Iterative Division

7/27/2019 Iterative Division

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ECE366

ComputerArchitecture

Instructor:ShantanuDutt

Departmento

fElectricalandCo

mputerEn

gineering

UniversityofIllinoisatChicago

LectureNotes#

13

CO

MPUTERARITH

METIC:

Ite

rativeDivisionTechniques

cShantanuDutt,UIC

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2/15

2

DivisionBas

ics

Rad

ix

divisionisess

entiallyatrial-and-errorprocess,inw

hichthenext

quo

tientbitischosenfrom

Exa

mple:

Binarydivisionismuchsim

pler,sincethenextquotientbitiseithera0

or1

dependin

gonwh

etherthepartialremainderislessthanor

greater

than

/equaltothedivisor,respectively

Inte

gerdivision:Give

n2integers

the

dividendand

thedivisor,we

wan

ttoobtainanintegerquotient

and

anintegerremaind

er

,s.t.

Inte

gerFPdivision:G

iven2integers

thedividendand

thedivisor,


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wewanttoobtainafloating-point(FP)o

rrealquotient

s.t.

We

willtalkaboutdivisionofunsignedn

umbersfirst

cShantanuDutt,UIC

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4/15

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Div

isionSHR-DivisorMethod

Startsubtractionof

fromleftmostp

ositionof

;SHR

fornext

subtractioneveryitera

tion

Pen

cil-paperdivisionexam

ple:

cShantanuDutt,UIC

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5/15

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Division

SHL-Partial-Rem

ainderMethod

Inst

eadofshiftingthe

divisorrightby1

bit,thepartialrem

aindercanbe

shif

tedleftbyonebit

Exa

mple:

cShantanuDutt,UIC

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6/15

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DivisionHandling

with0s

inHigh-OrderBits

Ano

therExam

ple:Somehigherorderbitsofthe

-bitdivis

or

are0s:

Not

ethat

isstoredasa4-bit#(0010)in

thecom

puter.Thetypeofman-

ualadjustmentdonein

theaboveexampleofconverting4-bitsubtractions

into

2-bitones(ingeneral,

-bitsubtractionsto

-bitones,

where

)

isnotpossibleinacom

puter

cShantanuDutt,UIC

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7/15

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DivisionHandl

ing

with0sinH

igh-OrderBits(contd.)

Problemwith

havinghi

gherorderbitsas

0s:

Twowaystotacklethepr

oblem:

Method1:Shift

totheleftby

bits(untilitsMSBisa1),dothe

divisionfor

stepsfo

raninteger

(mo

restepsforaFP

)

Examp

le:

cShantanuDutt,UIC

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8/15

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DivisionHandl

ing

with0sinH

igh-OrderBits(contd.)

Problemwith

havinghi

gherorderbitsas

0s:

Method2:Augment

totheleftby

bitsthatareall0s.Perform

divisionfor

stepsforan

integer

andmorestepsforaFP

Examp

le:

Thisis

thetypeofdivisionalgorithmusedinacomputer

cShantanuDutt,UIC

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9/15

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DivisionACa

veat

Examp

leforneedin

gane

xtrabittotheleftofthedividendtocatcha1ona

shiftleft(requiredwheno

riginally

hasmorebitsthan

(e.g.,dividingan

8-bitnumberbya6-bitone):

ThusneedanextrabitinACistocatcha1outofACwhentheMSBsofthe

partial

remainder

and

areboth1sbut

!


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Thead

der/subtracterthusalsoneedstobe

bits;seeb

lockdiagram

ofdividergivennext.

NOTE:

Theextrabitisneverrequiredwhenboth

and

havethesame

number

ofbitstostart

with(thisdoesno

tcountthe

bitsthatare

addedlaterto

).

cShantanuDutt,UIC

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11/15

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Divis

ionCompleteAl

gorithm#1

Sim

ple

restoringdivision

algorithm:

Accum

ulator

Q

M

Cout

1

1

1

17

17bitAdd/Sub

DividendD

DivisorV

XOR

1

7

17

17

Serial

divisionfor16bitunsignednumbers

T

FromControlUnit

FromCo

ntrolUnit

1.Tocompute

"

,store

inthe

register,

inthe

#

registerand

0in

AC(Accumulator

),and0in

$

(holdsthequotientbitbeforetheshift

left).Notethat

isa

n

-bitregister,w

hile#

and%

&

are

'

( -bit

ones,with0sinitially

intheir'

( th

bits.

Rep

eatthefollowings

teps

times:

2.ShiftAC-Q-Tregister

combinationleft1bit(initially,this

hastheeffect

ofa

ddingon

0s

totheleftofthedividend)

3.Perform

%

&

%

&

(subtractdivisorfrom

'

( -bitpartialremainder

)


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4.Ifth

eresultisnegative,set

$

,otherw

iseset

$

5.Iftheresultofstep2

isnegative,restor

etheoldvalueof

ACbydoin

g

%

&

%

&

NOTE:

Noticesimilaritybetweenhardware

forA&Smultiplic

ationanddi-

vision.

Thesamehardwarewithadditiona

lcontrolcanbeu

sedforboth

purposes.

cShantanuDutt,UIC

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13/15

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Divis

ionCompleteAl

gorithm#2

Non-re

storingdivisionalgorithm:Anextraadditionstepisneededinthe

previou

salgorithmtorestoretheoldvalueof%

&

whenthere

sultofasub-

tractionisnegative.Thisstepiseliminatedinnon-restoringdivision:

1.Sam

einitializationasbefore.

Rep

eatthefollowings

teps

times:

2.ShiftAC-Q-Tregister

combinationleft1

bit

3.IfcurrentQ[0]is0andthisisnotthe1st

iteration

then

perform%

&

%

&

else

perform%

&

%

&

4.Ifth

eresultisnegative,set

$

,otherw

iseset

$

Thisworksbecause:Let

bethecontentso

fregisterAC-Qat

thebeginning

ofthe

)

thiteration.AfteraSHLandsubtrac

tion,0

isco

mputed.

Inrestoringdivision:Ifth

isresult(

0

)is-ve,restorationandthenan-

othersubtractioninthenextiteration

givesus1

.


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Innon-restoringdivision:Iftheresultofthepreviousiteration

(0

)is

-ve,we

donotrestorebutinthe

'

)

(thiterationweperform

aSHLtoget

1

0

andthenaddto

get1

.This

givesusthesameresultinAC-Q

asinre

storingdivision,andbasedonthissa

meresultwedeter

minethenext

bitoft

hequotient.Thus

non-restoringdivisionwill

giveusthesamequo-

tientas

restorin

gdivision

,thoughfaster.Th

usnon-restoringd

ivisionworks

correctlysincerestoringd

ivisioniscorrect.

cShantanuDutt,UIC

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15/15

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Twos

ComplementDivision(contd.)

The

rearenosimpledivisionalgorithmsfor2scomplement#sunlikethe

caseformultiplication

Thisisprimarilyduetothedifficultyins

electingthequotientbitsinsuch

awaythatithastthec

orrect+veor-ve2

scom

plementrep

resentation

Thu

stheapproach

gen

erallyfollowedis

tonegatea-ve

or

,perform

divisionandthennega

tethequotientifo

nlyoneof

or

was-ve

Am

oreefficientdivisionalgorithmfor

2scom

plement#sistheSRT

met

hod(forSweeney,

RobertsonandTocher,itsinventors);seeRef.text

#1

(HenneseyandPatterson)forthisalg

orithmifinterested

cShantanuDutt,UIC

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Iterative Division

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