Chapter VIII

Chapter VIIICase 8. The Playfair cipher. This is the English military field cipher; as the method is published in English military manuals and as it is a cipher of proven reliability, it may be met with in general cipher work. The Playfair cipher operates with a key word; two letters are substituted for each two letters of the text.The Playfair cipher may be recognized by the following points: (a) It is a substitution cipher, (b) it always contains an even number of letters, (c) when the cipher is divided into groups of two letters each, no group consists of the repetition of the same letter asSSorBB, (d) there will be recurrence of pairs throughout the message, following in a general way, the frequency table of digraphs of pairs, (e) in short messages there may be recurrence of cipher groups representing words or even phrases, and these will always be found in long messages.In preparing a cipher by this method, a key word is chosen by the correspondents. A large square, divided into twenty-five smaller squares, is constructed as shown below and the letters of the key word are written in, beginning at the upper left hand corner. If any letter recurs in the key word, it is only used on the first occurrence. The remaining letters of the alphabet are used to fill up the square. It is customary to considerIandJas one letter in this cipher and they are written together in the same square.If the key word chosen isLEAVENWORTH, then the square would be constructed as follows:LEAVNWORTHBCDFGIJKMPQSUXYZThe text of the message to be sent is then divided up into groups of two letters each, and equivalents are found for each pair.Every pair of letters in the square must be: Either (1) in the same vertical line. Thus in the above example each letter is represented in cipher by that which stands next below it, and the bottom letter by the top one of the same column; for instance,TYis represented byFV.Or (2) in the same horizontal line. Each letter in this case is represented by that which stands next on its right, and the letter on the extreme right by that on the extreme left of the same horizontal line with it; for instanceRHis represented byTW.Or (3) at opposite corners of a rectangle. Each letter of the pair is represented by the letter in the other corner of the rectangle in the same horizontal line with it; for instanceTSis represented byWY.If, on dividing the letters of the text into pairs, it is found that a pair consists of the same letter repeated, a dummy letter, asX,Y, orZ, should be introduced to separate the similar letters.If the message to be sent were “The enemy moves at dawn,” it would be divided into pairs:THEXENEMYMOVESATDAWNand enciphered:HWAUALAKXPTELUVRMRHLThe message is then broken up into groups of five letters for transmission.To decipher such a cryptogram, (knowing the key word), the receiver divides it into pairs, andfrom his table finds the equivalent of these pairs, taking the letter immediately above each, when they are in the same vertical line; those immediately on the left, when in the same horizontal line; and those at opposite angles of the rectangle when this is formed.It is evident, from the foregoing description, that any letter of the plain text may be represented in cipher by one of five letters, viz: The one next below it and the other four letters in the same horizontal line with it in the square. Take, for example, the letterDof the plain text, in combination with each of the other letters of the alphabet. We have, using the keyLEAVENWORTH:DADBDCDEDFDGDHDIDKDLDMDNDODPDQDRDSDTDUDVDWDXDYDZMRFCFDCAFGFBGRBMCMBAMXGACRFMGMMDBXFRCXFABRMAFXGXThis givesDrepresented byBCFGM44844times,and, connected with these five letters representingD,we haveARDMXBCG55245111times.Note that these letters are those of the vertical column containingDplus the lettersB,CandG, of the horizontal line containingD.Lieut. Frank Moorman, U. S. Army, has developed a method for determining the letters which make up the key word in a Playfair cipher. In the first place, a key word necessarily contains vowels in the approximate proportion of two vowels to three consonants and it is also likely that a key word will contain other common letters. This key word is placed in the first row or rows. Now if a table is made, showing what letters in the cipher occur with every letter, it will be found that the letters having the greatest number of other letters in combination with them are very likely to be letters of the keyword, or in other words, letters occurring in the first or second lines. An example will make this clear:MessageDB FN EX TZ MF TO VB QB QT OB XA OF PR TZ EQ RH QK QV DX OK AB PR QI EL TV KE EX XS FS BP WD BO BY BF RO EA BO RH QK QV TX GU EL AB TH TR XN ON EA AY XH BO HN EX BS HR QB ZM SE XP HF GZ UG KC BD PO EA AY XH BO XP HF KR QI AB PR QI EL BX FZ BI SE FX PB RA PR QI WC BR XD YG TB QT EA AY XH BO HN EX BS HR QB PR QI EL BX BT HB QB NF SI SE BX NU XP BU RB XB QR OX BA TB RH BP WD RP RO GU GX QR SE ZY OX BA EL AX CW BY BA SX RK RO PR HB OP BD PI CN OX EM RP KR XT EL AX CW EQ FZ SX EL RH RO PR HB UX DA SE XN ZN GU EL BX FS DG DB TB ZL VE RH BO RQ.From this message, we make up the following table, considering the letters of each pair:First Letters of PairsABCDEFGHIKLMNOPQRSTUVWXYZA31411B3231141311C11D221E151F12111G1111H513I1151K121L81M11N112121O6141P2123Q2R122721S221T121U131V2W2X1514113211Y51Z212From this table we pick out the lettersB,E,F,O,R,T,X, as tentative letters of the key word on account of the variety of other letters with which they occur. As there are but two vowels for seven letters, we will addAto the list on account of its occurrences withB,D,E,R, andX. This leaves the letters for the bottom lines of the square as follows:........CDGHIJKLMNPQSUVWYZReferring to the table again we find the most frequent combination to beEL, occurring 8 times, with no occurrence ofLE. Now,THis the commonest pair in plain text, andHTis not common. The fact thatHoccurs in the same horizontal line withLand thatEandTare probably in the key, will lead us to putEin the first line overHandTin the first line overL, so as to makeELequalTH.The next most frequent combination isPRoccurring 7 times, withRPoccurring twice. In the square as partially arranged,PRequalsM_orN_orQ_orI_. We may eliminate all these exceptN_, and thisN_could only beNOorNA, so that we will put, tentatively theRin the second line overHand theOandAin the same line overIJ. We have then:.E..T.RAOCDGHIJKLMNPQSUVWYZLet us now check this by picking out the combinations beginning withELand seeing if the table will solve them. We find,ELTV,ELAB,ELBXFZ,ELBXBT,ELAXCWBY,ELAXCWEQ,ELRH,ELBXFS. Now, on the assumption that the letter afterELrepresentsE, we have it represented byAthree times,Bthree times,Ronce andTonce. This requiresthatAandBbe put in the same horizontal line withE, sinceTis already there, andRis tentatively underE.The combinationELTVnow equalsTHEZ. If theTwere moved one place to the left, it would beTHEY, a more likely combination, but this requires theLto be moved one place to the left also, by puttingIorKin the key word and taking outO,RorXand returning it to its place in the alphabetical sequence. The most frequent pairs containingOareB Osix times,R Ofour times, andO Xthree times. Now these pairs equal respectivelyE N,E SandH E, ifOis put betweenNandPin the fourth line. We will therefore cease to consider it as a letter of the the key word. The combinationELABcan only beTHE_on the assumption thatAis the first letter to the right ofE. The combinationELBXoccurs three times. If it representsTHE_, theBmust be the first letter of the first line and theXmust now be placed underEwhere theRwas tentatively put. We can getTHE_out ofELRHby puttingRin the first line or leaving it where it is, but the preponderance of theBXcombination should suggest the former alternative.A new square showing these changes will look like this:BEATR.X...GH.LMNOPQSUVWYXAsIput in the space underBwill give the wordBEATRIXand as a vowel is clearly necessary there, we will so use theIJand leaveKbetweenHandL. This leavesC,DandFto be placed. It appeared atfirst thatFwas in the key but if it is in the second line, in proximity to the letters of the first line, it will give the same indications. Completing the square then, we haveBEATRIJXCDFGHKLMNOPQSUVWYZWith this square, the message is deciphered without difficulty.