5 Variable Karnaugh Map Example o1r5i

Five Variable Maps

Five Variable Maps d

Two methods: 1

1) 32 squares

This map for e=0 a

Think of two maps on top of each other

1

1

1

1 1 0 1

b

1 1 1 1

b

1 1 e 1

b

c

d

0

This map for e=1

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

a

1

1 1

c

2) Enter variables (letters) on the 16 square map This is the same as putting a “1” on the e=0 map This is the same as putting a “1” on the e=1 map Can circle Cannot circle

e 1

e 1

d

e

1 1

a

e e

1

1 1

c

Best for where “e” has simple relations: “e” is in only a few squares or “e” is in almost all the squares. Printed; 11/02/04 Modified; February 11, 2004

Slide 46

Department of Electronics, Carleton University © John Knight

Digital Circuits p. 91

Five Variable Maps

Five Variable Maps

Five Variable Maps

d

d

1

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

1

a

1

e=0

c

d

1

For complicated functions using the double map is usually easier.

b a

1

1

Use of variable entered maps

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

1 b

e=0

OK 1 b

a

e=1

b a

1

1

58.• PROBLEM Plot the variable entered map on the right on the 5-variable map on its left.

1

c

1 c

e=1

d

b a

d

e=0

c

e

d a b

e=1


1 1 e 1

1 e

b

c

Variable entered map

a

Printed; 11/02/04 Modified; February 11, 2004

c

d

c

Comment on Slide 46


Five Variable Maps

Five Variable Maps

Five Variable Maps Method 1: Dual 4-Variable maps

d 1

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

e=0

F = e(b·d) for circles only on the e=1 map are ANDed with e

1

1

b

1 1 1 1

b

c

d

0

+ e(cd)

e=1

Method 2: Variable Entered Maps

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

1

+ acd + ab

for circles on both maps don’t mention e.

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

a

1 1 0 1

a

1

1 1

c

1 e 1 e or e 1 d

1 e or 1 e e 1

F = e(b·d)

1 1 e 1

e

+ acd + ab

for circles containing e are ANDed with e

a

+ e(cd)

1 1

1

b

c

for circles containing e are ANDed with e


Slide 47 onSlide63 Comment



Five Variable Maps

Five-Variable Maps

Five-Variable Maps The “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

d

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

b

Plot F on the 5 variable map on the right.

a

d

Plot F on the variable entered map on the right.

e=0

60.• PROBLEM (based on the last problem) Circle the 5 variable map and reduce F to 12 letters.

d

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

b a b


c

Variable entered map

a

e=1


c

c

Comment on Slide 47


Five Variable Maps

Five Variable Maps

Five Variable Maps John’s solution

Method 1: Dual maps

Tom’s solution

d

d 1 e=0

1 1

d d

d

e=0

d d

1

1

a

1

d

d b

e=0 a

1

1

1

1

e=1

1

d

1

1

1 1

a

b

e=1

c

Method 2: Variable entered

1

0

a

1

d

b

d

0

e=1

1

1

c

d 1

0

b

1

1

c 1

d d

d

1

c F = e(abd) + bd + abcd

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

d

d/e 1 d/0 e/d b e

d/e 1 d/0 e/d e a

e/d 1

e/d 1

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

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


Slide 48


Five Variable Maps


Five Variable Maps With Don’t Cares

Five Variable Maps With Don’t Cares Method 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 maps This 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 maps A third method for 5 variables would be to use a “/” if the symbols differed between the levels.

1/ 0

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.

a

d/1 1 d/0 1/d b 1/d 1

c Method 3 map



Comment on Slide 48


Multiple Output Maps

Five Variable Maps

Multiple Output Maps Two or more outputs Same inputs

d

Find the circuits for F and G Need two maps Need two circuits

a

1 1 1 1

1 1 1 1

Often one can share some gates

Map of F

We optimized maps individually got one common gate abd. size measures 29 letters (literals) 37 gate inputs 11 gates

d

a

1 1 1 1

1 1 1

1 c


b

d

a c d b c d a b d a b d a b c

F = a·c·d+bcd

b

a

1

d

a

1

1 1

1

b

c d

a

G=b·c·d+acd



G

a b c

1 1 1 c

+u+abd+abc 13 letters

b

b c d a c d a d b

3 letters

1

1

Map of G c

u=abd F

1

c

1 1

1 1 1

+u+abc+acd 13 letters Slide 49


Five Variable Maps With Don’t Cares

Circuit With Two Outputs Multiple Outputs The 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 . • In the next slide, common were agressively sought out. Circuit complexity There are several ways to estimate the size of the circuit. The same measures also estimate power dissipation which 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 one does a AND/OR to NAND/NOR conversion. • The number of gates. • The number of gate inputs. This its 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 usually considered the best estimate. Note these are relative estimates; used for estimating if one circuit bigger than another. An exact estimate is usually not needed since modern logic is quite inexpensive. It is like estimating the cost of something as $10.00 and quibbiling whether it was really $9.98.



