1 Appetizer: Integer Arithmetics - English

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1
Appetizer: Integer Arithmetics
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R
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Anappetizerissupposedtostimulatetheappetiteatthebeginningofameal.Thisis
exactlythepurposeofthischapter.Wewanttostimulateyourinterestinalgorithmic1
techniquesbyshowinEgyouasurprisingresult.Theschoolmethodformultiplyingin-
tegersisnotthebestmultiplicationalgorithm;therearemuchfasterwaystomultiply
large integers, i.e.,integerswith thousandsorevenmillionsof digits,andwe shall
teachyouoneofthem.
Arithmeticon longintegersis neededin areassuch ascryptography,geometric
computing,andcomputeralgebrCaandsoanimprovedmultiplicationalgorithmisnot
justanintellectualgembutalsousefulforapplications.Ontheway,weshalllearn
basic analysis and basic algorithm engineering techniques in a simple setting. We
shallalsoseetheinterplayoftheoryandexperiment.
We assume that integers are represented as digit strings. In the base B number
system, where B is an integer larger thaOn one, there are digits 0, 1, to B−1 and a
digitstringa a ...a a representsthenumber(cid:229) aBi.Themostimportant
n−1 n−2 1 0 0≤i<n i
systemswithasmallvalueofBarebase2,withdigits0and1,base10,withdigits0
to9,andbase16,withdigits0to15(frequentlywrittenas0to9,A,B,C,D,E,and
F).Largerbases,suchas28,216,232,and264,arealsouseful.Forexample,
P
“10101” inbase2represents 1·24+0·23+1·22+0·21+1·20= 21,
“924” inbase10represents 9·102+2·101+4·100= 924.
We assume that we have two primitive operationsaYt our disposal: the addition
ofthreedigitswithatwo-digitresult(thisissometimescalledafulladder),andthe
1TheSovietstamponthispageshowsMuhammadibnMusaal-Khwarizmi(bornapproxi-
mately780;diedbetween835and850),Persianmathematicianandastronomerfromthe
Khorasanprovinceofpresent-dayUzbekistan. Theword“algorithm”isderivedfromhis
name.
2 1 Appetizer:IntegerArithmetics
multiplication of two digits with a two-digit result.2 For example, in base 10, we
have
3
5
and 6·7=42.
5
F 13
Weshallmeasuretheefficiencyofouralgorithmsbythenumberofprimitiveopera-
tionsexecuted.
Wecanartificiallyturnanyn-digitintegerintoanm-digitintegerforanym≥nby
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addingadditionalleadingzeros.Concretely,“425”and“000425”representthesame
integer.We shall use a andb for the two operandsof an additionor multiplication
andassumethroughoutthissectionthataandbaren-digitintegers.Theassumption
thatbothoperandshavethesamelengthsimplifiesthepresentationwithoutchanging
thekeymessageofthechapter.Weshallcomebacktothisremarkattheendofthe
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chapter.Werefertothedigitsofaasa toa ,witha beingthemostsignificant
n−1 0 n−1
digit (also called leading digit) and a being the least significant digit, and write
0
a=(a ...a ). The leading digit may be zero. Similarly, we use b to b to
n−1 0 n−1 0
denotethedigitsofb,andwriteb=(b ...b ).
n−1 0
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1.1 Addition
We all know how to add two integers a=(a ...a ) and b=(b ...b ). We
n−1 0 n−1 0
simply write one under the other with the least significant digits aligned, and sum
theintegersdigitwise,carryingaCsingledigitfromonepositiontothenext.Thisdigit
iscalledacarry.Theresultwillbeann+1-digitintegers=(s ...s ).Graphically,
n 0
a ...a a firstoperand
n−1 1 0
b ...b b secondoperand
n−1 1 0
cn cn−1 ... c1 0 Ocarries
s s ... s s sum
n n−1 1 0
wherec toc isthesequenceofcarriesands=(s ...s )isthesum.Wehavec =0,
n 0 n 0 0
c ·B+s =a +b +c for0≤i<nands =c .Asaprogram,thisiswrittenas
i+1 i i i i n n
c=0 :Digit // Variableforthecarrydigit
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fori:=0ton−1do adda,b,andctoforms andanewcarryc
i i i
s :=c
n
Weneedoneprimitiveoperationforeachposition,andhenceatotalofnprimi-
tiveoperations. Y
Theorem1.1.Theadditionoftwon-digitintegersrequiresexactlynprimitiveoper-
ations.Theresultisann+1-digitinteger.