“It is very frequently neces(x)sary to employ ciphers and they have for many centuries been employed in the relations betwe(x)en governments, for com(x)munication betwe(x)en com(x)manders and their subordinates and particularly betwe(x)en governments and their agents in foreign countries; there are many cases in history where the capture of a message not in cipher has made the captors of the message victorious in their military movements.”It will be seen that the method of Lieut. Moorman enabled us to pick out six letters of the key word out of eight letters chosen tentatively. The reason for the appearance ofFhas already been noted; the letterOoccurred with many other letters because it happened to remain in the same line withNandSand to be underH. It thus was likely to represent any of these three letters which occur very frequently in any text.Two-character Substitution CiphersCase9.—Two-character substitution ciphers. In ciphers of this type, two letters, numerals, or conventional signs, are substituted for each letter of the text. There are many ways of obtaining the charactersto be substituted but, in general, these ciphers may be considered as special varieties of Case 6 or Case 7. The ciphers which come under this case are not well suited to telegraphic correspondence because the cipher message will contain twice as many letters as the plain text. However they are so used; an example is at hand in which two numerals are substituted for each letter and this makes transmission by telegraph very slow.Case 9 can be recognized by some or all of the following points; the number of characters in the cipher is always an even number; often only a few, say five to ten, of the letters of the alphabet appear; either a frequency table for pairs of the cipher text resembling the normal single letter frequency table can be made, or groups of four letters will show a regular recurrence, from which the cipher can be solved as in Case 7.Case9a.—MessageRNTGN RAAGR NARNA GTGRA TGAAN NANGG RARAT NAANR NNNRN AAAGG AANGR NGGNN NRNAA AANRA TNANN NGGRN RNNRG TTGRG TGGRN ARNTG NNART GGRNR GRNNT GTGAA NNARN ARNRT TGAGG GAAAA NANNA RNAGA NGNAT NNNATThis message contains 160 letters and it will be noted that the only letters used areA,G,N,RandT.We may expect a simple two-letter substitution cipher at once. It will simplify the work if we divide the cipher into groups of two letters and then, if we find there are 26 or less recurring groups, to assign an arbitrary letter to each group and work out the cipher by the method of Case 6.RN TG NR AA GR NA RN AG TG RA TG AA NN AN GG RA RA TN AA NR NN NR NA AA GG AA NG RN GG NN NR NA AA AN RA TN AN NN GG RN RN NR GT TG RG TG GR NA RN TG NN AR TG GR NR GR NN TG TG AA NN AR NA RN RT TG AG GG AA AA NA NN AR NA GA NG NA TN NN ATWith arbitrary letters substituted, we haveA B C D E F A G B H B D I J K H H L D C I C F D K D M A K I C F D J H L J I K A A C N B O B E F A B I P B E C E B B D I P F A Q B G K D D F I P F R M F L I SNow, preparing a frequency table, with note of prefixes and suffixes we have:FrequencyPrefixSuffixA71111111FMKAFFBGKACBQB101111111111AGHNOAPIBQCHDOEIEBDGC6111111BDIIAEDIFFNED9111111111CBLFKFBKDEICKMJIDFE41111DBBCFFCIF811111111ECCEPDPMADDAAIRLG211ABBKH41111BKJHBHLLI9111111111DCKJBEDFLJCCKPBPPJ3111IDLKHIK511111JDAIGHDIADL3111HHFDJIM211DRAFN11CBO11BBP3111IIIBFFQ11ABR11FMS11IA brief study of this table and the distribution in the cipher leads to the conclusion thatB,FandCare certainly vowels and are, if the normal frequency holds, equal toE,O, andAorI. SimilarlyDandIare consonants and we may take them asNandT.Iis taken asTbecause of the combinationIP(=possiblyTH) occurring three times. The next letter in order of frequency isA; it is certainly a consonant and may be taken asRon the basis of its frequency. Let us now try these assumptions on the first two lines of the message. We haveREAN_OR_E_ENT_____NATAON_N_IIIThis is clearly the wordREINFORCEMENTSand, using the letters thus found, the rest of the line becomesAMMUNITIONAND. We have then the following letters determined:Arbitrary lettersABCDEFGHIJKLMPlain TextREINFOCMTSAUDIf these be substituted we have for the message:REINFORCEMENTS AMMUNITION AND RATIONS MUST ARRI_E _EFORE T_E FIFTEENT_ OR _E CANNOT _O_D OUT_.From this the remainder of the letters are determined:Arbitrary lettersNOPQRSPlain textVBHWLXNow let us substitute the two-letter groups for the arbitrary letters:Arbitrary lettersKOGMBEPCRHDFAJILNQSTwo-letter groupsGGRGAGNGTGGRARNRGARAAANARNANNNTNGTRTATPlain textABCDEFHILMNORSTUVWXIt is evident that the cipher was prepared with the letters of the wordGRANTchosen by means of a square of this kind:GRANTGABCDERFGHIKALMNOPNQRSTUTVWXYZThusTG=E,AN=S, etc., as we have already found.Case9-bMessage195049295831232528154418452815204811504122521153455849134124502855252659331952225245113215621558414328613612652945565015234245585063455420191550185311211541582811241745534554205950255245413215334920485018152364An examination of the groups of two numerals each which make up this message, shows that we have 11 to 36 and 41 to 65 with eleven groups missing. Now the 11 to 36 combination is a very familiar one in numeral substitution ciphers (See Case 6-c) and itwill be noted that 41 to 66 would give us a similar alphabet. Let us make a frequency table in this form:GroupFrequencyGroupFrequency11111114111111121421131431144411511111111145111111111164617147181114811191114911120111150111111112115122115211112311153111241154112511155126156127572811111581111129115911306031161132116213311631346413565136166Each of these tables looks like the normal frequency table except for the position of 20 and 50 which should representT, by all our rules, and should be apparently 30 and 60. But suppose we put the alphabet and corresponding numerals in this form:12345678901 or 4ABCDEFGHIJ2 or 5KLMNOPQRST3 or 6UVWXYZThen A=11 or 41, J=10 or 40 and T=20 or 50 as we found. Using the above alphabet, the message may easily be read. Note that this cipher is made up of ten characters only, the Arabic numerals.Case9c—Message115625467625422944321949294015142321721129797031154924213511742414787576462524445143254845317974253340554615127573227945162748151170423519441378252149251476455315483421267215254075161125784546422174154952197929701524214329254449331970187531407925482945514914117321171554An examination of this message shows it to consist of forty-four different two-figure groups running from 11 to 79. Let us prepare a frequency table of these groups.GroupFrequency1111111112113114111115111111111161117118119111120211111111221231241111251111111111126127128291111113031111321331134135113637383940111141421114311441111451111146111147481111491111115051115215315415515615758597011117172117311741117511117611177781117911111We at once note the resemblance between the frequency tables for the groups 11 to 19 and 21 to29; for the groups 30 to 36 and 50 to 56; and for the groups 40 to 49 and 70 to 79. Also the groups 11 to 19 and 21 to 29 have a frequency fitting well with the normal frequency table of the lettersAtoI; the groups 41 to 49 and 71 to 79 have a frequency fitting well with the normal frequency table of the lettersKtoS; and the groups 31 to 36 and 51 to 56 have a frequency fitting well with the normal frequency table of the lettersUtoZ. We haveJandTunaccounted for, but note what occurred in Case 9-b and that 40 and 70 would correspond well withTif they followed respectively 49 and 79. We may now make up a cipher table as follows:12345678901 or 2ABCDEFGHIJ4 or 7KLMNOPQRST3 or 5UVWXYZand this table will solve the cipher message.In ciphers coming under case 9-b and 9-c, it is not uncommon to assign some of the unused numbers such as 85, 93, etc., to whole words in common use or to names of persons or places. In case such groups are found, the meaning must be guessed at from the context; but if many messages in the same cipher are available, the meaning of these groups will soon be obtained. The appearance of such odd groups of figures in a message does not interfere materially with the analysis, and it will be apparent at once on deciphering the message that they represent whole words instead of letters.

