Case7-b.MessageDDLRMERGLMUJTLLCHERSLSOEESMEJUZJIMUDAEESDUTDBGUGPNRCHOBEQEIEOOACDEIOOGCOLJLPDUVMIGIYXQQTOTDJCPJOISLYDUASIUPFNEAECOBOESHOBETNDQXUCYLUQOYEHYDULXPEQFIXZEPDCNZENELQMJTSQECFIEARNDNETSCFIFQSETDDNPUUZHQCDTXQIRMERGLXBEIQRXJFBSQDLDSVIXUMTBAEQEBYLECOIYCUDQTPYSVOQBLULYROYHEFMOYMUYROYMUEQBLVUBREYGHYTQCMUBREQTOFVSDDUDAFFSCEBSVTIOYETCLQXDVNLQXYTSIMZULXBAXQRECVTDETGOBCCUYFTTNXLUNEFSIVIJRZHSBYLLTSIOn the preliminary determination, we have the following count of letters out of a total of 385:A8L23J9E38N11Q22I19R14V9O21S20X13U24T21Z6Total110Total89Total5928%23%15%Every letter exceptKandWoccurs at least six times. We may say then that it is a substitution cipher, Spanish text, and certainly not Case 4, 5 or 6. We will now analyze it for recurring pairs or groupsto determine, if it be Case 7, how many alphabets were used. The following is a complete list of such recurring groups and pairs with the number of letters intervening and the factors thereof. In work of this kind, the groups of three or more letters are always much more valuable than single pairs. For example, the groups,HOBE,OYMU,RMERGLandUBREshow, without question, that six alphabets were used. It is not necessary, as a rule, to make a complete list like the following:AE74=2×37IE110=2×5×11RE50=2×5×5AE120=2×2×2×3×5IM302=2×151RMERGL198=2×3×3×11BE88=2×2×2×11IO250=2×5×5×5SC132=2×2×3×11CD132=2×2×3×11IX78=2×3×13SD262=2×131CFI12=2×2×3LY158=2×79SI230=2×5×23CH36=2×2×3×3JT150=2×3×5×5SI34=2×17CO42=2×3×7LL367 No factorsSI264=2×2×2×3×11CO126=2×3×3×7LQ164=2×2×41SI12=2×2×3CU114=2×3×19LQX6=2×3SL78=2×3×13DD186=2×3×31LU124=2×2×31SQ54=2×3×3×3DD116=2×2×29LU110=2×5×11SV27=3×3×3DE285=5×57LU234=2×3×3×13SV63=3×3×7DL218=2×109LX66=2×3×11SV90=2×3×3×5DN14=2×7LXB132=2×2×3×11TD47 No factorsDQ120=2×2×2×3×5LY158=2×79TD165=3×5×11DU36=2×2×3×3ME22=2×11TD96=2×2×2×2×2×3DU24=2×2×2×3MU24=2×2×2×3TN239 No factorsDU38=2×19MU240=2×2×2×2×3×5TS14=2×7DU165=3×5×11MU18=2×3×3TS156=2×2×3×13EA30=2×3×5ND47 No factorsTSI50=2×5×5EB78=2×3×13NE48=2×2×2×2×3UBRE12=2×2×3EC180=2×2×3×3×5NE18=2×3×3UD60=2×2×3×5ECO126=2×3×3×7NE192=2×2×2×2×2×2×3UDA270=2×3×3×3×5EES14=2×7OB6=2×3UL114=2×3×19EF105=3×5×7OB234=2×3×3×13ULX198=2×3×3×11EI8=2×2×2OE93=3×31UY89 No factorsEI152=2×2×2×19OI144=2×2×2×2×3×3UZ162=2×3×3×3×3EQ88=2×2×2×11OO7 No factorsVI148=2×2×37EQ264=2×2×2×3×11OY6=2×3VT33=3×11EQ44=2×2×11OY46=2×23XQ114=2×3×19EQE176=2×2×2×2×11OYMU6=2×3XQ144=2×2×2×2×3×3ER12=2×2×3PD75=3×5×5XU99=3×3×11ES78=2×3×13QBL24=2×2×2×3YE184=2×2×2×23ET135=3×3×5QC95=5×19YL106=2×53ET9=3×3QE108=2×2×3×3×3YL144=2×2×2×2×3×3ET54=2×3×3×3QE68=2×2×17YM6=2×3ET31 No factorsQR132=2×2×3×11YRO12=2×2×3HE245=5×7×7QTO210=2×3×5×7ZE6=2×3HOBE66=2×3×11QX198=2×3×3×11ZH183=3×61Out of one hundred and one recurring pairs we have fifty with the factors 2×3=6; out of twelve recurring triplets, nine have these factors; and the four recurring groups of four or more letters all have these factors. The percentages are respectively 49.5%, 75% and 100% and we may be certain from this that six alphabets were used. But, before the six frequency tables are made up, there is one more point to be considered; why are there so many recurring groups which do not have six as a