PART2. Theinsecurityofamonoalphabeticcodeisduetothefactthateachtimeagivenletteroccursintheoriginalmessage,itisencodedusingthesameletterintheencrypted
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1 Parabola Volume 36, Issue 3(2000) Cryptography PART2 RodJames 1 InthepreviousissueofParabola 2 wesawhowtoencode(andalsohowtobreak!) monoalphabetic ciphers(i.e. we replace each letter of the alphabet by some other letter everytimeitoccursinthemessage).wenowlookatsomemorecomplexcodes. Polyalphabetic Codes Theinsecurityofamonoalphabeticcodeisduetothefactthateachtimeagivenletteroccursintheoriginalmessage,itisencodedusingthesameletterintheencrypted text. Thus,ifthemessageislongenough,thecodecanbebrokenbycomparingthe frequency of each letter in the encypted text with frequencies of letters in most messages. For example, in the earlier article, we were able to recognize the encryption of theletterseandtbecausetheywerethemostfrequentandthefrequencyofthepair ZWsuggestedthatitwastheencryptionofTH. To overcome this insecurity, we(mentally) write the message in a number of columns and use different(monoalphabetic) codes for each column. The easiest way to do this istochooseaword(calledthekeyword),writethemessageinthesamenumberof columnsasthenumberoflettersinthekeywordandthenusea Caesarcipher by adding to each column the number corresponding to that letter of the keyword. For example,ifwechosethewordcatasthekeywordthenwewouldwritethemessage in3columnsandadd3(mod26)tothefirstcolumn,1(mod26)tothesecondcolumnand20(mod26)tothethirdcolumn.thusvenividiviciwouldbeencryptedas follows: VEN IVI DIV ICI which would then be written add 3,1,20 (mod 26) Y F H L W C G J P L D C. YFH LWC GJP LDC Atfirstsight,itappearsasthoughthiscodewouldbeimpossibletobreak.However, polyalphabetic codes can also be broken by using a little more ingenuity: all we need do is discover how many columns there are and then break the monoalphabetic code 1 RodJamesisalecturerinMathematicsatUNSWandistheeditorofParabola. 2 SeeParabolaVol.36No.2 1
2 for each column. For example, imagine that the following was confiscated from a(very silly and not very smart) student during an examination: CTAIF GBJGN QRUWF QNEZX ENQRN QVLMD ONFNR BHRFN QEYNV NRFNQ RMJZY JFNQR UBFCP AGNAN where the letters are grouped in fives for convenience only. Note that the sequence NQRoccurs4times,startingatpositions10,22,49and58,andsotheyare12,27and 9lettersapart.Sincetheselastthreenumbersareallmultiplesof3,thissuggeststhat acodewasusedwithakeywordoflengthamultipleof3.sowewritethelettersin triples(rather than columns for economy of space) as follows: CTA IFG BJG NQR UWF QNE ZXE NQR NQV LMD ONF NRB HRF NQE YNV NRFNQRUBFCPAGNA N NownoticethatthepairNQ(letters14and17ofthealphabet)occurs6timesinthe first 2 columns suggesting that this represents TH(letters 20 and 8 of the alphabet) intheoriginalmessage.since14 20(mod26)is20and17 8is 9,trysubtracting20 (mod26)fromthefirstcolumnand9(mod26)fromthesecondcolumn. Theresult (with decrypted letters in lower case) is ika owg hagthranfwee foe th R thv rd D uef ti B nif the eev tifthr saz eaf thr asf ig A mea t. ItisnoweasytoseethatthekeywordwasTIMandthemessagewas Iknowthattheanswerforthethirdquestionisthree. Itisthesameastheassignment. Public Key Cryptography Allofthecodeswehaveconsideredsofarrelyonsecrecyfortheirsecurity:ifaspy whoknowsthecoderevealsittoanyone,thentheycandecipheranymessageandthe code becomes useless. However, in 1976, Diffie and Helman suggested a new approach inwhicheveryoneknowsthekeyrequiredtoencodeamessage(andsocandoso),but asecondkeyisneededtodecodemessages. Thisdecodingkeyisonlyprovidedto thosewithauthoritytodecode,anditisimpossible(atleastinareasonableamountof time)tofindthedecodingkeyfromaknowledgeoftheencodingkey. Oneoftheearliest,andstillwidelyused,public-keycodeswascreatedin1977by Rivest, Shamir and Adleman, now referred to as RSA-coding. To understand the RSAcode,rememberthat,ifwehave nsymbols(e.g. n = 26forthealphabet),thenthe Caesarcipherisanexampleofan additioncipher m m + k(mod n), where kisthekey.alsoitwassuggestedinthepreviousissuethata multiplication cipher m km(mod n) couldbeusedprovidedthat kiscoprimeto n.amoresophisticated cipherwouldbea power cipher m m k (mod n), andamessageencodedby such a cipher can be decoded using the following two results: 2
