* Click on the problem number to see the answer*

*This exam is due on Wednesday, March 15th. The exam totals 100 points. Please show all work as answers alone are not sufficient. All work submitted must be your own.*

ibaid oldwf muneg ueaiw atycr ganth ymdea hrdii nyess tirsc ogaos aneth tadfa wgshi brlrs hipea edeoa eteom aeapn eucnp bnvey lptee lpeab biewr ersno ahdhf toodd ratee odlby rvhec gudus oeelt rcosa wawem oevhn tlhda wseed owevs fjrlt haeet lusso eylhe oshod lsbir wtofa placx eaedy iorlx rssei nvrex smpab ceoox

a) Calculate the Index of Coincidence and speculate on whether or not this might be a polyalphabetic cipher.

b) Supposing that the cipher is either a transposition or a monoalphabetic substitution (and I am not claiming that it is either of these), examine the frequency distribution and decide which of these two choices is the most likely.

c) Determine the plaintext if you know that the message starts with "I may not be ..." (a plaintext attack).

d) Explain how the ciphertext was formed.

**2**. Suppose that n = 215629 in the **RSA Cryptosystem**, and it has been revealed that (n) =
214684. Use this information to find the factorization of n by solving a quadratic equation.

**3**. Suppose the **Affine Cipher** is implemented in **Z**_{126}.

a) Determine the number of possible keys.

b) If the key is (23,7) (i.e., the encryption function is x (23x + 7) mod 126) determine the decryption function corresponding to this key.

**4**. An unknown multiplexed Feedback Shift Register system produces the following bit stream: 0 0 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 ... .

a) Find the linear equivalence of this FSR.

b) Find an a **LFSR** and a starting state for it which will produce the same repeating portion.

**5**. Here is an example of **RSA** ciphertext:

11437 6198 16611 2405 18636 2679 12205 24142 6375 2134 16611 2405 9529 7260 7834 15094 4667 24027 762 5878 5206 16683 5359 10888 4168 3536 23229 20351 15580 6704 7977 374 6525 4287 14402 527 12887 21628 11884 9402 15470 1339 10420 18051 23125 7747 135 22007 20049 9984 13199 15176 1379 8313 19574 7989 22869 406 10057 21758 3918 23991 14237 7989 3947 19529 15728 5601 3527 7200 7601 13282 21160 6291 15994 7785 8982 3045 6596 16796 4663 2405 20302 11929 17125 14533 21001 8351 11571 22082 11040 8687 6704 3330 5630 19650 13024The public modulus is n = 24637 and the public encryption exponent is

a) Factor n by trial and error.

b) Compute the private decryption exponent *d* using the extended Euclidean algorithm.

c) Decrypt the ciphertext, obtaining the plaintext as a sequence of elements of **Z**_{24637} . Use the square-and-multiply algorithm for modular exponentiation.

d) Translate the plaintext into English text, given that each element of **Z**_{24637} represents three alphabetic characters, as in the following examples:

CAT 2 × 26

ZZZ 25 × 26

a) The 260 letters of this encrypted message are distributed as follows: a(24), b(8), c(7), d(15), e(37), f(5), g(5), h(13), i(12), j(1), k(0), l(13), m(5), n(10), o(20), p(7), q(0), r(15), s(18), t(14), u(6), v(5), w(9), x(4), y(7), z(0). Using these numbers as the f_{i} and n = 260, we calculate the Index of Coincidence:

and obtain the value I = 0.060618... . As this is close to the value of 0.065, which is the Index for English, we should assume that the cipher is** not **polyalphabetical (which would give a value closer to 0.038.)

b) The frequency of vowels {a,e,i,o,u} is 38.1%, the frequency of {e, t, a, o, n} is 40.0%, and the frequency of {j, k, q, x, z} is 1.9%. The corresponding values for natural English are 38.6%, 45.1% and 1.1% respectively. This is fairly strong evidence that the cipher is a **transposition** rather than a monoalphabetical substitution.

c) The plaintext is from "The Diary of Samuel Marchbanks", by Robertson Davies, Clarke Irwin, 1947.

