3. Coding theory 3.1. Basic concepts

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1 3. CODING THEORY 1 3. Coding theory 3.1. Basic concepts In this chapter we will discuss briefly some aspects of error correcting codes. The main problem is that if information is sent via a noisy channel, then the received message contains errors which need to be detected and corrected. The lectures are given by a non native speaker of English. Any mispronunciation due to his accent will be corrected by your brain to the correct English word. The idea is to assign at each symbol a codeword. If there are not to many errors introduced during the transmission then the received word might not be a codeword but near to a unique codeword which, hopefully, was the original one. One way of decoding YES and NO is by YES= and NO= The sequence is not a codeword and is likely to mean NO (there were two errors) rather then YES (there were five errors). A q-ary code C is a given set of sequences of symbols where each symbol is chosen from a set F of q elements. The set F is called the alphabet. If q = 2, the code will be called binary. We limit ourself to the situation where F is a field of order q. IF n q will denote the set of all ordered n-tuples a = a 1 a 2... a n with a i IF q. The elements of IF n q are called vectors and n is called the length of a. Observe that the set IF n q has q n elements. A q-ary code C of length n is a subset of IF n q. We say that a code C of length n is s error detecting if changing up to s digits to a codeword does not produce a codeword. We say that a code C of length n is t error correcting if from a given string of length n which differs on at most t places from some codeword one can deduce the codeword. The Hamming distance between two vectors x and y of IF n q is the number of places in which they differ. It is denoted by d(x, y). In IF 5 2 we have d(00111, 11001) = 4. In IF 4 3 we have d(0122, 1220) = 3. Lemma The Hamming distance is a distance function, that is it satisfies the three conditions i.) d(x, y) = 0 if and only if x = y. ii.) d(x, y) = d(y, x), for all x, y IF n q. iii.) d(x, y) d(x, z) + d(z, y), for all x, y, z IF n q. The first two conditions are easy to verify. The third, known as the triangle inequality, is verified as follows. Note that if u, v IF n q, then d(u, v) is the minimum number of changes of digits required to change u into v. But we can change x into y by first changing x into z and then z into y. Hence d(x, y) d(x, z) + d(z, y), The problem of decoding we consider is as follows. Suppose a codeword x, unknown to us, has been transmitted to us and we have received the vector y. This vector y may have been distorted by

2 3. CODING THEORY 2 noise. It seem reasonable to decode y as that code word z such that d(z, y) is as small as possible. This is called nearest decoding. Given a code C the minimum distance, denoted d(c), is the smallest distances between distinct code words. That is, d(c) = min{d(x, y) x, y C, x y}. Lemma Let C be a code. (i) One can detect up to s errors in any codeword if d(c) s + 1. (ii) One can correct up to t errors in any codeword if d(c) 2t + 1. Suppose d(c) s + 1. Suppose a codeword x is transmitted and s or fewer errors are introduced. Then the received vector cannot be a different codeword and so the errors can be detected. Suppose d(c) 2t + 1. Suppose a codeword x is transmitted and the vector y is received in which t or fewer errors have occurred, so d(x, y) t. Suppose z is a codeword with d(z, y) t. Then d(x, z) d(x, y) + d(y, z) 2t, hence z = x. So x is the nearest codeword to y. Corollary Let C be a code with minimum distance d. Then up to d 1 errors can be detected and up to (d 1)/2 errors can be corrected. We have d s + 1 if and only if s d 1, and d 2t + 1 if and only if t (d 1)/2. Let YES=1111 and NO=0000, then one error can be corrected but the two errors in 1001 can t be corrected, it is unclear if the correct word is YES or NO. A (n, M, d)-code is a code of length n, containing M codewords and having minimum distance d. A good (n, M, d)-code has small n (for fast transmission), large M (for a large variety of messages) and large d (to correct many errors). These are conflicting aims. One of the main problems is to optimize one parameter given the other two. The usual version of the problem is to find the largest code M, given the the length n and minimum distance d. Note that M q n. The q-ary repetition code of length n is defined as follows C = {a 1 a 2... a n a 1 =... = a n, a 1 IF q }. It has length n, q codewords and minimal distance n. Note that any q-ary code with minimum distance n can have at most q codewords as any two codewords must differ in all positions. Two q-ary codes are called equivalent if one can be obtained from the other by a combination of operations of the following type (A) a permutation of the position of the code; (B) a permutation of the symbols appearing in a fixed position. The code C = {00100, 00011, 11111, 11000} is equivalent to D = {00000, 01101, 10110, 11011}. Indeed first permute 0 1 in the third position and then interchange positions 2 and 4.

