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Minimal Polynomials. The minimal polynomial of element of is the monic polynomial h(x) of least degree in GF p [x] such that h() = 0 when evaluated in Notation: the minimal polynomial of is denoted m (x). Minimal Polynomials.
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Minimal Polynomials • The minimal polynomial of element of is the monic polynomial h(x) of least degree in GFp[x] such that h() = 0 when evaluated in • Notation: the minimal polynomial of is denoted m(x)
Minimal Polynomials TheoremLet f(x) be a monic polynomial with coefficients in GFp and let Then f(x) is the minimal polynomial of if and only if 1. f() = 0 when evaluated in 2. f(x) is an irreducible polynomial over GFp. Proof. First we show, by contradiction, that m(x) is irreducible. Thus suppose m(x) = a(x)b(x) where neither a(x) nor b(x) is a constant. Then degree of each of a(x) and b(x) is less than the degree of m(x). Also, a()b() = m() = 0 Since a() and b() are elements of the field , a() = 0 or b() = 0 In either case, is a zero of a polynomial of degree less than that of m(x) But this is not possible since m(x) is the minimal polynomial of . We may thus conclude that m(x) must, in fact, be irreducible.
Minimal Polynomials TheoremLet f(x) be a monic polynomial with coefficients in GFp and let Then f(x) is the minimal polynomial of if and only if 1. f() = 0 when evaluated in 2. f(x) is an irreducible polynomial over GFp. Proof continued Now suppose that f(x) is an irreducible polynomial in GFp[x] and that f() = 0. Then the degree of m(x) is the degree of f(x) Write f(x) = q(x) m(x) + r(x) where r(x) is either 0 or has degree < deg m(x) Then r(x) = f(x) - q(x) m(x), hence r() = f() - q() m() = 0. If r(x) is not zero, it is a polynomial of lower degree than m(x) Since would then be a root of r(x), this is not possible Thus r(x) = 0, so f(x) = q(x) m(x). Since f(x) is irreducible and m(x) is nonconstant, q(x) must be a constant Since f(x) and m(x) are both monic, q(x) = 1 and thus f(x) = m(x).
Finding Minimal Polynomials • Recall that we showed that, for every element of , and hence • Therefore is a root of the polynomial • If we were to factor the above polynomial as a product of irreducible polynomials, then would have to be a root of one of the factors • By our previous theorem, that irreducible factor would be the minimal polynomial of . • In the case p = 2, the polynomial takes on the form since -x = +x mod 2. • Example: using the Maple command Factor(x^8 + x) mod 2,we get (x^3+x+1)*(1+x)*(x^3+x^2+1)*x • Thus in GF8, x is the minimal polynomial of 0; x+1 is the minimal polynomial of 1; and the minimal polynomial of every other element is either x3 + x + 1 or x3 + x2 + 1
Minimal Polynomials of Elements of GF8 • Recall that the nonzero elements of GF8 form a group under multiplication, denoted GF8* • By definition, the order of an element of a group is the least power n such that an = 1. • By Lagrange’s theorem, the order of an element must divide the cardinality of the group • Thus, the order of an element of GF8*must divide 7, hence is either 1 or 7 • The only element of order one is the multiplicative identity 1 and thus every other element has order 7 • Suppose we use the polynomial x3 + x + 1 to construct GF8 • Then there is an element of GF8, such that 3 + + 1 = 0 (namely, x renamed ) • Since 1, every nonzero element of GF8 is a power of . • Since 3 + + 1 = 0 and x3 + x + 1 is irreducible, the minimal polynomial of is x3 + x + 1.
Minimal Polynomials of Elements of GF8 • We can use a simple trick to find minimal polynomials in GF8 (and other powers of 2). • If u is an element of GF2, then u2 = u (there are only two cases!) • Moreover, if and are elements of for any positive integer n, then ( + )2 = 2 + 2 + 2 = 2 + 2 • Recall that, inGF8 = GF2 /(x3 + x + 1), there is an element having minimal polynomial x3 + x + 1 such that every other nonzero element is a power of • Since 0 = 02 = ( 3 + + 1)2 = (3)2 + 2 + 1 = (2)3 + 2 + 1, we see that x3 + x + 1 is also the minimal polynomial of 2 • Similarly, x3 + x + 1 is the minimal polynomial of 4 • The fun stops here, since 8 = .