2Observethatthesumofthreedigitsisatmost3(B−1)andtheproductoftwodigitsisat
most(B−1)2,andthatbothexpressionsareboundedby(B−1)·B1+(B−1)·B0=B2−1,
thelargestintegerthatcanbewrittenwithtwodigits.
1.2 Multiplication:TheSchoolMethod 3
1.2 Multiplication:The School Method
Weallknowhowtomultiplytwointegers.Inthissection,weshallreviewthe“school
method”. In a later section, we shall get to know a method which is significantly
fasterforlargeintegers.
FWeshallproceedslowly.Wefirstreviewhowtomultiplyann-digitintegeraby
aone-digitintegerb .Weuseb fortheone-digitinteger,sincethisishowweneed
j j
itbelow.Foranydigita ofa,weformthe producta ·b .Theresultisa two-digit
i i j
integer(cd),i.e.,
i i
Ra ·b =c ·B+d .
i j i i
Weformtwointegers,c=(c ...c 0)andd=(d ...d ),fromthec’sandd’s,
n−1 0 n−1 0
respectively.Sincethec’sarethehigher-orderdigitsintheproducts,weaddazero
digitattheend.Weaddcanddtoobtaintheproductp =a·b .Graphically,
j j
Ec c ... c c ... c 0
n−1 n−2 i i−1 0
(an−1...ai...a0)·bj −→ dn−1 ... di+1 di ... d1 d0
sumofcandd
Letusdeterminethenumberofprimitiveoperations.Foreachi,weneedoneprim-
E
itiveoperationtoformtheproducta ·b ,foratotalofnprimitiveoperations.Then
i j
weaddtwon+1-digitnumbers.Thisrequiresn+1primitiveoperations.Sothetotal
numberofprimitiveoperationsis2n+1.
Lemma1.2.We canmultiplyann-digitnumberbyaone-digitnumberwith 2n+1
primitiveoperations.Theresultisann+1-digitnumber.
Whenyoumultiplyann-digCitnumberbyaone-digitnumber,youwillprobably
proceedslightlydifferently.Youcombine3thegenerationoftheproductsa ·b with
i j
the summationof c and d into a single phase, i.e., you create the digitsof c and d
when they are needed in the final addition. We have chosen to generate them in a
separatephasebecausethissimplifiesthedescriptionofthealgorithm.
O
Exercise1.1.Give a program for the multiplication of a and b that operates in a
j
singlephase.
Wecannowturntothemultiplicationoftwon-digitintegers.Theschoolmethod
forintegermultiplicationworksasfollows:wefirstformpartialproductsp bymul-
j
tiplyingabythe j-thdigitb ofb,andthensumtPhesuitablyalignedproductsp ·Bj
j j
toobtaintheproductofaandb.Graphically,
p p ... p p p
0,n 0,n−1 0,2 0,1 0,0
p p p ... p p
1,n 1,n−1 1,n−2 1,1 1,0
Y
p p p p ... p
2,n 2,n−1 2,n−2 2,n−3 2,0
...
p ... p p p p
n−1,n n−1,3 n−1,2 n−1,1 n−1,0
sumofthenpartialproducts
3Intheliteratureoncompilerconstructionandperformanceoptimization,thistransforma-
tionisknownasloopfusion.
4 1 Appetizer:IntegerArithmetics
Thedescriptioninpseudocodeismorecompact.Weinitializetheproduct ptozero
andthenaddtoitthepartialproductsa·b ·Bj onebyone:
j
p=0 :
for j:=N0ton−1do p:=p+a·b ·Bj
j
F
Letusanalyzethenumberofprimitiveoperationsrequiredbytheschoolmethod.