3 3. CODING THEORY 3 If a code is displayed as an M n-matrix whose rows are the code words, then the operations of type (A) correspond to a permutation of the columns of the matrix. The operations of type (B) correspond to re-labeling the symbols in a given column. Observe that the operations do no change the hamming-distance between two codewords. Hence does not change the minimum distance The codes and column and then to the third are equivalent, first apply to the second Lemma Any q-ary (n, M, d)-code over an alphabet containing 0 is equivalent to a q-ary (n, M, d)-code containing Choose a codeword a 1 a 2 a n and sequentially apply the operations of type (B) of the form 0 a j if a j 0 to the code. We will denote with 0 the all zero codeword. From now on we will assume that 0 C. The weight of codeword x, denoted by w(x), is the number of non zero entries. That is, w(x) = d(x, 0). Lemma Suppose that C is a binary code of length n. Let x, y C with w(x) and w(y) even, then d(x, y) is even too. Let x = x 1 x 2 x n and y = y 1 y 2 y n. Let A = {i x i = 1, 1 i n} and B = {i y i = 1, 1 i n}. Then A = w(x), B = w(y) and d(x, y) = (A B)\A B. Hence d(x, y) = w(x) + w(y) 2 A B, which is even. Theorem Suppose d is odd. A binary (n, M, d)-code exists if and only if a binary (n + 1, M, d + 1)- code exists. Suppose C a binary (n + 1, M, d + 1)-code, with d odd. Choose two codewords x and y at minimal distance, that is d(x, y) = d + 1. Choose a position in which x and y differ and delete this position from all codewords. The codewords obtained from x and y are now at distance d and any two codewords will differ in at least d positions. Hence the resulting code is a (n, M, d)-code. Suppose D is a binary (n, M, d)-code, with d odd. For a codeword x D, with x = x 1 x 2 x n we define ˆx = x 1 x 2 x n x n+1, where x n+1 w(x) (mod 2). Let ˆD be the code of length n + 1 defined as follows ˆD = {ˆx x D}. Clearly d d( ˆD) d + 1. Observe that w(ˆx) is even for all ˆx ˆD. By the previous lemma the distance between two codewords will be even. Hence d( ˆD) will be even too. Hence ˆD is a binary (n + 1, M, d + 1)-code. Consider the Fano plane, that is the geometry with 7 points and 7 lines such that any two points are on a unique line and any two lines meet in a unique point. Let A be the incidence matrix of the Fano plane (rows indexed by the points, columns by the lines) and B the matrix obtained from A by swapping 0 1. Let C be the code consisting of the rows of A, rows of B, the all 0 vector and the all 1 vector. The codewords are

4 3. CODING THEORY 4 { rows of A and rows of B This code has minimum distance 3. Hence is a binary (7, 16, 3)-code. The all zero vector and the first row of A are at minimal distance and differ in the first entry. Deleting the first entry of all vectors gives a binary (6, 16, 2)-code Linear Codes In this section V (n, q) denotes the vector space of dimension n over the field IF q, where vectors will be written as row-vectors. Thus V (n, q) = {(x 1, x 2,..., x n ) x 1, x 2,... x n IF q } A vector x = (x 1, x 2,..., x n ) will often simply written as x 1 x 2 x n. A linear code C is a subspace of V (n, q). That is 1. (0,..., 0) C; 2. If x, y C, then x + y C; 3. If x C and a IF q, then ax C. If C has dimension k, then the linear code C is called a [n, k]-code. If it has minimum distance d it is also called a [n, k, d]-code. So a q-ary [n, k, d]-code is a q-ary (n, q k, d)-code. If C is a linear code, then the weight of C, denoted by w(c), is the smallest of the weights of a non-zero code words. That is, w(c) = min {w(x) x C, x 0}. Lemma Let x, y V (n, q). Then d(x, y) = w(x y). The vector x y has a non zero entry precisely on those places where x and y differ. Lemma Let C be a linear code. Then d(c) = w(c). Let x and y be two codewords with d(x, y) = d(c). Then d(c) = d(x, y) = w(x y) w(c), since x y C. On the other hand, for some codeword z we have w(c) = w(z) = d(z, 0) d(c), since 0 C. Hence d(c) = w(c) A k n matrix whose rows form a basis of a linear [n, k]-code is called a generator matrix for the code.

