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### hw2soln

Course: M E 435, Fall 2008
School: Wisconsin
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Word Count: 1639

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435 ECE/CS/MATH 1) Homework 2 Solutions Consider repeated affine encryptions. Let the first encryption be e1(x)=ax+b and the second encryption be e2(x)=cx+d where a and c are in Z26* and b and d are in Z26. We first apply e1, then e2. A typical letter x is encrypted to e2 (e1 ( x)) = c(ax + b) + d = cax + cb + d = cax + (cb + d ) Now, note that since a and c are in Z26* their product ac is as well. Further, cb+d...

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435 ECE/CS/MATH 1) Homework 2 Solutions Consider repeated affine encryptions. Let the first encryption be e1(x)=ax+b and the second encryption be e2(x)=cx+d where a and c are in Z26* and b and d are in Z26. We first apply e1, then e2. A typical letter x is encrypted to e2 (e1 ( x)) = c(ax + b) + d = cax + cb + d = cax + (cb + d ) Now, note that since a and c are in Z26* their product ac is as well. Further, cb+d is some element of Z26. Overall, then, the repeated encryption is simply equal to another affine encryption, namely eequivalent ( x) = (ca) x + (cb + d ) = rx + s where r is in Z26* and s is in Z26. There is no advantage (with respect to decryption difficulty) to applying repeated affine encryptions. Now we consider repeated Hill ciphers. A valid Hill cipher encrypts a block of plaintext into a block of ciphertext by the relation Ap=c where A is the encryption matrix with determinant in ZN*, p is a vector representing the plaintext block and c is a vector representing the ciphertext block. N is the number of symbols in the symbol alphabet. First, we consider the case where the block size is the same. Let the first encryption be e1(v)=A1v and the second be e2(v)=A2v. We first apply e1, then e2 as above. A typical block v is encrypted as e2 (e1 (v)) = A2 ( A1v) = ( A2 A1 )v Clearly this is equivalent to another cipher on the same block size. The encrypting matrix is A1A2. The question remains, however, whether the determinant of this new encrypting matrix is in ZN*. A fundamental result from linear algebra tells us that if A1 and A2 are square matrices of the same size (they are) then det(A1A2)=det(A1)det(A2). Since det(A1) and det(A1) are in ZN* so is their product, and the resulting encryption is equivalent to a single Hill cipher of the same size as A1 or A2. Now we generalize the result to the case where the block size is different. Suppose the first encryption has block size m and the second has block size n. I claim that the equivalent system has block size lcm(m,n) where lcm denotes the least common multiple. The argument for this is similar to the argument made for repeated Vigenere encryptions except now the procedure is multiplication by a matrix instead of addition by a keyword. For example, suppose we have m=2 and n=3, and the encryption matrices are as follows a b Am = c d Notice that these are equivalent to a b c d 0 0 Am ' = 0 0 0 0 0 0 00 00 ab cd 0 0 0 0 0 0 0 0 0 0 0 0 a b c d e h k An ' = 0 0 0 f i l 0 0 0 g0 j0 m0 0e 0h 0k 0 0 0 f i l 0 0 0 g j m e An = h k f i l g j m And again employing some matrix algebra we note that since the determinants of Am and An are values such that the matrices are invertible (i.e. the determinants are relatively prime to the modulus of the symbol alphabet, and thus invertible modulo the modulus) then the block matrices Am and An are also invertible. By the above argument for square matrices, we see the system is equivalent to another Hill cipher with block length lcm(m,n). 2) Rather than test all possible keys we try to find a smarter approach. Lets tabulate the letter frequencies: ABCDEFGH I JKL MNOPQRSTUVWXYZ 2 0 2 2 064 3030 0 1 0 0 1 2 0 110 0 0 010 We expect the most frequent letters to be E and T. Suppose e(E)=F and e(T)=G. We get a system of two equations and two unknowns: a(4)+b=5 and a(19)+b=6. Solving, we find that a(15)=1. Thus, a=7 and b=3 is the most likely key. Trying this (see posted m file) we find the plain text is MEET ME AFTER MIDNIGHT IN THE ALLEY where the spaces were inserted for readability. 3) Solving the cryptoquote should be easier than solving the cipher in problem 2 since we benefit from the spaces between the words. We again tabulate the letter frequencies: ABCDEFGH I JKL MNOPQRSTUVWXYZ 1 0 4 9 105 3037 2 0 4 4 4 1 0 030 7 1 202 From this, we might assume that E encrypted to D since D is most frequent. Continuing in this fashion (with some occasional mistakes) we decrypt the quote to a limited fortune is no excuse for deficiency in neatness. - charlotte *ilman where the * is still unknown. Since I did not recognize the authors name, I created a table to indicate the reverse mapping and which letters are unused. ABCDEFGH I J KL MNOPQRSTUVWXYZ SEH TF RIM OCAX L NYU D Unused BGJ KLPQVWZ Performing Google a search of Charlotte *ilman with the * replaced by each of the unused letters yielded one hit for Charlotte Wilman and numerous hits for Charlotte Gilman (a poet). Concluding that the author is likely the poet, we claim that the quote is A limited fortune is no excuse for deficiency in neatness. - Charlotte Gilman And the reverse mapping is ABCDEFGH I J KL MNOPQRSTUVWXYZ G SEH TF RIM OCAX L NYU D 4) In order to reduce the amount of work needed on this problem, we recall from problem 1 that for square matrices A1 and A2, det(A1A2)=det(A1)det(A2). We know that the Hill matrix must have determinant in Z26* so if the encryption is done on a letter matrix with determinant in Z26*, the ciphertext matrix must also have determinant in Z26*. This allows us to discard some possibilities immediately. We tabulate the digram frequencies LM = 3 QE = 1 TX = 4 YE = 1 AG = 2 CT = 1 UI = 1 EW = 2 NC = 1 LZ = 1 UA = 1 IS = 1 PZ = 1 YV = 2 AP = 2 (note here that YVAP appears twicewe might be able to use this info) GQ = 1 WY = 1 AX = 2 FT = 1 CJ = 1 MS = 1 QC = 1 AD = 1 AG = 1 DX = 1 NX = 1 SN = 1 PJ = 1 QS = 2 RI = 1 MH = 1 NO = 1 CV = 1 FV = 1 Rather than try combinations, I resorted to an exhaustive search (using Matlab and some good guessing) to find THE KING WAS IN HIS COUNTING HOUSE COUNTING OUT HIS MONEY THE QUEEN WAS IN THE PARLOUR EATING BREAD AND HONEY Z The spaces were inserted for readability. Notice the dummy Z at the end of the phrase to create an even length for encryption. The encrypting matrix was found to be 4 13 H encryption = 11 9 And the corresponding decrypting matrix is 23 13 H decryption = 21 16 Notice that the ciphertext YVAP corresponds to plaintext TING a hard (albeit unlikely) guess to make a priori. See the posted m file for my code for decryption. I hypothesized that the first word of the plaintext was THE and the program outputs only those decryptions matching the criteria. A more robust algorithm is shown in the commented-out code, where I search for the occurrence of THE and AND anywhere in the supposed plaintext. 5) There are several ways to approach this problem. Two solutions will be presented here. Vector Approach Let P=(p0, p1, , p25) represent a vector in 26-dimensional space (actually in the hypercube [0,1]26) . The coordinate pi is simply the probabilities of occurrence of the ith letter in plaintext. For sufficiently long texts we can suppose that the letter probabilities approach the empirically observed English letter frequencies. Thus the entries of Q should be close in value to ...

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