a painless guide to crc error detection algorithms.docx

a painless guide to crc error detection algorithms.docx

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apainlessguidetocrcerrordetectionalgorithms

目录

原文网址http:

//www.repairfaq.org/filipg/LINK/F_crc_v3.html2

2.Introduction:

ErrorDetection2

3.TheNeedForComplexity4

4.TheBasicIdeaBehindCRCAlgorithms5

5.Polynomial（多项式）Arithmetic7

Chapter6BinaryArithmeticwithNoCarries10

Chapter7AFullyWorkedExample13

Chapter8ChoosingAPoly16

Chapter9）AStraightforwardCRCImplementation17

Chapter10ATable-DrivenImplementation19

Chapter11ASlightlyMangled（损坏的）Table-DrivenImplementation23

Chapter12）"Reflected"Table-DrivenImplementations27

Chapter13）"Reversed"Polys29

Chapter14）InitialandFinalValues30

Chapter15）DefiningAlgorithmsAbsolutely30

Chapter16）AParameterizedModelForCRCAlgorithms31

Chapter17）ACatalogofParameterSetsforStandards33

Chapter18）AnImplementationoftheModelAlgorithm35

Chapter19）RollYourOwnTable-DrivenImplementation35

Chapter20）GeneratingALookupTable36

Chapter21）Summary37

Chapter22）Corrections37

Chapter23）Glossary37

Chapter24）References38

Chapter25）ReferencesIHaveDetectedButHaven'tYetSighted39

原文网址http:

//www.repairfaq.org/filipg/LINK/F_crc_v3.html

2.Introduction:

ErrorDetection

Theaimofanerrordetectiontechniqueistoenablethereceiverofamessagetransmittedthroughanoisy（error-introducing）channeltodeterminewhetherthemessagehasbeencorrupted.Todothis,thetransmitterconstructsavalue（calledachecksum）thatisafunctionofthemessage,andappendsittothemessage.Thereceivercanthenusethesamefunctiontocalculatethechecksumofthereceivedmessageandcompareitwiththeappendedchecksumtoseeifthemessagewascorrectlyreceived.Forexample,ifwechoseachecksumfunctionwhichwassimplythesumofthebytesinthemessagemod256（i.e.modulo256）,thenitmightgosomethingasfollows.Allnumbersareindecimal.

Message:

6234

Messagewithchecksum:

623433

Messageaftertransmission:

627433

Intheabove,thesecondbyteofthemessagewascorrupted（变坏）from23to27bythecommunicationschannel.However,thereceivercandetectthisbycomparingthetransmittedchecksum（33）withthecomputerchecksumof37（6+27+4）.Ifthechecksumitselfiscorrupted,acorrectlytransmittedmessagemightbeincorrectlyidentifiedasacorruptedone.However,thisisasafe-sidefailure.Adangerous-sidefailureoccurswherethemessageand/orchecksumiscorruptedinamannerthatresultsinatransmissionthatisinternallyconsistent.Unfortunately,thispossibilityiscompletelyunavoidableandthebestthatcanbedoneistominimizeitsprobabilitybyincreasingtheamountofinformationinthechecksum（e.g.wideningthechecksumfromonebytetotwobytes）.

Othererrordetectiontechniquesexistthatinvolveperformingcomplextransformationsonthemessagetoinjectitwithredundant（冗长的）information.However,thisdocumentaddressesonlyCRCalgorithms,whichfallintotheclassoferrordetectionalgorithmsthatleavethedataintact（原封不动的）andappendachecksumontheend.i.e.:

3.TheNeedForComplexity

Inthechecksumexampleintheprevioussection,wesawhowacorruptedmessagewasdetectedusingachecksumalgorithmthatsimplysumsthebytesinthemessagemod256:

Message:

6234

Messagewithchecksum:

623433

Messageaftertransmission:

627433

Aproblemwiththisalgorithmisthatitistoosimple.Ifanumberofrandomcorruptionsoccur,thereisa1in256chancethattheywillnotbedetected.Forexample:

Message:

6234

Messagewithchecksum:

623433

Messageaftertransmission:

820533

Tostrengthenthechecksum,wecouldchangefroman8-bitregistertoa16-bitregister（i.e.sumthebytesmod65536insteadofmod256）soastoapparentlyreducetheprobabilityoffailurefrom1/256to1/65536（why？

因为结果有65536种可能性，不变只占其中一种）.Whilebasicallyagoodidea,itfailsinthiscasebecausetheformulausedisnotsufficiently"random";withasimplesummingformula,eachincomingbyteaffectsroughlyonlyonebyteofthesummingregisternomatterhowwideitis.Forexample,inthesecondexampleabove,thesummingregistercouldbeaMegabyte（兆字节）wide,andtheerrorwouldstillgoundetected.Thisproblemcanonlybesolvedbyreplacingthesimplesummingformulawithamoresophisticatedformulathatcauseseachincomingbytetohaveaneffectontheentirechecksumregister.

Thus,weseethatatleasttwoaspectsarerequiredtoformastrongchecksumfunction:

WIDTH

Aregisterwidthwideenoughtoprovidealowa-prioriprobabilityoffailure（e.g.32-bitsgivesa1/2^32chanceoffailure）.

CHAOS

Aformulathatgiveseachinputbytethepotentialtochangeanynumberofbitsintheregister.

Note:

Theterm"checksum"waspresumablyusedtodescribeearlysummingformulas,buthasnowtakenonamoregeneralmeaningencompassingmoresophisticatedalgorithmssuchastheCRCones.TheCRCalgorithmstobedescribedsatisfythesecondconditionverywell,andcanbeconfiguredtooperatewithavarietyofchecksumwidths.