Chapter VIII

Case 8. The Playfair cipher. This is the English military field cipher; as the method is published in English military manuals and as it is a cipher of proven reliability, it may be met with in general cipher work. The Playfair cipher operates with a key word; two letters are substituted for each two letters of the text.The Playfair cipher may be recognized by the following points: (a) It is a substitution cipher, (b) it always contains an even number of letters, (c) when the cipher is divided into groups of two letters each, no group consists of the repetition of the same letter asSSorBB, (d) there will be recurrence of pairs throughout the message, following in a general way, the frequency table of digraphs of pairs, (e) in short messages there may be recurrence of cipher groups representing words or even phrases, and these will always be found in long messages.In preparing a cipher by this method, a key word is chosen by the correspondents. A large square, divided into twenty-five smaller squares, is constructed as shown below and the letters of the key word are written in, beginning at the upper left hand corner. If any letter recurs in the key word, it is only used on the first occurrence. The remaining letters of the alphabet are used to fill up the square. It is customary to considerIandJas one letter in this cipher and they are written together in the same square.If the key word chosen isLEAVENWORTH, then the square would be constructed as follows:LEAVNWORTHBCDFGIJKMPQSUXYZThe text of the message to be sent is then divided up into groups of two letters each, and equivalents are found for each pair.Every pair of letters in the square must be: Either (1) in the same vertical line. Thus in the above example each letter is represented in cipher by that which stands next below it, and the bottom letter by the top one of the same column; for instance,TYis represented byFV.Or (2) in the same horizontal line. Each letter in this case is represented by that which stands next on its right, and the letter on the extreme right by that on the extreme left of the same horizontal line with it; for instanceRHis represented byTW.Or (3) at opposite corners of a rectangle. Each letter of the pair is represented by the letter in the other corner of the rectangle in the same horizontal line with it; for instanceTSis represented byWY.If, on dividing the letters of the text into pairs, it is found that a pair consists of the same letter repeated, a dummy letter, asX,Y, orZ, should be introduced to separate the similar letters.If the message to be sent were “The enemy moves at dawn,” it would be divided into pairs:THEXENEMYMOVESATDAWNand enciphered:HWAUALAKXPTELUVRMRHLThe message is then broken up into groups of five letters for transmission.To decipher such a cryptogram, (knowing the key word), the receiver divides it into pairs, andfrom his table finds the equivalent of these pairs, taking the letter immediately above each, when they are in the same vertical line; those immediately on the left, when in the same horizontal line; and those at opposite angles of the rectangle when this is formed.It is evident, from the foregoing description, that any letter of the plain text may be represented in cipher by one of five letters, viz: The one next below it and the other four letters in the same horizontal line with it in the square. Take, for example, the letterDof the plain text, in combination with each of the other letters of the alphabet. We have, using the keyLEAVENWORTH:DADBDCDEDFDGDHDIDKDLDMDNDODPDQDRDSDTDUDVDWDXDYDZMRFCFDCAFGFBGRBMCMBAMXGACRFMGMMDBXFRCXFABRMAFXGXThis givesDrepresented byBCFGM44844times,and, connected with these five letters representingD,we haveARDMXBCG55245111times.Note that these letters are those of the vertical column containingDplus the lettersB,CandG, of the horizontal line containingD.Lieut. Frank Moorman, U. S. Army, has developed a method for determining the letters which make up the key word in a Playfair cipher. In the first place, a key word necessarily contains vowels in the approximate proportion of two vowels to three consonants and it is also likely that a key word will contain other common letters. This key word is placed in the first row or rows. Now if a table is made, showing what letters in the cipher occur with every letter, it will be found that the letters having the greatest number of other letters in combination with them are very likely to be letters of the keyword, or in other words, letters occurring in the first or second lines. An example will make this clear:MessageDB FN EX TZ MF TO VB QB QT OB XA OF PR TZ EQ RH QK QV DX OK AB PR QI EL TV KE EX XS FS BP WD BO BY BF RO EA BO RH QK QV TX GU EL AB TH TR XN ON EA AY XH BO HN EX BS HR QB ZM SE XP HF GZ UG KC BD PO EA AY XH BO XP HF KR QI AB PR QI EL BX FZ BI SE FX PB RA PR QI WC BR XD YG TB QT EA AY XH BO HN EX BS HR QB PR QI EL BX BT HB QB NF SI SE BX NU XP BU RB XB QR OX BA TB RH BP WD RP RO GU GX QR SE ZY OX BA EL AX CW BY BA SX RK RO PR HB OP BD PI CN OX EM RP KR XT EL AX CW EQ FZ SX EL RH RO PR HB UX DA SE XN ZN GU EL BX FS DG DB TB ZL VE RH BO RQ.From this message, we make up the following table, considering the letters of each pair:First Letters of PairsABCDEFGHIKLMNOPQRSTUVWXYZA31411B3231141311C11D221E151F12111G1111H513I1151K121L81M11N112121O6141P2123Q2R122721S221T121U131V2W2X1514113211Y51Z212From this table we pick out the lettersB,E,F,O,R,T,X, as tentative letters of the key word on account of the variety of other letters with which they occur. As there are but two vowels for seven letters, we will addAto the list on account of its occurrences withB,D,E,R, andX. This leaves the letters for the bottom lines of the square as follows:........CDGHIJKLMNPQSUVWYZReferring to the table again we find the most frequent combination to beEL, occurring 8 times, with no occurrence ofLE. Now,THis the commonest pair in plain text, andHTis not common. The fact thatHoccurs in the same horizontal line withLand thatEandTare probably in the key, will lead us to putEin the first line overHandTin the first line overL, so as to makeELequalTH.The next most frequent combination isPRoccurring 7 times, withRPoccurring twice. In the square as partially arranged,PRequalsM_orN_orQ_orI_. We may eliminate all these exceptN_, and thisN_could only beNOorNA, so that we will put, tentatively theRin the second line overHand theOandAin the same line overIJ. We have then:.E..T.RAOCDGHIJKLMNPQSUVWYZLet us now check this by picking out the combinations beginning withELand seeing if the table will solve them. We find,ELTV,ELAB,ELBXFZ,ELBXBT,ELAXCWBY,ELAXCWEQ,ELRH,ELBXFS. Now, on the assumption that the letter afterELrepresentsE, we have it represented byAthree times,Bthree times,Ronce andTonce. This requiresthatAandBbe put in the same horizontal line withE, sinceTis already there, andRis tentatively underE.The combinationELTVnow equalsTHEZ. If theTwere moved one place to the left, it would beTHEY, a more likely combination, but this requires theLto be moved one place to the left also, by puttingIorKin the key word and taking outO,RorXand returning it to its place in the alphabetical sequence. The most frequent pairs containingOareB Osix times,R Ofour times, andO Xthree times. Now these pairs equal respectivelyE N,E SandH E, ifOis put betweenNandPin the fourth line. We will therefore cease to consider it as a letter of the the key word. The combinationELABcan only beTHE_on the assumption thatAis the first letter to the right ofE. The combinationELBXoccurs three times. If it representsTHE_, theBmust be the first letter of the first line and theXmust now be placed underEwhere theRwas tentatively put. We can getTHE_out ofELRHby puttingRin the first line or leaving it where it is, but the preponderance of theBXcombination should suggest the former alternative.A new square showing these changes will look like this:BEATR.X...GH.LMNOPQSUVWYXAsIput in the space underBwill give the wordBEATRIXand as a vowel is clearly necessary there, we will so use theIJand leaveKbetweenHandL. This leavesC,DandFto be placed. It appeared atfirst thatFwas in the key but if it is in the second line, in proximity to the letters of the first line, it will give the same indications. Completing the square then, we haveBEATRIJXCDFGHKLMNOPQSUVWYZWith this square, the message is deciphered without difficulty.