircle. It can do double duty if this term, WXYZ,is given a single circle on each map.(

The square matches an isolated square on another map

The pair matches another single partner square.Map “e” has such a square. WXYZ has one possible partner.

Map “g” has such a square. W·XYZ has one possible partner.

On all the other maps where WXYZ =1, it has the samepartner (maps d and g), or is covered already (maps f and c).This means it is not a likely candidate for a single circle.

11

1

1

e

X

W

Z

Y

d dddd d

Circle both partners.Then expand to cover the “d”s.

11

1 1

11 1

g

X

W

Z

Y

d dddd dThe partner works on map “d”, and “b” has another partner.

Circle both partners and expand the circle to cover the “d”s.

There is no example in the BCD display where such term shouldhave only a single circle.

Comment on Slide 54




1 1 111

1

11111

1 1 1 1 1

11 11111

d gfe

X

W

Y

X

W

Z

Y

X

W

Y

X

W

Y

1 1 1 1 1 1 1 1 1 111111111

11 1 1 1 1b ca

X

W

Y

X

W

Y

X

W

Y

1

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

Final fill in

11

1

1

e

X

W

Y

3 new AND terms needed

Add terms to cover them

One new term is unique XY

The others (K, L) are reused.

1 1 111

1

11111

1 1 1 1 1

11 11111

d gfe

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

1d d d

ddd

d d dd

dd

d d dd

dd

d d dd

dd

11

1

1

e

X

W

Z

Y

K

1 1 1 1 1 1 1 1 1 111111111

11 1 1 1 1b ca

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

d d dd

dd

d d dd

dd

d d dd

dd

LK

L

Uncovered bits identified as “1”

Slide 55



Display Driver, Final Fill In


Squares that are left

1 1 111

1

11111

1 1 1 1 1

11 111111

d gfe

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

1 1 1 1 1 1 1 1 1 111111111

11 1 1 1 1b ca

XW

Z

Y

XW

Z

Y

XW

Z

Yd d

d d

d ddd

ddd d

d dddd d

d dddd d

d dddd d

d dddd d

d dddd d

These squares have little chance of sharing by circling a smaller than optimum circle.

Example: Assume one did not do maps a, b and c, only d, e, f, and g.However keep your eyes open.

Then one would do a two square circle for d, e and f

1 1 111

1

11111

1 1 1 1 1

11 111111

d gfe

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

d dddd d

d dddd d

d dddd d

d dddd d

For all seven maps, the largest circles appear to be optimum.

If only 4 maps were optimized

However another circling, using smaller circlesmight still be optimum.

1 1 1 111

1 1

b

XW

Z

Yd dddd d

N

R

L

K64.• PROBLEM

Reducing the size of L to L=XYZand adding reusable term R makesall multi-letter terms reusable.

Find the number of letters, gateinputs and gates.

L L

Comment on Slide 55




K M 1PN

K

1ML1K

K 1 W W W

11 PXX1P

d gfe

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

K 1 K 1 X 1 X Y 1 ZX111JLJN

11 1 1 1 1b ca

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

d d dd

dd

Form Equations

1

e

Label AND terms with letters

J = YZ

L = Y·Z

N = XYZ

M = XY

P = YZ

a = J + W + K + N

c = Z + Y + X

d = N + M + P + K

b = J + L + X

e = K + P

f = L + W + X

g = W + M + P + XY

37 letters (literals)

14 gates

38 gate inputs

K = X·Z

Size measures

If one term covers squarereplace “1” by letter

If several terms cover squareleave as “1”.

Only term not reused.

c = f + Z d = e + N + MUsing:

35 letters 14 gates 36 gate inputs

Slide 56


Multiple Output Maps Display Drivers, Forming Equations

Display Drivers, Forming Equations

J = YZ

L = Y·ZN = XYZ

M = XY

P= YZ

Q M 1PN

Q

1MQPQ

Q 1 W 1 W

11 PXXXXP

d gfe

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

X

W

Z

Y

Q 1 M 1 X 1 X Y Y JX1X1JLJN

W1 1 X Y Yb ca

XW

Z

Y

XW

Z

Y

XW

Z

Y

a = J + W+ Q + M

c = J + Y + Xd = N + M+ P + Q

b = J + L + X

e = Q + Pf = Q + W+ Xg = W+ M + P + X·Y

14+24 literals = 3814 gates16 and gate inputs + 23 or = 39

One way of forming equations is to put a letter like J,K L ... in the square covered by a circle.The one can write the OR inputs for the equation of the map by writing down the letters.

