Question 1. Decrypt the following cryptotexts: (a) The given ciphered text is encoded using Bacon's cipher. From AAAAB AAAAA AAABA ABBBA ABBAB AAABA ABAAA ABBBB AABBB AABAA BAAAB, we obtain BACONCIPHER (b) The given ciphered text is a anagram. There are more possible ways how to decode the given text. For example LESTER SANDERS HILL orHANDLIST RESELLERS. (c) The given ciphered word is encoded using Four Square cipher. We obtain ITWASINVEMTEDBYFELIXDELASTELLE after decoding the given ciphered word using the Four Square decoding algorithm with keys being four and square. Question 2. (a) ('-.19(A) = {A.l + 19) mod 26 = (0.7 + 19) mod 26 = 19 mod 26 = 19 = T C7,i9(F) = (F.7+ 19) mod 26 = (5.7 + 19) mod 26 = 54 mod 26 = 2 = C e7jl9(7) = (1.7 + 19) mod 26 = (8.7 + 19) mod 26 = 75 mod 26 = 23 = X e7.i9(JV) = (JV.7 + 19) mod 26 = (13.7 + 19) mod 26 = 110 mod 26 = 6 = G e7A9(E) = (E.7 + 19) mod 26 = (4.7 + 19) mod 26 = 47 mod 26 = 21 = V Ciphertexxt - TCCXGV (b) 7"1 mod 26 = 15 d7,ig = (15.(1? - 19)) mod 26 = (15.(7- - 19)) mod 26 = = 2 = C d7,ig = (15.(T - 19)) mod 26 = (15.(19 - 19)) mod 26 = 0 = ■A d7,19 = (15.(V - 19)) mod 26 = (15.(21 - 19)) mod 26 = 4 = --E d7,,y = (15.(P - 19)) mod 26 = (15.(15 - 19)) mod 26 = 18 = S dr,i9 = (15.(1 - - 19)) mod 26 = = (15.(8- 19)) mod 26 = 17 = R Plaintext - CAESAR Question 3. (3 points) What is the number of possible keys and the unicity distance of an affine cipher if the following modulus is used: For a modulus of 26, the number of possible keys is n = (^(26) -26 = 12- 26 = 312. where y is the Eulcr's totient function, as we can have ^(26) different values of a co-prime to 26 and 26 different values of b. 1. 30 n = p(30) • 30 = 8 • 30 = 240 U = = |"!2§f2l = [2.471 = 3 The number of possible keys is 240, the unicity distance is 3. 2. 31 w = p(31)-31 = 30-31 = 930 U = ft; = = [3.081 = 4 The number of possible keys is 930, the unicity distance is 4. Question 4. (a) i. Since the keyspace is basically the set of possible values of a, it always holds that \K\ = n — 1, because 0 K. That implies that \K\ < \M\, which means that this encryption function is not perfectly secure. It's also obvious that Pr[P — 0|C = 0] = 1 (zero always gets encrypted as zero) and assuming that n > 1, Pr[P = 0] ^ 1, therefore Pr[P = Q\C = 0] £ Pr[P = Q\. ii. Since the keys (b) are uniformly distributed and \K\ = n, Pr[K = b] = ^. Pr\C = c] = A, because c = i + &(modn), x and b are independent and b is uniformly distributed: E P[X = c-b(modn)] = E P[X = x] = l. Pr\C = c\P = x] = Pr[K = c-x (modn)] = Pr[P = x\C = c] = Pr[P^]'n = pr[p = x] This means that it is a perfectly secure cryptosystem, iii. Here the key consists of