I may not be able to grow flowers, but my garden produces just as many dead leaves, old overshoes, pieces of rope, and bushels of dead grass as anybody's, and today I bought a wheelbarrow to help in clearing it up. I have always loved and respected the wheelbarrow. It is the one wheeled vehicle of which I am perfect master. xxxx

d)** The plaintext was written, by rows, into a 10 × 26 array and the last row was padded by 4 nulls (the x's). The ciphertext was obtained by reading out by columns, starting at the leftmost column.**

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Answer to Problem 2

In the RSA Cryptosystem, n = pq, the product of two primes. For this n, (n) = (p-1)(q-1). So, we have (n) = pq - p -q +1 and q = n/p. Thus, we may write
(n) = n -p -n/p + 1 and by multiplying through by p and putting all the terms on one side of the equation we get: p^{2} + ((n) -n -1)p + n = 0. By substituting the known values, we have: p^{2} -946p + 215629 = 0. This quadratic equation can be solved with the use of the quadratic formula to get the values** p = 383 or 563**. Since their product is 215629, these are the two values for p and q.

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Answer to Problem 3

a) The encryption rule for an affine cipher mod n is given by e(x) = ax + b (mod n). The keys of the system are the pairs (a,b) where b may be any integer in **Z**_{n} but a must be relatively prime to n. Therefore, the number of keys is n(n). Since n = 126 = 2(3^{2})(7), we have (126) = 126(1/2)(2/3)(6/7) = 36 (by the formula given in class), and so the number of keys is (126)(36) = **4536**.

b) With the encryption rule being e(x) = 23x + 7 (mod 126), to find the decryption rule we must solve the modular equation y = 23x + 7 (mod 126) for x. Thus, x = 23^{-1}(y - 7) (mod 126). To find 23^{-1} (mod 126) we can use the extended Euclidean algorithm:

23 = 2(11) + 1, t

11 = 11(1) + 0, t

t

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Answer to Problem 4

a) Since this sequence does not come from a *linear* feedback shift register, it will become periodic only after some initial (non-periodic) string. Indeed, by examining the sequence, we see that after the first two 0's, the string 1 0 0 0 1 0 1 starts to repeat. The repeating portion is given by the polynomial S^{(7)}(x) = 1 + x^{4} + x^{6}. To obtain the linear equivalence, we find the minimal characteristic polynomial, m(x), for this string. We know that the reciprocal polynomial m^{*}(x) = (1 + x^{7})/ gcd(S^{(7)}(x), 1 + x^{7}). To calculate the gcd, we use the Euclidean algorithm:

x

x

x

b) The LFSR with characteristic polynomial m(x) = x^{4} + x^{2} + x + 1 is:

.

And, with starting state **1 0 0 0** it will produce the repeating portion of the sequence. (Answers to the starting state may vary if you used a different repeating portion of the sequence, but the LFSR will be the same.)

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Answer to Problem 5

a) After a few minutes of trial and error factoring, the prime factors of 24637 are found to be **71** and **347**.

b) From a) we calculate that (24637) = (70)(346) = 24220. The decryption exponent *d* = 3^{-1}(mod 24220), and we can use the extended Euclidean algorithm to find it:

3 = 3(1) + 0, t

t

c) We need to take each number in the message and raise it to the 16147 power mod 24637. If doing this by hand, with a calculator or programming it yourself, you will need to use the square and multiply algorithm. Alternatively, you could use a symbolic algebra package such as Derive, Maple or Mathematica to do the calculations (the square and multiply algorithm is incorporated in these packages). The results are:

294 | 1467 | 5958 | 12389 | 8914 | 16605 | 6820 | 13468 | 8867 | 9501 |

5958 | 12389 | 7818 | 344 | 2994 | 12669 | 3328 | 11613 | 2760 | 9293 |

16310 | 14577 | 1459 | 341 | 1728 | 8645 | 5460 | 12970 | 4849 | 13030 |

16316 | 5902 | 12855 | 7813 | 8061 | 16722 | 7555 | 5109 | 3056 | 15373 |

3335 | 550 | 13211 | 2764 | 2154 | 5114 | 5116 | 7520 | 3198 | 8801 |

13068 | 13643 | 37 | 3205 | 513 | 4860 | 4753 | 2952 | 5765 | 8068 |

8974 | 11719 | 2801 | 4860 | 173 | 2827 | 9972 | 12636 | 1724 | 9270 |

8223 | 10036 | 7646 | 2717 | 2068 | 1008 | 12667 | 5749 | 572 | 16593 |

13226 | 12389 | 4557 | 2770 | 5768 | 5909 | 10003 | 12706 | 10246 | 13065 |

4160 | 1548 | 13030 | 11861 | 1538 | 502 | 8967 |

d) Translating these numbers back to triples of letters gives the following :

This quote is from Beutelspacher,Alice lives in New York City and Bob lives in Los Angeles. They are recently divorced and communicate, when they do it at all, only by telephone. Now, they have to decide who should get an antique table they have jointly inherited. They agree to toss a coin. Somehow, Alice and Bob mus find a way of tossing the coin without suspecting each other of cheating.x

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