Comment on Slide 49



Multiple Outputs, Finding Common

Multiple Outputs, Finding Common d

Try to share • Identify common squares on both maps • Circle common even if the individual maps allow larger circles

a

d

1 1 1 1

• Check, sometimes this doesn’t help.

1 1 1 1

b

a

1

1 1

1

1 1 1

1

c

c

Map of F

Map of G

b

• Here changed 3-input AND to 4-input and removed another 3-input. size measures Prev This slide slide 29 letters (literals) 29 33 gate inputs 37 9 gates 11

d

d

a

1 1

1 1 1

w=abcd u=abd c d

d

1 a

abc +abc+abc 11 gate inputs


G

F

F =v+w+u

1 c

a b c

a b c a b c

b

11 gate inputs

v=a·b·c·d

a b

a d b

d

1 1

c

1

1 1

1 1 1

1

b c

b

a a

G=v+w+u +acd+ab·c acd 11 gate inputs

c

Slide 50




Multiple Outputs

Multiple Outputs Collecting the u+v+w would reduce the number of letters and gate inputs, but will increase the number of gates. However the total logic is clearly reduced. F = abc + abc + x G

= acd + ab·c + x

x = a·b·c·d + abcd + abd

(7 letters, 9 inputs, 3 gates) (7 letters, 9 inputs, 3 gates) (11 letters, 14 inputs, 4 gates))

Total: 25 letters, 10 gates, 32 gate inputs 61.• PROBLEM Find the Σ of Π expressions with minimal logic for the two-output circuit E,F.

Z

Z

YZ WX 00 01 11 10

YZ WX 00 01 11 10

W X Y Z

Soln has 5 gates.

01

d 1

11

d

10

1

00

E F

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

W

00

d E

1

d

1 1

1

d Y

X

01

1

1

1

11

d

d

1

W 10

F

X

Y

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

Z

YZ WX 00 01 11 10 00 01

1 1

11

W 10 1

1 1 1

00 01

X

d

W

1 1

E=


YZ WX 00 01 11 10

1 1

Y

11 10

1

YZ WX 00 00

1

d

1 1

01

X W

11 10

1

Z 01 11 10

1 1 1 1 1 1 1 1 1 1 d

Y

F=


G=

Y

Comment on Slide 50



7-Segment Display Driver,

7-Segment Display Driver, 01

0001 0101 1101 1001

11

YZ WX 00

10

0011 0010 0111 0110 1111 1110 1011 1010

BCD Digits in binary

Design Driver Logic

Z Y X W

LOGIC

YZ WX 00 00 0000 01 0100 11 1100 10 1000

b a f g e d c

a f

01 11

10

00 b

g

01 11

e

Design this logic

c d

10

Decimal digits displayed

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

Generate Maps Choose segment “a” find all the squares where “a” is lit. Repeat for “b”, “c”, . . .

Digits with “d” lit


Digits with “a” lit

Digits with “b” lit

Digits with “c” lit

Digits with “e” lit

Digits with “f” lit

Digits with “g” lit

Slide 51




Seven-Segment Display Driver

Seven-Segment Display Driver Design Example The slide above, shows the bars in a seven segment display such as is used in many automotive dashboard displays, 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 the proper signals to light the 7-segment display on leads a, b, c, d, e, f, and g. Binary-coded decimal (BCD) digits only go from 0 to 9. The other numbers, 10 through 15 will never be received as inputs. Utilize this fact in your solution. 63.• PROBLEM The digits on the right have a revised form for 1, 7, 6 and 9.

YZ WX\

Derive the equations for the display drivers. Keep the same notation for all items that do not change. Minimize the equations for multiple outputs, as is done in the next few pages. If you have new , the letters H, Q, S, T, U and V have not been used.

