The following was enciphered with a Vigenere cipher. Find the key and decipher i
ID: 3624795 • Letter: T
Question
The following was enciphered with a Vigenere cipher. Find the key and decipher it. Code to compute Index of Coincidence (IC). You may work on part of it by hand (like which IC to choose). You may reuse work from question 1 here once you have found the period and split the alphabets accordingly. Show your work in detail.TSMVM MPPCW CZUGX HPECP RFAUE IOBQW PPIMS FXIPC TSQPK SZNUL OPACR DDPKT SLVFW ELTKR GHIZS FNIDF ARMUE NOSKR GDIPH WSGVL EDMCM SMWKP IYOJS TLVFA HPBJI RAQIW HLDGA IYOUX (spaces don’t mean beginning or end of word)
Explanation / Answer
8. First, we test to be sure this is a Vigenère cipher. We compute the frequency counts of each letter, and from that get the index of coincidence (IC), which is 0.040. Using Figure 9–4, this indicates a polyalphabetic cipher, so we are justified in assuming a Vigenère cipher. We next look for repetitions. The following table summarizes them: repetition begins at ends at interval length factorization of interval length ts 0 40 40 2 2 2 5 ts 0 59 59 59 sm 1 105 104 2 2 2 13 pp 6 30 24 2 2 2 3 pc 7 38 31 31 hp 15 120 105 3 5 7 fa 21 79 58 2 29 fa 21 118 97 97 ue 23 83 60 2 2 3 5 pi 31 109 78 2 3 13 ms 33 104 71 71 sf 34 74 40 2 2 2 5 ip 37 92 55 5 11 ts 40 59 19 19 pk 43 57 14 2 7 lvf 61 116 55 5 11 krg 68 88 20 2 2 5 fa 79 118 39 3 13 iyo 110 135 25 5 5 Of these, the interval lengths have the factor 2 appear 9 times, 3 appears 5 times, 4 appeears 6 times, 5 appears 7 times, 6 appears 3 times, 7 appears 2 times, and 8 and 10 appear 4 times each. Given the IC indicates a key length of greater than 2, we will first try a key of length 5, as that is the most frequent factor greater than 2. Splitting the ciphertext into 5 parts, we compute the IC for each alphabet. They are: alphabet #1: 0.042 alphabet #3: 0.050 alphabet #5: 0.48 alphabet #2: 0.066 alphabet #4: 0.066 Of these, alphabets 2 and 4 have an IC indicating a key length of 1, alphabets 3 and 5 have an IC indicating a key length of between 2 and 3, and alphabet 1 has an IC indicating a key length of between 5 and 10. Given the short ciphertext, we will accept that our hypothesis of a key length of 5 is worth pursuing, and proceed accordingly. We next lay out the frequencies for each alphabet: a b c d e f g h i j k l mn o p q r s t u v wx y z alphabet #1: 1 0 1 1 2 2 2 3 3 0 0 0 1 1 1 1 0 2 3 3 0 0 1 0 0 0 alphabet #2: 1 0 0 3 0 1 0 1 0 0 0 4 1 1 2 5 0 1 3 0 0 0 0 1 2 2 alphabet #3: 2 2 0 1 1 0 1 0 5 0 0 0 3 1 2 2 2 0 1 1 1 2 1 0 0 0 alphabet #4: 0 0 4 1 0 2 2 0 1 2 4 0 1 0 0 3 1 0 0 0 4 2 0 0 0 1 alphabet #5: 2 0 1 0 2 1 0 1 1 0 1 2 2 0 0 2 0 3 3 1 0 0 4 2 0 0 First consider the second alphabet. The frequencies in the middle (4 1 1 2 5 0 1 3), representing counts for the letters l through s, are similar to the frequency counts expected at the beginning of the unshifted alphabet. So, let us assume that the second alphabet maps A into L. In the fourth alphabet, the frequencies from w to z, and a and b, are 0 0 0 1 0 0, and are similar to the frequency counts expected at the end of the unshifted alphabet. So, let us assume the fourth alphabet maps A into C. For the first alphabet, the frequencies from the letter on match the frequency counts expected at the end of the unshifted alphabet, so let us assume the first alphabet maps A into A. We now substitute back into the ciphertext. As in the book, the bold letters are plaintext and the unbolded letters are the ciphertext: THMTM MEPAW COUEX HEEAP RUASE IDBOW PEIKS FMINC THQNK SONSL OEAAR DSPIT SAVDW EATIR GWIXS FCIBF AGMSE NDSIR GSINH WHGTL ESMAM SBWIP INOHS TAVDA HEBHI RPQGW HADEA INOSX The first word is either “THE” or “THAT”. But if it’s “THAT”, then the third alphabet is unshifted, and the frequency counts do not match those of the unshifted alphabet. If it’s “THE”, on the other hand, then E maps into M, meaning A maps into I, and the frequency counts are closer to those of the unshifted alphabet than if the first word were “THAT”. Adopting as a working hypothesis that the third alphabet maps A into I, and updating the text as before, we have: THETM MEHAW COMEX HEWAP RUSSE IDTOW PEAKS FMANC THINK SOFSL OESAR DSHIT SANDW EALIR GWAXS FCABF AGESE NDKIR GSANH WHYTL ESEAM SBOIP INGHS TANDA HETHI RPIGW HAVEA INGSX At this point, the cipher falls apart. The “W” at the end of the second block is obviously “S”, so the fifth alphabet maps A into E. The plaintext is: THETI MEHAS COMET HEWAL RUSSA IDTOS PEAKO FMANY THING SOFSH OESAN DSHIP SANDS EALIN GWAXO FCABB AGESA NDKIG GSAND WHYTH ESEAI SBOIL INGHO TANDW HETHE RPIGS HAVEW INGST So, the key is ALICE, and the plaintext is: ‘The time has come,’ the Walrus said, ‘To speak of many things: Of shoes—and ships—and sealing wax— Of cabbages—and kings— And why the sea is boiling hot— And whether pigs have wings.’ This is a part of a poem from Through the Looking Glass (and yes, I know it should be “talk”, not “speak”, in the second line!)
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