“It is very frequently neces(x)sary to employ ciphers and they have for many centuries been employed in the relations betwe(x)en governments, for com(x)munication betwe(x)en com(x)manders and their subordinates and particularly betwe(x)en governments and their agents in foreign countries; there are many cases in history where the capture of a message not in cipher has made the captors of the message victorious in their military movements.”It will be seen that the method of Lieut. Moorman enabled us to pick out six letters of the key word out of eight letters chosen tentatively. The reason for the appearance ofFhas already been noted; the letterOoccurred with many other letters because it happened to remain in the same line withNandSand to be underH. It thus was likely to represent any of these three letters which occur very frequently in any text.Two-character Substitution CiphersCase9.—Two-character substitution ciphers. In ciphers of this type, two letters, numerals, or conventional signs, are substituted for each letter of the text. There are many ways of obtaining the charactersto be substituted but, in general, these ciphers may be considered as special varieties of Case 6 or Case 7. The ciphers which come under this case are not well suited to telegraphic correspondence because the cipher message will contain twice as many letters as the plain text. However they are so used; an example is at hand in which two numerals are substituted for each letter and this makes transmission by telegraph very slow.Case 9 can be recognized by some or all of the following points; the number of characters in the cipher is always an even number; often only a few, say five to ten, of the letters of the alphabet appear; either a frequency table for pairs of the cipher text resembling the normal single letter frequency table can be made, or groups of four letters will show a regular recurrence, from which the cipher can be solved as in Case 7.Case9a.—MessageRNTGN RAAGR NARNA GTGRA TGAAN NANGG RARAT NAANR NNNRN AAAGG AANGR NGGNN NRNAA AANRA TNANN NGGRN RNNRG TTGRG TGGRN ARNTG NNART GGRNR GRNNT GTGAA NNARN ARNRT TGAGG GAAAA NANNA RNAGA NGNAT NNNATThis message contains 160 letters and it will be noted that the only letters used areA,G,N,RandT.We may expect a simple two-letter substitution cipher at once. It will simplify the work if we divide the cipher into groups of two letters and then, if we find there are 26 or less recurring groups, to assign an arbitrary letter to each group and work out the cipher by the method of Case 6.RN TG NR AA GR NA RN AG TG RA TG AA NN AN GG RA RA TN AA NR NN NR NA AA GG AA NG RN GG NN NR NA AA AN RA TN AN NN GG RN RN NR GT TG RG TG GR NA RN TG NN AR TG GR NR GR NN TG TG AA NN AR NA RN RT TG AG GG AA AA NA NN AR NA GA NG NA TN NN ATWith arbitrary letters substituted, we haveA B C D E F A G B H B D I J K H H L D C I C F D K D M A K I C F D J H L J I K A A C N B O B E F A B I P B E C E B B D I P F A Q B G K D D F I P F R M F L I SNow, preparing a frequency table, with note of prefixes and suffixes we have:FrequencyPrefixSuffixA71111111FMKAFFBGKACBQB101111111111AGHNOAPIBQCHDOEIEBDGC6111111BDIIAEDIFFNED9111111111CBLFKFBKDEICKMJIDFE41111DBBCFFCIF811111111ECCEPDPMADDAAIRLG211ABBKH41111BKJHBHLLI9111111111DCKJBEDFLJCCKPBPPJ3111IDLKHIK511111JDAIGHDIADL3111HHFDJIM211DRAFN11CBO11BBP3111IIIBFFQ11ABR11FMS11IA brief study of this table and the distribution in the cipher leads to the conclusion thatB,FandCare certainly vowels and are, if the normal frequency holds, equal toE,O, andAorI. SimilarlyDandIare consonants and we may take them asNandT.Iis taken asTbecause of the combinationIP(=possiblyTH) occurring three times. The next letter in order of frequency isA; it is certainly a consonant and may be taken asRon the basis of its frequency. Let us now try these assumptions on the first two lines of the message. We haveREAN_OR_E_ENT_____NATAON_N_IIIThis is clearly the wordREINFORCEMENTSand, using the letters thus found, the rest of the line becomesAMMUNITIONAND. We have then the following letters determined:Arbitrary lettersABCDEFGHIJKLMPlain TextREINFOCMTSAUDIf these be substituted we have for the message:REINFORCEMENTS AMMUNITION AND RATIONS MUST ARRI_E _EFORE T_E FIFTEENT_ OR _E CANNOT _O_D OUT_.From this the remainder of the letters are determined:Arbitrary lettersNOPQRSPlain textVBHWLXNow let us substitute the two-letter groups for the arbitrary letters:Arbitrary lettersKOGMBEPCRHDFAJILNQSTwo-letter groupsGGRGAGNGTGGRARNRGARAAANARNANNNTNGTRTATPlain textABCDEFHILMNORSTUVWXIt is evident that the cipher was prepared with the letters of the wordGRANTchosen by means of a square of this kind:GRANTGABCDERFGHIKALMNOPNQRSTUTVWXYZThusTG=E,AN=S, etc., as we have already found.Case9-bMessage195049295831232528154418452815204811504122521153455849134124502855252659331952225245113215621558414328613612652945565015234245585063455420191550185311211541582811241745534554205950255245413215334920485018152364An examination of the groups of two numerals each which make up this message, shows that we have 11 to 36 and 41 to 65 with eleven groups missing. Now the 11 to 36 combination is a very familiar one in numeral substitution ciphers (See Case 6-c) and itwill be noted that 41 to 66 would give us a similar alphabet. Let us make a frequency table in this form:GroupFrequencyGroupFrequency11111114111111121421131431144411511111111145111111111164617147181114811191114911120111150111111112115122115211112311153111241154112511155126156127572811111581111129115911306031161132116213311631346413565136166Each of these tables looks like the normal frequency table except for the position of 20 and 50 which should representT, by all our rules, and should be apparently 30 and 60. But suppose we put the alphabet and corresponding numerals in this form:12345678901 or 4ABCDEFGHIJ2 or 5KLMNOPQRST3 or 6UVWXYZThen A=11 or 41, J=10 or 40 and T=20 or 50 as we found. Using the above alphabet, the message may easily be read. Note that this cipher is made up of ten characters only, the Arabic numerals.Case9c—Message115625467625422944321949294015142321721129797031154924213511742414787576462524445143254845317974253340554615127573227945162748151170423519441378252149251476455315483421267215254075161125784546422174154952197929701524214329254449331970187531407925482945514914117321171554An examination of this message shows it to consist of forty-four different two-figure groups running from 11 to 79. Let us prepare a frequency table of these groups.GroupFrequency1111111112113114111115111111111161117118119111120211111111221231241111251111111111126127128291111113031111321331134135113637383940111141421114311441111451111146111147481111491111115051115215315415515615758597011117172117311741117511117611177781117911111We at once note the resemblance between the frequency tables for the groups 11 to 19 and 21 to29; for the groups 30 to 36 and 50 to 56; and for the groups 40 to 49 and 70 to 79. Also the groups 11 to 19 and 21 to 29 have a frequency fitting well with the normal frequency table of the lettersAtoI; the groups 41 to 49 and 71 to 79 have a frequency fitting well with the normal frequency table of the lettersKtoS; and the groups 31 to 36 and 51 to 56 have a frequency fitting well with the normal frequency table of the lettersUtoZ. We haveJandTunaccounted for, but note what occurred in Case 9-b and that 40 and 70 would correspond well withTif they followed respectively 49 and 79. We may now make up a cipher table as follows:12345678901 or 2ABCDEFGHIJ4 or 7KLMNOPQRST3 or 5UVWXYZand this table will solve the cipher message.In ciphers coming under case 9-b and 9-c, it is not uncommon to assign some of the unused numbers such as 85, 93, etc., to whole words in common use or to names of persons or places. In case such groups are found, the meaning must be guessed at from the context; but if many messages in the same cipher are available, the meaning of these groups will soon be obtained. The appearance of such odd groups of figures in a message does not interfere materially with the analysis, and it will be apparent at once on deciphering the message that they represent whole words instead of letters.