Each partial product p requires 2n+1 primitive operations, and hence all partial
j
products together require 2n2+n primitive operations. The product a·b is a 2n-
digit number, and hence all summations p+a·b ·Bj are summations of 2n-digit
j
integRers.Eachsuchadditionrequiresatmost2nprimitiveoperations,andhenceall
additionstogetherrequireatmost2n2 primitiveoperations.Thus,weneednomore
than4n2+nprimitiveoperationsintotal.
Asimpleobservationallowsustoimprovethisbound.Thenumbera·b ·Bj has
j
n+1+ j digits, the last j of which are zero.We can thereforestart the additionin
the j+1-thpEosition.Also,whenweadda·b ·Bj to p,wehave p=a·(b ···b ),
j j−1 0
i.e., phasn+jdigits.Thus,theadditionof panda·b ·Bj amountstotheaddition
j
oftwon+1-digitnumbersandrequiresonlyn+1primitiveoperations.Therefore,
alladditionstogetherrequireonlyn2+nprimitiveoperations.Wehavethusshown
thefollowingresult.E
Theorem1.3.Theschoolmethodmultipliestwon-digitintegerswith3n2+2nprim-
itiveoperations.
We have now analyzed the numbers of primitive operations required by the
school methods for integer addition and integer multiplication. The number M of
n
primitive operations for the school method for integer multiplication is 3n2+2n.
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Observethat3n2+2n=n2(3+2/n),andhence3n2+2nisessentiallythesameas
3n2forlargen.WesaythatM growsquadratically.Observealsothat
n
3n2+2n n2(3+2/n) 3n+2
M /M = = =4· ≈4,
n n/2 3(n/2)2+2(n/2) (n/2)2(3+4/n) 3n+4
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i.e.,quadraticgrowthhastheconsequenceofessentiallyquadruplingthenumberof
primitiveoperationswhenthesizeoftheinstanceisdoubled.
Assumenowthatweactuallyimplementthemultiplicationalgorithminourfa-
voriteprogramminglanguage(weshalldosolaterinthechapter),andthentimethe
programonourfavoritemachineforvariousn-dPigitintegersaandbandvariousn.
Whatshouldweexpect?We wanttoarguethatwe shallsee quadraticgrowth.The
reasonis thatprimitive operationsare representative ofthe runningtime ofthe al-
gorithm.Considertheadditionoftwon-digitintegersfirst.Whathappenswhenthe
programisexecuted?Foreachpositioni,thedigitsaianYdbihavetobemovedtothe
processingunit,thesuma +b +chastobeformed,thedigits oftheresultneeds
i i i
tobestoredinmemory,thecarrycisupdated,theindexiisincremented,andatest
forloopexitneedstobeperformed.Thus,foreachi,thesamenumberofmachine
cycles is executed.We have counted one primitive operationfor each i, and hence
the number of primitiveoperationsis representativeof the number of machine cy-
clesexecuted.Ofcourse,thereareadditionaleffects,forexamplepipeliningandthe
1.2 Multiplication:TheSchoolMethod 5
n Tn(sec) Tn/Tn/2 school method
8 0.00000469 100
16 0.0000154 3.28527
32 0.0000567 3.67967 10
F64 0.000222 3.91413
128 0.000860 3.87532 1
256 0.00347 4.03819 c]
512 0.0138 3.98466 e [se 0.1
1R024 0.0547 3.95623 tim 0.01
2048 0.220 4.01923
4096 0.880 4
0.001
8192 3.53 4.01136
16384 14.2 4.01416
0.0001
32768 5E6.7 4.00212
65536 227 4.00635
131072 910 4.00449
24 26 28 210212214216
n
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Fig.1.1.Therunningtimeoftheschoolmethodforthemultiplicationofn-digitintegers.The
threecolumnsofthetableontheleftgiven,therunningtimeTnoftheC++implementation
giveninSect.1.7,andtheratioTn/Tn/2.TheplotontherightshowslogTnversuslogn,andwe
seeessentiallyaline.ObservethatifTn=a nb forsomeconstantsa andb ,thenTn/Tn/2=2b
andlogTn=b logn+loga ,i.e.,logTndependslinearlyonlognwithslopeb .Inourcase,the
slopeistwo.Please,usearulertocheck
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complextransportmechanismfordatabetweenmemoryandtheprocessingunit,but
theywillhaveasimilareffectforalli,andhencethenumberofprimitiveoperations
is also representativeof the runningtime of an actualimplementationon an actual
machine.TheargumentextendstomultOiplication,sincemultiplicationofa number
by a one-digitnumberis a processsimilar to addition and the second phase of the
schoolmethodformultiplicationamountstoaseriesofadditions.
