EmilSekerinski,McMasterUniversity,WinterTerm16/17COMPSCI1MD3IntroductiontoProgramming

"Thereareonly10typesofpeopleintheworld:thosewhounderstandbinaryandthosewhodon't"

Integeraisanapproximatesquarerootofnifa2 ≤ n < (a+1)2

Onewaytocomputethesquarerootisbylinearsearch(exhaustivesearch):def linearSqrt(n): a = 0 while (a+1)*(a+1) <= n: a = a+1 return a

>>> linearSqrt(27)5

Statement a

A 0

B 1

B 2

B 3

B 4

B 5

A:

B:

Traceforinputn=27:

def linearSqrt (n): a = 0 while (a+1)*(a+1) <= n: a = a+1 return a

A:

Aisexecutedexactlyatimes,so:

linearSqrt(4) = 2 →2timeslinearSqrt(8) = 2 →2timeslinearSqrt(9) = 3 →3timeslinearSqrt(10) = 3 →3times

Hence,Aisexecuted√ntimes(√hereintegersquareroot)

Supposewehavea,bsuchthattherootisbetweenaandb:0 ≤ a < b and a2 ≤ n < b2

Werepeatedlyreplaceeitheraorbby(a+b)//2suchthataboveholds,untila+1 == b;thenaistheapproximatesquarerootwhile a+1 != b: c = (a+b)//2 if c*c <= n: a = c else: b = cHowdowefindsuitableinitialvaluesforaandb?

Statement a b c

C 0 8 4

D 4 8 4

D 4 8 6

E 4 6 6

C 4 6 5

D 5 6 5

C:D:E:

Traceforn=27,a=0,b=8:

Forawetake0.Forb,thesmallestvaluesatisfying0 ≤ a < b is1,whichwemultiplyby2untilbsatisfiesa2 ≤ n < b2Howoftenaretheloopbodiesexecuted?

a,b:=0,1

a+1≠b

c*c≤n

a:=c b:=c

+

–

+ –

c:=(a+b)//2

def binarySqrt(n): a, b = 0, 1 while b*b <= n: b = 2*b while a+1 != b: c = (a+b)//2 if c*c <= n: a = c else: b = c return a

0≤a<ba2≤n<b2

b*b≤n+

–

b:=2*b

a+1=ba2≤n<(a+1)2

Thefirstloopsetsb = 2k > √nafterkexecutions.Thesecondloophalvestheintervalb - aateveryexecution,untila+1 = b,hencealsotakeskexecutions.Henceeachlooptakesk = log2 bexecutions,sok ≈ log2 √n

def binarySqrt(n): a, b = 0, 1 while b*b <= n: b = 2*b while a+1 != b: c = (a+b)//2 if c*c <= n: a = c else: b = c return a

Statement a b c

A 0 1

B 0 2

B 0 4

B 0 8

C 0 8 4

… … … …

Traceforn=27:

A:

B:

C:D:

E:

Atriple(a,b,c)ofintegersisPythagoreanifa2+b2=c2Asimplewaytofindsuchtriplesisbybruteforce:enumerateallvaluesofa,b,candcheckiftheyformaPythagoreanTripledef printPythagoreanTriples1(n): for a in range(1, n+1): for b in range(1, n+1): for c in range(1, n+1): if a**2+b**2 == c**2: print(a, b, c)printPythagoreanTriples1doesnotreturnaresult,buthasaside-effect,printingonthescreen.Itisaprocedure(method)ratherthanafunctioninthemathematicalsense

3

4

5

def printPythagoreanTriples1(n): for a in range(1, n+1): for b in range(1, n+1): for c in range(1, n+1): if a*a+b*b == c*c: print(a, b, c)

A:

Eachofa,b,ctakendifferentvalues,inallcombinations,soAisexecutedonn*n*n=n3combinationsintotalHowcanweimprovethat?

Ratherthangoingoverallvaluesofc,wecalculatecfroma,bandcheckifitisanintegerdef printPythagoreanTriples2(n): for a in range(1, n+1): for b in range(1, n+1): c2 = a*a+b*b c = binarySqrt(c2) if c <= n and c2 == c*c: print(a, b, c)Bothaandbtakendifferentvaluesinallcombinations,soAisexecutedonn*n=n2combinationsintotalCanweimprovefurther?