factor? The answer is that one or more of the alphabets is repeated in each cycle; that is, a key word of the formHAVANAhas been used. If this were the key word, the second, fourth and sixth alphabets would be the same. We will see later that in this example the second and sixth alphabets are the same and this introduces the great number of recurring groups without the factor 6.We will now proceed to make a frequency table for each alphabet. As the message is written in thirty columns, we take the first, seventh, thirteenth, etc., as constituting the first alphabet; the second, eighth, fourteenth, etc., as constituting the second alphabet and so on. The prefix and suffix letter is noted for each occurrence of each letter. The importance of this will be appreciated when the form of the frequency tables is examined. None bears any resemblance to the normal frequency table except that each is evidently a mixed up alphabet. Thenumbers after “Prefix” and “Suffix” refer to the alphabet to which these belong, for convenience in future reference.
Case7-b.MessageDDLRMERGLMUJTLLCHERSLSOEESMEJUZJIMUDAEESDUTDBGUGPNRCHOBEQEIEOOACDEIOOGCOLJLPDUVMIGIYXQQTOTDJCPJOISLYDUASIUPFNEAECOBOESHOBETNDQXUCYLUQOYEHYDULXPEQFIXZEPDCNZENELQMJTSQECFIEARNDNETSCFIFQSETDDNPUUZHQCDTXQIRMERGLXBEIQRXJFBSQDLDSVIXUMTBAEQEBYLECOIYCUDQTPYSVOQBLULYROYHEFMOYMUYROYMUEQBLVUBREYGHYTQCMUBREQTOFVSDDUDAFFSCEBSVTIOYETCLQXDVNLQXYTSIMZULXBAXQRECVTDETGOBCCUYFTTNXLUNEFSIVIJRZHSBYLLTSIOn the preliminary determination, we have the following count of letters out of a total of 385:A8L23J9E38N11Q22I19R14V9O21S20X13U24T21Z6Total110Total89Total5928%23%15%Every letter exceptKandWoccurs at least six times. We may say then that it is a substitution cipher, Spanish text, and certainly not Case 4, 5 or 6. We will now analyze it for recurring pairs or groupsto determine, if it be Case 7, how many alphabets were used. The following is a complete list of such recurring groups and pairs with the number of letters intervening and the factors thereof. In work of this kind, the groups of three or more letters are always much more valuable than single pairs. For example, the groups,HOBE,OYMU,RMERGLandUBREshow, without question, that six alphabets were used. It is not necessary, as a rule, to make a complete list like the following:AE74=2×37IE110=2×5×11RE50=2×5×5AE120=2×2×2×3×5IM302=2×151RMERGL198=2×3×3×11BE88=2×2×2×11IO250=2×5×5×5SC132=2×2×3×11CD132=2×2×3×11IX78=2×3×13SD262=2×131CFI12=2×2×3LY158=2×79SI230=2×5×23CH36=2×2×3×3JT150=2×3×5×5SI34=2×17CO42=2×3×7LL367 No factorsSI264=2×2×2×3×11CO126=2×3×3×7LQ164=2×2×41SI12=2×2×3CU114=2×3×19LQX6=2×3SL78=2×3×13DD186=2×3×31LU124=2×2×31SQ54=2×3×3×3DD116=2×2×29LU110=2×5×11SV27=3×3×3DE285=5×57LU234=2×3×3×13SV63=3×3×7DL218=2×109LX66=2×3×11SV90=2×3×3×5DN14=2×7LXB132=2×2×3×11TD47 No factorsDQ120=2×2×2×3×5LY158=2×79TD165=3×5×11DU36=2×2×3×3ME22=2×11TD96=2×2×2×2×2×3DU24=2×2×2×3MU24=2×2×2×3TN239 No factorsDU38=2×19MU240=2×2×2×2×3×5TS14=2×7DU165=3×5×11MU18=2×3×3TS156=2×2×3×13EA30=2×3×5ND47 No factorsTSI50=2×5×5EB78=2×3×13NE48=2×2×2×2×3UBRE12=2×2×3EC180=2×2×3×3×5NE18=2×3×3UD60=2×2×3×5ECO126=2×3×3×7NE192=2×2×2×2×2×2×3UDA270=2×3×3×3×5EES14=2×7OB6=2×3UL114=2×3×19EF105=3×5×7OB234=2×3×3×13ULX198=2×3×3×11EI8=2×2×2OE93=3×31UY89 