Minimal Polynomials of Elements of GF8 • You can verify that x3 + x2 + 1 is the minimal polynomial of 3 = + 1, and from that it is also the minimal polynomial of 6 and 12 = 5 • Thus, we have accounted for all elements of GF8. • The above techniques are generalized and justified by the following sequence of lemmas
The Powers Lemma Lemma P1If p is a prime and 1 i p-1, then p 0 Lemma P2Let p be a prime and let f(x), g(x) be elements of GFp[x]. Then (f(x) + g(x))p = f(x)p + g(x)p mod p Lemma P3Let p be a prime and let h(x) GFp[x] be given by anxn+an-1xn-1++a1x+a0Then (h(x))p = anxpn+an-1xp(n-1)++a1xp+a0 mod p = h(xp).
The Squaring Lemma LemmaSuppose h(x) GFp[x] and . Then, when evaluated in , for every integer i. CorollarySuppose h(x) GF2[x] and . Then if is a zero of h(x), so is for every positive integer i.
Error-Control Codes • Error-control codes are designed to deal with low-probability errors in communications over a data channel • Shannon’s Model: • Assumption: channel is noisy, so that changes will be made with some positive probability between the encoder and the decoder • Idea: add enough redundant information so that errors can be detected and/or corrected Sender Encoder Channel Decoder Receiver
Error-Control Codes • Example: triple-repetition code • Each binary bit is repeated three times and a simple majority rule is used for decoding • Thus the word 01101 would be encoded as 000111111000111 • Transmission and decoding:Original 0 1 1 0 1 Encoded 000 111 111 000 111 Received 010 111 110 000 101 Decoded 0 1 1 0 1
Binary Symmetric Channels • Triple repetition codes are very inefficient in that a great deal of extra bits are inserted • We will explore more efficient codes • Assumption: binary symmetric channel The message will be a stream of zeros and ones, the only errors are bit reversals and the probability of 0 changing to 1 is the same as that of 1 changing to 0 • In particular, if the probability of a change is p, then the probability of no change is 1-p • The four possible events in a binary symmetric channel no error error
Hamming Distance • Encoders will work on blocks of binary words u = (u1,u2,…,un) where each ui is either 0 or 1 DefinitionThe weight w(u) of a word u = (u1,u2,…,un) is the number of nonzero components of u. • The weight of word u = (01101000) is 3 DefinitionThe Hamming distance d(u,v) between binary words u = (u1,u2,…,un) and v = (v1,v2,…,vn) is the number of coordinates in which the words differ. DefinitionThe sum u+v of binary words is the vector sum modulo 2. Example(0,0,1,0,1,1,0,1,0,0,1) + (1,1,1,0,1,1,0,1,0,0,0) = (1,1,0,0,0,0,0,0,0,0,1)
Hamming Distance • Note: 1d(u,v) = w(u+v) 2 uu = w(u)(vector dot product) Theorem If u, v and w are codewords, then (i) d(u,u) = 0 (ii) d(u,v) = d(v,u) (iii) d(u,v) d(u,w) + d(w,v)
Maximum Likelihood Decoding • Suppose the minimum distance between codewords is 3. • Then if a single error occurs, the result is not a codeword and there is exactly one codeword at distance 1 from the given word • Thus, we can correct the error by choosing the closest codeword • This is known as maximum likelihood decoding
Hamming Codes • The (7,4)-Hamming code uses three parity bits for a 4-bit message • The encoder uses matrix multiplication with the following matrix • The codeword v for the 4-bit message word u is given by v = uH • Note that the first 4 bits of v are exactly the bits of u; bits 5, 6, and 7 are parity bits
Hamming Codes • Example: suppose the message word is u = (0 0 1 1) Then the codeword is given by • The codeword v for the 4-bit message word u is given by v = uH • Note that the first 4 bits of v are exactly the bits of u; bits 5, 6, and 7 are parity bits
Hamming Codes • What is the minimum distance between codewords in the (7,4) Hamming code? • Every codeword is a linear combination of rows of H and the rows have weights 3,3,3 and 4, respectively. • The sum of any two rows is at least 3: 2 from the first four columns and at least one from the parity columns • Similarly, the sum of any three of the rows is at least 3 and the sum of all four rows is 7 • Thus, the (7,4) Hamming code can correct 1 error • See the text for the decoding method
Hamming Codes HW • Pages 125-126, # 2, 4, 8
Binary BCH Codes • The idea behind BCH codes is to use polynomials over a field to represent plaintext words • Encoding is then done by multiplying by a fixed polynomial • Thus a binary plaintext word arar-1 a0 is represented by the polynomial a(x) = arxr + ar-1xr-1 + + a1x + a0 • If g(x) is the encoding polynomial, then we write the product a(x)g(x) = cnxn + cn-1xn-1 + + c1x + c0, and the code word for the original binary plaintext is the word cncn-1 c0 • The choice of the field and the polynomial g(x) are the key to the error-correcting capabilities of the code. • The methodology is captured in the fundamental theorem by discovered and proved by Bose and Chaudhury and independently by Hocquenghem.