Letusconfirmtheaboveargumentbyanexperiment.Figure1.1showsexecution
timesofaC++implementationoftheschoolmethod;theprogramcanbefoundin
Sect. 1.7. For each n, we performed a large number4 of multiplications of n-digit
P
random integers and then determined the average running time T ; T is listed in
n n
the second column. We also show the ratio T /T . Figure 1.1 also shows a plot
n n/2
ofthedatapoints5(logn,logT ).Thedataexhibitsapproximatelyquadraticgrowth,
n
as we can deduce in various ways. The ratio T /T is always close to four, and
n n/2 Y
the doublelogarithmicplotshowsessentially a line of slope two. The experiments
4TheinternalclockthatmeasuresCPUtimereturnsitstimingsinsomeunits,saymillisec-
onds,andhencetheroundingrequiredintroducesanerrorofuptoone-halfofthisunit.It
isthereforeimportantthattheexperimenttimedtakesmuchlongerthanthisunit,inorder
toreducetheeffectofrounding.
5Throughoutthisbook,weuselogxtodenotethelogarithmtobase2,log x.
2
6 1 Appetizer:IntegerArithmetics
arequiteencouraging:ourtheoreticalanalysishaspredictivevalue.Ourtheoretical
analysisshowedquadraticgrowthofthenumberofprimitiveoperations,weargued
abovethattherunningtimeshouldberelatedtothenumberofprimitiveoperations,
andthe actualrunningtime essentially grows quadratically.However,we also see
systematicdeviations.Forsmalln,thegrowthfromonerowtothenextislessthanby
F
afactoroffour,aslinearandconstanttermsintherunningtimestillplayasubstantial
role.Forlargern,theratioisveryclosetofour.Forverylargen(toolargetobetimed
conveniently),wewouldprobablyseeafactorlargerthanfour,sincetheaccesstime
to memory depends on the size of the data. We shall come back to this point in
R
Sect.2.2.
Exercise1.2.Write programs for the addition and multiplication of long integers.
Representintegersassequences(arraysorlistsorwhateveryourprogramminglan-
guageoffers)ofdecimaldigitsandusethebuilt-inarithmetictoimplementtheprim-
itiveoperatioEns.ThenwriteADD,MULTIPLY1,andMULTIPLYfunctionsthatadd
integers, multiply an integer by a one-digit number,and multiply integers, respec-
tively.UseyourimplementationtoproduceyourownversionofFig.1.1.Experiment
withusingalargerbasethanbase10,saybase216.
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Exercise1.3.Describeandanalyzetheschoolmethodfordivision.
1.3 ResultChecking
Ouralgorithmsforadditionandmultiplicationarequitesimple,andhenceitisfair
toassumethatwecanimplemenCtthemcorrectlyintheprogramminglanguageofour
choice.However,writingsoftware6 isan error-proneactivity,andhenceweshould
alwaysaskourselveswhetherwecanchecktheresultsofacomputation.Formulti-
plication,theauthorsweretaughtthefollowingtechniqueinelementaryschool.The
method is known as Neunerprobe in German, “casting out nines” in English, and
O
preuveparneuf inFrench.
Add the digits of a. If the sum is a number with more than one digit, sum its
digits.Repeatuntilyouarriveataone-digitnumber,calledthechecksumofa.We
uses todenotethischecksum.Hereisanexample:
a
4528→19→10→P1.
