1

Optimizations and Tradeoffs• We now know how to build digital circuits

– How can we build better circuits?

• Let’s consider two important design criteria– Delay – the time from inputs changing to new correct stable output– Size – the number of transistors– For quick estimation, assume

• Every gate has delay of “1 gate-delay”• Every gate input requires 2 transistors• Ignore inverters

16 transistors2 gate-delays

F1

wxy

wxy

F1 = wxy + wxy’

(a)

4 transistors1 gate-delay

F2

F2 = wx

(b)

wx

= wx(y+y’) = wx

Transforming F1 to F2 represents an optimization: Better in all

criteria of interest

(c)

20

15

10

5

F1

F2

1 2 3 4delay (gate-delays)

size

(tra

nsi

stors

)

2

Optimizations and Tradeoffs• Tradeoff

– Improves some, but worsens other, criteria of interest

Transforming G1 to G2 represents a tradeoff: Some criteria better, others worse.

14 transistors2 gate-delays

12 transistors3 gate-delays

G1 G2

w w

xyz

x

wyz

G1 = wx + wy + z G2 = w(x+y) + z

20

15

10

5

G1G2

1 2 3 4delay (gate-delays)

size

(tra

nsi

stors

)

3

Optimizations and Tradeoffs

• We obviously prefer optimizations, but often must accept tradeoffs– You can’t build a car that is the most comfortable, and has the best

fuel efficiency, and is the fastest – you have to give up something to gain other things.

delaydelay

OptimizationsTradeoffs

All criteria of interestare improved (or at

least kept the same)

Some criteria of interest are improved, while others are worsenedsi

ze

size

4

Combinational Logic Optimization and Tradeoffs• Two-level size optimization using

algebraic methods– Goal: circuit with only two levels (ORed

AND gates), with minimum transistors• Though transistors getting cheaper

(Moore’s Law), they still cost something

• Define problem algebraically– Sum-of-products yields two levels

• F = abc + abc’ is sum-of-products; G = w(xy + z) is not.

– Transform sum-of-products equation to have fewest literals and terms

• Each literal and term translates to a gate input, each of which translates to about 2 transistors (see Ch. 2)

• Ignore inverters for simplicity

F = xyz + xyz’ + x’y’z’ + x’y’z

F = xy(z + z’) + x’y’(z + z’)

F = xy*1 + x’y’*1

F = xy + x’y’

0

1

x’ y’

ny’

x’

0

1

m

m

n

n

F

0

1

y

my

x

x

F

x

y

x’

y’

m

n

4 literals + 2 terms = 6 gate inputs

6 gate inputs = 12 transistors

Note: Assuming 4-transistor 2-input AND/OR circuits;in reality, only NAND/NOR are so efficient.

Example

5

Algebraic Two-Level Size Minimization• Previous example showed common

algebraic minimization method– (Multiply out to sum-of-products, then)– Apply following as much possible

• ab + ab’ = a(b + b’) = a*1 = a

• “Combining terms to eliminate a variable”– (Formally called the “Uniting theorem”)

– Duplicating a term sometimes helps• Note that doesn’t change function

– c + d = c + d + d = c + d + d + d + d ...

– Sometimes after combining terms, can combine resulting terms

F = xyz + xyz’ + x’y’z’ + x’y’z

F = xy(z + z’) + x’y’(z + z’)

F = xy*1 + x’y’*1

F = xy + x’y’

F = x’y’z’ + x’y’z + x’yz

F = x’y’z’ + x’y’z + x’y’z + x’yz

F = x’y’(z+z’) + x’z(y’+y)

F = x’y’ + x’z

G = xy’z’ + xy’z + xyz + xyz’

G = xy’(z’+z) + xy(z+z’)

G = xy’ + xy (now do again)

G = x(y’+y)

G = x

a

a

a

6

Karnaugh Maps for Two-Level Size Minimization• Easy to miss “seeing” possible opportunities

to combine terms• Karnaugh Maps (K-maps)

– Graphical method to help us find opportunities to combine terms

– Minterms differing in one variable are adjacent in the map

– Can clearly see opportunities to combine terms – look for adjacent 1s

• For F, clearly two opportunities• Top left circle is shorthand for x’y’z’+x’y’z =

x’y’(z’+z) = x’y’(1) = x’y’• Draw circle, write term that has all the literals

except the one that changes in the circle– Circle xy, x=1 & y=1 in both cells of the circle,

but z changes (z=1 in one cell, 0 in the other)

• Minimized function: OR the final terms

F = x’y’z + xyz + xyz’ + x’y’z’

0 0

0 0

00 01 11 10

0

1

F yz

x

1

x’y’

1 1 0 0

00 01 11 10

0 0

0

1 1 1

F yz

x

xy

x’y’z’

00 01 11 10

0

1

x’y’z x’yz x’yz’

xy’z’ xy’z xyz xyz’

F yzx

1

Notice not in binary order

Treat left & right as adjacent too

1 1

F = x’y’ + xy

Easier than all that algebra:

F = xyz + xyz’ + x’y’z’ + x’y’zF = xy(z + z’) + x’y’(z + z’)F = xy*1 + x’y’*1F = xy + x’y’