No factorsEI152=2×2×2×19OI144=2×2×2×2×3×3UZ162=2×3×3×3×3EQ88=2×2×2×11OO7 No factorsVI148=2×2×37EQ264=2×2×2×3×11OY6=2×3VT33=3×11EQ44=2×2×11OY46=2×23XQ114=2×3×19EQE176=2×2×2×2×11OYMU6=2×3XQ144=2×2×2×2×3×3ER12=2×2×3PD75=3×5×5XU99=3×3×11ES78=2×3×13QBL24=2×2×2×3YE184=2×2×2×23ET135=3×3×5QC95=5×19YL106=2×53ET9=3×3QE108=2×2×3×3×3YL144=2×2×2×2×3×3ET54=2×3×3×3QE68=2×2×17YM6=2×3ET31 No factorsQR132=2×2×3×11YRO12=2×2×3HE245=5×7×7QTO210=2×3×5×7ZE6=2×3HOBE66=2×3×11QX198=2×3×3×11ZH183=3×61Out of one hundred and one recurring pairs we have fifty with the factors 2×3=6; out of twelve recurring triplets, nine have these factors; and the four recurring groups of four or more letters all have these factors. The percentages are respectively 49.5%, 75% and 100% and we may be certain from this that six alphabets were used. But, before the six frequency tables are made up, there is one more point to be considered; why are there so many recurring groups which do not have six as a factor? The answer is that one or more of the alphabets is repeated in each cycle; that is, a key word of the formHAVANAhas been used. If this were the key word, the second, fourth and sixth alphabets would be the same. We will see later that in this example the second and sixth alphabets are the same and this introduces the great number of recurring groups without the factor 6.We will now proceed to make a frequency table for each alphabet. As the message is written in thirty columns, we take the first, seventh, thirteenth, etc., as constituting the first alphabet; the second, eighth, fourteenth, etc., as constituting the second alphabet and so on. The prefix and suffix letter is noted for each occurrence of each letter. The importance of this will be appreciated when the form of the frequency tables is examined. None bears any resemblance to the normal frequency table except that each is evidently a mixed up alphabet. Thenumbers after “Prefix” and “Suffix” refer to the alphabet to which these belong, for convenience in future reference.
Case7-b.MessageDDLRMERGLMUJTLLCHERSLSOEESMEJUZJIMUDAEESDUTDBGUGPNRCHOBEQEIEOOACDEIOOGCOLJLPDUVMIGIYXQQTOTDJCPJOISLYDUASIUPFNEAECOBOESHOBETNDQXUCYLUQOYEHYDULXPEQFIXZEPDCNZENELQMJTSQECFIEARNDNETSCFIFQSETDDNPUUZHQCDTXQIRMERGLXBEIQRXJFBSQDLDSVIXUMTBAEQEBYLECOIYCUDQTPYSVOQBLULYROYHEFMOYMUYROYMUEQBLVUBREYGHYTQCMUBREQTOFVSDDUDAFFSCEBSVTIOYETCLQXDVNLQXYTSIMZULXBAXQRECVTDETGOBCCUYFTTNXLUNEFSIVIJRZHSBYLLTSIOn the preliminary determination, we have the following count of letters out of a total of 385:A8L23J9E38N11Q22I19R14V9O21S20X13U24T21Z6Total110Total89Total5928%23%15%Every letter exceptKandWoccurs at least six times. We may say then that it is a substitution cipher, Spanish text, and certainly not Case 4, 5 or 6. We will now analyze it for recurring pairs or groupsto determine, if it be Case 7, how many alphabets were used. The following is a complete list of such recurring groups and pairs with the number of letters intervening and the factors thereof. In work of this kind, the groups of three or more letters are always much more valuable