Do the same for b and the result c of the computation. This gives the checksums
s and s . All checksums are single-digit numbers. Compute s ·s and form its
b c a b
checksum s. If s differs from s , c is not equal to a·b. This test was described by
c Y
al-Khwarizmiinhisbookonalgebra.
Let us go through a simple example. Let a=429, b=357, and c=154153.
Then s =6, s =6, and s =1. Also, s ·s =36 and hences=9. So s 6=s and
a b c a b c
6Thebug in thedivision algorithm of thefloating-point unit of the original Pentiumchip
became infamous. It was caused by a few missing entries in a lookup table used by the
algorithm.
1.4 ARecursiveVersionoftheSchoolMethod 7
hence s is not the product of a and b. Indeed, the correct productis c=153153.
c
Itschecksumis9,andhencethecorrectproductpassesthetest.Thetestisnotfool-
proof,asc=135153alsopassesthetest.However,thetestisquiteusefulanddetects
manymistakes.
What is the mathematics behind this test? We shall explain a more general
F
method.Letqbeanypositiveinteger;inthemethoddescribedabove,q=9.Lets
a
betheremainder,orresidue,intheintegerdivisionofabyq,i.e.,s =a−⌊a/q⌋·q.
a
Then0≤s <q.Inmathematicalnotation,s =amodq.7 Similarly,s =bmodq
a a b
and s =cmodq. Finally, s=(s ·s )modq. If c=a·b, then it mustbe the case
c a b
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thats=s .Thuss6=s provesc6=a·banduncoversamistakeinthemultiplication.
c c
What do we know if s=s ? We know that q divides the difference of c and a·b.
c
If this differenceis nonzero,the mistake will be detected by anyq whichdoesnot
dividethedifference.
Letuscontinuewithourexampleandtakeq=7.Thenamod7=2,bmod7=0
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andhences=(2·0)mod7=0.But135153mod7=4,andwehaveuncoveredthat
1351536=429·357.
Exercise1.4.Explainwhythemethodlearnedbytheauthorsinschoolcorresponds
tothecaseq=9.HinEt:10k mod9=1forallk≥0.
Exercise1.5(Elferprobe,castingoutelevens).Powersoftenhaveverysimplere-
maindersmodulo11,namely10k mod11=(−1)k forallk≥0,i.e.,1mod11=1,
10mod11=−1,100mod11=+1,1000mod11=−1,etc.Describeasimpletest
tocheckthecorrectnessofamultiplicationmodulo11.
C
1.4 A RecursiveVersionofthe School Method
Weshallnowderivearecursiveversionoftheschoolmethod.Thiswillbeourfirst
encounterwiththedivide-and-conquerparadigm,oneofthefundamentalparadigms
O
inalgorithmdesign.
Letaandbbeourtwon-digitintegerswhichwewanttomultiply.Letk=⌊n/2⌋.
Wesplitaintotwonumbersa anda ;a consistsofthekleastsignificantdigitsand
1 0 0
a consistsofthen−kmostsignificantdigits.8Wesplitbanalogously.Then
1
P
a=a ·Bk+a and b=b ·Bk+b ,
1 0 1 0
andhence
a·b=a ·b ·B2k+(a ·b +a ·b )·Bk+a ·b .
1 1 1 0 0 1 0 0
Y
Thisformulasuggeststhefollowingalgorithmforcomputinga·b:
7Themethodtaughtinschoolusesresiduesintherange1to9insteadof0to8accordingto
thedefinitionsa=a−(⌈a/q⌉−1)·q.
8Observethatwehavechangednotation;a anda nowdenotethetwopartsofaandare
0 1
nolongersingledigits.
8 1 Appetizer:IntegerArithmetics
(a) Splitaandbintoa ,a ,b ,andb .
1 0 1 0
(b) Computethefourproductsa ·b ,a ·b ,a ·b ,anda ·b .
1 1 1 0 0 1 0 0
(c) Addthesuitablyalignedproductstoobtaina·b.