A:

Both(3,4,5)and(4,3,5)areprinted,whichisunnecessary.Wecanrestricta,b,csuchthat0<a≤b<c≤ndef printPythagoreanTriples(n): for a in range(1, n): for b in range(a, n): c2 = a*a+b*b c = binarySqrt(c2) if c <= n and c2 == c*c: print(a, b, c)Ifa=1,thenbtakesn-1values,ifa=2,thenn-2values,etc.Intotal(n-1)+(n-2)+…+1=n(n-1)/2=n2/2-n/2Comparedtothepreviousversion,Aisexecutedonlyhalfasoften

A:

Ifweknowhowmanystepsaprogramtakesdependingontheinput,wecanusethistopredicttheexecutiontimeThiscanbedonewithoutknowingdetailsoftheprocessorandcompilationtomachinelanguage!

log2nn

√n

nnlog2nn2n3

Modificationwithoutprintingforbettermeasurement:def countPythagoreanTriples1(n): k = 0 for a in range(1, n+1): for b in range(1, n+1): for c in range(1, n+1): if a*a+b*b == c*c: k = k+1 return kEachofa,b,ctakendifferentvalues,inallcombinations,soAisexecutedonn*n*n=n3combinationsintotalForn=200:2003executionsofA;forn=400:4003executions

(4003/2003)=(400/200)3=8Forn=400:8timeslonger,8tsec

A:

Modificationwithoutprintingforbettermeasurement:def countPythagoreanTriples2(n): k = 0 for a in range(1, n+1): for b in range(1, n+1): c2 = a*a+b*b c = binarySqrt(c2) if c <= n and c2 == c*c: k = k+1 return kBothaandbtakendifferentvaluesinallcombinations,soAisexecutedonn*n=n2combinationsintotalForn=200:2002executionsofA;forn=400:4002executionsForn=400:4timeslonger,4tsec,whenignoringbinarySqrt

A:

Modificationwithoutprintingforbettermeasurement:def countPythagoreanTriples(n): k = 0 for a in range(1, n): for b in range(a, n): c2 = a*a+b*b c = binarySqrt(c2) if c <= n and c2 == c*c: k = k+1 return kIfa=1,thenbtakesn-1values,ifa=2,thenn-2values,etc.Intotal(n-1)+(n-2)+…+1=n(n-1)/2=n2/2-n/2

A:

Forlargen,n/2isnegligible:((8002/2)/(4002/2))=4Forn=800:4timeslonger,4tsec,whenignoringbinarySqrt

1/2+1/2==1.01/3+1/3+1/3==1.01/4+1/4+1/4+1/4==1.01/5+1/5+1/5+1/5+1/5==1.01/6+1/6+1/6+1/6+1/6+1/6==1.01/7+1/7+1/7+1/7+1/7+1/7+1/7==1.01/8+1/8+1/8+1/8+1/8+1/8+1/8+1/8==1.01/9+1/9+1/9+1/9+1/9+1/9+1/9+1/9+1/9==1.01/10+1/10+1/10+1/10+1/10+1/10+1/10+1/10+1/10+1/10==1.01.0+2.0==3.00.1+0.2==0.3

Thefixedpointrepresentationofafractionalnumberconsistsofthesignificantdigitswithapointatafixedposition.Fordecimalandbinarywith2integerand2fractionaldigits:2*101+0*100+5*10-1+3*10-2 1*21+0*20+1*2-1+1*2-2

=2o+0+.5+.03 =2+0+.5+.25=20.53 =2.75

2 0. 5 3

101 100 10-1 10-2

1 0. 1 1

21 20 2-1 2-2

Howmanybinarydigitsareneededtorepresent0.1?

=1/16+1/32=3/32=.09375

=51/512=.099609375

=819/8192=.0999755859375

. 0 0 0 1 1

2-1 2-2 2-3 2-4 2-5

. 0 0 0 1 1 0 0 1 1

2-1 2-2 2-3 2-4 2-5 2-6 2-7 2-8 2-9

. 0 0 0 1 1 0 0 1 1 0 0 1 1

2-1 2-2 2-3 2-4 2-5 2-6 2-7 2-8 2-9 2-10 2-11 2-12 2-13

Infinitelymanybinarydigitsareneeded:aftertheinitialdigit0,thedigits0011keeprepeatingDependingonthebase,certainrationalnumberscannotbewrittenwithfinitelymanydigits,e.g.1/3indecimal,1/10inbinary

WIKIPEDIA

3/4=.11base2 è finitebinaryrepresentation1/3=.01010101…base2 è nofinitebinaryrepresentationFollowingtheIEEEstandard,Pythonuses52binarydigits,approximately16decimaldigits.Standardoutputgivesonlythefirst16decimaldigits,eveniftheconversionfrombinaryresultsinmoredigits.Onlyanapproximatevalueisprinted!>>> 1/3, 1/10>>> format(1/3, ".60f"), format(1/10, ".60f").60f:printfloatingpointnumberwith60digitsafter.

. ? ? ? ? ?