The Playfair cipher may be recognized by the following points: (a) It is a substitution cipher, (b) it always contains an even number of letters, (c) when the cipher is divided into groups of two letters each, no group consists of the repetition of the same letter asSSorBB, (d) there will be recurrence of pairs throughout the message, following in a general way, the frequency table of digraphs of pairs, (e) in short messages there may be recurrence of cipher groups representing words or even phrases, and these will always be found in long messages.

In preparing a cipher by this method, a key word is chosen by the correspondents. A large square, divided into twenty-five smaller squares, is constructed as shown below and the letters of the key word are written in, beginning at the upper left hand corner. If any letter recurs in the key word, it is only used on the first occurrence. The remaining letters of the alphabet are used to fill up the square. It is customary to considerIandJas one letter in this cipher and they are written together in the same square.

If the key word chosen isLEAVENWORTH, then the square would be constructed as follows:

LEAVNWORTHBCDFGIJKMPQSUXYZ

The text of the message to be sent is then divided up into groups of two letters each, and equivalents are found for each pair.

Every pair of letters in the square must be: Either (1) in the same vertical line. Thus in the above example each letter is represented in cipher by that which stands next below it, and the bottom letter by the top one of the same column; for instance,TYis represented byFV.

Or (2) in the same horizontal line. Each letter in this case is represented by that which stands next on its right, and the letter on the extreme right by that on the extreme left of the same horizontal line with it; for instanceRHis represented byTW.

Or (3) at opposite corners of a rectangle. Each letter of the pair is represented by the letter in the other corner of the rectangle in the same horizontal line with it; for instanceTSis represented byWY.

If, on dividing the letters of the text into pairs, it is found that a pair consists of the same letter repeated, a dummy letter, asX,Y, orZ, should be introduced to separate the similar letters.

If the message to be sent were “The enemy moves at dawn,” it would be divided into pairs:

THEXENEMYMOVESATDAWNand enciphered:HWAUALAKXPTELUVRMRHL

The message is then broken up into groups of five letters for transmission.

To decipher such a cryptogram, (knowing the key word), the receiver divides it into pairs, andfrom his table finds the equivalent of these pairs, taking the letter immediately above each, when they are in the same vertical line; those immediately on the left, when in the same horizontal line; and those at opposite angles of the rectangle when this is formed.

It is evident, from the foregoing description, that any letter of the plain text may be represented in cipher by one of five letters, viz: The one next below it and the other four letters in the same horizontal line with it in the square. Take, for example, the letterDof the plain text, in combination with each of the other letters of the alphabet. We have, using the keyLEAVENWORTH:

DADBDCDEDFDGDHDIDKDLDMDNDODPDQDRDSDTDUDVDWDXDYDZMRFCFDCAFGFBGRBMCMBAMXGACRFMGMMDBXFRCXFABRMAFXGX

This givesDrepresented byBCFGM44844times,

and, connected with these five letters representingD,we haveARDMXBCG55245111times.