5 3. CODING THEORY 5 The code constructed from the Fano plane is a [7, 4]-code with generator matrix G = The notion of equivalence between linear codes is slightly different from the one defined before. The second operation on the codewords is more restrictive. Two linear q-ary codes are called equivalent if the one can be obtained from the other by combining operations of the following types (A.) permutations of the positions of the code; (B.) multiplication of the symbols appearing in a fixed position by a non-zero element of IF q. Theorem Two k n matrices generate equivalent linear [n, k]-codes over IF q if one matrix can be obtained from the other by a sequence of operations of the following types: (R1) Permutation of rows. (R2) Multiplication of a row by a non-zero scalar. (R3) Addition of a scalar multiple of one row to another row. (C1) Permutation of columns. (C2) Multiplication of a column by a non-zero scalar. The row operations (R1), (R2) and (R3) preserve the subspace C and simply replace one basis of C by another. The operations (C1) and (C2) convert the generator matrix to one of an equivalent code. Theorem Let G be a generator matrix of an [n, k]-code. Then by performing operations of the type (R1), (R2), (R3), (C1) and (C2), G can be transformed into standard type [I k A] where I k is the identity matrix and A a k (n k) matrix. Note that n k. Using row operations (R1), (R2) and (R3) we can transform G into reduced echelon form and then use (C1) to move the columns containing the pivots to the left. On the vector space V (n, q) we define an inner-product. Let x, y V (n, q) with x = x 1 x 2 x n and y = y 1 y 2 y n, then the inner-product of x and y is defined by x y = x 1 y 1 + x 2 y x n y n. If x y = 0, then the vectors are called orthogonal. For a matrix A, the transposed will be denoted by A T. Let C be a [n, k]-code. The dual code C is defined as the set of vectors of V (n, q) that are orthogonal to every codeword of C, that is C = {x V (n, q) x y = 0 for all y C}.

6 3. CODING THEORY 6 Lemma Suppose C is a [n, k]-code with generator matrix G. Let x V (n, q). Then x C if and only if xg T = 0. That is, if and only if x is orthogonal to all the rows of G. Note that the rows of G generate the code. So a vector is orthogonal to all codewords in C if and only if it is orthogonal to each row of G. Theorem Suppose C is a [n, k]-code. Then C is a linear [n, n k]-code and (C ) = C. This can be shown using arguments of linear algebra. Let C be an [n, k]-code. A parity-check matrix H for C is a generator matrix of C. Thus H is an (n k) n matrix satisfying GH T = O k,n k, where O k,n k is the all 0 matrix. A parity-check matrix H is said to be in standard form if H = [B I n k ]. Since (C ) = C it follows that the code C can also be described via a parity-check matrix, as C = {x V (n, q) xh T = 0}. Theorem Let C be an [n, k]-code with generator matrix G = [I k A] in standard form. Then a parity check-matrix for C is H = [ A T I n k ]. Indeed the matrix H has the right size and [I k A]([ A T I n k ]) T = Coding and syndrome decoding of linear codes Let C be an [n, k]-code over IF q with generator matrix G. We identify messages with vectors of V (k, q). Suppose H is a parity check matrix for an [n, k]-code C. Then for any vector y V (n, q) the row vector S(y) = yh T, which is of length n k, is called the syndrome of y. Observe that S(y) = 0 if and only if y C. Let z V (n, k), then the C-coset of z is defined as z + C = {z + x x C}. Lemma Suppose C is an [n, k]-code. Then (i) every vector of V (n, q) is in some C-coset. (ii) every coset contains exactly q k vectors. (iii) two cosets are either disjoint or coincide. Straight forward. The lemma tells us that there are q n k different C-cosets in V (n, k). In the following lemma we will see that the syndrome is a way to decide quickly if two vectors give rise to the same coset. Lemma Let x, y V (n, k). Then x + C = y + C if and only if S(x) = S(y).