Revised display

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



Comment on Slide 51



7-Segment Display Driver,

Maps for 7-Segment Display Driver

Z

1 1

Digits with “a” lit

a

Z

1 1

X

1

1 1

1

d d d d d d 1

b

d

e

Y

1 1

X

1 1

1 1 1

d d d d 1 1 d d

W

c

Y

Z

1

1

1

1

d d d d d d 1

W

1

d d d d 1 1 d d

W

Y

1 X

Z

1 1 1

Z

1

W

1

d d d d 1 1 d d

W

Transfer lit segment maps to Karnaugh maps

Z

1 1

X

Y

Z

1 1 1

1

1

1

1

d d d d 1 1 d d

W

Y

f

Y

1

X

d d d d 1 1 d d

W

X

g

X

Y

Minimization 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. Z

1

1

1 1

d d d d d d 1 d

1 X

1

1

1

1

d d d d d d 1

W

Y

e


Z

Z

Z

1 W

Expand circles to include “d”

X W

1 1 1

1

d d d d 1 1 d d

X

1

1

1

X

d d d d 1 1 d d

W

g

Y

f

Y

1

Y Slide 52




BCD Display

BCD Display Typical Minimization Procedure This should work fairly well as as a general procedure, but a clever person may find more efficient procedures for certain problems. Z

1. Locate isolated “1”s. 1 1 1 These are “1”s in a square 1 1 that cannot be grouped with d d W any other square except 1 d d d Y possibly a “d” squares. Isolated 2. Circle these isolated “1”s and expand the circle to include any “d”s.

Z

Z

1 X W

1

1

1

1

d d

1 e

X W

1

Y

Z

1 1

1

1

1

X W

1 f

1

1

1

d d d d 1 1 d d

Y

Isolated pair

g

X

Y

3. Locate isolated pairs of “1”s in which neither “1” can be paired with another any other square. 4. Circle this pair and expand the circle to include any “d”s.

Solution to Prob 62.• See also prob 74.• . L=YZ+WX·Y+WXZ E=W· Y+L

F=WYZ+L

Z

00 01

1 1

11

W 10 1

YZ WX 00 01 11 10

1 1

00 01

X

d

G=WX+Z+XY

Z

YZ WX 00 01 11 10

W

1 1 Y


11 10

1

1 1 1

1

1 1

d

YZ WX 00 00

Y

01

X W

11 10

1

Z 01 11 10

1 1 1 1

1 1 1 1 1 1 d

20 letters, 27 gate inputs, 11 gates X

Y


Comment on Slide 52



Maps for 7-Segment Display Driver ,

Maps for 7-Segment Display Driver

,

Minimization Look for half-map circles (one letter )

Example: f needs only one AND gate for three circles

These do not require an AND gate. Hence they can always be circled without loss of potential gate sharing. On maps “b” and “c”, circle W is redundant

X

Z

1

Z

1 1

a Z

1

1 1

d d d d d d 1 d

1 1

X W

b

e

Y

1 1

X

1 1

1

1

1

1

X

c

Y

Y

1 1 1

1

1

X

1

1

1

d d d d 1 1 d d

W

Y

f

Y

X

Z

d d d d 1 1 d d

W

1 1 1

d d d d 1 1 d d

W

Z

d d d d d d 1

W

1

d d d d 1 1 d d

Y

1 X

Z

1 1 1

Z

1 W

1

d d d d 1 1 d d

W

f

Y Z

1

1

W

g

X

Y

The green (light) circles are repeats of half-map circles. A common error is to add the dashed circle W to the “b” map and on the “c” map. They are not needed.


Slide 53




BCD Display

Maps for 7-Segment Display Driver Minimization (continued) Circles that cover half the map These are representend by a single letter and are particularly good. Since they only contain a single letter, they do not need an AND gate. The input can feed directly into the OR gate. There is no advantage to sharing these between maps because there is no hardware to share. 5. Locate all circles which, with “d”s if needed, cover half of a map. 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 by other circles. The only new squares they cover contain “d’s and hence are useless. Remove such circles. The “b” and “c” maps have such useless circles. Z 1 1 W

a 1

1 1

1 W

d d d d 1 d d d


Y

1

W

W

e

1 1

X W

1 1

1 1 1

d d d d 1 1 d d

Y Z

b

d d d d 1 d d

1

d d d d 1 1 d d

1 1

Z

1 1 1

Y Z

1 X

1 1

X

d d d d 1 1 d d

Z

1

Z 1 1

c

Y Z

1

1 1

X

Y


W

1

1

1

d d d d 1 1 d d f

X

Y

1

X W

1

1

1

d d d d 1 1 d d g

X

Y

Comment on Slide 53



Maps for 7-Segment Display Driver,

Maps for 7-Segment Display Driver , One partner squares.