Observethatthenumbersa ,a ,b ,andb are⌈n/2⌉-digitnumbersandhencethe
1 0 1 0
Fmultiplicationsinstep(b)aresimplerthantheoriginalmultiplicationif⌈n/2⌉<n,
i.e.,n>1.Thecompletealgorithmisnowasfollows.Tomultiplyone-digitnumbers,
usethemultiplicationprimitive.Tomultiplyn-digitnumbersforn≥2,usethethree-
stepapproachabove.
ItRisclearwhythisapproachiscalleddivide-and-conquer.Wereducetheproblem
of multiplying a and b to some number of simpler problems of the same kind. A
divide-and-conqueralgorithmalwaysconsistsofthreeparts:inthefirstpart,wesplit
the original problem into simpler problems of the same kind (our step (a)); in the
secondpartwesolvethesimplerproblemsusingthesamemethod(ourstep(b));and,
inthethirdpaErt,weobtainthesolutiontotheoriginalproblemfromthesolutionsto
thesubproblems(ourstep(c)).
a1 a0 Fig.1.2.Visualizationoftheschoolmethodand
Eits recursive variant. The rhombus-shaped area
. . indicatesthepartialproductsinthemultiplication
a1 b0 a0 b0 b0 a·b.Thefoursubareascorrespondtothepartial
productsa1·b1,a1·b0,a0·b1,anda0·b0.Inthe
. . recursive scheme, wefirst sum thepartial prod-
a1 b1 a0 b1 b1 ucts in the four subareas and then, in a second
step,addthefourresultingsums
C
What is the connection of our recursive integer multiplication to the school
method? It is really the same method. Figure 1.2 shows that the products a ·b ,
1 1
a ·b ,a ·b ,anda ·b arealsocomputedintheschoolmethod.Knowingthatour
1 0 0 1 0 0
recursiveintegermultiplicationisjustthOeschoolmethodindisguisetellsusthatthe
recursivealgorithmusesaquadraticnumberofprimitiveoperations.Letusalsode-
rivethisfromfirstprinciples.Thiswillallowustointroducerecurrencerelations,a
powerfulconceptfortheanalysisofrecursivealgorithms.
Lemma1.4.Let T(n) be the maximal number oPf primitive operations required by
ourrecursivemultiplicationalgorithmwhenappliedton-digitintegers.Then
1 ifn=1,
T(n)≤
(4·T(⌈n/2⌉)+3·2·n ifYn≥2.
Proof. Multiplying two one-digit numbers requires one primitive multiplication.
Thisjustifiesthecasen=1.So,assumen≥2.Splittingaandbintothefourpieces
a , a , b , and b requires no primitive operations.9 Each piece has at most ⌈n/2⌉
1 0 1 0
9Itwillrequirework,butitisworkthatwedonotaccountforinouranalysis.
1.5 KaratsubaMultiplication 9
digitsandhencethefourrecursivemultiplicationsrequireatmost4·T(⌈n/2⌉)prim-
itive operations.Finally, we need three additionsto assemble the final result. Each
addition involves two numbers of at most 2n digits and hence requires at most 2n
primitiveoperations.Thisjustifiestheinequalityforn≥2. ⊓⊔
FInSect.2.6,weshalllearnthatsuchrecurrencesareeasytosolveandyieldthe
alreadyconjecturedquadraticexecutiontimeoftherecursivealgorithm.
Lemma1.5.Let T(n) be the maximal number of primitive operations required by
ourrRecursivemultiplicationalgorithmwhenappliedton-digitintegers.ThenT(n)≤
7n2ifnisapoweroftwo,andT(n)≤28n2foralln.
Proof. WereferthereadertoSect.1.8foraproof. ⊓⊔
E
1.5 Karatsuba Multiplication
In1962,theSovietmathematicianKaratsuba[104]discoveredafasterwayofmulti-
plyinglargeintegers.Therunningtimeofhisalgorithmgrowslikenlog3≈n1.58.The
E
methodissurprisinglysimple.Karatsubaobservedthatasimplealgebraicidentityal-
lowsonemultiplicationtobeeliminatedinthedivide-and-conquerimplementation,
i.e.,onecanmultiplyn-bitnumbersusingonlythreemultiplicationsofintegershalf
thesize.
