课程性质︰数学系选修 课程教师︰陈君明 开课学院：理学院 开课系所︰数学系 考试日期（年月日）︰2018 年 5 月 8 日 考试时限（分钟）：15:30 ~ 18:20 (不延长) 注意事项：考试时会发一张标准期考作答卷当作计算纸，答案则是填在题本的最後一页。 试题 : Cryptography Midterm Exam 2018/05/08 Part I (3 points each) 1. Which can NOT be the number of elements of a Galois field? A. 21 B. 23 C. 25 D. 27 E. None of the above 2. How many irreducible polynomials of degree 9 over GF_2? Hint: Factor x^(2^9) - x. A. 53 B. 55 C. 57 D. 59 E. None of the above 3. Suppose α ∈ GF_8 is a root of x^3 + x + 1. Whose minimal polynomial is x^3 + x^2 + 1? A. α^2 B. α^3 + α^2 C. α^4 D. α^4 + α^2 E. None of the above 4. Which group is isomorphic to the Klein-4 group (Z_8*, *)? A. (Z_4, +) B. (Z_5*, *) C. (Z_10*, *) D. (Z_12*, *) E. None of the above 5. Which ideal is NOT the same as the principal ideal <6> in Z? A. <-6, 42> B. <36, 48> C. <18, -24> D. <30, 60, 72> E. None of the above 6. Which is NOT a finalist of the AES (Advanced Encryption Standard) competition? A. Rijndael B. Serpent C. Twofish D. RC4 E. None of the above 7. Which mode of operation has no error propagation? That is, a transmission error of one bit causes a decryption error of only one bit. A. ECB B. CBC C. OFB D. CFB E. None of the above 8. Let mi's and ci's be plaintext and ciphertext blocks respectively. With a decryption algorithm d and a key k, which is the decryption operation of CBC mode for i > 1? A. mi = dk(ci ⊕ m{i-1}) B. mi = dk(ci) ⊕ c{i-1} C. mi = dk(ci ⊕ c{i-1}) D. mi = dk(ci) ⊕ m{i-1} E. None of the above 9. Which can NOT replace the square matrix in the affine transformation constructing the S-box of AES to become a new block cipher? A. [10001110] B. [11010110] C. [10001001] D. [01010010] [01000111] [01101011] [11000100] [00101001] [10100011] [10110101] [01100010] [10010100] [11010001] [11011010] [00110001] [01001010] [11101000] [01101101] [10011000] [00100101] [01110100] [10110110] [01001100] [10010010] [00111010] [01011011] [00100110] [01001001] [00011101] [10101101] [00010011] [10100100] E. None of the above 10. In a Feistel cipher, every encryption round consists of Li = R{i-1} and A. Ri = R{i-1} ⊕ f(L{i-1}, ki) B. Ri = R{i-1} ⊕ f(R{i-1}, ki) C. Ri = L{i-1} ⊕ f(R{i-1}, ki) D. Ri = L{i-1} ⊕ f(L{i-1}, ki) E. None of the above Part II (3 points each) * Euclidean Algorithm * GCD(80837, 88603) = ___. * a = ___ and b = ___ is the pair of integers satisfying 17 a + 43 b = 1 where a is the least positive one. * The least positive solution to the equation 85 x ≡ 10 (mod 43) is ___. * Consider the multiplicative group G = (Z_15*, * mod 15). * The order of the group, |G| = ___. * G has 4 elements of order 4: 2, 8, 13, and ___. * G has 3 subgroups with 2 elements: {1, 4}, {1, 14}, and ___. * G has 3 subgroups with 4 elements: {1, 2, 4, 8}, {1, 4, 11, 14}, and ___. * Consider the quotient ring Rm = GF_5[x] / <x^2 + 4x + m> where m ∈ GF_5. * For any specific value of m, the number of the elements in Rm is ___. * Rm is a field for m = 2 and m = ___. * The multiplicative inverse of x + 2 in R_2 (for m = 2) is ___. * To show that x^2 + 4x + 2 is a primitive polynomial over GF_5, it suffices to verify x^8≠1 and x^n≠1 in R^2 where n = ___. * Complete the table for DES, Triple DES (3DES), and AES: _______________________________________________________________ | | | | DES | Triple DES | AES | | Key Length (bits) | 56 | 112 | 168 | 128 192 256 | | Block Length (bits) | ????? | 128 | | Number of Rounds | 16 | 48 | ??? 