2-1 2-2 2-3 2-4 …

.5 .25 .125 .0625 …

Withafinitenumberofdigits,arithmeticoperationswillleadtoroundingerrors.Sometimetheerrorgetscancelled,sometimesnot>>> 1/3+1/3+1/3, 1/6+1/6+1/6+1/6+1/6+1/6>>> format(1/3+1/3+1/3, ".60f"), ...Asaconsequence,fractionalnumbersshouldneverbecomparedforequality:

a == bshouldbecome

abs(b-a) ≤ εHowever,εmustnotbetoosmall!

def x_intersect(f, a, b, eps): "f(a) ≤ 0 ≤ f(b), a ≤ b" while b-a > eps: m = (a+b)/2 if f(m) <= 0: a = m else: b = m return a, b

InPython,functionsaredata;theycanbepassedaroundasanyotherdata>>> x_intersect(f1, 0, 100, 1e-8)>>> x_intersect(f2, 0, 100, 1e-8)

>>> x_intersect(f1, 0, 100, 1e-15)>>> x_intersect(f1, 0, 100, 1e-16) # interrupt

def f1(x): return x*x-4

def f2(x): return x*x-2

>>> type(f1)

Afloatingpointnumberconsistsofafraction(mantissa)withthesignificantdigitsandanexponent.Forexample,fordecimalnumbers:

(1.963,3)=1.963*103=1963FollowingtheIEEEstandard,Pythonstoresthefractionwith52bitsinnormalizedform(leading1before.)andtheexponentwith11bits(withrange-1022to1023):

(1.f)*2e

Thelargestpositivenumberis(1.11…base2)*21023=1.7976931348623157*10308

Thesmallestpositivenumberis1.0*2-1022=2.2250738585072014*10-308

MostcomputersfollowtheIEEEstandard:floating-pointcomputationwillhavethesameresultsacrosscomputers.Severalformatsexist,with8bytesbeingwidelyused:Thenumberofbitsofthefractionlimitstheprecision.Thenumberofbitsintheexponentlimitstherange.Arithmeticoperationsonfloatmayleadtoalossofsignificantdigits:>>> 10000000000000000+1>>> 10000000000000000.0+1

52bitsfraction11bitsexponent1bitsign

+,-onfloatarethe"dangerous"operations!

def quadraticEquationSolution(a, b, c): d = math.sqrt(b*b-4*a*c) return (-b+d)/(2*a), (-b-d)/(2*a)

>>> quadraticEquationSolution(1, -3, -4)>>> quadraticEquationSolution(1, -2e8, 1)However,0.0isnotasolutionofx2-2e8*x+1=0:b*b-4*a*c=(-2e8)*(-2e8)-4*1*1=4e16-4=4e16Therefore,d=sqrt(4e16)=2e8and:(-b+d)/(2*a)=(-(-2e8)+2e8)/(2*1)=2e8(-b-d)/(2*a)=(-(-2e8)-2e8)/(2*1)=0Howcanweavoidsucherrors?

!

x1/2=−b± b2 −4ac

2a

lossofsignificantdigits

SolutionsofthequadraticequationarerelatedbyVieta'sformula:Giventhesolutionwithlargerabsolutevalue,thesmallercanbecomputedusingVieta,avoidingthelossofsignificantdigitsdef quadraticEquationSolutionPlus(a, b, c): d = math.sqrt(b*b-4*a*c) x1 = -(b+d)/(2*a) if b>=0 else (d-b)/(2*a) x2 = c/(x1*a) return x1, x2

quadraticEquationSolutionPlus(1, -2e8, 1)

x1/2=−b± b2 −4ac

2a

x1x2=ca

WIKIPEDIA

ArithmeticwiththePythondecimallibrarywillproducethesameerrorsascalculationsbyhand;bydefault,28fractionaldigitsarekept;theprecisioncanbechanged:>>> Decimal(1)/Decimal(3), Decimal(1)/Decimal(10)>>> Decimal('0.1')+Decimal('0.2')==Decimal('0.3')Howfastisdecimalarithmetic?def timeDecimalAddition(): from time import time start = time() for i in range(100000): x = Decimal(0) for j in range(10): x = x+Decimal(1)/Decimal(10) return time()-start

python.org

ThePythonlibraryfractionsstoresrationalnumbersa/bwithnumeratoraanddenominatorbasapair(a, b).Thismakescalculationswith+,-,*,/precise.>>> Fraction(1)/Fraction(3) == Fraction(1, 3)Conversionbetweenfloat,Decimal,Fractionrevealsdifferencesinrepresentation:>>> Decimal(0.1) >>> Fraction(0.1)

1.  Avoidfloat,useintegerinstead:computewith¢rather$,mmratherthanm

2.  Usedecimalorfractionsstandardlibraryinstead:slower,particularlyforverylarge/smallnumbers

3.  Usealibraryforintervalarithmeticwithfloat:givessafelowerandupperbounds,takestwiceasmuchmemory/time

4.  Useaproblem-specificlibrarywithorcheckliteratureforalgorithmswithknownnumericalproperties(e.g.solvingdifferentialequations)

5.  Usesymboliccomputationasincomputeralgebrasystemsinsteadofaprogramminglanguage

6.  Whenusingfloat,nevercompareforequality;checkplausibilityofresult