It is very hard to find the optimal circling in a problem of this size.

The equations.

Q = X·Y·Z

To avoid confusion, leave a 1 in squares which are covered by several circles.All letters must appear at least once, or the circle they represent is redundant.

Terms which have only one input like X, do not require a special letter, and we give them the name ofthe input variable.

These maps show a solution which is suboptimal, but you will probably havedifficulty improving without comparing answers.

Comment on Slide 56

Factoring and Multiplying Out Two canonical forms


Factoring and Multiplying Out

Two canonical formsThese form can represent any Boolean function.

Sum of Products (Σ of Π)abc + ae + ace + abd + . . .

NAND-NAND logic

Product of Sum (Π of Σ) Dual of Σ of Π(a+b+c)(a+e)(a+c+e)(a+b+d)( . . .

NOR-NOR logic

Factoringtransforms Σ of Π → Π of Σ

ab + cad → (a + c)(b + a + d)

Multiplying outtransforms Π of Σ → Σ of Π

(a + c)(b + a + d) → ab + cad

Three Methods of factoring1. Use the 2nd distributive law

x + ab = (x + a)(x + b)

2. Take the dual, multiply out and take the dual again.

3. Plot F on a map and use DeMorgan’s Law

eac

cab

dab

ae

eac

cab

dab

ae

NOR

NAND

Σ of Π(NAND-NAND)

(NOR-NOR)Π of Σ

Slide 57


Factoring and Multiplying Out Display Drivers, Forming Equations

Factoring and Multiplying Out

Why factor?Often the factored form is about the same complexity as the Σ of Π form, but sometimes it can be much simpler.

Example where the factored form has half the letters and just over half the gate inputs.

a·c·d + a·b·c + abc + acd = (a + c)(b + d)( a + c)

Three methods of factoringUsing (D2)

This is the straightforward way, unfortunately it uses the “unfamiliar” distributive law which makes the algebraharder for many people.

Using duality and (D1)

This is algebraically just as difficult as the previous method. However using the more familiar (D1) makes iteasier for most people.

Using Karnaugh maps

This is quite easy, but is more complex for over five input variables.

Comment on Slide 57

Factoring and Multiplying Out Method 1: Factoring Using (D2) YZ + U =


Method 1: Factoring Using (D2) YZ + U = (Y + U)(Z + U)

Review

bx + xy + yb

(b + x)(x + y)(y + b)

ExampleABC + X

(AB + X) (C + X)

(A + X)(B + X)(C + X)

ExampleBC + AD

(BC + A) (BC + D)

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

ExampleBC + BD

(BC + B) (BC + D)

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

1 (C + B)(B + D) (C + D)

Can reduce further using Consensus (C2)

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

= (C + B)(D + B)

BC + BD = (B + C)(B + D)

= bx + yb

= (b + x)(y + b)

Consensus Rules

Use (D2)

Use (D2) again

Use (D2) again

Use (D2) again, twice

Get the Swap Rule

= (B + C)(C + D) (D + B)

ABC + X= (A + X)(B + X)(C + X)

Get extended (D2)

Slide 58


Factoring and Multiplying Out Factoring Using (D2)

Factoring Using (D2)Review of the thorems that will be used.

The extended (D2) The dual is the extended (D1)

ABC + X = (A + X)(B + X)(C + X) (A + B + C)X = AX + BX + CX

Consensus The dual Consensus

bx + xy + yb = bx + yb (b + x)(x + y)(y + b) = (b + x)(y + b)

Swap rule (Sw) The dual of the Swap rule

BC + BD = (B + C)(B + D) (B + C)(B + D) = BC + BD

65.• PROBLEM

What can you say about the dual of the Swap rule?If you can’t say anything, substitute X for W and X for W.

66.• PROBLEM

Factor AB + BC + CA

Comment on Slide 58

Factoring and Multiplying Out Method 1: Factoring Using (D2) YZ + U =


Using D2 alone always works, but long!

