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Unformatted text preview: University of California San Diego ECE 259A: Solutions to Problem Set #4
1. A binary polynomial g x generates a linear cyclic code of length 15 if and only if it is a divisor of x15 1 over GF 2 . Using the factorization
¡ ¤ ¤ ¤ ¤ ¡ ¤ ¤ ¡ ¤ ¤ ¡ ¤ ¤ ¡ ¤ £ ¡ ¡ ¢ 1
¢ ¢ x x
£ x we conclude that there are 2 5 32 binary linear cyclic codes of length 15, including the trivial n cyclic codes 0 and IF2 . Note that the factorization of x 15 1 itself is not essential to answer this question. All one needs to know is the total number of factors, which may be easily found by writing out the cyclotomic cosets.
!¡ ¡ £ ¡ ¡ £ ¤ ¤ ¤ which means that 1 α 0 is a root of g x . Therefore, the minimal polynomial of α 0 divides g x . Yet the minimal polynomial of α 0 is x 1 over any ﬁeld. (b) The allone vector is described by the polynomial 1 x x n 1 xn 2 x 1. In n any ﬁeld we have x 1 x 1 1 x . Since n and q are relatively prime, all the roots of xn 1 are distinct and therefore 1 α 0 is not a root of 1 x . Thus if 1 x then α 0 cannot be a root of g x . On the other hand, since all the nonzero powers of α are roots of 1 x , if α 0 is not a root of g x then necessarily g x 1 x and hence 1 x . ¡ ¢ ¢ ¤ ¤ "¤ ¤ £ %¡ ¢ ¡ ¡ ¢ £ ¢ £ ¡ 3. (a) Any factor of x 8 1 over GF 3 generates a cyclic code of length 8. Thus it would sufﬁce to show that g x divides x 8 1 over GF 3 . Indeed,
¡ ¤ ¤ ¤ ¤ ¡ ¤ ¤ £ ¤ £ 1
¢ 2 x which may be obtained by long division of polynomials, for instance. (b) From the above, the paritycheck polynomial for is
¤ ¤ ¤ ¤ £ (¡ ) 0¡ ¢ £ (¡ hx
) 1¡ 1 gx 2 4. If all the roots of g1 x are also the roots of g 2 x , then it is obvious that g 1 x g2 x . Therefore if c x is any polynomial of degree n 1, then g 2 x c x implies g1 x c x . In other words, cx 2 implies c x 1. ¡ £ 7¡ &¡ !¡ 5. implies x n 1 g x 1 . As the code is also cyclic (a) Since the code is reversible, g x xn 1 g x 1 implies xr g x 1 . Now deg xr g x 1 deg g x r, and since r g x 1 must be a scalar multiple g x is the unique monic polynomial of degree r in , x of g x . This completes the proof for binary codes. gr
¤
1x where we have used the fact that g x is monic. Subsituting x 0 in the above expression, we see that β g0 . Further, equating the coefﬁcients of xr yields g2 1. Since clearly 0 β 0, we must have β g 0 1. Exercise: show that β 1 if and only if g 1 0.
@£ C¡ £ £ £ ¡ A B£ £ £ @£ ¡ ¤ ¤ "¤ ¤ £ ¤ g1 x g0 ¤ 9¤ xr r1 β g 0 xr g 1 xr 1 ¡ ¡ £ 3¡ (b) For nonbinary codes g x for some nonzero β GF q . In other words, gr
1x β xr g x 1 1 ¤ ¡ ¤ £ 8¡ ¡ ' ¡ ¤ ' ¡ ¤ £ ©¡ ¡ ' ¡ £ 3¡ Writing h x 2 in reverse order gives g
¢ 4 ¡ &5¡ ¡ x
¡ x8 x5 2 x5 h x ¡ x8 x8 x3 1 x5 2 x3 2 x2 x 2 2 x3 2 x2 x 2 x3 x2 2. 1 x5 2 x4 &¡ ¡ ¡ ¡ £ $¡ ' ¡ £ ¡ ¡ ¤ #"¤ £ ¤ g0 g1 gr g1 ¡ c1 0. 