The BCH Theorem Theorem 90 (page 243)Construct using a degree-n, irreducible polynomial q(x) GFp[x]. Let be a primitive element of , let t be a positive integer and letmi(x) be the minimal polynomial of i for i = 1, 2, … , 2t. Define g(x) to be the least common multiple of m1(x), m2(x), … , m2t(x) and let k = deg( g(x) ). Then the minimum weight of the codewords corresponding to the polynomials in {a(x)g(x) : a(x) is a plaintext polynomial of degree at most pn-k-2 }is at least d = 2t. Thus, for such codewords, at least t errors can be corrected. Notation: the polynomial g(x) above is called a generator for the code.
Properties of BCH Codes • The way we constructed the generator polynomial has some consequences which are important in decoding received code polynomials • Since g(x) is the least common multiple of the minimal polynomials of the first 2t powers of , we know that g(i) = 0 for i = 1, 2, …, 2t. • That means that if we encode a plaintext polynomial a(x) by setting c(x) = a(x)g(x), we also know that c(i) = 0 for i = 1, 2, …, 2t • Suppose codeword c(x) is transmitted over a line and the received codeword is r(x). • No errors if r(x) mod g(x) = 0 and then we can decode a(x) = r(x)/g(x) • If there were errors, then the error polynomial is defined to be e(x) = r(x) – c(x) • If we compute e(x), we can recover c(x) as r(x) + e(x) • Note that e(i) = r(i) – c(i) = r(i) for i = 1, 2, …, 2t . • The values e(), e(2), e(3), …, e(2t) can be used to decode the received message with at most t errors.
BCH Code Example • We will use GF16 to construct a double-error-correcting code. • Let q(x) = x4 + x + 1 and construct GF16 as GF2 / q(x) • Let be a root of q(x) in GF16 so that 4 + + 1 = 0 and hence 4 = + 1. • Since the multiplicative order of 1 must divide |GF16 -{0}| = 15, the order must be 3, 5 or 15. • Since 3 1 and 5 = 4 = (+ 1) = 2 + 1, the order of is 15 • Thus every nonzero element of GF16 is a power of • Now we know that x4 + x + 1 is the minimal polynomial of and hence is the minimal polynomial of 2, 4 and 8 . • Thus m1(x) = m2(x) = m4(x) = m8(x) = q(x) = x4 + x + 1. • Since q(x) has degree 4, it must be the case that q(x) = (x-)(x-2 )(x-4)(x-8),which can be verified by carrying out the multiplication in GF16.
BCH Code Example • Since we want to build a code capable of correcting at least 2 errors, we need the minimal polynomial of the first 4 powers of . • We already have m1(x), m2(x) and m4(x), so we need m3(x). • We know that m3(x) is the minimal polynomial of 3, 6, 12 and 24 = 9 • Thus m3(x) is divisible by the polynomial (x-3)(x-6)(x-12) (x-9) = x4 + (3 + 6 + 9 + 12)x3 + (9 + 12 + 1 + 1 + 3 + 6)x2 + (3 + 6 + 9 + 12)x + 1 = x4 + (3 + 6 + 9 + 12)x3 + (3 + 6 + 9 + 12)x2 + (3 + 6 + 9 + 12)x + 1 Since 3 + 6 + 9 + 12 = 3 + (3 + 2) + (3 + ) + (3 + 2 + + 1) = 1 We have that m3(x) is divisible by x4 + x3 + x2 + x + 1 But this polynomial is in GF2[x] and has 3 as a root. Since m3(x) is the polynomial in GF2[x] of least degree that has 3 as a root, m3(x) = x4 + x3 + x2 + x + 1
BCH Code Example • Now that we have found the minimal polynomials of the first 4 powers of , we need to compute their least common multiple • Since all the polynomials are irreducible, the lcm is just the product of the distinct polynomials in the list: g(x) = (x4 + x + 1)(x4 + x3 + x2 + x + 1)= x8 + x7 + x6 + x4 + 1 • Since the degree of g(x) is 8, the plaintext polynomial must have degree less than or equal to 24 – deg(g(x)) – 2 = 16 – 8 – 2 = 6 • In particular, if the plaintext word is a6a5a4a3a2a1a0 = 1011001, then a(x) = x6 + x4 + x3 + 1 and a(x)g(x) = x14 + x13 + x10 + x9 + x3 + 1 and hence the codeword is transmitted as 110011000001001