Z

1 1 W

Z

1 1

1

1

1 1

W

d d d d d d 1 d

b

W

Y

1

1

1

1

X

d d d d d d 1 e

1 1

X

1 1

1 1 1

d d d d 1 1 d d

W

c

Y

Z

1 X

1

d d d d 1 1 d d

W

Z

1

Z

1 1 1

Y

a 1

X

d d d d 1 1 d d

Z

1 1

1 1 1

1

1

X

1

1

1

d d d d 1 1 d d

W

Y

f

Y

Squares with only one partner: (See arrows

Y

Z

d d d d 1 1 d d

W

X

g

X

Y

)

“d”s don't count as partners. 1 1

Single circle if: • Square is isolated on another map. • Square has a different single partner on another map, and both partners other circles. Else circle both partners. They are unlikely to be shareable.

1 1

1 1 1 G H 3 AND gates

1 1 1 G H 4 AND gates

1

1

There is no example in the BCD display maps.

Expand circles to include “d”s; lighter (blue) circles. Printed; 11/02/04 Modified; February 11, 2004

Slide 54




BCD Display

Display Driver, One Partner Squares Minimization (continued) Squares that have one partner These squares can be encircled with one and only one other square. “d”s don’t count as a partner. There are two cases, depending on what is at the same position on the other maps The square matches an isolated square on another map Map “a” has such a square. It can be given a single circle or a double square circle. Looking at map “d” one sees it has to have a single circle. It can do double duty if this term, WXYZ, is given a single circle on each map.(

Z 1

W

Z

1 1

1

1

The pair matches another single partner square. Map “e” has such a square. WXYZ has one possible partner. On all the other maps where WXYZ =1, it has the same partner (maps d and g), or is covered already (maps f and c). This means it is not a likely candidate for a single circle. Circle both partners. Then expand to cover the “d”s. Map “g” has such a square. W ·XYZ has one possible partner. The partner works on map “d”, and “b” has another partner. Circle both partners and expand the circle to cover the “d”s.

1

X

d d d d 1 1 d d a

1

W

Y

d d d d 1 d d Z

Z

1

1 1

d d d d 1 d d e

1 1

X W

e

Y

1

1

1

d d d d 1 1 d d g

Y Z

1

W

X

d d d d 1 d d

Z

1

X

Y

d

1

W

1 1

1

Y

1 1

X W

1

1

1

d d d d 1 1 d d g

X

Y

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



Comment on Slide 54




Maps for 7-Segment Display Driver, Final fill in

1

Uncovered bits identified as “1” Add to cover them

1 1

1

1

1 1

1 W

d d d d d d 1 d

1 X

b 1

1

1

1

d d d d d d 1

W

Y

e

X

One new term is unique XY

1

The others (K, L) are reused.

W

Z

1

1

1 1

1 W

d d d d d d 1 d

W

d d d d d d 1 e


L

1

Y

1

1

X

X

1

Y

g

Y

Z

1 1 1

1

1 1

X

1 1

1 1 1

X

d d d d 1 1 d d

W

c

Y

Y

Z

1 1 1

1

1

X

1

1

1

X

d d d d 1 1 d d

W

Y

f

Y

X

d d d d 1 1 d d

W

d d d d 1 1 d d

W

1

1

1 1

X

Y

1

b

1

X

c

Z

1

1 1 1

d d d d 1 1 d d

W

d d d d 1 1 d d

W

Z

K

1 1

X

Y

a

K

X

1 1

Z

L 1

d d d d 1 1 d d

1 1

Y

1 1

f

Y

1 1

1

d d d d 1 1 d d

W

Z

1

1 1 1

d d d d 1 1 d d

W

Y

a

Z

1 1

X

d d d d 1 1 d d

W

3 new AND needed 1

1

g

Y

Slide 55




BCD Display

Display Driver, Final Fill In Minimization (continued) Squares that are left These squares have little chance of sharing by circling a smaller than optimum circle. However keep your eyes open. Example: Assume one did not do maps a, b and c, only d, e, f, and g. Then one would do a two square circle for d, e and f Z

