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Fields: Defns

Fields: Defns. “Closed”: a,b in F  a+b , a.b in F Properties: Commutative: a+b = b+a , a.b = b.a Associative: a+(b+c )=( a+b)+c , a.(b.c ) = ( a.b).c Distributive: a.(b+c )= a.b+a.c a+0=0+a=a, a.1=1.a=a a+(-a)=0, a.a -1 =1. Facts about fields.

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Fields: Defns

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  1. Fields: Defns • “Closed”: a,b in F a+b, a.b in F • Properties: • Commutative: a+b=b+a, a.b=b.a • Associative: a+(b+c)=(a+b)+c, a.(b.c) = (a.b).c • Distributive: a.(b+c)=a.b+a.c • a+0=0+a=a, a.1=1.a=a • a+(-a)=0, a.a-1=1

  2. Facts about fields • Examples: Q, R, C, P(x)/Q(x) if P(x),Q(x) in F(x),… • Non-examples: Z, P(x) in F(x), … • Algebraically closed: C • roots of P(x) in C(x) must be in C (Fundamental theorem of algebra) • Not algebraically closed: C • roots of P(x) in R(x) may not be in C

  3. Q1. “Useful facts” about finite F • Characteristic: • Finite (else infinite field) • Prime (else exist non-zero a,bs.t. a.b = 0) • Closed set under + and scalar ., other props “Must be” n copies of set of characteristic p. • Let the set (“group”) generated by powers of a be H. Then all sets of the form aH have the same size and are disjoint (bijection). Hence |H| divides |F|. Hence… • Eg: 3 in F7, but not 2.

  4. Q2. Prime-order fields • (a+b)mod(p), (a.b)mod(p) • … • -a = p-a, a-1 = a|F|-1 (why?) • Hint: Binomial theorem, mod p,… • Keep dividing P(x) by (x-ri). Not closed eg: x2+x+1 over F2

  5. Q2. Prime-order fields (contd.) • a±b (a±b)mod(p), cost O(log(p)) • a.b (a.b)mod(p), cost O(log2(p)) (why?) • ab (ab)mod(p), cost O(log3(p)) (generate a, a2 ,a4,… in time O(log3(p)), then multiply subset also in time O(log3(p)) ) • logabHARD (brute force, O(p.poly(log(p)) • a/ba. b-1 • mb+np=1 (Euclid’s algorithm, find m) O(…?) • b|F|-1 ,cost O(log3(p))

  6. Q3.Prime-power-order fields • Analogue • a≅a(x) (with coeffs from Fp) • p≅p(x) (prime≅“irreducible” (no factors)) • … • If p(x) irreducible, consider F(x)(modp(x))… • Eg: x2+1 no solutions over R, but over C=R(x)/(x2+1)… • Bits…

  7. Q4. Linear algebra over finite fields • Yes • Yes • Yes • Yes • No. Example: (1 1) over F2. • No. • Yes • Yes • Yes

  8. S-Z Lemma (easy case) • If P(x) has degree d, then at most n roots. • Pra in F(P(a) = 0) ≤d/q • If P(x1,x2,…,xk) has degree d, then • Pra1,a2,…,ak in F(P(a1,a2,…,ak) = 0) ≤d/q • (Proof by Induction) • degree(x2y5+x4y4) = 8 by definition

  9. Q5. Rank of random matrices • m/q • mxm matrix M=(xij). • Det(M) polynomial of degree m • (1-q-n) (1-q-n+1)…(1-q-n+m+1)≥(1-q-n+m+1)m ≥1-mq-n+m+1 If n>(1+ε)m, ≈1-mq-mε

  10. Q6. BEC(p) • Prev question, q=2, R=…? • Approx pn bits erased • Complexity • Encoding time = O(n2) (Why?) • Decoding time = O(n3) (Why?) • Storage O(n2) • Design time O(n2)

  11. Q7. Prop. of Linear codes • x=Gm, 0=Hx • No. GT and T’H, for any invertible T, T’ • [G -I].[HT IT]T =[0] • x,y in C means (x-y) in C (why?) • Complexity: • Encoding: O(n2) • BSC(p) decoding: O(exp(n)) (naïve)

  12. Q8. Linear GV codes • Let xi be codeword with “low” weight d= dmin. • # codewords of weight at most d ~2nH(d) • PrG(Gx≠0 for all x of low wt) < (2nH(d). 2-n). 2-nR • Probabilistic method…

  13. Q9. Singleton Bound n qn-d+1≤qnR d-1 n-d+1

  14. Q10. Reed-Solomon encoding • Determinant(Vandermonde matrix) = ri distinct, q≥n. nR (x-x’) (m-m’) 0 m=m’ nR=n-dmin = n-nR=dmin

  15. 11. q-BSC(p) • Say q=2m, • Append (say) m’ = m1/2 zeroes to each packet. • Detect errors (w.p. ~ 2m’). • Use erasure code to decode. • Random vs. worst-case noise • Naïve: O(n2), O(n3), O(n), O(n) • (Can “cleverly” do O(n.log(n)), O(n.log(n)), O(1), O(1) – how?)

  16. 12. Reed-Solomon decoding • Note • xi = M(ri). • Define “error-locator polynomial” E(ri)= • Define q(r,y) = E(r)(y-M(r)) • q(ri,yi)=0 (why?) • E(ri)yi=E(ri)M(ri)=T(ri) (definition) • T(.) of degree k+t-1 in r, and E(ri) of degree t, hence # unknown coefficients k+2t+1 ≤ n, linear transform • Not unique (null-space), but only interested in T(r)/E(r). • This unique since T(ri)E’(ri)yi=T’(ri)E(ri)yi. • If yi= 0, then T(ri)=T’(ri) • If yi≠ 0, then T(ri)/E(ri)=T’(ri)/E’(ri) • Degree of M(r) = T(t)/E(r) at most k-1, hence must be equal.

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