4.TheBasicIdeaBehindCRCAlgorithms

Wheremightwegoinoursearchforamorecomplexfunctionthansumming?

Allsortsofschemesspringtomind.Wecouldconstruct（建造）tablesusingthedigitsofpi,orhash（散列）eachincomingbytewithallthebytesintheregister.Wecouldevenkeepalargetelephonebookon-line,anduseeachincomingbytecombinedwiththeregisterbytestoindexanewphonenumberwhichwouldbethenextregistervalue.Thepossibilitiesarelimitless.

However,wedonotneedtogosofar;thenextarithmeticstepsuffices（足够）.Whileadditionisclearlynotstrongenoughtoformaneffectivechecksum,itturnsoutthatdivisionis,solongasthedivisorisaboutaswideasthechecksumregister.

ThebasicideaofCRCalgorithmsissimplytotreatthemessageasanenormousbinarynumber,todivideitbyanotherfixedbinarynumber,andtomaketheremainderfromthisdivisionthechecksum.Uponreceiptofthemessage,thereceivercanperformthesamedivisionandcomparetheremainderwiththe"checksum"（transmittedremainder）.

Example:

Supposethethemessageconsistedofthetwobytes（6,23）asinthepreviousexample.Thesecanbeconsideredtobethehexadecimalnumber0617whichcanbeconsideredtobethebinarynumber0000-0110-0001-0111.Supposethatweuseachecksumregisterone-bytewideanduseaconstantdivisorof1001,thenthechecksumistheremainderafter0000-0110-0001-0111isdividedby1001.Whileinthiscase,thiscalculationcouldobviouslybeperformedusingcommongardenvariety32-bitregisters,inthegeneralcasethisismessy.Soinstead,we'lldothedivisionusinggood-'ollongdivisionwhichyoulearnedinschool（remember?

）.Exceptthistime,it'sinbinary:

...0000010101101=00AD=173=QUOTIENT

____-___-___-___-

9=1001）0000011000010111=0617=1559=DIVIDEND

DIVISOR0000.,,....,.,,,

----.,,....,.,,,

0000,,....,.,,,

0000,,....,.,,,

----,,....,.,,,

0001,....,.,,,

0000,....,.,,,

----,....,.,,,

0011....,.,,,

0000....,.,,,

----....,.,,,

0110...,.,,,

0000...,.,,,

----...,.,,,

1100..,.,,,

1001..,.,,,

====..,.,,,

0110.,.,,,

0000.,.,,,

----.,.,,,

1100,.,,,

1001,.,,,

====,.,,,

0111.,,,

0000.,,,

----.,,,

1110,,,

1001,,,

====,,,

1011,,

1001,,

====,,

0101,

0000,

----

1011

1001

====

0010=02=2=REMAINDER

Indecimalthisis"1559dividedby9is173witharemainderof2".

Althoughtheeffectofeachbitoftheinputmessageonthequotientisnotallthatsignificant,the4-bitremaindergetskickedaboutquitealotduringthecalculation,andifmorebyteswereaddedtothemessage（dividend）it'svaluecouldchangeradically（完全的）againveryquickly.Thisiswhydivisionworkswhereadditiondoesn't.

Incaseyou'rewondering,usingthis4-bitchecksumthetransmittedmessagewouldlooklikethis（inhexadecimal）:

06172（wherethe0617isthemessageandthe2isthechecksum）.Thereceiverwoulddivide0617by9andseewhethertheremainderwas2.

5.Polynomial（多项式）Arithmetic

WhilethedivisionschemedescribedintheprevioussectionisveryverysimilartothechecksummingschemescalledCRCschemes,theCRCschemesareinfactabitweirder,andweneedtodelve（钻研）intosomestrangenumbersystemstounderstandthem.

ThewordyouwillhearallthetimewhendealingwithCRCalgorithmsistheword"polynomial".AgivenCRCalgorithmwillbesaidtobeusingaparticularpolynomial,andCRCalgorithmsingeneralaresaidtobeoperatingusingpolynomialarithmetic.Whatdoesthismean?

Insteadofthedivisor,dividend（message）,quotient（商）,andremainder（asdescribedintheprevioussection）beingviewedaspositiveintegers,theyareviewedaspolynomialswithbinarycoefficients.Thisisdonebytreatingeachnumberasabit-stringwhosebitsarethecoefficientsofapolynomial.Forexample,theordinarynumber23（decimal）is17（hex）and10111binaryandsoitcorrespondstothepolynomial:

1*x^4+0*x^3+1*x^2+1*x^1+1*x^0

or,moresimply:

x^4+x^2+x^1+x^0

Usingthistechnique,themessage,andthedivisorcanberepresentedaspolynomialsandwecandoallourarithmeticjustasbefore,exceptthatnowit'sallcluttered（弄乱）upwithXs.Forexample,supposewewantedtomultiply1101by1011.Wecandothissimplybymultiplyingthepolynomials:

（x^3+x^2+x^0）（x^3+x^1+x^0）

=（x^6+x^4+x^3

+x^5+x^3+x^2

+x^3+x^1+x^0）=x^6+x^5+x^4+3*x^3+x^2+x^1+x^0

Atthispoint,togettherightanswer,wehavetopretendthatxis2andpropagatebinarycarriesfromthe3*x^3yielding:

x^7+x^3+x^2+x^1+x^0

It'sjustlikeordinaryarithmeticexceptthatthebaseisabs