0 ¡ ¨ £ ©¡ ¡ ¡ ¡ ¡ 6¡ ¡ ¢ ¡ £©¡ ¡ ¡ ¡ 2. (a) If x c0 1 is a factor of g x then g 1 c1 cn 1 f1g1 0, and thereforee for all c x , we have Since g x , we have ¢ ¦ x15 1 x2 1 x4 1 x4 x3 1 x4 x3 x2 x 1 ¥ § ¨ (c) The part follows directly from (a). If γ is a root of g x then 0 g γ βγ r g γ 1 , and hence γ 1 is also a root of g x . To establish , note that by deﬁnition c x iff n 1 c x 1 evaluated at any root γ of g x yields cγ 0 for any root γ of g x . Now x γ n 1 c γ n 1 , which must be 0 since γ 1 is also a root of g x . Hence x n 1 c x 1 . (d) The conjugates of any nth root of unity γ are γ , γ q , γ q . . ., where the exponents can be 1 mod n it follows that γ 1 is a conjugate of γ , and reduced modulo n. Since q m hence the minimal polynomial of γ has both γ and γ 1 as roots. Furthermore, any factor of xn 1 is a product of such minimal polynomials. Hence any generator polynomial of a cyclic coide satisﬁes the conditions of (c). 1 and det V1 1, which corre6. The proof is by induction on m. For m 1, we have V1 sponds to an empty product. For m 2, we have det V2 α2 α1 as required. Now assume that det Vm 1 has the desired form, and deﬁne the polynomial D x as follows:
P
1 1 1 1 1
2 Using the last column to expand the foregoing determinant, we can write
¤ ¤ ¤ ¤ £ Q¡ Dx dm 1x m1 dm 2x m2 d1 x j 1i j 1 by induction hypothesis. Evaluating D x at α 1 , α2 , . . . , αm 1 , we see that D αi 0 for all these elements, since in each case we have two identical columns in the determinant. Thus, we have found m 1 roots of D x , and since deg D x m 1 these are all the roots of D x . Hence, D x splits into linear factors over as follows:
¡
1 i1 Now,
£ Q¡ £ Q¡ where the last equality follows from (1). It is obvious that the above expression is nonzero, and hence Vm is nonsingular, if and only if α 1 , α2 , . . . αm are all distinct.
£ W¡ £ £ YXW¡ £ W¡ 7. By deﬁnition c x iff c α of 1. Consider the polynomial
¢ £ ©¡ We show next that f x . Since f x x b 1 xn 1, any power of α which is not a root of xb 1 is a root of f x . Consider α i for i 1, 2, . . . a 1. Clearly α i b α bi 1, since bi is strictly less that n and α is primitive. Hence α , α 2 , . . . α a 1 are all roots of f x , and f x . Since the weight of f x is a, we have d a δ.
&¡ @£ ¡ £ ¡ ¢ ¢ £ 3¡ £ £ ¢ 4 ¡ ¡ !¡ ¡ ¢ ¤ ¤ ¤ ¡ ¤ ¡ £ W¡ ¢ ) 0¡ fx 1 1 xn xb xb a1 c α2 c αa 1 0, where α is a primitive nth root
a2 xb S R R i1 j 1i j 1 xb 1 ¡ £ Q¡ 1 ¢ ¢ det Vm D αm dm αm αi ∏
R ∏∏ m1 ¢ R £ V¡ Dx dm m1 ∏ x αi m1 m αi αj ¡ £ T¡ ¡ ¢ S R £ %¡ R £ Q¡ U ¡ £ 1 1 ¢ ¡ dm det Vm ∏∏ for some coefﬁcients d 0 , d1 , . . . , dm 1. Note that
m 2m 1 αi αj P 1 1 xm 1 P m α1 m α2 m αm P P . . . P P P P P P P P P P £ I¡ Dx P P P P 1 α1 2 α1 3 α1 . . . 1 α2 2 α2 3 α2 . . . 1 αm 2 αm 3 αm . . . 1 x x2 x3 . . . d0 (1) ¡ 5¡ &¡ ¡ £ 3¡ £ %¡ £ P ¡ ¡ ¢ ¡ £ 3¡ H G£ F ¡ § D ¡ ¡ £ ¢ ¡ E £ P P P P P P 6¡ ¡ ¡ ¢ ¡ ¡ ¢ £ %¡ D 8. Denote β α r . Since gcd r, n 1, we have o β o α gcd o α , r n. Thus β is b s for some s. It follows that the zeros also a primitive nth root of unity, and therefore α β of may be written as α b , α b r , α b 2r , . . . , α b δ 2 r βs , βs 1 , βs 2 , . . . , βs δ 2 and the minimum distance of is at least δ by the BCH bound.
S £ ¡ ¡ S S ) 1¡ £ £ `¡ c £ bS a S £ 6¡ S £ ...
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This note was uploaded on 10/12/2009 for the course COMM ECE259A taught by Professor Prof.alexandervardy during the Fall '08 term at San Diego.
 Fall '08
 Prof.AlexanderVardy

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