7 3. CODING THEORY 7 We have x+c = y +C if and only if x y C. But x y C if and only if (x y)h T = 0. Which holds if and only if xh T = yh T. That is, if and only if S(x) = S(y). Let x + C be a coset. Among the vectors in x + C we choose a vector z such that the weight is minimal. If the choice of a vector of minimal weight in the coset is not unique we choose z to be one of them. We have w(z) w(y) for all y x + C. The vector z is called the coset leader of the coset z + C. Observe that if w(c) = d, then any non-zero vector of C has at least d non-zero entries. So if y is a vector with w(y) < d 2, then any vector of y + C different from y has at least d 2 non zero entries. hence y is a coset leader of y + C. Given a [n, k]-code we first produce a table of coset leaders 0, a 1, a 2,..., a s, where s = q n k 1, together with their syndromes S(0), S(a 1 ), S(a 2 ),..., S(a s ). This table is called the syndrome lookup table. Encoding A message x = x 1 x 2 x k is encoded by multiplying x on the right with G, that is, the encoded message is xg. Note that this is a vector of length n and is a linear combination of the rows of G, hence a code word. To obtain the message x from a given codeword y, one needs to solve the system of linear equations xg = y. In case G is in standard form the encoding gets even simpler. Since G is of the form [I k A] and x = x 1 x 2 x k, we have xg = y 1 y 2 y k y k+1 y n, where y i = x i for 1 i k and y k+j = a ij x j for 1 j n k. The digits y 1 y 2 y k are called the message digits and the digits y k+1 y n are the check digits. The check digits are added to the message to give protection against noise. Note that y k+1 y n = xa. Syndrome decoding The decoding procedure is as follows: 1. For a received vector y we calculate the syndrome S(y) = yh T. 2. In the syndrome look-up table locate S(y) and the corresponding coset leader z. 3. Correct y as y z. 4. y z is a code word, compute the corresponding message from it. Suppose a codeword x is sent through a noisy channel and the vector received is y. We define the error vector e to be e = y x. Now we have y = x + e. The decoding problem now becomes given y to find e and subtract it from y to obtain x. Note that e y + C, for x is a codeword. Hence e z + C as y + C = z + C. If the error was small then the weight w(e) is small and so it is likely that e = z and thus y z = x. In particular, if w(e) < w(c) 2, then e is the vector of smallest weight in e + C, thus e is the coset leader of the coset of syndrome S(e) = S(y). Hence e = z, and so the procedure corrects up to w(c) 1 2 errors. If G is in standard form, then the message is the first k digits of y z Binary Hamming codes In this subsection we briefly discuss binary Hamming codes. It should be noted that these codes also exists over any finite field but we will only study the case where the field is IF 2, the field with two elements.

8 3. CODING THEORY 8 Let r be a positive integer, and H be the r (2 r 1) matrix whose columns are the distinct non-zero vectors of V (r, 2). The code having H as parity-check matrix is called the binary Hamming code and is denoted by Ham(r, 2). Since the columns can be taken in any order, the code Ham(r, 2) is any of a number of equivalent codes. Let r = 2, then if we choose choose H and G in standard form: ( ) H = and G = ( ) Hence Ham(2, 2) is just the binary triple repetition code. Theorem The binary Hamming code Ham(r, 2), with r 2, is a [2 r 1, 2 r 1 r]-code and has minimum distance 3, hence is single-error correcting. By definition the dual code Ham(r, 2) is a [2 r 1, r]-code, hence Ham(2, r) is a [2 r 1, 2 r 1 r]-code. Since Ham(2, r) is a linear code the minimum weight is equal to the minimum distance. We show that there are no codewords of weight 1 or 2, but there are codewords of weight 3. Suppose that x is a codeword of weight 1. Then x = , with an 1 in the i-th place. Since xh T = 0, we have that the i-th column of H is the all-zero vector, a contradiction. Suppose that x is a codeword of weight 2. Then x = , with an 1 in the i-th and j-th place. Since xh T = 0, we have that the sum of the i-th and j-the column of H is the all-zero vector. Since these are vectors of V (r, 2) it means that i-th and j-the column of H are the same, a contradiction. Remains to show that there are codewords of weight 3. We can suppose that the first 3 rows of H T are 0 001, and Showing that is a codeword. The code has k = 2 r r 1, and n = 2 r 1, so there are 2 n k = 2 r cosets. The vectors of weight at most one of V (n, 2) are coset leaders, since there are n + 1 = 2 r of them they are all the coset leaders. The syndrome of the vector e j = , with a 1 on place j is H T, which is the transpose of the j-th column of H. If the columns are arranged in order of increasing binary numbers (that is the j-the column of H is the binary representation of j), then the syndrome of the coset leader e j is the binary representation of j. The decoding algorithm becomes very nice: 1. For a received vector y we calculate the syndrome S(y) = yh T. 2. If S(y) = 0, then y is a codeword and no error was made. 3. If S(y) 0, then S(y) gives the binary representation of the position in which the error was made, and so the error can be corrected. Consider Ham(3, 2) with parity-check matrix H = If y = , then S(y) = 110, which is 6. The error is in place 6 and we correct y to the codeword ,