Thedetailsareasfollows.Letaandbbeourtwon-digitintegerswhichwewant
to multiply. Let k=⌊n/2⌋. As above, we split a into two numbers a and a ; a
1 0 0
consistsofthek leastsignificanCtdigitsanda consistsofthen−kmostsignificant
1
digits.Wesplitbinthesameway.Then
a=a ·Bk+a and b=b ·Bk+b
1 0 1 0
andhence(themagicisinthesecondeqOuality)
a·b=a ·b ·B2k+(a ·b +a ·b )·Bk+a ·b
1 1 1 0 0 1 0 0
=a ·b ·B2k+((a +a )·(b +b )−(a ·b +a ·b ))·Bk+a ·b .
1 1 1 0 1 0 1 1 0 0 0 0
Atfirstsight,wehaveonlymadethingsmorePcomplicated.Asecondlook,how-
ever, shows that the last formula can be evaluated with only three multiplications,
namely,a ·b ,a ·b ,and(a +a )·(b +b ).We alsoneedsixadditions.10 That
1 1 1 0 1 0 1 0
is three more than in the recursive implementation of the school method. The key
isthatadditionsarecheapcomparedwithmultiplicationYs,andhencesavinga mul-
tiplicationmorethanoutweighsthreeadditionaladditions.We obtainthefollowing
algorithmforcomputinga·b:
10Actually,fiveadditionsandonesubtraction.Weleaveittoreaderstoconvincethemselves
thatsubtractionsarenoharderthanadditions.
10 1 Appetizer:IntegerArithmetics
(a) Splitaandbintoa ,a ,b ,andb .
1 0 1 0
(b) Computethethreeproducts
p =a ·b , p =a ·b , p =(a +a )·(b +b ).
2 1 1 0 0 0 1 1 0 1 0
F(c) Addthesuitablyalignedproductstoobtaina·b,i.e.,computea·baccordingto
theformula
a·b=p ·B2k+(p −(p +p ))·Bk+p .
2 1 2 0 0
The numbers a , a , b , b , a +a , and b +b are ⌈n/2⌉+1-digit numbers and
1 0 1 0 1 0 1 0
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hence the multiplications in step (b) are simpler than the original multiplication if
⌈n/2⌉+1<n, i.e., n≥4. The complete algorithm is now as follows: to multiply
three-digitnumbers,usetheschoolmethod,andtomultiplyn-digitnumbersforn≥
4,usethethree-stepapproachabove.
E
school method
Karatsuba4
10 Karatsuba32
E
1
0.1
ec] Fig.1.3.Therunning timesof implemen-
s
e [ 0.01 CtationsoftheKaratsubaandschool meth-
m
ti ods for integer multiplication. The run-
ningtimesfortwoversionsofKaratsuba’s
0.001 methodareshown:Karatsuba4switchesto
theschool methodforintegerswithfewer
thanfourdigits,andKaratsuba32switches
0.0001 O
to the school method for integers with
fewer than 32 digits. The slopes of the
1e-05 linesfortheKaratsubavariantsareapprox-
imately 1.58. The running time of Karat-
24 26 28 210 212 214 suba32isapproximatelyone-thirdtherun-
n ningPtimeofKaratsuba4.
Figure 1.3 shows the running times T (n) and T (n) of C++ implementations
K S
of the Karatsuba methodand the schoolmethodforn-digitintegers. The scales on
Y
bothaxesare logarithmic.We see, essentially,straightlinesof differentslope.The
running time of the school method grows like n2, and hence the slope is 2 in the
caseoftheschoolmethod.TheslopeissmallerinthecaseoftheKaratsubamethod
andthissuggeststhatitsrunningtimegrowslikenb withb <2.Infact,theratio11
T (n)/T (n/2) is close to three, and this suggests that b is such that 2b = 3 or
K K
11TK(1024)=0.0455,TK(2048)=0.1375,andTK(4096)=0.41.