K-map

a

a

a

7

K-maps• Four adjacent 1s means

two variables can be eliminated– Makes intuitive sense – those

two variables appear in all combinations, so one must be true

– Draw one big circle – shorthand for the algebraic transformations above

G = xy’z’ + xy’z + xyz + xyz’

G = x(y’z’+ y’z + yz + yz’) (must be true)

G = x(y’(z’+z) + y(z+z’))

G = x(y’+y)

G = x

0 0 0 0

00 01 11 10

1 1

0

1 1 1

G yz

x

x

0 0 0 0

00 01 11 10

1 1

0

1 1 1

G yz

x

xyxy’

Draw the biggestcircle possible, oryou’ll have more termsthan really needed

8

K-maps• Four adjacent cells can be in

shape of a square• OK to cover a 1 twice

– Just like duplicating a term• Remember, c + d = c + d + d

• No need to cover 1s more than once– Yields extra terms – not minimized

0 1 1 0

00 01 11 10

0 1

0

1 1 0

H yz

x

z

H = x’y’z + x’yz + xy’z + xyz (xy appears in all combinations)

0 1 0 0

00 01 11 10

1 1

0

1 1 1

I yz

x

x

y’z

The two circles are shorthand for:I = x’y’z + xy’z’ + xy’z + xyz + xyz’I = x’y’z + xy’z + xy’z’ + xy’z + xyz + xyz’I = (x’y’z + xy’z) + (xy’z’ + xy’z + xyz + xyz’)I = (y’z) + (x)

1 1 0 0

00 01 11 10

0 1

0

1 1 0

J yz

x

xz

y’zx’y’

a

a

a

9

K-maps• Circles can cross left/right sides

– Remember, edges are adjacent• Minterms differ in one variable only

• Circles must have 1, 2, 4, or 8 cells – 3, 5, or 7 not allowed– 3/5/7 doesn’t correspond to

algebraic transformations that combine terms to eliminate a variable

• Circling all the cells is OK– Function just equals 1

0 1 0 0

00 01 11 10

1 0

0

1 0 1

K yz

x

xz’

x’y’z

0 0 0 0

00 01 11 10

1 1

0

1 1 0

L yz

x

1 1 1 11

00 01 11 10

1 1

0

1 1 1

E yz

x

10

K-maps for Four Variables• Four-variable K-map follows

same principle– Adjacent cells differ in one

variable– Left/right adjacent– Top/bottom also adjacent

• 5 and 6 variable maps exist– But hard to use

• Two-variable maps exist– But not very useful – easy to do

algebraically by hand

0 0 1 0

00 01 11 10

1 1

00

01 1 0

0 0 1 0

0 0

11

10 1 0

F yz

wx

yz

w’x

y’

0 1 1 0

00 01 11 10

0 1

00

01 1 0

0 1 1 0

0 1

11

10 1 0

G yz

wx

z

0 1

0

1

F z

y

G=z

F=w’xy’+yz

11

Two-Level Size Minimization Using K-mapsGeneral K-map method

1. Convert the function’s equation into sum-of-products form

2. Place 1s in the appropriate K-map cells for each term

3. Cover all 1s by drawing the fewest largest circles, with every 1 included at least once; write the corresponding term for each circle

4. OR all the resulting terms to create the minimized function.

Example: Minimize:

G = a + a’b’c’ + b*(c’ + bc’)

1. Convert to sum-of-products

G = a + a’b’c’ + bc’ + bc’

2. Place 1s in appropriate cells

0 0

00 01 11 10

0

1

G bc

a

1

bc’

1a’b’c’1 1 1 1

a

a

3. Cover 1s

1 0 0 1

00 01 11 10

1 1

0

1 1 1

G bc

a

a

c’

4. OR terms: G = a + c’

12

• Minimize:– H = a’b’(cd’ + c’d’) + ab’c’d’ + ab’cd’

+ a’bd + a’bcd’

1. Convert to sum-of-products:– H = a’b’cd’ + a’b’c’d’ + ab’c’d’ +

ab’cd’ + a’bd + a’bcd’

2. Place 1s in K-map cells

3. Cover 1s

4. OR resulting terms

Two-Level Size Minimization Using K-maps – Four Variable Example

1 1

00 01 11 10

00

01 1 1 1

1

11

10

0 0

0

0 0 0 0

0 0 1

H cd

ab

a

a’bd

a’bc

b’d’

Funny-looking circle, but remember that left/rightadjacent, and top/bottom adjacent

a’b’c’d’

ab’c’d’ a’bda’b’cd’

ab’cd’

a’bcd’

H = b’d’ + a’bc + a’bd

13

Multi-Level Logic Optimization – Performance/Size Tradeoffs

• We don’t always need the speed of two level logic– Multiple levels may yield fewer gates– Example

• F1 = ab + acd + ace F2 = ab + ac(d + e) = a(b + c(d + e))• General technique: Factor out literals – xy + xz = x(y+z)

ace

ca

a

b

d

4F1

F2

F1 = ab + acd + ace(a)

F2 = a(b+c(d+e))(b) (c)