Review of Rules

xy + x = x

xy + x = y + x

u + yz =

u + xyz =(u + x)(u + y)(u + z)

(u + y)(u + z)Example:ABC + BCD + BF

(B + CD + F)(B + AC)

Simplify (S)

Distributive (D2)

Extended Distributive (D2)

Swap (Sw)

by + bz = (b + y)(b + z)

Shortcut method:(a) Look for common variables and use (D1)

xa + xbc = x(a + bc)(b) Look for complemented variables and use (Sw)

xa + xbc = (x + a)(x + bc)(c) Do simplifications as soon as possible

xa + a = a, x + xy = x + y

BAC + B(CD + F)

Find common variablesChoose B over C

commute Use (D1) with B

Find complimented variablesUse (Sw) with B

(B + F + CD)(B + AC)Ready to use (D2)commute

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

(B + C)(B + F + C)= (B + C)(F + C)

This can be reduced

If one expanded using (D2) alone, get(B + A)(F + A + B)(F + A + C)(F + A + D)(B + F + C)(B + F + D)(B + C)(F + C) ugh!

Use (D2) twice

(b + x)(b)=(b + x)

(b + x)(b) = b

(d) Then use (D2)

Reduction (R)

see page below

Slide 59


Factoring and Multiplying Out Shortcuts for Factoring

Shortcuts for FactoringReason for choosing B over C

There are two common factors B and C.If one factors out C first get C(AB + BD) + BFThen one can only use (Sw) on part of the expression to give C(A + B)(B + D) +BF

This requires about 6 applications of (D2) to get the final result.

Details of the extra reduction

(B + C)(B + F + C)

= C +B(B + F) (D2) (c + x)(c + y) = c + xy

= C + BB + BF (D1)

= (C + B)(C + F) (D2)

The long expression using (D2) alone, except for simplifications

ABC + BCD + BF

= (A + BCD)(B + BCD)(C + BCD) + BF (D2)

= (A + BCD)(B + CD)(C) + BF (Simplification) xy + x = y + x, z + uz = z

= (BF + A + BCD)(BF + B + CD)(BF + C) (extended D2)

= (BF + A + BCD)(F + B + CD)(BF + C) (Simplification) xy + x = y + x

= (B + A + BCD)(F + A + BCD) (F + B + C)(F + B + D) (B + C)(F + C) Use (D2) three times

= (B + A)(F + A + BCD) (F + B + C)(F + B + D) (B + C)(F + C) (Simplification) z + uz = z

= (B + A)(F + A + B)(F + A + C) (F + A + D) (F + B + C)(F + B + D) (B + C)(F + C) (extended D2)

After completing the simplification, you can see why we like the short cuts.

Comment on Slide 59



Factoring Examples Using Shortcut Methods

67.• PROBLEM:Factor AC + ABD + ABE + A·CDE Five factors, four have 3 letters, one has 2.

Factor WXY + W·XZ + WYZ + WYZ

WXY + WYZ + W·XZ + WYZ

W(XY + YZ) + W(XZ + YZ)

[(W +(Z +Y)(Z + X)][W +Y(X + Z)]

[W +(XZ + YZ)][W +(XY + YZ)]

(W + Z + Y)(W + Z + X) (W + Y)(W + X + Z)

Associative law

(D1) used twice

Swap: w(a) + w(b) = (w + b)(w + a)

Swap on left; (D1) on right

Finally we have to use (D2).

Swap D1

D2 D2

Factor AB·D + ACD + A·C + ABD

A(B·D + CD) + A(C + BD)

[A + (C + BD)][A + (B·D + CD)]

[A + (C + B)(C + D)][A + (B + D)(C + D)]

[A + (C + B)][A + (C + D)][A + (B + D)][A + (C + D)]

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

(D1) on both sides

swap

(D2) on left; swap on right

(D2) on left; (D2) on right

Clean up brackets

Comment on Slide 59



Example of Multiplying Out

(A + B)(C + D)(F + G + H)

=

All letters are different, no simplification possible

Use (D1)

rewrite

With all the letters different, there is no way to simplify.

= (AC + AD + BC + BD)(F + G + H)

(AC + AD + BC + BD) · F

+ (AC + AD + BC + BD) · G

+ (AC + AD + BC + BD) · H+ (AC + AD + BC + BD) · H

=

ACF + ADF + BCF + BDF

+ ACG + ADG + BCG + BDG

+ ACH + ADH + BCH + BDH

Use (D1)

The expressions get long rapidly.Using (D1) always works, it is easy on the brain, but hard on the pencil.Also the simplifications must be done by other means.