than single pairs. For example, the groups,HOBE,OYMU,RMERGLandUBREshow, without question, that six alphabets were used. It is not necessary, as a rule, to make a complete list like the following:AE74=2×37IE110=2×5×11RE50=2×5×5AE120=2×2×2×3×5IM302=2×151RMERGL198=2×3×3×11BE88=2×2×2×11IO250=2×5×5×5SC132=2×2×3×11CD132=2×2×3×11IX78=2×3×13SD262=2×131CFI12=2×2×3LY158=2×79SI230=2×5×23CH36=2×2×3×3JT150=2×3×5×5SI34=2×17CO42=2×3×7LL367 No factorsSI264=2×2×2×3×11CO126=2×3×3×7LQ164=2×2×41SI12=2×2×3CU114=2×3×19LQX6=2×3SL78=2×3×13DD186=2×3×31LU124=2×2×31SQ54=2×3×3×3DD116=2×2×29LU110=2×5×11SV27=3×3×3DE285=5×57LU234=2×3×3×13SV63=3×3×7DL218=2×109LX66=2×3×11SV90=2×3×3×5DN14=2×7LXB132=2×2×3×11TD47 No factorsDQ120=2×2×2×3×5LY158=2×79TD165=3×5×11DU36=2×2×3×3ME22=2×11TD96=2×2×2×2×2×3DU24=2×2×2×3MU24=2×2×2×3TN239 No factorsDU38=2×19MU240=2×2×2×2×3×5TS14=2×7DU165=3×5×11MU18=2×3×3TS156=2×2×3×13EA30=2×3×5ND47 No factorsTSI50=2×5×5EB78=2×3×13NE48=2×2×2×2×3UBRE12=2×2×3EC180=2×2×3×3×5NE18=2×3×3UD60=2×2×3×5ECO126=2×3×3×7NE192=2×2×2×2×2×2×3UDA270=2×3×3×3×5EES14=2×7OB6=2×3UL114=2×3×19EF105=3×5×7OB234=2×3×3×13ULX198=2×3×3×11EI8=2×2×2OE93=3×31UY89 No factorsEI152=2×2×2×19OI144=2×2×2×2×3×3UZ162=2×3×3×3×3EQ88=2×2×2×11OO7 No factorsVI148=2×2×37EQ264=2×2×2×3×11OY6=2×3VT33=3×11EQ44=2×2×11OY46=2×23XQ114=2×3×19EQE176=2×2×2×2×11OYMU6=2×3XQ144=2×2×2×2×3×3ER12=2×2×3PD75=3×5×5XU99=3×3×11ES78=2×3×13QBL24=2×2×2×3YE184=2×2×2×23ET135=3×3×5QC95=5×19YL106=2×53ET9=3×3QE108=2×2×3×3×3YL144=2×2×2×2×3×3ET54=2×3×3×3QE68=2×2×17YM6=2×3ET31 No factorsQR132=2×2×3×11YRO12=2×2×3HE245=5×7×7QTO210=2×3×5×7ZE6=2×3HOBE66=2×3×11QX198=2×3×3×11ZH183=3×61Out of one hundred and one recurring pairs we have fifty with the factors 2×3=6; out of twelve recurring triplets, nine have these factors; and the four recurring groups of four or more letters all have these factors. The percentages are respectively 49.5%, 75% and 100% and we may be certain from this that six alphabets were used. But, before the six frequency tables are made up, there is one more point to be considered; why are there so many recurring groups which do not have six as a factor? The answer is that one or more of the alphabets is repeated in each cycle; that is, a key word of the formHAVANAhas been used. If this were the key word, the second, fourth and sixth alphabets would be the same. We will see later that in this example the second and sixth alphabets are the same and this introduces the great number of recurring groups without the factor 6.We will now proceed to make a frequency table for each alphabet. As the message is written in thirty columns, we take the first, seventh, thirteenth, etc., as constituting the first alphabet; the second, eighth, fourteenth, etc., as constituting the second alphabet and so on. The prefix and suffix letter is noted for each occurrence of each letter. The importance of this will be appreciated when the form of the frequency tables is examined. None bears any resemblance to the normal frequency table except that each is evidently a mixed up alphabet. Thenumbers after “Prefix” and “Suffix” refer to the alphabet to which these belong, for convenience in future reference.
