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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:

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Presentation on theme: "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:"— Presentation transcript:

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,b s.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 F 7, 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-r i ). Not closed eg: x 2 +x+1 over F 2

5 Q2. Prime-order fields (contd.) a±b  (a±b)mod(p), cost O(log(p)) a.b  (a.b)mod(p), cost O(log 2 (p)) (why?) a b  (a b )mod(p), cost O(log 3 (p)) (generate a, a 2,a 4,… in time O(log 3 (p)), then multiply subset also in time O(log 3 (p)) ) log a b  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(log 3 (p))

6 Q3.Prime-power-order fields Analogue – a ≅ a(x) (with coeffs from F p ) – p ≅ p(x) (prime ≅ “irreducible” (no factors)) … If p(x) irreducible, consider F(x)(mod p(x))… – Eg: x 2 +1 no solutions over R, but over C=R(x)/(x 2 +1)… Bits…

7 Q4. Linear algebra over finite fields Yes No. Example: (1 1) over F 2. No. Yes

8 S-Z Lemma (easy case) If P(x) has degree d, then at most n roots. – Pr a in F (P(a) = 0) ≤d/q If P(x 1,x 2,…,x k ) has degree d, then – Pr a1,a2,…,ak in F (P(a1,a2,…,ak) = 0) ≤d/q (Proof by Induction) – degree(x 2 y 5 +x 4 y 4 ) = 8 by definition

9 Q5. Rank of random matrices m/q – mxm matrix M=(x ij ). – 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(n 2 ) (Why?) – Decoding time = O(n 3 ) (Why?) – Storage O(n 2 ) – Design time O(n 2 )

11 Q7. Prop. of Linear codes x=Gm, 0=Hx – No. GT and T’H, for any invertible T, T’ – [G -I].[H T I T ] T =[0] x,y in C means (x-y) in C (why?) Complexity: – Encoding: O(n 2 ) – BSC(p) decoding: O(exp(n)) (naïve)

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

13 Q9. Singleton Bound n n-d+1 d-1 q n-d+1 ≤q nR

14 Q10. Reed-Solomon encoding nR (m-m’)(x-x’) = n-nR=d min nR=n-d min 0 m=m’ Determinant(Vandermonde matrix) = r i distinct, q≥n.

15 11. q-BSC(p) Say q=2 m, – Append (say) m’ = m 1/2 zeroes to each packet. – Detect errors (w.p. ~ 2m’). – Use erasure code to decode. Random vs. worst-case noise Naïve: O(n 2 ), O(n 3 ), 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 – x i = M(r i ). – Define “error-locator polynomial” E(r i )= – Define q(r,y) = E(r)(y-M(r)) – q(r i,y i )=0 (why?) – E(r i )y i =E(r i )M(r i )=T(r i ) (definition) – T(.) of degree k+t-1 in r, and E(r i ) 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(r i )E’(r i )y i =T’(r i )E(r i )y i. If y i = 0, then T(r i )=T’(r i ) If y i ≠ 0, then T(r i )/E(r i )=T’(r i )/E’(r i ) Degree of M(r) = T(t)/E(r) at most k-1, hence must be equal.


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