3 Theorem1 If a, baretwocoprimeintegers,thenthereareintegers x, ysuchthat ax + by = 1. Theorem2 If aisanintegerand φ(n)isthenumberofintegersbetween 0and nwhichare coprimeto n,then a φ(n) = 1 (mod n). Thus,if kisthekeyforapowercipherand kiscoprimeto φ(n),thenthereareintegers x, ysuchthat kx + φ(n)y = 1 andso m = m 1 = m kx+φ(n)y = m kx m φ(n)y = m kx 1 = m kx (mod n). Sodecodingconsistsofraisingtheencodedmessage c = m k c c x = (m k ) x = m. Wewillwrite k 1 (mod φ(n))forthis x. Examples tothe x thpower: 1. Suppose we were encoding a message which included spaces between words, commasandfullstops(aswellasthe26lettersofthealphabet)byreplacing A by 1, Bby 2,..., Zby26,spaceby27,commaby28andfullstopby0,andthen using m m 3 (mod 29). Sinceallofthenumbers 1, 2,..., 28arecoprimeto29, φ(29) = 28 and = 1. So 3 1 (mod 28) = 19anddecodingconsistsof c c 19 (mod 29). 2.Supposeweonlyusedthe26lettersofthealphabet(replacinglettersbynumbers asinexample1)andthenused m m 5 (mod 26) The set of numbers between 0 and 26 which are coprime to 26 is {1, 3, 5, 7, 9, 11, 15, 17, 19, 21,23, 25} andso φ(26) = 12. Also So φ(26) = 12 = and 5 = = = 5 2(12 2 5) = (mod 12) and 5 1 (mod 12) = 5. Decodingconsistsof c c 5 (mod 26). 3
4 Ingeneral,if pand qareprimenumbers,then φ(p) = p 1and φ(pq) = pq 1 (numberofmultiplesof q +numberofmultiplesof p) = pq 1 (p + q) = (p 1)(q 1). Nowtheimportantfeatureofnumbersoftheform n = pqisthatitisimpossibleto calculate φ(n) unless you know p and q. Currently there is no way of quickly calculatingthefactorsoflargenumbers(evenbycomputer)andsoamessagewhichwas encodedby m m k (mod n)cannotbedecodedunlessthefactorsof nareknown, eventhough nand kareknown. Thusacodeofthisformissafeevenifeveryone knows nand kprovidedtheydonotknow pand q(or,equivalently φ(n))andso,we cancreateanrsacode,bydoingthefollowing: 1.choose2randomlargeprimenumbers p, q(say100digits),find n = pqandcalculate φ(n) = (p 1)(q 1); 2.choosearandomnumber kbetween1and φ(n)whichiscoprimeto φ(n)andfind l = k 1 (mod φ(n)) usingtheorem1. Thepublickeyisthenthepair (n, k)andthesecretkeyis l(where,nowthat lhasbeen found,wecanforget p, qand φ(n)). Example3 Theproblemwithexamples1and2isthattheletter Acanalwaysbe recognisedsince 1 k = 1.Soweusethefollowingreplacements: 0 0, 1 1, space 2, A 3,...,Z 28, Nowsupposethat p = 31and q = 47.Then n = pq = 1457(public)and φ(n) = = Ifthepublickeyis k = 7,thenthesecretkeyis 7 1 (mod 1380) = Wemustkeep p, qand φ(n)secret(orjustforgetthemonce 7 1 isfound). IfanyonewantedtoencodetheletterY,theywouldusethefollowing: Y (mod 1457) = 914 = c. If(andonlyif)someoneknewthesecretkey,theycoulddecodethisasfollows: This example is quite unrealistic since (mod 1457) = 27 Y. (a)1457iseasilyfactoredandhencesomeonecaneasilybreakit; (b)itisalsosimplya29lettermonoalphabeticsubstitutionandsoiseasilybrokenby the statistical methods above. 4
5 Wecouldimproveon(b)byencoding,andthenenciphering,pairsofletters (a, b)by writing (a, b) as 29a + b, asinthefollowingexample then UP (23, 18) = (mod 1457) = c. However,eventhisisstillonlya29 2 = 841lettersimplealphabeticsubstitution. Inreallife, pand qarechosentohaveabout100digitseachandso nhasabout200 digits. This allows us to break the message up into 75-character chunks of integers and applyrsatoeachchunkinturn.thisissolargethatstatisticalmethodsareuseless for breaking it. Also RSA is slow, since calculating powers mod n for large n needs special computerpackagesandlotsofarithmetic.hencersaisrarelyusedinreallifetosendthe whole message. What usually happens is: 1. Arandomlygeneratesalargetemporarykey k, 2. Aencryptsthekey kusingrsa, 3. Aencryptsthemessage mto cusing k, 4. Asendsboth candtheencryptedkey kto B, 5. Bfinds kusingrsadecryption. 6. Bthenusesthis ktodecodethemessage. Thekey kcannotbefoundbystatitisticalmeanssinceadifferentkeyisgeneratedfor each message. Ifyouareinterestedinlearningmore,youcanstartwith TheCodeBook bysimon Singh(published by Fourth Estate) or visit the RSA webside 5
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