Comment on Slide 60

Factoring and Multiplying Out Multiplying Out


Multiplying Out

Needed to change Π of Σ to Σ of Π( - - - )( - - - )( - - - ) → ( ) + ( ) + ( )

Multiplying out is the dual operation of factoring.

Multiplying Out Uses (D1)

F BA

D

C

BCD

ABCF

B

A

D

C

BCD

BF

ACD

ACF

ABC

ACD

Example:(B + F + C)(B + C)(B + F + D)(B + A)

= (BB + BC + FB + FC + CB + C)(BB + BA + FB + FA + DB + DA)

= FBBA + FBFB + FBFA + FBDB + FBDA

0

= (FB + C)(BA + FB + FA + DB + DA)

0

+ CBA + CFB + CFA + CDB + CDA)

0

= BF + ABC + ACF + BCD + ACD

= BF + ABC + BCD

x +xy = x

x +xy = x Use D1 twice

Use D1

Put letters in order

Use map

Slide 60



Example of Multiplying Out

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

= (A + D)(A + B) · (A + C + D )(A + C + D) · (A + B + C)

= (AB + A·D) · [A(C + D) + A(C + D )] · (A + B + C)

= {AB [A(C + D) + A(C + D )] + A·D)[A(C + D) + A(C + D )]} · (A + B + C)

= {AB·C + ABD + A·D·C + A·D)} · (A + B + C)

= {AB(C + D) + A·D(C + D)} · (A + B + C)

BA

D

CAB·C

ABD

A·C·D

{AB·C + ABD + A·D)} · A+ {AB·C + ABD + A·D)} · B+ {AB·C + ABD + A·D)} · C ABDC + A·DC

AB·C + ABD ++ A·DB

ABD

=

= AB·C + ABD + ABD + ACD

=

Arrange terms to ready to use Swap

Use Swap twice

Use (D1)

Use (D1)

xy + x = x

Use (D1)

xy + x = x

Collect terms

Check on mapthat there is no

more simplification

Comment on Slide 60

Factoring and Multiplying Out Multiplying Out


Using D1 alone eventually works, but long!

Example:

Shortcut method:(a) Look for common variables and use (D2)

(x + a)(x + b) = x + bc(b) Look for complemented variables and use (Sw)

(x + a)(x + bc) = xa + xbc

(c) Do simplifications as soon as possiblexa + a = a, x + xy = x + y

Find common variables

Use (Sw) with B

Use (D2) twice

(d) Then use (D1)

Review of Rules

xy + x = x

xy + x = y + x

u + yz =

u + xyz =(u + x)(u + y)(u + z)

(u + y)(u + z)

Simplify (S)

Distributive (D2)

Extended Distributive (D2)

Swap (Sw)

by + bz = (b + y)(b + z)

(b + x)(b)=(b + x)

(b + x)(b) = b

Reduction (R)

(e) Loop back and try again until done.

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

= (B + F + C)(B + F + D) · (B + A)(B + C)

= (B + F + CD) · (B + AC)

= BAC + B(F + CD)

= ABC + BF + BCD

Finally use (D1)

Go back and look at Slide 60

Slide 61



Multiplying Out

68.• PROBLEM

Multiply out. Remember to check for obvious simplifications before starting.

(W + Y + Z)(X + Y + Z)(W + X)(W + Z)(X + Y + Z)(W + X + Z)

69.• PROBLEM

Multiply out to get four terms of three letters each. The answer should be very symmetric on an AB\CDKarnaugh map.

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

Comment on Slide 61

Factoring and Multiplying Out Multiplying Out, Short Cut Method


Multiplying Out, Short Cut Method

B

A

D

C

CD

BCD

Example:(A + C + D)(B + D)(A + B + C)(C + D)(A + B + D)

= CD + ABD + BCD

0

(x)(x+y) = x

x +xy = x

Rearrange to use (D2) and Swap

Always check for obvious simplifications

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

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

(A + C + BD) (CD + BD)

(A + C + BD) · CD

+ (A + C + BD) · BD=

Use (D2) and Swap

ACD + CD + BCD

+ ABD + CBD + BDBD

ABD

=Use (D1)

simplifications

Check map for further

= CD + ABD + BC

Collect terms

Slide 62


Factoring and Multiplying Out Multiplying Out, Short Cuts

Multiplying Out, Short CutsSimplify before you start

You always want to look for (x + ... + y )( x + y) =(x + y)

Reduction and consensus are much harder to spot. Here

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

= [C + (A + D)(D)] Using (D2)

= [C + AD]

= (C + A)(C + D)

Further (C + A)(A + B + C) = (C + A)

Hence the expression could be reduced before starting to

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

However, it is sometimes easier just to use (D2) and Swap.