Case7-b.MessageDDLRMERGLMUJTLLCHERSLSOEESMEJUZJIMUDAEESDUTDBGUGPNRCHOBEQEIEOOACDEIOOGCOLJLPDUVMIGIYXQQTOTDJCPJOISLYDUASIUPFNEAECOBOESHOBETNDQXUCYLUQOYEHYDULXPEQFIXZEPDCNZENELQMJTSQECFIEARNDNETSCFIFQSETDDNPUUZHQCDTXQIRMERGLXBEIQRXJFBSQDLDSVIXUMTBAEQEBYLECOIYCUDQTPYSVOQBLULYROYHEFMOYMUYROYMUEQBLVUBREYGHYTQCMUBREQTOFVSDDUDAFFSCEBSVTIOYETCLQXDVNLQXYTSIMZULXBAXQRECVTDETGOBCCUYFTTNXLUNEFSIVIJRZHSBYLLTSIOn the preliminary determination, we have the following count of letters out of a total of 385:A8L23J9E38N11Q22I19R14V9O21S20X13U24T21Z6Total110Total89Total5928%23%15%Every letter exceptKandWoccurs at least six times. We may say then that it is a substitution cipher, Spanish text, and certainly not Case 4, 5 or 6. We will now analyze it for recurring pairs or groupsto determine, if it be Case 7, how many alphabets were used. The following is a complete list of such recurring groups and pairs with the number of letters intervening and the factors thereof. In work of this kind, the groups of three or more letters are always much more valuable than single pairs. For example, the groups,HOBE,OYMU,RMERGLandUBREshow, without question, that six alphabets were used. It is not necessary, as a rule, to make a complete list like the following:AE74=2×37IE110=2×5×11RE50=2×5×5AE120=2×2×2×3×5IM302=2×151RMERGL198=2×3×3×11BE88=2×2×2×11IO250=2×5×5×5SC132=2×2×3×11CD132=2×2×3×11IX78=2×3×13SD262=2×131CFI12=2×2×3LY158=2×79SI230=2×5×23CH36=2×2×3×3JT150=2×3×5×5SI34=2×17CO42=2×3×7LL367 No factorsSI264=2×2×2×3×11CO126=2×3×3×7LQ164=2×2×41SI12=2×2×3CU114=2×3×19LQX6=2×3SL78=2×3×13DD186=2×3×31LU124=2×2×31SQ54=2×3×3×3DD116=2×2×29LU110=2×5×11SV27=3×3×3DE285=5×57LU234=2×3×3×13SV63=3×3×7DL218=2×109LX66=2×3×11SV90=2×3×3×5DN14=2×7LXB132=2×2×3×11TD47 No factorsDQ120=2×2×2×3×5LY158=2×79TD165=3×5×11DU36=2×2×3×3ME22=2×11TD96=2×2×2×2×2×3DU24=2×2×2×3MU24=2×2×2×3TN239 No factorsDU38=2×19MU240=2×2×2×2×3×5TS14=2×7DU165=3×5×11MU18=2×3×3TS156=2×2×3×13EA30=2×3×5ND47 No factorsTSI50=2×5×5EB78=2×3×13NE48=2×2×2×2×3UBRE12=2×2×3EC180=2×2×3×3×5NE18=2×3×3UD60=2×2×3×5ECO126=2×3×3×7NE192=2×2×2×2×2×2×3UDA270=2×3×3×3×5EES14=2×7OB6=2×3UL114=2×3×19EF105=3×5×7OB234=2×3×3×13ULX198=2×3×3×11EI8=2×2×2OE93=3×31UY89 No factorsEI152=2×2×2×19OI144=2×2×2×2×3×3UZ162=2×3×3×3×3EQ88=2×2×2×11OO7 No factorsVI148=2×2×37EQ264=2×2×2×3×11OY6=2×3VT33=3×11EQ44=2×2×11OY46=2×23XQ114=2×3×19EQE176=2×2×2×2×11OYMU6=2×3XQ144=2×2×2×2×3×3ER12=2×2×3PD75=3×5×5XU99=3×3×11ES78=2×3×13QBL24=2×2×2×3YE184=2×2×2×23ET135=3×3×5QC95=5×19YL106=2×53ET9=3×3QE108=2×2×3×3×3YL144=2×2×2×2×3×3ET54=2×3×3×3QE68=2×2×17YM6=2×3ET31 No factorsQR132=2×2×3×11YRO12=2×2×3HE245=5×7×7QTO210=2×3×5×7ZE6=2×3HOBE66=2×3×11QX198=2×3×3×11ZH183=3×61Out of one hundred and one recurring pairs we have fifty with the factors 2×3=6; out of twelve recurring triplets, nine have these factors; and the four recurring groups of four or more letters all have these factors. The percentages are respectively 49.5%, 75% and 100% and we may be certain from this that six alphabets were used. But, before the six frequency tables are made up, there is one more point to be considered; why are there so many recurring groups which do not have six as a