1

If only 4 maps were optimized

1 1

1 W

Z

Z

1

d d d d 1 d d

1

1

X W

d d d d d d 1

Y

d

e

1 X W

1

1

1 1

d d d d 1 d d d

1 1

X W

Y

d d d d 1 d d

1 1

X W

1 1

1 1 1

d d d d 1 1 d d c

N

Z

1 X

Y


W

Y

1 1

1

1

d d d d 1 1 d d f

Y

X

Z

1

1

e

1

Y

b

L

1 W

X

Y

g

d d d d 1 1 d d

Y

1 X

1

d d d d 1 1 d d

Z

1 1 1

Z

1 W


a

L

W

1

1

Z 1

d d d d 1 1 d d

Z

1

X

Y

f

1 1

1 W

1

R

1

L

1

d d d d 1 1 d d

Z

K

Find the number of letters, gate inputs and gates.

1 1

Y

For all seven maps, the largest circles appear to be optimum. However another circling, using smaller circles might still be optimum. 64.• P ROBLEM Reducing the size of L to L=XYZ and adding reusable term R makes all multi-letter reusable.

Z

1

1

1

X W

1

1

1

d d d d 1 1 d d g

X

Y

Comment on Slide 55




Maps for 7-Segment Display Driver, Form Equations

Z

Label AND with letters If one term covers square replace “1” by letter If several cover square leave as “1”.

K

W

N W

K

a

W

d d d d d d K

Y

b

d

W

1

L

P

1

X

d d d d d d K

Y

e

Y 1

X W

1 1

Z 1 X

d d d d 1 1 d d c

Y

Z

K X

X

d d d d 1 1 d d

Z

M 1 P

Z

X 1 J

1 L

X

d d d d 1 1 d d

Z

K

Z

1 N J

W

Y

Z

M 1 X X

X

d d d d 1 W d d

X W

P

d d d d W W d d

Y

f

Y

1 1

g

J = YZ

M = XY

a=J+W+K+N

e=K+P

N = XYZ

b=J+L+X

f=L+W+X

L = Y·Z

P = YZ

c=Z+Y+X

g = W + M + P + XY

d=N+M+P+K

Only term not reused.

Size measures 37 letters (literals)

Using:

38 gate inputs

c=f+Z

35 letters


d=e+N+M

14 gates

36 gate inputs Slide 56



X

Y

K = X·Z

14 gates

X


Display Drivers, Forming Equations

Display Drivers, Forming Equations 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. 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. which have only one input like X, do not require a special letter, and we give them the name of the input variable.

It is very hard to find the optimal circling in a problem of this size. These maps show a solution which is suboptimal, but you will probably have difficulty improving without comparing answers. Z Q

W

W


J = YZ

W

Q

Q Y

N = XYZ

L = Y·Z

P = YZ

M = XY

Q = X·Y·Z

X

J 1 X

Y Y 1 X

X

X

X W

Q

1

Y Y c

Y

Y Z

M 1 X

a= J + W+ Q+ M b=J+L+X c=J+Y+X d = N+ M+ P + Q e=Q+P f = Q+ W+ X g = W + M + P + X·Y


X

X

1

X W

W f

Y

X

W

1 X b Q

P

e

1 J

Z

P W

d

The equations.

X

X

Z

1 X L

a Y Z M 1 P

N

M

1 W

Z

Q

Z

1 N J

Y

1

P

X

1 W g

Y

14+24 literals = 38 14 gates 16 and gate inputs + 23 or = 39

Comment on Slide 56


Factoring and Multiplying Out

Two canonical forms

Factoring and Multiplying Out Two canonical forms

c b

a

a e a c e a b d

These 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 NAND

(a+b+c)(a+e)(a+c+e)(a+b+d)( . . . NOR-NOR logic

Factoring

Σ of Π

transforms Σ of Π → Π of Σ

(NAND-NAND)

ab + cad → (a + c)(b + a + d)

Multiplying out

transforms Π of Σ → Σ of Π (a + c)(b + a + d) → ab + cad

Three Methods of factoring

c

b

a

a e a e c a d b

1. Use the 2nd distributive law x + ab = (x + a)(x + b) 2. Take the dual, multiply out and take the dual again.

NOR

Π of Σ (NOR-NOR)

3. Plot F on a map and use DeMorgan’s Law Printed; 11/02/04 Modified; February 11, 2004



Slide 57


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 factoring Using (D2) This is the straightforward way, unfortunately it uses the “unfamiliar” distributive law which makes the algebra harder for many people. Using duality and (D1) This is algebraically just as difficult as the previous method. However using the more familiar (D1) makes it easier for most people. Using Karnaugh maps This is quite easy, but is more complex for over five input variables.