9 3. CODING THEORY Cyclic codes A code C is called cyclic if 1.) C is a linear code; 2.) any cyclic shift of a codeword is also a codeword, that is, if a 0 a 1 a n 1 is a codeword, then so is a n 1 a 0 a 1 a n 2. The code constructed from the Fano plane is cyclic. When considering cyclic codes we number the coordinate positions 0, 1,..., n 1. This because we will identify the codewords with polynomials. Let F = IF q and f(x) = X n 1 F [X]. We will consider the ring R n = F [X]/(f), a ring of q n elements. In R n we have that X n = 1. A vector a 0 a 1 a n 1 V (n, q) will be identified with the polynomial a 0 + a 1 X + + a n 1 X n 1. Adding vectors in V (n, q) now corresponds to adding the corresponding polynomials in R n. The cyclic shift a 0 a 1 a n 1 a n 1 a 0 a 1 a n 2 corresponds to multiplying the polynomial with X. Lemma A code C in R n is a cyclic code if and only if C satisfies the following two conditions (i) if a(x), b(x) C, then a(x) + b(x) C. (ii) if a(x) C and r(x) R n, then r(x)a(x) C. Suppose C is a cyclic code in R n. Then C is linear so (i) holds. Let a(x) C. Since the multiplication with X corresponds to a cyclic shift the multiplication with X i corresponds to a cyclic shift in i positions and since C is cyclic we have that X i a(x) C. Since C is linear we also have λx i a(x) C, for all λ IF q. If r(x) = r 0 + r 1 X + + r n 1 X n 1, then r(x)a(x) = r 0 a(x) + r 1 Xa(x) + + r n 1 X n 1 a(x). Since each summand is in C we have that (ii) holds. Suppose (i) and (ii) hold. Taking r(x) a scalar implies that the code is linear. Taking r(x) = X shows that C is cyclic. There is an easy way to describe and construct cyclic codes. Let f(x) R n, we define the set of all multiples of f(x) in R n. f(x) = {r(x)f(x) r(x) R n } Theorem For any f(x) R n, the set f(x) is a cyclic code: it is called the cyclic code generated by f(x). This follows immediately from the previous lemma. The following theorem shows that all cyclic codes can be constructed that way. Theorem Let C be a non-zero cyclic code in R n. Then (i) there exists a unique monic polynomial g(x) of smallest degree in C; (ii) g(x) = C (iii) in F [X] the polynomial g(x) is a divisor of X n 1

10 3. CODING THEORY 10 (i) Suppose g(x) and h(x) are both monic polynomials in C of smallest degree. Then g(x) h(x) C. If g(x) h(x), then the degree of g(x) h(x) is smaller then that of g(x) and a suitable multiple of g(x) h(x) is monic and will be in C, a contradiction. Thus g(x) = h(x). (ii) Suppose a(x) C. View a(x) as a polynomial in F [X]. Then a(x) = q(x)g(x) + r(x), with deg(r(x)) < deg(g(x)) or r = 0. Since C is cyclic we have r(x) C, and by the minimality of the degree of g we have r = 0. (iii) By the division algorithm X n 1 = q(x)g(x) + r(x), with deg(r(x)) < deg(g(x)) or r = 0. In R n we have r(x) = q(x)g(x), hence r(x) C. By the minimality of the degree of g(x) we have r(x) = 0. The code constructed form the Fano plane is cyclic. The polynomial of minimal degree is 1+X 2 +X 3, corresponding to the fourth row of A. It is the generating polynomial. The rows of A are obtained by multiplying with monomials. The all one vector is obtained by multiplying the polynomial by itself and the rows of B are obtained by multiplying with X + 1 to obtain the last row from B and then with monomials to obtain the other rows of B. In IF 2 [X], we have X 3 1 = (X 1)(X 2 + X + 1). Thus there are 4 cyclic codes in V (3, 2): Generator Corresponding polynomial Code in R 3 code in V (3, 2) 1 all of R 3 all of V (3, 2) X + 1 {0, 1 + X, X + X 2, 1 + X 2 } {000, 110, 011, 101} X 2 + X + 1 {0, 1 + X + X 2 } {000, 111} X 3 1 {0} {000} The monic polynomial of least degree is called the generator polynomial of C. Theorem Suppose C is a cyclic code with generator polynomial g(x) = g 0 + g 1 X + + g r X r of degree r. Then g 0 0, dim(c) = n r and a generator matrix for C is g 0 g 1 g 2 g r g 0 g 1 g 2 g r g 0 g 1 g 2 g r g 0 g 1 g 2 g r Suppose g 0 = 0, then X n 1 g(x) = X 1 g(x) is a codeword of degree r 1, contradicting the minimality of the degree of g(x). The n r rows of the matrix are linear independent, as the matrix is in echelon form, and are code words. It suffices to show that they generate the code. Let a(x) be a codeword, we may assume that deg(a(x)) < n. Since a(x) = g(x)q(x), for some q(x) R n and deg(a(x)) < n, this equality also folds in IF q [X]. It follows that deg(q(x)) < n r.