factor? The answer is that one or more of the alphabets is repeated in each cycle; that is, a key word of the formHAVANAhas been used. If this were the key word, the second, fourth and sixth alphabets would be the same. We will see later that in this example the second and sixth alphabets are the same and this introduces the great number of recurring groups without the factor 6.We will now proceed to make a frequency table for each alphabet. As the message is written in thirty columns, we take the first, seventh, thirteenth, etc., as constituting the first alphabet; the second, eighth, fourteenth, etc., as constituting the second alphabet and so on. The prefix and suffix letter is noted for each occurrence of each letter. The importance of this will be appreciated when the form of the frequency tables is examined. None bears any resemblance to the normal frequency table except that each is evidently a mixed up alphabet. Thenumbers after “Prefix” and “Suffix” refer to the alphabet to which these belong, for convenience in future reference.
Case7-b.MessageDDLRMERGLMUJTLLCHERSLSOEESMEJUZJIMUDAEESDUTDBGUGPNRCHOBEQEIEOOACDEIOOGCOLJLPDUVMIGIYXQQTOTDJCPJOISLYDUASIUPFNEAECOBOESHOBETNDQXUCYLUQOYEHYDULXPEQFIXZEPDCNZENELQMJTSQECFIEARNDNETSCFIFQSETDDNPUUZHQCDTXQIRMERGLXBEIQRXJFBSQDLDSVIXUMTBAEQEBYLECOIYCUDQTPYSVOQBLULYROYHEFMOYMUYROYMUEQBLVUBREYGHYTQCMUBREQTOFVSDDUDAFFSCEBSVTIOYETCLQXDVNLQXYTSIMZULXBAXQRECVTDETGOBCCUYFTTNXLUNEFSIVIJRZHSBYLLTSIOn the preliminary determination, we have the following count of letters out of a total of 385:A8L23J9E38N11Q22I19R14V9O21S20X13U24T21Z6Total110Total89Total5928%23%15%Every letter exceptKandWoccurs at least six times. We may say then that it is a substitution cipher, Spanish text, and certainly not Case 4, 5 or 6. We will now analyze it for recurring pairs or groupsto determine, if it be Case 7, how many alphabets were used. The following is a complete list of such recurring groups and pairs with the number of letters intervening and the factors thereof. In work of this kind, the groups of three or more letters are always much more valuable than single pairs. For example, the groups,HOBE,OYMU,RMERGLandUBREshow, without question, that six alphabets were used. It is not necessary, as a rule, to make a complete list like the following:AE74=2×37IE110=2×5×11RE50=2×5×5AE120=2×2×2×3×5IM302=2×151RMERGL198=2×3×3×11BE88=2×2×2×11IO250=2×5×5×5SC132=2×2×3×11CD132=2×2×3×11IX78=2×3×13SD262=2×131CFI12=2×2×3LY158=2×79SI230=2×5×23CH36=2×2×3×3JT150=2×3×5×5SI34=2×17CO42=2×3×7LL367 No factorsSI264=2×2×2×3×11CO126=2×3×3×7LQ164=2×2×41SI12=2×2×3CU114=2×3×19LQX6=2×3SL78=2×3×13DD186=2×3×31LU124=2×2×31SQ54=2×3×3×3DD116=2×2×29LU110=2×5×11SV27=3×3×3DE285=5×57LU234=2×3×3×13SV63=3×3×7DL218=2×109LX66=2×3×11SV90=2×3×3×5DN14=2×7LXB132=2×2×3×11TD47 No factorsDQ120=2×2×2×3×5LY158=2×79TD165=3×5×11DU36=2×2×3×3ME22=2×11TD96=2×2×2×2×2×3DU24=2×2×2×3MU24=2×2×2×3TN239 No factorsDU38=2×19MU240=2×2×2×2×3×5TS14=2×7DU165=3×5×11MU18=2×3×3TS156=2×2×3×13EA30=2×3×5ND47 No factorsTSI50=2×5×5EB78=2×3×13NE48=2×2×2×2×3UBRE12=2×2×3EC180=2×2×3×3×5NE18=2×3×3UD60=2×2×3×5ECO126=2×3×3×7NE192=2×2×2×2×2×2×3UDA270=2×3×3×3×5EES14=2×7OB6=2×3UL114=2×3×19EF105=3×5×7OB234=2×3×3×13ULX198=2×3×3×11EI8=2×2×2OE93=3×31UY89 