Comment on Slide 62

Factoring and Multiplying Out Method 2: Factoring Using Duality


Method 2: Factoring Using Duality

Recall Factoring Methods:1. Use (D1)

Includes short cuts where one looks for (D2) and Swap first.

2. Use Duality to change factoring to the easier multiplying out.

3. Use Karnaugh Maps and DeMorgan’s Law.

Factoring Using DualitySteps:

(1) The expression to factor is Σ of ΠExample: AB + BC + ACD

(2) Take its dual to get Π of Σ.

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

(3) Multiply out the dual to get Σ of Π again

BA + BC + BD + AC

(3a) The dual identity is(A + B)(B + C)(A + C + D) = BA + BC + BD + AC

(4) Taking the dual back gives a valid identitywith the desired Π of Σ.

AB + BC + ACD = (B + A)(B + C)(B + D)(A + C)

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

= BA + BC + BD +ACA + ACC + ACD

= BA + BC + BD + AC

(A + B)(B + C)(A + C + D)(D2)

(D1)

Multiply Out Details

Algebra of one is the

Multiplying out is based on (D1).

Easier than factoring based on (D2).

dual of the other

Slide 63


Factoring and Multiplying Out Changing Factoring into Multiplying Out

Changing Factoring into Multiplying OutFactoring a Σ of Π is Coverted to Multiplying Out its Dual

We take a factoring problem which is confusing, because factoring is based on (D2). This law is not a normalalgebraic law and is harder to work with.

In the dual space, the dual expression is already factored. The problem is transformed into multiplying out,which is based on the first distributive law (D1). (D1) is more familiar, and hence multiplying out is usuallyeasier than factoring.

Multiplying out in the dual space does not give the answer. One take the dual of the answer. This will then bethe factored form of the original expression.

70.• PROBLEM

Show thatF = A·B·C·D + ABCD +ABCD + ABCD

takes only 8 letters or 12 gate inputs in factored form.

Comment on Slide 63

Factoring and Multiplying Out Factoring Using Duality


Factoring Using Duality

B

A

D

C

Example:

A·BD + AB·D + ABC + AC

Rearrange to use (D2)

Take dual

Use Swap

Check on map

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

= [A + (B + D)(B + C)] [A + (B + D)C]

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

Use (D2)

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

= A (B + D)C) + A(B·C + BD)

= A·BC + A·DC + AB·C + ABD

Use Swap

Use (D1)

(-25% if you say these are equal)

A·BC

AB·C

ABD= A·BC + A·DC + AB·C + ABD

A·DCTake dual

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

equalequ

al

OK

Slide 64


Factoring and Multiplying Out Factoring In the Dual Space

Factoring In the Dual SpaceExample

71.• PROBLEM

Factor EF·D + ECD + E·C + EFD

A·BC + ACD + ABC + BCD

BA

D

CAB·C

ABC

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

To Factor

Note the excess of C and C

Use (D2)

more simplifications

Take the dual

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

rearrange for (D2)= (C + B + A)(C + B + A)(C + A + D)(C + B + D)

= [C + (B + A)(B + A)] · [(C + (A + D)(B + D)]

= C(A + D)(B + D) + C(B + A)(B + A)Use Swap

= C(AB + D) + C(BA + BA)

= CAB + CD + C·BA + CBA

Take the dual ABC

CD

Use Swap

Use (D1)

Check on map forOK

Comment on Slide 64

Factoring and Multiplying Out Method 3: Factoring Using Karnaugh


Method 3: Factoring Using Karnaugh Maps

Steps:(1) Given F = ( Σ of Π expression)

(2) Plot it on a map.

(3) Make a map for F,It has “1” where F had “0”

(4) Circle the F map

(5) Write out the equation for F

(6) Invert F using DeMorgan’s lawto get F as Π of Σ

F = A·BC + AD + ABC + BCD B

C

D

11 1

1

A1

11

1

Map of F

B

C

D

11

1

1

A

11

11

Map of F

B

C

D

11

1

1

A

11

11

Map of FF = C·D + ABC + AB·D + A·B·C

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

Slide 65



Method 3: Factoring Using Karnaugh Maps• This method is probably the easiest, and least error prone, for up to 4 variables. Above 4, it gets hard.