Comment on Slide 57



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

Method 1: Factoring Using (D2) Example

YZ + U = (Y + U)(Z + U) Get extended (D2)

ABC + X

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

Use (D2) (AB + X) (C + X)

Use (D2) again Review Consensus Rules

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

Example

bx + xy + yb

BC + AD

= bx + yb

Use (D2) again (BC + A) (BC + D)

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

Use (D2) again, twice

= (b + x)(y + b)

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

Can reduce further using Consensus (C2)

Example

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

BC + BD

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

= (C + B)(D + B)

Get the Swap Rule

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


BC + BD = (B + C)(B + D) Department of Electronics, Carleton University © John Knight


Slide 58


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



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

Using D2 alone always works, but long!

Review of Rules

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 (d) Then use (D2)

Example:

Find common variables Choose B over C Use (D1) with B

ABC + BCD + BF commute

BAC + B(CD + F)

Find complimented variables Use (Sw) with B

Simplify (S) xy + x = x

xy + x = y + x (b + x)(b)=(b + x) Distributive (D2) u + yz = (u + y)(u + z) Extended Distributive (D2) u + xyz = (u + x)(u + y)(u + z) Swap (Sw) by + bz = (b + y)(b + z)

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

(b + x)(b) = b

Reduction (R)

Ready to use (D2)

This can be reduced (B + C)(B + F + C) = (B + C)(F + C)

(B + F + CD)(B + AC) Use (D2) twice (B + F + C)(B + F + D)(B + A)(B + C)

see page below

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) Printed; 11/02/04 Modified; February 11, 2004

ugh! Slide 59




Shortcuts for Factoring

Shortcuts for Factoring Reason for choosing B over C There are two common factors B and C. If one factors out C first get C(AB + BD) + BF Then 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,

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

z + uz = z

Use (D2) three times

(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 Factor

WXY + W·XZ + WYZ + WYZ Associative law WXY + WYZ + W·XZ + WYZ (D1) used twice W(XY + YZ) + W(XZ + YZ) Swap: w(a) + w(b) = (w + b)(w + a) [W +(XZ + YZ)][W +(XY + YZ)] D1 Swap

Swap on left; (D1) on right

[(W +(Z +Y)(Z + X)][W +Y(X + Z)] D2 D2 (W + Z + Y)(W + Z + X) (W + Y)(W + X + Z) Factor

Finally we have to use (D2).

AB·D + ACD + A·C + ABD (D1) on both sides A(B·D + CD) + A(C + BD) swap [A + (C + BD)][A + (B·D + CD)]

(D2) on left; swap on right

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

(D2) on left; (D2) on right

[A + (C + B)][A + (C + D)][A + (B + D)][A + (C + D)] Clean up brackets (A + C + B)(A + C + D)(A + B + D)(A + C + D) 67.• PROBLEM: Factor AC + ABD + ABE + A·CDE


Five factors, four have 3 letters, one has 2.



Comment on Slide 59



Example of Multiplying Out All letters are different, no simplification possible

(A + B)(C + D)(F + G + H) = (AC + AD + BC + BD)(F + G + H)

Use (D1) rewrite

(AC + AD + BC + BD) · F =

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

=

ACF + ADF + BCF + BDF + ACG + ADG + BCG + BDG

Use (D1)

+ ACH + ADH + BCH + BDH

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



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)

Example: (B + F + C)(B + C)(B + F + D)(B + A) Use D1 twice

x +xy = x

0 0 = (BB + BC + FB + FC + CB + C)(BB + BA + FB + FA + DB + DA)

BCD D

ACF

Use D1 BF

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

F

0 = FBBA + FBFB + FBFA + FBDB + FBDA

B

A

ABC C

ACD

+ CBA + CFB + CFA + CDB + CDA)

BCD

Put letters in order

D

= BF + ABC + ACF + BCD + ACD Use map

F

= BF + ABC + BCD

B A

ABC C

ACD


Slide 60





Example of Multiplying Out (A + D)(A + C + D)(A + B)(A + B + C)(A + C + D)

Arrange to ready to use Swap

= (A + D)(A + B) · (A + C + D )(A + C + D) · (A + B + C) Use Swap twice = (AB + A·D) · [A (C + D) + A(C + D )] · (A + B + C) Use (D1) = {AB [A (C + D) + A(C + D )] + A·D)[A (C + D) + A(C + D )]} · (A + B + C) Use (D1) = {AB(C + D) + A·D(C + D)} · (A + B + C) xy + x = x

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

Use (D1)

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

D

xy + x = x

= AB·C + ABD + ABD + ACD


ABD

Collect Check on map that there is no more simplification


A · C ·D B

A

AB·C

ABD C

Comment on Slide 60



Multiplying Out

Using D1 alone eventually works, but long!