No factorsEI152=2×2×2×19OI144=2×2×2×2×3×3UZ162=2×3×3×3×3EQ88=2×2×2×11OO7 No factorsVI148=2×2×37EQ264=2×2×2×3×11OY6=2×3VT33=3×11EQ44=2×2×11OY46=2×23XQ114=2×3×19EQE176=2×2×2×2×11OYMU6=2×3XQ144=2×2×2×2×3×3ER12=2×2×3PD75=3×5×5XU99=3×3×11ES78=2×3×13QBL24=2×2×2×3YE184=2×2×2×23ET135=3×3×5QC95=5×19YL106=2×53ET9=3×3QE108=2×2×3×3×3YL144=2×2×2×2×3×3ET54=2×3×3×3QE68=2×2×17YM6=2×3ET31 No factorsQR132=2×2×3×11YRO12=2×2×3HE245=5×7×7QTO210=2×3×5×7ZE6=2×3HOBE66=2×3×11QX198=2×3×3×11ZH183=3×61Out of one hundred and one recurring pairs we have fifty with the factors 2×3=6; out of twelve recurring triplets, nine have these factors; and the four recurring groups of four or more letters all have these factors. The percentages are respectively 49.5%, 75% and 100% and we may be certain from this that six alphabets were used. But, before the six frequency tables are made up, there is one more point to be considered; why are there so many recurring groups which do not have six as a factor? The answer is that one or more of the alphabets is repeated in each cycle; that is, a key word of the formHAVANAhas been used. If this were the key word, the second, fourth and sixth alphabets would be the same. We will see later that in this example the second and sixth alphabets are the same and this introduces the great number of recurring groups without the factor 6.We will now proceed to make a frequency table for each alphabet. As the message is written in thirty columns, we take the first, seventh, thirteenth, etc., as constituting the first alphabet; the second, eighth, fourteenth, etc., as constituting the second alphabet and so on. The prefix and suffix letter is noted for each occurrence of each letter. The importance of this will be appreciated when the form of the frequency tables is examined. None bears any resemblance to the normal frequency table except that each is evidently a mixed up alphabet. Thenumbers after “Prefix” and “Suffix” refer to the alphabet to which these belong, for convenience in future reference.
Case7-b.
Message
DDLRMERGLMUJTLLCHERSLSOEESMEJUZJIMUDAEESDUTDBGUGPNRCHOBEQEIEOOACDEIOOGCOLJLPDUVMIGIYXQQTOTDJCPJOISLYDUASIUPFNEAECOBOESHOBETNDQXUCYLUQOYEHYDULXPEQFIXZEPDCNZENELQMJTSQECFIEARNDNETSCFIFQSETDDNPUUZHQCDTXQIRMERGLXBEIQRXJFBSQDLDSVIXUMTBAEQEBYLECOIYCUDQTPYSVOQBLULYROYHEFMOYMUYROYMUEQBLVUBREYGHYTQCMUBREQTOFVSDDUDAFFSCEBSVTIOYETCLQXDVNLQXYTSIMZULXBAXQRECVTDETGOBCCUYFTTNXLUNEFSIVIJRZHSBYLLTSI
On the preliminary determination, we have the following count of letters out of a total of 385:
A8L23J9E38N11Q22I19R14V9O21S20X13U24T21Z6Total110Total89Total5928%23%15%
Every letter exceptKandWoccurs at least six times. We may say then that it is a substitution cipher, Spanish text, and certainly not Case 4, 5 or 6. We will now analyze it for recurring pairs or groupsto determine, if it be Case 7, how many alphabets were used. The following is a complete list of such recurring groups and pairs with the number of letters intervening and the factors thereof. In work of this kind, the groups of three or more letters are always much more valuable than single pairs. For example, the groups,HOBE,OYMU,RMERGLandUBREshow, without question, that six alphabets were used. It is not necessary, as a rule, to make a complete list like the following:
