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MRLAQ EDUEQ QWGKI LFMFE XZYXA QXGJH FMXKM QWRLA LGWCL SOLMX RLWPI OCVWL SKNIS IMFES JUVAR MFEXZ MJHTC RGRVM RLSZS MREFW XZGRY RLWPI OMYDB SFJCT AQ.
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To recover the key and decrypt the message, we can make use of the fact that the ciphertext is composed of three simple substitutions. To accomplish t,his, we tabulate the letter frequencies for the sets
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SO= {co,c:~, cg,.
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. . }, 51 = {Q, c 4 , ~ 7 , . . . }, and 52 =
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where c, is the ith ciphertext letter. Tables 1.4, 1.5, and 1.6, respectively.
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Doing so, we obtain the results in
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Table 1.4: Letter Frcquericics in S o Letter II R Q U K F E Y J M L G P C N I Z W B Freauencv110 4 3 1 2 3 2 3 3 4 2 2 4 1 I 1 1 1
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Table 1.5: Letter Frequencies in S 1 Letter IL v E w I M x Q K S H R Y C A Frequency 16 5 4 2 4 4 7 1 1 6 1 2 1 1 1 From the S ciphertext in Table 1.4, we might reasonably guess that o ciphertext R corresponds to plaintext E. T, N, 0, R, I. A or S. which gives 11s
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1.4 SELECTED CLASSIC C R Y P T 0 TOPICS
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Table 1.6: Letter Frequencies in
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W M A D Q G L F Z K C 0 X S J T Y Letter Frequency 6 4 5 2 1 3 3 5 4 2 1 3 1 2 1 2 1
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candidate values for ko, namely ko E {13,24,4,3,0,9,17,25}. Similarly, for 1 ciphertext X niight correspond to plaintext E, T, N, 0, R, I, A or S, set S , from which we obtain likely values for k l , and from set Sz, ciphertext W likely correspond to plaintext E, T, N, 0,R, I,A or S. The corresponding likely keyword letters are tabulated in Table 1.7. Table 1.7: Likely Keyword Letters
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ko N Y E D A J R Z
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k1 T E K J G P
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k2 S
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The conibinations of likely keyword letters in Table 1.7 yield 83 = 2 putative keywords. By testing each of these putative keyword on the first few letters of the ciphertext, we can easily determine which, if any, is the actual keyword. For this example, we find that ( k o ,k l , k2) = (24,4,18), which corresponds t o YES, and the original plaintext is
THE TRUTH IS ALWAYS SOMETHING THAT IS TOLD, NOT SOMETHING THAT IS KNOWN. IF THERE WERE NO SPEAKING OR WRITING, THERE WOULD BE NO TRUTH ABOUT ANYTHING. THERE WOULD ONLY BE WHAT IS.
This attack provides a significant shortcut, as conipared t o trying all possible 263 M 214 keywords. Knowing the length of the keyword used in a Vigenkre cipher helps greatly in the cryptanalysis. If the keyword is known, and the message is long enough, we can simply perform letter frequency counts on the associated sets of ciphertext t o begin solving for the plaintext. However, it is not so obvious how to determine the length of an unknown keyword. Next, we consider two methods for approximating the length of the keyword in a Vigenhre cipher.
CLASSICCZPHER,S
Friederich W. Kasiski (1805-1881) was a major in the East Prussian infantry regiment and the author of the cryptologic text Die Geheimschriflen und die Dechiger-kunst. Kasiski developed a test (amazingly, known as the Kasiski Test), that can sonietimes be used to find the length of a keyword used in a cipher such as the Vigenkre. It relies on the occasional coincidental alignment of letter groups in plaintext with the keyword. To attack a periodic cipher using the Kasiski Test, we find repeated letter groups in the ciphertext arid tabulate the separations between them. The greatest common divisor of these separations (or a divisor of it) gives a possible length for the keyword. For example, suppose we encrypt the plaintext
THECHILDISFATHEROFTHEMAN
with a Vigenkre cipher using the keyword POETRY.The resulting ciphertest is
IVIVYGARMLMYIVIKFDIVIFRL.
Notice that the second Occurrence of the ciphertext letters IVI begins exactly 12 letters after the first, and the third occurrence of IVI occurs exactly six letters after the second. Therefore, it is likely that the length of the keyword is a divisor of six. In this case, the keyword length is exactly six.