Review of Rules

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 possible xa + a = a, x + xy = x + y (d) Then use (D1) (e) Loop back and try again until done.

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

Find common variables = (B + F + C)(B + F + D) · (B + A)(B + C) Use (D2) twice = (B + F + CD) · (B + AC) Use (Sw) with B = BAC + B(F + CD) Finally use (D1)

Simplify (S) xy + x = x

(b + x)(b) = b

Reduction (R) xy + x = y + x (b + x)(b)=(b + x) Distributive (D2) u + yz = (u + y)(u + z) Extended Distributive (D2) u + xyz = (u + x)(u + y)(u + z) Swap (Sw) by + bz = (b + y)(b + z)

= ABC + BF + BCD Go back and look at Slide 60




Slide 61



Multiplying Out 68.• PROBLEM Multiply out. 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 of three letters each. The answer should be very symmetric on an AB\CD Karnaugh map. (B + C)(A + B + C)(A + C + D)(A + B + C)(A + C + D)



Comment on Slide 61



Multiplying Out, Short Cut Method

Multiplying Out, Short Cut Method Example: (A + C + D)(B + D)(A + B + C)(C + D)(A + B + D) (x)(x+y) = x Always check for obvious simplifications

= (A + C + D)(B + D)(A + B + C)(C + D)(A + B + D) Rearrange to use (D2) and Swap

= (A + C + D)(A + C + B)(B + D)(C + D) Use (D2) and Swap

(A + C + BD) (CD + BD) x +xy = x

=

(A + C + BD) · CD

ACD + CD + BCD

=

+ (A + C + BD) · BD

0 + ABD + CBD + BDBD

Use (D1)

Collect

= CD + ABD + BCD

CD D

Check map for further

BCD

simplifications

= CD + ABD + BC

ABD

B

A C




Slide 62


Multiplying Out, Short Cuts

Multiplying Out, Short Cuts Simplify 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



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 Duality Multiply Out Details

Steps: (1) The expression to factor is Σ of Π Example:

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

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

(2) Take its dual to get Π of Σ.

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

(A + B)(B + C)(A + C + D) (3) Multiply out the dual to get Σ of Π again BA + BC + BD + AC

= 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 identity with the desired Π of Σ.



Multiplying out is based on (D1). Easier than factoring based on (D2).

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


(D1)

Algebra of one is the dual of the other Slide 63


Changing Factoring into Multiplying Out

Changing Factoring into Multiplying Out Factoring 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 normal algebraic 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 usually easier than factoring. Multiplying out in the dual space does not give the answer. One take the dual of the answer. This will then be the factored form of the original expression. 70.• PROBLEM Show that F = A·B·C ·D + ABCD +ABCD + ABCD takes only 8 letters or 12 gate inputs in factored form.



Comment on Slide 63



Factoring Using Duality

Factoring Using Duality Example: (-25% if you say these are equal)

A ·BD + AB ·D + ABC + AC Take dual

(A + B + D)(A + B + D)(A + B + C)(A + C) Rearrange to use (D2) = (A + B + D)(A + B + C)(A + B + D)(A + C) Use (D2)

equal

= [A + (B + D)(B + C)] [A + (B + D)C] Use Swap equal

= A (B + D)C) + A(B + D)(B + C) Use Swap = A (B + D)C) + A(B ·C + BD) Use (D1) = A·BC + A ·DC + AB ·C + ABD

A·BC

= A ·BC + A ·DC + AB ·C + ABD OK

ABD

Check on map

B

A

Take dual

C

AB ·C

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


D

Slide 64



A ·DC


Factoring In the Dual Space

Factoring In the Dual Space Example A ·BC + ACD + ABC + BCD Take the dual

To Factor

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

Note the excess of C and C rearrange for (D2)