Note that these letters are those of the vertical column containingDplus the lettersB,CandG, of the horizontal line containingD.

Lieut. Frank Moorman, U. S. Army, has developed a method for determining the letters which make up the key word in a Playfair cipher. In the first place, a key word necessarily contains vowels in the approximate proportion of two vowels to three consonants and it is also likely that a key word will contain other common letters. This key word is placed in the first row or rows. Now if a table is made, showing what letters in the cipher occur with every letter, it will be found that the letters having the greatest number of other letters in combination with them are very likely to be letters of the keyword, or in other words, letters occurring in the first or second lines. An example will make this clear:

Message

DB FN EX TZ MF TO VB QB QT OB XA OF PR TZ EQ RH QK QV DX OK AB PR QI EL TV KE EX XS FS BP WD BO BY BF RO EA BO RH QK QV TX GU EL AB TH TR XN ON EA AY XH BO HN EX BS HR QB ZM SE XP HF GZ UG KC BD PO EA AY XH BO XP HF KR QI AB PR QI EL BX FZ BI SE FX PB RA PR QI WC BR XD YG TB QT EA AY XH BO HN EX BS HR QB PR QI EL BX BT HB QB NF SI SE BX NU XP BU RB XB QR OX BA TB RH BP WD RP RO GU GX QR SE ZY OX BA EL AX CW BY BA SX RK RO PR HB OP BD PI CN OX EM RP KR XT EL AX CW EQ FZ SX EL RH RO PR HB UX DA SE XN ZN GU EL BX FS DG DB TB ZL VE RH BO RQ.

From this message, we make up the following table, considering the letters of each pair:

First Letters of Pairs

ABCDEFGHIKLMNOPQRSTUVWXYZA31411B3231141311C11D221E151F12111G1111H513I1151K121L81M11N112121O6141P2123Q2R122721S221T121U131V2W2X1514113211Y51Z212

From this table we pick out the lettersB,E,F,O,R,T,X, as tentative letters of the key word on account of the variety of other letters with which they occur. As there are but two vowels for seven letters, we will addAto the list on account of its occurrences withB,D,E,R, andX. This leaves the letters for the bottom lines of the square as follows:

........CDGHIJKLMNPQSUVWYZ

Referring to the table again we find the most frequent combination to beEL, occurring 8 times, with no occurrence ofLE. Now,THis the commonest pair in plain text, andHTis not common. The fact thatHoccurs in the same horizontal line withLand thatEandTare probably in the key, will lead us to putEin the first line overHandTin the first line overL, so as to makeELequalTH.

The next most frequent combination isPRoccurring 7 times, withRPoccurring twice. In the square as partially arranged,PRequalsM_orN_orQ_orI_. We may eliminate all these exceptN_, and thisN_could only beNOorNA, so that we will put, tentatively theRin the second line overHand theOandAin the same line overIJ. We have then:

.E..T.RAOCDGHIJKLMNPQSUVWYZ

Let us now check this by picking out the combinations beginning withELand seeing if the table will solve them. We find,ELTV,ELAB,ELBXFZ,ELBXBT,ELAXCWBY,ELAXCWEQ,ELRH,ELBXFS. Now, on the assumption that the letter afterELrepresentsE, we have it represented byAthree times,Bthree times,Ronce andTonce. This requiresthatAandBbe put in the same horizontal line withE, sinceTis already there, andRis tentatively underE.

The combinationELTVnow equalsTHEZ. If theTwere moved one place to the left, it would beTHEY, a more likely combination, but this requires theLto be moved one place to the left also, by puttingIorKin the key word and taking outO,RorXand returning it to its place in the alphabetical sequence. The most frequent pairs containingOareB Osix times,R Ofour times, andO Xthree times. Now these pairs equal respectivelyE N,E SandH E, ifOis put betweenNandPin the fourth line. We will therefore cease to consider it as a letter of the the key word. The combinationELABcan only beTHE_on the assumption thatAis the first letter to the right ofE. The combinationELBXoccurs three times. If it representsTHE_, theBmust be the first letter of the first line and theXmust now be placed underEwhere theRwas tentatively put. We can getTHE_out ofELRHby puttingRin the first line or leaving it where it is, but the preponderance of theBXcombination should suggest the former alternative.

A new square showing these changes will look like this:

BEATR.X...GH.LMNOPQSUVWYX

AsIput in the space underBwill give the wordBEATRIXand as a vowel is clearly necessary there, we will so use theIJand leaveKbetweenHandL. This leavesC,DandFto be placed. It appeared atfirst thatFwas in the key but if it is in the second line, in proximity to the letters of the first line, it will give the same indications. Completing the square then, we have

BEATRIJXCDFGHKLMNOPQSUVWYZ

With this square, the message is deciphered without difficulty.

“It is very frequently neces(x)sary to employ ciphers and they have for many centuries been employed in the relations betwe(x)en governments, for com(x)munication betwe(x)en com(x)manders and their subordinates and particularly betwe(x)en governments and their agents in foreign countries; there are many cases in history where the capture of a message not in cipher has made the captors of the message victorious in their military movements.”

It will be seen that the method of Lieut. Moorman enabled us to pick out six letters of the key word out of eight letters chosen tentatively. The reason for the appearance ofFhas already been noted; the letterOoccurred with many other letters because it happened to remain in the same line withNandSand to be underH. It thus was likely to represent any of these three letters which occur very frequently in any text.

Two-character Substitution CiphersCase9.—Two-character substitution ciphers. In ciphers of this type, two letters, numerals, or conventional signs, are substituted for each letter of the text. There are many ways of obtaining the charactersto be substituted but, in general, these ciphers may be considered as special varieties of Case 6 or Case 7. The ciphers which come under this case are not well suited to telegraphic correspondence because the cipher message will contain twice as many letters as the plain text. However they are so used; an example is at hand in which two numerals are substituted for each letter and this makes transmission by telegraph very slow.Case 9 can be recognized by some or all of the following points; the number of characters in the cipher is always an even number; often only a few, say five to ten, of the letters of the alphabet appear; either a frequency table for pairs of the cipher text resembling the normal single letter frequency table can be made, or groups of four letters will show a regular recurrence, from which the cipher can be solved as in Case 7.

Two-character Substitution Ciphers

Case9.—Two-character substitution ciphers. In ciphers of this type, two letters, numerals, or conventional signs, are substituted for each letter of the text. There are many ways of obtaining the charactersto be substituted but, in general, these ciphers may be considered as special varieties of Case 6 or Case 7. The ciphers which come under this case are not well suited to telegraphic correspondence because the cipher message will contain twice as many letters as the plain text. However they are so used; an example is at hand in which two numerals are substituted for each letter and this makes transmission by telegraph very slow.Case 9 can be recognized by some or all of the following points; the number of characters in the cipher is always an even number; often only a few, say five to ten, of the letters of the alphabet appear; either a frequency table for pairs of the cipher text resembling the normal single letter frequency table can be made, or groups of four letters will show a regular recurrence, from which the cipher can be solved as in Case 7.