BCH Decoding • Now suppose we received the message 11101000101111 • The corresponding to the polynomial is r(x) = x14 + x13 + x12 + x10 + x6 + x4 + x3 + x2 + 1 • Suppose a single error occurred at bit position k • Then r(x) = c(x) + xk. Why? • If the original bit at position k was 0, then there is no xk in c(x) • Changing that 0 to a 1 corresponds to adding the term xk. • If the original bit at position k was 1, then xk is a term of c(x) and the coefficient of xk in r(x) is 0. • Since addition is done mod 2, if we add xk to c(x), the xk term will have coefficient 2; when we reduce modulo 2, that term disappears • In other words, the kth bit has changed to the correct value: 0 • Since addition is done mod 2, we can recover c(x) by adding xk to r(x) • Of course, we first must find the value k, which is to say the position where the bit reversal occurred during transmission
Single-error Correction • We can find the bit error position k if we can find k. • r() = c() + k = 0 + k = k = (14 + 13 + 12 + 10 + 6 + 4 + 3 + 2 + 1 mod(q()) ) mod 2 = 3 + 2 + • Checking the list of powers of , we see that 3 + 2 + = 11 • Thus k = 11 and we recover the codeword that was sent as c(x) = r(x) + x11 = x14 + x13 + x12 + x11 + x10 + x6 + x4 + x3 + x2 + 1 • Dividing c(x) by g(x), we get a(x) = x6 + x3 + x2 + 1 • The corresponding bit string is 000000001001101
Single-error Correction HW Problem 2, page 257
Correcting Two Errors • If errors occur at two bit positions, say j and k, then the error polynomial will be e(x) = xj + xk. • Obviously, decoding in this situation is more complicated than in the case of a single error. • As before, we have r() = c() + e() = 0 + j + k = j + k • Our goal will be to find j + k and j+k by computing the polynomial (xj +1)(xk +1) = x2j+k + x(j + k) + 1 • After that we express j + k as a power of • This information will be enough to determine j and k • For this example, suppose r(x) = c(x) + e(x) = x14 + x11 + x10 + x5 + x +1 and assume that exactly two errors occurred.
Correcting Two Errors • Since g(x) is the least common multiple of the first 2t = 4 powers of , we know that g() = g(2) = g(3) = g(4) = 0 • We then obtain the following: r() = c() + e() = a()g() + e() = a()0 + j + k = j + k r(2) = c(2) + e(2) = a(2)g(2) + e(2) = a(2)0 + 2j + 2k = 2j + 2k r(3) = c(3) + e(3) = a(3)g(3) + e(3) = a(3)0 + 3j + 3k = 3j + 3k r(4) = c(4) + e(4) = a(4)g(4) + e(4) = a(4)0 + 4j + 4k = 4j + 4k • From this we have j + k = r() = 14 + 11 + 10 + 5 + +1 • Since we are evaluating the expression in GF16, we must reduce the expression mod 4 + +1 and also mod 2 • Then, since we are interested in finding the exponents i and j, we express each of the above values as a power of
Correcting Two Errors • Carrying out the evaluations and reductions described on the previous slide, we get: j + k = 2 + 1 = 8 2j + 2k = 3j + 3k= 3 + 2 + = 11 4j + 4k =2
Correcting Two Errors • We have derived the following information j + k= 82j + 2k= 3j + 3k= 114j + 4k = 2 • Now for the algebraic “insight” when computing mod 2: (j + k) j+k + (2j + 2k) (j + k) = 3j + 3k (2j + 2k) j+k + (3j + 3k) (j + k) = 4j + 4k • We want to solve for j+kand j + k, so let u = j+kand v = j + k • Substituting these values and the values calculated above, the equations become 8u + v = 11 u + 11v = 2
Correcting Two Errors • It will be useful to express the equations 8u + v = 11 u + 11v = 2 in terms of matrices: Convert the above to the augmented matrix: Now we can solve the equations by row operations
Correcting Two Errors • Solving for j+k and j + k : From the above, we have j + k = 8 and j+k = 9
Correcting Two Errors • We have j + k = 8 and j+k = 9 • We know that j+k = 9, so we check the possible values of j,k: • Therefore, we see that j = 4 and k = 5 and thus the corrected code polynomial is c(x) = r(x) + e(x) = (x14 + x11 + x10 + x5 + x +1) + x4 + x5 = x14 + x11 + x10 + x4 + x +1 a(x) = c(x)/g(x) = x6 + x5 + x +1
Double Error Correcting HW • The polynomial x4 + x + 1 was used to construct GF16 and used to construct a double-error correcting code as in the slides. The polynomial r(x) = x13 + x11 + x9 + x8 + x7 + x2 + 1 with at most 2 errors. Find the plaintext polynomial that was sent.