12 14 | | Number of Different S-box(s) | 8 | 1 | |_______________________________________________________________| * A simple PRNG, Linear Congruential Generator, is generated by a recursive formula S{i+1} = A * S{i} + B mod m, where A, B, and the seed S{0} are kept secret. Suppose m = 31, and the first three outputs S{1} = 18, S{2} = 4, and S{3} = 17 are obtained, then S{4} = ___ and S{5} = ___. All the following output Si's are predictable, which's very bad for cryptographic applications. * The following reference code comes from the book "The Design of Rijndael" written by J. Daemen and V. Rijmen: typedef unsigned char word8; word8 Logtable[256] = { 0, 0, 25, 1, 50, 2, 26,198, 75,199, 27,104, 51,238,223, 3,100, 4,224, 14, 52,141,129,239, 76,113, 8,200,248,105, 28,193,125,194, 29,181,249,185, 39,106, 77,228,166,114,154,201, 9,120,101, 47,138, 5, 33, 15,225, 36, 18,240,130, 69, 53,147,218,142,150,143,219,189, 54,208,206,148, 19, 92,210,241, 64, 70,131, 56, 102,221,253, 48,191, 6,139, 98,179, 37,226,152, 34,136,145, 16,126,110, 72,195, 163,182, 30, 66, 58,107, 40, 84,250,133, 61,186, 43,121, 10, 21,155,159, 94,202, 78,212,172,229,243,115,167, 87,175, 88,168, 80,244,234,214,116, 79,174,233,213, 231,230,173,232, 44,215,117,122,235, 22, 11,245, 89,203, 95,176,156,169, 81,160, 127, 12,246,111, 23,196, 73,236,216, 67, 31, 45,164,118,123,183,204,187, 62, 90, 251, 96,177,134, 59, 82,161,108,170, 85, 41,157,151,178,135,144, 97,190,220,252, 188,149,207,205, 55, 63, 91,209, 83, 57,132, 60, 65,162,109, 71, 20, 42,158, 93, 86,242,211,171, 68, 17,146,217, 35, 32, 46,137,180,124,184, 38,119,153,227,165, 103, 74,237,222,197, 49,254, 24, 13, 99,140,128,192,247,112, 7}; word8 Alogtable[256] = { 1, 3, 5, 15, 17, 51, 85,255, 26, 46,114,150,161,248, 19, 53, 95,225, 56, 72, 216,115,149,164,247, 2, 6, 10, 30, 34,102,170,229, 52, 92,228, 55, 89,235, 38, 106,190,217,112,144,171,230, 49, 83,245, 4, 12, 20, 60, 68,204, 79,209,104,184, 211,110,178,205, 76,212,103,169,224, 59, 77,215, 98,166,241, 8, 24, 40,120,136, 131,158,185,208,107,189,220,127,129,152,179,206, 73,219,118,154,181,196, 87,249, 16, 48, 80,240, 11, 29, 39,105,187,214, 97,163,254, 25, 43,125,135,146,173,236, 47,113,147,174,233, 32, 96,160,251, 22, 58, 78,210,109,183,194, 93,231, 50, 86, 250, 21, 63, 65,195, 94,226, 61, 71,201, 64,192, 91,237, 44,116,156,191,218,117, 159,186,213,100,172,239, 42,126,130,157,188,223,122,142,137,128,155,182,193, 88, 232, 35,101,175,234, 37,111,177,200, 67,197, 84,252, 31, 33, 99,165,244, 7, 9, 27, 45,119,153,176,203, 70,202, 69,207, 74,222,121,139,134,145,168,227, 62, 66, 198, 81,243, 14, 18, 54, 90,238, 41,123,141,140,143,138,133,148,167,242, 13, 23, 57, 75,221,124,132,151,162,253, 28, 36,108,180,199, 82,246, 1}; /* The tables Logtable and Alogtable are used to perform multiplications in GF(256) */ word8 mul(word8 a, word8 b) { if (a && b) return Alogtable[(Logtable[a] + Logtable[b])%255]; else return 0; } GF256 is generated by m(x) = x^8 + x^4 + x^3 + x + 1 in AES. The above tables (20 entries in each row) are built by the primitive element x+1 of GF_2[x] / < m(x) > ~= GF_256. (~=: isomorphic). * (x + 1)^n mod m(x) = x^4 + x^2 + x, then n = ___. * (x + 1)^100 mod m(x) = ___ (a polynomial in GF_2[x] of degree < 8) * (x^5 + x^3)^(-1) mod m(x) = ___ (a polynomial in GF_2[x] of degree < 8) * Complete the subroutine computing patched multiplicative inverses in GF_256: word8 inverse(word8 a) { if (a) return Alogtable[ ___ ]; else return 0; } Part III (Write down all details of your work) 31. (3 points) Prove that the identity element e in a group G is unique. 32. (3 points) In the design of AES, the Galois field GF_256 ~= GF_2[x] / < x^8 + x^4 + x^3 + x + 1 > is represented with the polynomial basis {1, x, x^2, x^3, x^4, x^5, x^6, x^7}. The normal basis might be a better choice for hardware implementation. Show that the square operation of an element in GF_256 represented with the normal basis {x, x^2, x^(2^2), x^(2^3), x^(2^4), x^(2^5), x^(2^6), x^(2^7)} is executed as a rotation by one bit. 33. (4 points) To show that the Galois field GF_8 is unique up to isomorphism, construct an isomorphism between the quotient rings R1 = GF_2[x] / < x^3 + x^2 + 1 > and R2 = GF_2[x] / < x^3 + x + 1 >. That is, determine h(x) ∈ GF_2[x] such that the homomorphism f: R1 → R2 defined by f([0]) = [0], f([1]) = [1], and f([x]) = [h(x)] is one-to-one and onto, where [t] denotes the coset which t belongs to. Explain why your choice is correct. --

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