Case 9 can be recognized by some or all of the following points; the number of characters in the cipher is always an even number; often only a few, say five to ten, of the letters of the alphabet appear; either a frequency table for pairs of the cipher text resembling the normal single letter frequency table can be made, or groups of four letters will show a regular recurrence, from which the cipher can be solved as in Case 7.

Case9a.—MessageRNTGN RAAGR NARNA GTGRA TGAAN NANGG RARAT NAANR NNNRN AAAGG AANGR NGGNN NRNAA AANRA TNANN NGGRN RNNRG TTGRG TGGRN ARNTG NNART GGRNR GRNNT GTGAA NNARN ARNRT TGAGG GAAAA NANNA RNAGA NGNAT NNNATThis message contains 160 letters and it will be noted that the only letters used areA,G,N,RandT.We may expect a simple two-letter substitution cipher at once. It will simplify the work if we divide the cipher into groups of two letters and then, if we find there are 26 or less recurring groups, to assign an arbitrary letter to each group and work out the cipher by the method of Case 6.RN TG NR AA GR NA RN AG TG RA TG AA NN AN GG RA RA TN AA NR NN NR NA AA GG AA NG RN GG NN NR NA AA AN RA TN AN NN GG RN RN NR GT TG RG TG GR NA RN TG NN AR TG GR NR GR NN TG TG AA NN AR NA RN RT TG AG GG AA AA NA NN AR NA GA NG NA TN NN ATWith arbitrary letters substituted, we haveA B C D E F A G B H B D I J K H H L D C I C F D K D M A K I C F D J H L J I K A A C N B O B E F A B I P B E C E B B D I P F A Q B G K D D F I P F R M F L I SNow, preparing a frequency table, with note of prefixes and suffixes we have:FrequencyPrefixSuffixA71111111FMKAFFBGKACBQB101111111111AGHNOAPIBQCHDOEIEBDGC6111111BDIIAEDIFFNED9111111111CBLFKFBKDEICKMJIDFE41111DBBCFFCIF811111111ECCEPDPMADDAAIRLG211ABBKH41111BKJHBHLLI9111111111DCKJBEDFLJCCKPBPPJ3111IDLKHIK511111JDAIGHDIADL3111HHFDJIM211DRAFN11CBO11BBP3111IIIBFFQ11ABR11FMS11IA brief study of this table and the distribution in the cipher leads to the conclusion thatB,FandCare certainly vowels and are, if the normal frequency holds, equal toE,O, andAorI. SimilarlyDandIare consonants and we may take them asNandT.Iis taken asTbecause of the combinationIP(=possiblyTH) occurring three times. The next letter in order of frequency isA; it is certainly a consonant and may be taken asRon the basis of its frequency. Let us now try these assumptions on the first two lines of the message. We haveREAN_OR_E_ENT_____NATAON_N_IIIThis is clearly the wordREINFORCEMENTSand, using the letters thus found, the rest of the line becomesAMMUNITIONAND. We have then the following letters determined:Arbitrary lettersABCDEFGHIJKLMPlain TextREINFOCMTSAUDIf these be substituted we have for the message:REINFORCEMENTS AMMUNITION AND RATIONS MUST ARRI_E _EFORE T_E FIFTEENT_ OR _E CANNOT _O_D OUT_.From this the remainder of the letters are determined:Arbitrary lettersNOPQRSPlain textVBHWLXNow let us substitute the two-letter groups for the arbitrary letters:Arbitrary lettersKOGMBEPCRHDFAJILNQSTwo-letter groupsGGRGAGNGTGGRARNRGARAAANARNANNNTNGTRTATPlain textABCDEFHILMNORSTUVWXIt is evident that the cipher was prepared with the letters of the wordGRANTchosen by means of a square of this kind:GRANTGABCDERFGHIKALMNOPNQRSTUTVWXYZThusTG=E,AN=S, etc., as we have already found.

Case9a.—

Message

RNTGN RAAGR NARNA GTGRA TGAAN NANGG RARAT NAANR NNNRN AAAGG AANGR NGGNN NRNAA AANRA TNANN NGGRN RNNRG TTGRG TGGRN ARNTG NNART GGRNR GRNNT GTGAA NNARN ARNRT TGAGG GAAAA NANNA RNAGA NGNAT NNNAT

This message contains 160 letters and it will be noted that the only letters used areA,G,N,RandT.

We may expect a simple two-letter substitution cipher at once. It will simplify the work if we divide the cipher into groups of two letters and then, if we find there are 26 or less recurring groups, to assign an arbitrary letter to each group and work out the cipher by the method of Case 6.

RN TG NR AA GR NA RN AG TG RA TG AA NN AN GG RA RA TN AA NR NN NR NA AA GG AA NG RN GG NN NR NA AA AN RA TN AN NN GG RN RN NR GT TG RG TG GR NA RN TG NN AR TG GR NR GR NN TG TG AA NN AR NA RN RT TG AG GG AA AA NA NN AR NA GA NG NA TN NN AT

With arbitrary letters substituted, we have

A B C D E F A G B H B D I J K H H L D C I C F D K D M A K I C F D J H L J I K A A C N B O B E F A B I P B E C E B B D I P F A Q B G K D D F I P F R M F L I S

Now, preparing a frequency table, with note of prefixes and suffixes we have:

FrequencyPrefixSuffixA71111111FMKAFFBGKACBQB101111111111AGHNOAPIBQCHDOEIEBDGC6111111BDIIAEDIFFNED9111111111CBLFKFBKDEICKMJIDFE41111DBBCFFCIF811111111ECCEPDPMADDAAIRLG211ABBKH41111BKJHBHLLI9111111111DCKJBEDFLJCCKPBPPJ3111IDLKHIK511111JDAIGHDIADL3111HHFDJIM211DRAFN11CBO11BBP3111IIIBFFQ11ABR11FMS11I

A brief study of this table and the distribution in the cipher leads to the conclusion thatB,FandCare certainly vowels and are, if the normal frequency holds, equal toE,O, andAorI. SimilarlyDandIare consonants and we may take them asNandT.Iis taken asTbecause of the combinationIP(=possiblyTH) occurring three times. The next letter in order of frequency isA; it is certainly a consonant and may be taken asRon the basis of its frequency. Let us now try these assumptions on the first two lines of the message. We have

REAN_OR_E_ENT_____NATAON_N_III

This is clearly the wordREINFORCEMENTSand, using the letters thus found, the rest of the line becomesAMMUNITIONAND. We have then the following letters determined:

Arbitrary lettersABCDEFGHIJKLMPlain TextREINFOCMTSAUD

If these be substituted we have for the message:

REINFORCEMENTS AMMUNITION AND RATIONS MUST ARRI_E _EFORE T_E FIFTEENT_ OR _E CANNOT _O_D OUT_.