a and b (k =), therefore \K\ = n{n — 1) and assuming the keys are used with equal probability, Pr[K — k] — n^-i) • Since n is a prime and a,b < n, it holds for every a that ;jc n (and the encryption functions must be chosen uniformly), so there would be equal probability that any plaintext message p will be encrypted to any c, in other words, we need enough encryption functions so that any plaintext message p can get encrypted to any ciphertext message c. If we had less than n encryption functions, some of the ciphertext messages c would occur with higher probability. Question 5. (a) We have : C] = Wl © U)2 © «»3 © k, C2 = U>2 © IU3 © fc, c3 = tug © fc Taking the XOR. of cj and C2 we get cj © C2 = wi © tt>2 © *"3 © k © W2 © W3 © fc = w\. Also taking the XOR of C2 and C3 we get C2 © C3 = W2 © W3 © k © w-j © k = w2. Therefore Bob can recover the messages in the following way: '«'1 = ci © C2, w2 = c2 © C3, W3 = C3 © k (b) Malicious Kve can discover the length n of the messages and of the key respectively. She can also recover the binary messages w\ and u>2 using the above mentioned procedure. Question 6. UACVL GVVVUN YEZKJ MEXNG GMEEY VHCF.T Nl BPT NIPMI NKMOX HBSl'Y XTBRF XZFWM CI.PAB XJKOX YICKT HFEYX OGTSC MUKMQ ETNXS HU'VH TWULB SICYG YHJMVV ATOCW ZVVVA HCVUS XXTAE CLTTZ HINTS YMPME MNMVA AGIOY WFBT.I IIII1KL ONSEZ RVXRX KLEXN Ml'SSD CVETM QUCGT CQTPJ QNRLO SIGIS EEVPC \ 1 [\\ 1' NKYTR NHEMS XEPAT EXRBN 1 BI'Pl GIC'YD ZIIXH IGI.GK ICCI 11 i IHTSI' YPNCE BPTTR PEBPA YMUW1 LCPPP \BS.IW LLRXV LPDJS QGAYT W.1tce EBLLD ELJAP X 111 X 1! R\ 1 XAWSU VAEEO OA DTK SUMZF EEVBO IDSPL 1GGIN RPMAO UWYJM PI1LFK ITAII. RYUXH XNFPP WAEOM NIIFPC CYDVT GGDXX YULIT NBLXY XINMN HIMHM PTVGJ GIJUF TZEEY JAYOI ESBPC VWTI RZ1UP HBCIG BTNDF BZWVZ FLEHV SWMBK MEPMA ZEPPP mini:-, otyev UCTFZ GKTAE RFBXZ TSUOI GAMMP CLVKM SETSF XB1IOL IHFMH FPTGV SCjLlEW ZXJMI1 RGYDE II.HFX KSXXB GTNV'II UF.JDZ WEHVX VZGKS BNDYH EGUOX OIVHG BLSSI GGKGO IFTXVV VV1SOI GLZBL DRLCU TGYXl' XGZET GKTAE SYTSR EBEZF KLCWR RKLGB RKUTM UAUBB APKVI SYDTF JVAEP KMqCO YHVMS YTPSI" NTZGG RGKEV IQBPR PAJXP SSYVA TTKI.G WILCP DTBUA in iii.ii RFXMY WVKXD IIVRJX PNBMC VHBFG I.TRBY XIII.SK LGDAL III' 11 TGFZI. IMOGH ZTUMP XEHRX VAII.I RCETX ROUEI BMFUY WRVCG 1 U IP Willi RAIl.W PLVSV OSFDl WXJBS IEKVVI CHDFV MHTTI [T)CFR XEUIR HBBVW ILDJM XOMIT III IPG CFEMO XGGSB VFHZT KIDBH ICI.GF. D1IVHJ BSVRZ PLTI.F. DI.PTN ITETT VWVTN ICXQI VVUK BSAIO AARFN TDFSK SPXMH MIGIJI' I Mil I DHNFM SEFVA CDQLM V'XPHS GTXTM HVMAR PATNS ETEUX CEROK GGTCI VBNNE F.CYTF YRG.IB EOEEC RVVXXU WZMGP MGRHK SPEYB VWKGT AENYL MDATC DKZET JEAEE NRTBC VKMIG GBHOT WBBVM AXIUV TZVLK LCKAD GVTLA OXYOK RFYIA UTXXO IFR.1B SWITW BSMHB NTTXE TCMLV X.JTTU FZKWI CGI 111' TAEOC EWFYN CPVVQ GGTL1 X.1XTP EONTF. This is KASISKI method: substring distances of occurences distance TAE gcd([528, 638, 676. 