• It is very easy to incorporate don’t cares with this method.

72.• PROBLEM

Factor EF·D + ECD + E·C + EFD

Using a Karnaugh map and compare your answer with that from Problem 71.•

73.• PROBLEM

Factor ACD + BD + ABC + CD + ABD

Use a Karnaugh map and obtain the minimum Π of Σ expression.

74.• PROBLEM

Show that Prob 62.• has a slightly simpler solution if you find the Π of Σ expression.

Solution:to Prob 74.•

1

1

X

W

Z

Y

1 1

X

W

Z

Y

1

1

X

W

Z

Y

1 1

1d1

1

1 d

11

00

01

11

10

00 01 11 10WXYZ

00

01

11

10

00 01 11 10WXYZ

00

01

11

10

00 01 11 10WXYZ

1

d1

1 1

1

E=YZ+Q+P F=W·Z+Q G=X·Z+M

Q=M+XYZ P=WY

1 111

11

1

E=(Y+Z)Q·P F=(W+Z)Q

M=WXY·Z

G=(X+Z)M

20 letters, 23 gate inputs, 10 gates

X

W

Z

Y

1

1

X

W

Z

Y

W

Z

Y

11

11 1

1

1

d1

1

00

01

11

10

00 01 11 10WXYZ

00

01

11

10

00 01 11 10WXYZ

00

01

11

10

00 01 11 10WXYZ

1

d11

1

1 1 11

11

1 1

1

1

d

E= F= G=

Maps from Prob 62.•

Σ of Π solution had 20 letters, 27 gate inputs, and 11 gates.

Comment on Slide 65



Product of Sum Karnaugh Maps

Simplify the Π of Σ expression

(1) Find Inverse with DeMorgan.

(2) Make a map for F

(3) Simplify the F map

(4) Write out the equation for F

(5) Invert F using DeMorgan’s law

F = ABC + AC·D + BCD + B·D + AC·D B

C

D

1 1

1 1

A1

11

1

Map of F

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

11

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

B

C

D

1 1

1 1

A1

11

1

Map of F1

1

F = BC + AB·C + ABD + B·D

to get simplified F

The Σ of Π map for FCan be used as a Π of Σ map for F

Slide 66



Product of Sum Maps

75.• PROBLEM

Multiply out to get three terms of 2 of three letters, and one of two letters. Use a Karnaugh map.

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

00011110

0 1abc

Constructing an AND of ORs map

F = a + b

1

00011110

0 1abc

1

bc

00011110

0 1a

bc

F = c

bc

00011110

0 1abc

F = (a + b)(a + b)c

bc1 11

1 1

11

11

1

11 1

0 0

000

0

F = a + b

00

0 0

00

00

1

The “0”s are the important thing in Π of Σ maps.When one ANDs the three maps, the product map will have a “0” where any of the sum terms have a “0”.The other squares have “1”s.

11

If one multiplies out F = (a + b)(a + b)c, one gets F= abc + abc. 00011110

0 1abc

bc

0 0

000

01

1Which has exactly the same map except now one thinks about the“1”s instead of the “0”s.

Thus the map of F is not quite the Π of Σ map for F.However it can be used as that map.Further using DeMorgan’s theorem on the map of F, is far easierthan twisting ones mind around the Π of Σ maps such as the one above.

True Π of Σ map

Σ of Π map of F

00011110

0 1abc

bc

1 1

111

10

0

Σ of Π map of F

Product map

Comment on Slide 66

Method 3: Factoring Using Karnaugh


Slide 67


Method 3: Factoring Using Karnaugh

Solution problem 75

76.• PROBLEM (ACTUALLY SOLN TO 75)

Multiply out to get four terms of three letters each. Use a Karnaugh map.

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

B

C

D

11

11

A1

1111

1

F = (B + C)(A + B + C)(A + C + D)(A + B + C)(A + C + D)F = BC + ABC + AC + ABC + A CD

F

B

C

D

11

1

A 1

1

F

1

F = B D + AB C + A B C

Comment on Slide 63

Documents

5 Variable Karnaugh Map Example