From this the remainder of the letters are determined:

Arbitrary lettersNOPQRSPlain textVBHWLX

Now let us substitute the two-letter groups for the arbitrary letters:

Arbitrary lettersKOGMBEPCRHDFAJILNQSTwo-letter groupsGGRGAGNGTGGRARNRGARAAANARNANNNTNGTRTATPlain textABCDEFHILMNORSTUVWX

It is evident that the cipher was prepared with the letters of the wordGRANTchosen by means of a square of this kind:

GRANTGABCDERFGHIKALMNOPNQRSTUTVWXYZ

ThusTG=E,AN=S, etc., as we have already found.

Case9-bMessage195049295831232528154418452815204811504122521153455849134124502855252659331952225245113215621558414328613612652945565015234245585063455420191550185311211541582811241745534554205950255245413215334920485018152364An examination of the groups of two numerals each which make up this message, shows that we have 11 to 36 and 41 to 65 with eleven groups missing. Now the 11 to 36 combination is a very familiar one in numeral substitution ciphers (See Case 6-c) and itwill be noted that 41 to 66 would give us a similar alphabet. Let us make a frequency table in this form:GroupFrequencyGroupFrequency11111114111111121421131431144411511111111145111111111164617147181114811191114911120111150111111112115122115211112311153111241154112511155126156127572811111581111129115911306031161132116213311631346413565136166Each of these tables looks like the normal frequency table except for the position of 20 and 50 which should representT, by all our rules, and should be apparently 30 and 60. But suppose we put the alphabet and corresponding numerals in this form:12345678901 or 4ABCDEFGHIJ2 or 5KLMNOPQRST3 or 6UVWXYZThen A=11 or 41, J=10 or 40 and T=20 or 50 as we found. Using the above alphabet, the message may easily be read. Note that this cipher is made up of ten characters only, the Arabic numerals.

Case9-b

Message

195049295831232528154418452815204811504122521153455849134124502855252659331952225245113215621558414328613612652945565015234245585063455420191550185311211541582811241745534554205950255245413215334920485018152364

An examination of the groups of two numerals each which make up this message, shows that we have 11 to 36 and 41 to 65 with eleven groups missing. Now the 11 to 36 combination is a very familiar one in numeral substitution ciphers (See Case 6-c) and itwill be noted that 41 to 66 would give us a similar alphabet. Let us make a frequency table in this form:

GroupFrequencyGroupFrequency11111114111111121421131431144411511111111145111111111164617147181114811191114911120111150111111112115122115211112311153111241154112511155126156127572811111581111129115911306031161132116213311631346413565136166

Each of these tables looks like the normal frequency table except for the position of 20 and 50 which should representT, by all our rules, and should be apparently 30 and 60. But suppose we put the alphabet and corresponding numerals in this form:

12345678901 or 4ABCDEFGHIJ2 or 5KLMNOPQRST3 or 6UVWXYZ

Then A=11 or 41, J=10 or 40 and T=20 or 50 as we found. Using the above alphabet, the message may easily be read. Note that this cipher is made up of ten characters only, the Arabic numerals.

Case9c—Message115625467625422944321949294015142321721129797031154924213511742414787576462524445143254845317974253340554615127573227945162748151170423519441378252149251476455315483421267215254075161125784546422174154952197929701524214329254449331970187531407925482945514914117321171554An examination of this message shows it to consist of forty-four different two-figure groups running from 11 to 79. Let us prepare a frequency table of these groups.GroupFrequency1111111112113114111115111111111161117118119111120211111111221231241111251111111111126127128291111113031111321331134135113637383940111141421114311441111451111146111147481111491111115051115215315415515615758597011117172117311741117511117611177781117911111We at once note the resemblance between the frequency tables for the groups 11 to 19 and 21 to29; for the groups 30 to 36 and 50 to 56; and for the groups 40 to 49 and 70 to 79. Also the groups 11 to 19 and 21 to 29 have a frequency fitting well with the normal frequency table of the lettersAtoI; the groups 41 to 49 and 71 to 79 have a frequency fitting well with the normal frequency table of the lettersKtoS; and the groups 31 to 36 and 51 to 56 have a frequency fitting well with the normal frequency table of the lettersUtoZ. We haveJandTunaccounted for, but note what occurred in Case 9-b and that 40 and 70 would correspond well withTif they followed respectively 49 and 79. We may now make up a cipher table as follows:12345678901 or 2ABCDEFGHIJ4 or 7KLMNOPQRST3 or 5UVWXYZand this table will solve the cipher message.In ciphers coming under case 9-b and 9-c, it is not uncommon to assign some of the unused numbers such as 85, 93, etc., to whole words in common use or to names of persons or places. In case such groups are found, the meaning must be guessed at from the context; but if many messages in the same cipher are available, the meaning of these groups will soon be obtained. The appearance of such odd groups of figures in a message does not interfere materially with the analysis, and it will be apparent at once on deciphering the message that they represent whole words instead of letters.

Case9c—

Message

115625467625422944321949294015142321721129797031154924213511742414787576462524445143254845317974253340554615127573227945162748151170423519441378252149251476455315483421267215254075161125784546422174154952197929701524214329254449331970187531407925482945514914117321171554

An examination of this message shows it to consist of forty-four different two-figure groups running from 11 to 79. Let us prepare a frequency table of these groups.

GroupFrequency1111111112113114111115111111111161117118119111120211111111221231241111251111111111126127128291111113031111321331134135113637383940111141421114311441111451111146111147481111491111115051115215315415515615758597011117172117311741117511117611177781117911111

We at once note the resemblance between the frequency tables for the groups 11 to 19 and 21 to29; for the groups 30 to 36 and 50 to 56; and for the groups 40 to 49 and 70 to 79. Also the groups 11 to 19 and 21 to 29 have a frequency fitting well with the normal frequency table of the lettersAtoI; the groups 41 to 49 and 71 to 79 have a frequency fitting well with the normal frequency table of the lettersKtoS; and the groups 31 to 36 and 51 to 56 have a frequency fitting well with the normal frequency table of the lettersUtoZ. We haveJandTunaccounted for, but note what occurred in Case 9-b and that 40 and 70 would correspond well withTif they followed respectively 49 and 79. We may now make up a cipher table as follows:

12345678901 or 2ABCDEFGHIJ4 or 7KLMNOPQRST3 or 5UVWXYZ

and this table will solve the cipher message.

In ciphers coming under case 9-b and 9-c, it is not uncommon to assign some of the unused numbers such as 85, 93, etc., to whole words in common use or to names of persons or places. In case such groups are found, the meaning must be guessed at from the context; but if many messages in the same cipher are available, the meaning of these groups will soon be obtained. The appearance of such odd groups of figures in a message does not interfere materially with the analysis, and it will be apparent at once on deciphering the message that they represent whole words instead of letters.

Back to Index Next