= (C + B + A)(C + B + A)(C + A + D)(C + B + D) Use (D2) = [C + (B + A)(B + A)] · [(C + (A + D)(B + D)] Use Swap = C(A + D)(B + D) + C(B + A)(B + A) Use Swap = C(AB + D) + C(BA + BA)

CD

Use (D1)

= CAB + CD + C ·BA + CBA OK Take the dual

Check on map for more simplifications

D

ABC

B

ABC

A

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

AB·C

C

71.• PROBLEM Factor EF ·D + ECD + E ·C + EFD



Comment on Slide 64



Method 3: Factoring Using Karnaugh

Method 3: Factoring Using Karnaugh Maps D Steps: (1) Given F = ( Σ of Π expression) F = A·BC + AD + ABC + BCD

A

(2) Plot it on a map.

1

1 1 1 1 1 1

Map of F

B

1 C

D

(3) Make a map for F, It has “1” where F had “0” (4) Circle the F map

1 1 1 1

A

Map of F

1 1

1

B

1 C

D

(5) Write out the equation for F F = C ·D + ABC + AB ·D + A·B ·C

A

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

1 1 1 1

Map of F

1 1

1

B

1 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.• Z

Maps from Prob 62.• Z

Z

00 01

1 1

1 1

1

1 1

10

00 01

X

d

11

W

E=

Z

YZ WX 00 01 11 10

YZ WX 00 01 11 10

11

W 10 1

1 1 1

1

1 1

d

Y

YZ WX 00 00 01

X W

11 10

1

01 11 10

1 1 1 1

Y

F=

00

G=

1 1 1 1 1 1 d Y

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


01

1 1

11

W 10 1

1 1

00 01

X

d 1

Z

Z

YZ WX 00 01 11 10

YZ WX 00 01 11 10

11

W 10 1

1

1 1 1 1

YZ WX 00 01 11 10

1

00

1

d

1

01

X W

Y

Y

E=YZ+Q+P

F=W ·Z+Q

M=WXY ·Z

Q=M+XYZ

E=(Y+Z)Q ·P

10

1 1 1 1

1 1 1

1 1

1

d

X

Y

G=X·Z+M

P=WY

F=(W+Z)Q 20 letters, 23 gate inputs, 10 gates


11

1

G=(X+Z)M

Comment on Slide 65




Product of Sum Karnaugh Maps Simplify the Π of Σ expression F = (A + B + C)(A + C + D)(B + C + D)(B + D)(A + C + D)

D

(1) Find Inverse with DeMorgan.

1

F = ABC + AC·D + BCD + B ·D + AC·D

1 1

A

(2) Make a map for F

1 1

1 1

1

Map of F

B

1

1 C

(3) Simplify the F map

D

(4) Write out the equation for F

1

F = BC + AB ·C + ABD + B ·D

1 1

A

(5) Invert F using DeMorgan’s law to get simplified F

1 1

1 1

1

1

1

Map of F

B

C

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

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


Slide 66





Product of Sum Maps Constructing an AND of ORs map a bc 0 00 01 11 10

0 0 1 1

1

1 1 1 1

a bc 0 00 01 11 b 10

1 1 1 1

True Π of Σ map a bc 0

1

1 1 0 0

c

00 01 11 b 10

0 1 1 0

a bc 0

1

0 1 1 0

00 01 11 b 10

c

1

0 0 0 1 1 0 0 0

c

F=a+b F=a+b F=c F = (a + b)(a + b)c Product map 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 have a “0”. The other squares have “1”s. a bc 0

If one multiplies out F = (a + b)(a + b)c, one gets F= abc + abc. Which has exactly the same map except now one thinks about the “1”s instead of the “0”s. b

00 01 11 10

0 0

1

0

1 1 0

c

0 0

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

a bc 0 00 01 11 b 10

1

1 1 1 0 0 1 1 1

c

Σ of Π map of F

75.• PROBLEM Multiply out to get three 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)



Comment on Slide 66





Slide 67



Solution problem 75 F = (B + C)(A + B + C)(A + C + D)(A + B + C)(A + C + D) F = BC + ABC + AC + ABC + A CD

D

A

1 1 1

1 1

D

1

1 1 1 1 1 1

F

C

1

B

B

A 1

1 F

1 C F = B D + AB C + A B C

76.• PROBLEM (ACTUALLY SOLN TO 75) Multiply out to get four of three letters each. Use a Karnaugh map. (B + C)(A + B + C)(A + C + D)(A + B + C)(A + C + D)



Comment on Slide 63


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