AE74=2×37IE110=2×5×11RE50=2×5×5AE120=2×2×2×3×5IM302=2×151RMERGL198=2×3×3×11BE88=2×2×2×11IO250=2×5×5×5SC132=2×2×3×11CD132=2×2×3×11IX78=2×3×13SD262=2×131CFI12=2×2×3LY158=2×79SI230=2×5×23CH36=2×2×3×3JT150=2×3×5×5SI34=2×17CO42=2×3×7LL367 No factorsSI264=2×2×2×3×11CO126=2×3×3×7LQ164=2×2×41SI12=2×2×3CU114=2×3×19LQX6=2×3SL78=2×3×13DD186=2×3×31LU124=2×2×31SQ54=2×3×3×3DD116=2×2×29LU110=2×5×11SV27=3×3×3DE285=5×57LU234=2×3×3×13SV63=3×3×7DL218=2×109LX66=2×3×11SV90=2×3×3×5DN14=2×7LXB132=2×2×3×11TD47 No factorsDQ120=2×2×2×3×5LY158=2×79TD165=3×5×11DU36=2×2×3×3ME22=2×11TD96=2×2×2×2×2×3DU24=2×2×2×3MU24=2×2×2×3TN239 No factorsDU38=2×19MU240=2×2×2×2×3×5TS14=2×7DU165=3×5×11MU18=2×3×3TS156=2×2×3×13EA30=2×3×5ND47 No factorsTSI50=2×5×5EB78=2×3×13NE48=2×2×2×2×3UBRE12=2×2×3EC180=2×2×3×3×5NE18=2×3×3UD60=2×2×3×5ECO126=2×3×3×7NE192=2×2×2×2×2×2×3UDA270=2×3×3×3×5EES14=2×7OB6=2×3UL114=2×3×19EF105=3×5×7OB234=2×3×3×13ULX198=2×3×3×11EI8=2×2×2OE93=3×31UY89 No factorsEI152=2×2×2×19OI144=2×2×2×2×3×3UZ162=2×3×3×3×3EQ88=2×2×2×11OO7 No factorsVI148=2×2×37EQ264=2×2×2×3×11OY6=2×3VT33=3×11EQ44=2×2×11OY46=2×23XQ114=2×3×19EQE176=2×2×2×2×11OYMU6=2×3XQ144=2×2×2×2×3×3ER12=2×2×3PD75=3×5×5XU99=3×3×11ES78=2×3×13QBL24=2×2×2×3YE184=2×2×2×23ET135=3×3×5QC95=5×19YL106=2×53ET9=3×3QE108=2×2×3×3×3YL144=2×2×2×2×3×3ET54=2×3×3×3QE68=2×2×17YM6=2×3ET31 No factorsQR132=2×2×3×11YRO12=2×2×3HE245=5×7×7QTO210=2×3×5×7ZE6=2×3HOBE66=2×3×11QX198=2×3×3×11ZH183=3×61
Out of one hundred and one recurring pairs we have fifty with the factors 2×3=6; out of twelve recurring triplets, nine have these factors; and the four recurring groups of four or more letters all have these factors. The percentages are respectively 49.5%, 75% and 100% and we may be certain from this that six alphabets were used. But, before the six frequency tables are made up, there is one more point to be considered; why are there so many recurring groups which do not have six as a factor? The answer is that one or more of the alphabets is repeated in each cycle; that is, a key word of the formHAVANAhas been used. If this were the key word, the second, fourth and sixth alphabets would be the same. We will see later that in this example the second and sixth alphabets are the same and this introduces the great number of recurring groups without the factor 6.
We will now proceed to make a frequency table for each alphabet. As the message is written in thirty columns, we take the first, seventh, thirteenth, etc., as constituting the first alphabet; the second, eighth, fourteenth, etc., as constituting the second alphabet and so on. The prefix and suffix letter is noted for each occurrence of each letter. The importance of this will be appreciated when the form of the frequency tables is examined. None bears any resemblance to the normal frequency table except that each is evidently a mixed up alphabet. Thenumbers after “Prefix” and “Suffix” refer to the alphabet to which these belong, for convenience in future reference.