11 3. CODING THEORY 11 Let q(x) = q 0 +q 1 X + +q n r 1 X n r 1, then a(x) = q 0 g(x)+q 1 Xg(x)+ +q n r 1 X n r 1 g(x), which gives the desired linear combination. Let C be cyclic [n, k]-code with generator polynomial g(x), then x n 1 = g(x)h(x) for some monic polynomial h(x). Since g(x) has degree n k, h(x) has degree k. This polynomial h(x) is called the check polynomial of C. Theorem Suppose C is a cyclic code in R n with generator polynomial g(x) and check polynomial h(x). Then an element c(x) of R n is a codeword of C if and only if c(x)h(x) = 0. Observe that in R n we have g(x)h(x) = 0. If c(x) C, then c(x) = a(x)g(x), for some a(x) R n, hence c(x)h(x) = a(x)g(x)h(x) = 0. On the other hand, by the devision algorithm, c(x) = q(x)g(x) + r(x) with deg(r(x)) < deg(g(x)) or r(x) = 0. If c(x)h(x) = 0, then r(x)h(x) = 0, so viewed as polynomials in F [X], r(x)h(x) is a multiple of X n 1. If r(x) 0, then deg(r(x)h(x)) < n k+k = n, contradicting the fact that it is a multiple of X n 1. Hence r(x) = 0 and c(x) g(x). Theorem Suppose C is a cyclic [n, k]-code with check polynomial Then h(x) = h 0 + h 1 X + + h k X k. (i) a parity-check matrix for C is h k h k 1 h h k h k 1 h h k h k 1 h 0. (ii) C is a cyclic code generated by the polynomial h(x) = h k + h k 1 X + + h 0 X k. A polynomial c(x) = c 0 + c 1 X + + c n 1 X n 1 is a codeword if and only if c(x)h(x) = 0. In particular, in c(x)h(x) = 0 the coefficient in front of X k, X k+1,... X n must be zero. Hence c 0 c 1... c n must be orthogonal to h k h k 1 h and its cyclic shifts. This shows that the rows of the above matrix are codewords of C. Since h k 0 these rows are linear independent. Thus generate a subspace of C of dimension n k. But since the dimension of C is n k, these rows must be a basis of C. Hence it is a parity-check matrix. Observe that if the polynomial h(x) is a divisor of X n 1, then the generator matrix of h(x) is the above matrix, which is the generator matrix of C. So h(x) is the generator polynomial of C. We need to show that h(x) is a divisor of X n 1. Note that h(x) = X k h(x 1 ) and X n k g(x 1 ) are polynomials. Moreover (X k h(x 1 ))(X n k g(x 1 )) = X n (X n 1) = 1 X n, showing that h(x) is a divisor of X n 1. In IF 2 [X] we have that X 23 1 = (X 1)(X 11 + X 10 + X 6 + X 5 + X 4 + X 2 + 1)(X 11 + X 9 + X 7 + X 6 + X 5 + X + 1) is a factorization into irreducible polynomials. The cyclic code generated by X 11 + X 10 + X 6 + X 5 + X 4 + X is a [23, 12]-code with minimum distance 7. It is known as the binary Golay code.