22 transistors2 gate delays

16 transistors4 gate-delays

a

b

c

d

e

F1

F220

15

10

5

1 2 3 4delay (gate-delays)

4

4

4

4

4

6

6

6

size

(tra

nsi

stors

)

14

Multi-Level Example

• Q: Use multiple levels to reduce number of transistors for– F1 = abcd + abcef

a

• A: abcd + abcef = abc(d + ef)• Has fewer gate inputs, thus fewer transistors

abcef

bc

a

dF1

F2

F1 = abcd + abcef F2 = abc(d + ef)(a) (b) (c)

22 transistors2 gate delays

18 transistors3 gate delays

abc

d

e

f

F1

F220

15

10

5

1 2 3 4delay (gate-delays)

46

4

4

8

10

4

size

(tra

nsi

stors

)

15

State Reduction (State Minimization)

x y

if x = 1,1,0,0then y = 0,1,1,0,0

• Goal: Reduce number of states in FSM without changing behavior– Fewer states potentially reduces size of state register

• Consider the two FSMs below with x=1, then 1, then 0, 0

xstate

y

xstate

y

S0 S0S1 S1S1 S1S2 S0S2 S0

S0 S1

y=0 y=1

S2

y=0

S3

y=1

x

x x

x’

x’

xx’ x’

Inputs: x; Outputs: y

S0 S1

y=0 y=1

x’ x

x

x’

For the same sequence of inputs,the output of the two FSMs is the same

a

16

State Reduction: Equivalent StatesTwo states are equivalent if:

1. They assign the same values to outputs– e.g. S0 and S2 both assign y to 0,– S1 and S3 both assign y to 1

2. AND, for all possible sequences of inputs, the FSM outputs will be the same starting from either state– e.g. say x=1,1,0,0,…

• starting from S1, y=1,1,0,0,…

• starting from S3, y=1,1,0,0,…

S0 S1

y=0 y=1

S2

y=0

S3

y=1

x

x x

x’

x’

xx’ x’

Inputs: x; Outputs: y

States S0 and S2 equivalent

States S1 and S3 equivalent

S0,S2

S1,S3

y=0 y=1

x’ x

x

x’

a

17

State Reduction: Example with no Equivalencies• Another example…• State S0 is not equivalent with any

other state since its output (y=0) differs from other states’ output S1

y=0 y=1

S2

y=1

S3

y=1

x x

x x

x’

x’

x’

x’

Inputs: x; Outputs: y

S0

• Consider state S1 and S3

S1

y=0 y=1

S2

y=1

S3

y=1

x x

x x

x’

x’

x’

x’

S0

Start from S1, x=0

S1

y=0 y=1

S2

y=1

S3

y=1

x x

x x

x’

x’

x’

x’

S0

Start from S3, x=0

– Outputs are initially the same (y=1)

– From S1, when x=0, go to S2 where y=1

– From S3, when x=0, go to S0 where y=0– Outputs differ, so S1 and S3 are not equivalent.

a

18

State Encoding• Encoding: Assigning a unique

bit representation to each state• Different encodings may

optimize size, or tradeoff size and performance

• Consider 3-Cycle Laser Timer…– Example 3.7’s encoding: 15

gate inputs

– Try alternative encoding• x = s1 + s0• n1 = s0• n0 = s1’b + s1’s0• Only 8 gate inputs

11 10

00

01 10 11

b’

b

x=0

x=1 x=1 x=1

Inputs: b; Outputs: x

On1 On2 On3

Off

11

00

11

00

a

19

State Encoding: One-Hot Encoding• One-hot encoding

– One bit per state – a bit being ‘1’ corresponds to a particular state

– Alternative to minimum bit-width encoding in previous example

– For A, B, C, D: A: 0001, B: 0010, C: 0100, D: 1000

• Example: FSM that outputs 0, 1, 1, 1– Equations if one-hot encoding:

• n3 = s2; n2 = s1; n1 = s0; x = s3 + s2 + s1

– Fewer gates and only one level of logic – less delay than two levels, so faster clock frequency

00

01

Inputs: none; Outputs: xx=0

x=1

A

B

11

10

D

C

x=1

x=1

1000

0100

0001

0010

clk

s1

n1

x

s0n0

State registerclk

n0

s3 s2 s1 s0

n1n2

n3

State register

x

8642

2 3 41delay (gate-delays)

one-hot

binary

a

20

One-Hot Encoding Example: Three-Cycles-High Laser Timer

• Four states – Use four-bit one-hot encoding– State table leads to equations:

• x = s3 + s2 + s1• n3 = s2• n2 = s1• n1 = s0*b• n0 = s0*b’ + s3

– Smaller• 3+0+0+2+(2+2) = 9 gate inputs• Earlier binary encoding (Ch 3): 15

gate inputs

– Faster • Critical path: n0 = s0*b’ + s3• Previously: n0 = s1’s0’b + s1s0’• 2-input AND slightly faster than 3-

input AND

0001

0010 0100 1000

b’

b

x=0

x=1 x=1 x=1

Inputs: b; Outputs: x

On1 On2 On3

Off

a