803, 836]) - 1 IGG gcd([22, 41, 561, 792]) = 1 AIL gcd([22, 649, 924]) = 11 PPR gcd([22, 550, 770]) = 22 KLG gcd([352, 693, 759|) = 11 UBP gcd([154, 297, 836]) = 11 GCD of found distances is 11, I'm going to try this length and compute corresponding indices of correspondence: # This is algorithm how I obtain 11 columns below for column in range(0, 11): print( column ) for i in range(len( cryptotext ) ): if i '/, 11 == column: print(cryptotext[i]) 1. UVMMVHOXLLPLXESXLLMBXAXUBABPXXKDBBTBNVBNXGBLGHGJGLVVGTEWTUKV AGMALBLBZLHYBNAXEWVYKBHMAEZKRHUETULMGGWAPPABEZYHTLXGTTNPMGH 2. AEAHIBENOSADNYENHCONTEXAVESANRIARNNNBRCGIUFEGFGUOIBTSYWGLAHE LOIASCCEIRHSDERAIUTEMNHEEEUISRLNBWIIGISEOAETOTHDCISAPTEEUGO 3. CTOFQBEGPQJJUVEFFPMUIOBHFNUTUFDTFUBDMZCINOGDBEUFXTBSC.IIFJDB SZGOOIPITTF.I.IFSOOTSFBFFBCBPTUTJ.IEBPVUOUSSTOJOSVFIOUMSIEOSIL 4. VMUPBVCCHUNMWHVPXPCBWMGYHYYCFXBCXBLYC.ILCMXLLHYBBHEVCYASZWEP BBHAXCPGNDYMWBYUUWLLEUNNLLPMCCNAYALMBYUYFNCHAUMVCNMMUUCNSN 5. LCJWTPWYQLEPNZCPPTPRPPNTDZLNY.JMHDZPXHDEPYNOTPONPLLTMMPYOLTTT PLZAYYPTEFIWLZTEEZPEZYSTTLPTDMTPOPVAPJCDOSEHDOSCPSZPNDYTDS 6. GUYGRYETFWSOMECRTRHTNHNETNHDDYIKFPYHHAADHIRTTITSSTAUNOIIZNT CDTROGRGHSAALWSIIVDHFWETTDRAQLSNTKSRMETTIEWITIIEVEFCTCTEIE 7. VC.IVNRZPKZSMGTRGKPKNKFVOVKEVZWCZWIVVVEBZIVBNYGVSKVXKCIUCVXR VRWFKYZYRKDTRVRMFVJVKRFXZMKELVYTYVVFNRFFWTFKRKERVZELZFFKDT 8. WGMREGKJIXYIPSYYLGSIYPHXWMMTWVLEMYZRHEXIMHYVILGILWIMEIFXLSP VLMNTMIXXSTOXZRIRVSSLVVEWELIMXMIEIGZMVZJXEYLRMKOQREVGRXLSS 9. UTUGUJJQTJVTGIUDGCPPTCDGKOSGEKGTEUGJJNJIHGPKCGJGGVUQTFCQKHE VCPGWUUUVPNCVGCGJAQWGCAXJPGPVJPFVRTKVJGVJUNVFQWKGVVKGVJEPF 10. NRHKIBMNAMARMGXEWFEMRKTGGXNGHXEXXLKXBRKXMVLMKKGGDTVGZTUILLB TUBVUWPNAXXWBKEQBHGMVGBXTMBXXTMXUAXWAXKABXWABGIGGXBMMBXXLX 11. CPFEREERIHIWEIHIIEYIWSBSTHEDVDMTRISPSTOHEHTITIIKANOHENBUCUP IISTLIIIGIMOATTTPSCABRDSTCARMPTEMAITIAATESTFENOCGTROYHNTNIB Indices of correspondence are high, thus the key length is 11. All columns are encoded with Caesar cipher. We can use frequency analysis, where "E" most used letter in English and Vigenere table to decode. If "E" doesn't make sense, try other widely used letter. Collunm Index of correspondence 2 3 4 5 6 7 8 9 10 11 0.05708 0.06244 0.06316 0.06278 0.07574 0.06248 0.07118 0.06057 0.07515 0.07324 0.06823 Collunm Most frequent letters Corresponding row in Vigenere table 2 3 4 5 6 7 8 9 10 11 B,X.L E Y,T,H A B U,X F Y, C P G X I, E, T I, E, T V E,A,P R E C T E,A,P BABBAGES SUCCESSFUL CRYPTANALYSIS OF THE VIGENERE CIPHER WAS PROBABLY ... 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