Erasure Codes. Erasure Code: Example. Example Make polynomial, P ( - PowerPoint PPT Presentation

Erasure Codes. Erasure Code: Example. Example Make polynomial, P ( x ) = a 2 x 2 + a 1 x + a 0 Satellite n packet message. So send n + k ! with P ( 1 ) = 1, P ( 2 ) = 4, P ( 3 ) = 4. Send message of 1,4, and 4. Modulo 7 to accommodate at least 6 packets. Make polynomial, P ( x ) = a 2 x 2 + a 1 x + a 0 ··· n + k 1 2 Linear equations: ······ with P ( 1 ) = 1, P ( 2 ) = 4, P ( 3 ) = 4. How? P ( 1 ) = a 2 + a 1 + a 0 ≡ 1 ( mod 7 ) Lose k packets. Lagrange Interpolation. P ( 2 ) = 4 a 2 + 2 a 1 + a 0 ≡ 4 ( mod 7 ) Linear System. ··· n + k 1 2 P ( 3 ) = 2 a 2 + 3 a 1 + a 0 ≡ 4 ( mod 7 ) ······ Work modulo 5. 6 a 1 + 3 a 0 = 2 ( mod 7 ) , 5 a 1 + 4 a 0 = 0 ( mod 7 ) P ( x ) = x 2 ( mod 5 ) P ( 1 ) = 1 , P ( 2 ) = 4 , P ( 3 ) = 9 = 4 ( mod 5 ) a 1 = 2 a 0 ( mod 7 ) a 0 = 2 ( mod 7 ) a 1 = 4 ( mod 7 ) a 2 = 2 ( mod 7 ) GPS device Any n packets is enough! P ( x ) = 2 x 2 + 4 x + 2 Send ( 0 , P ( 0 )) ... ( 5 , P ( 5 )) . 6 points. Better work modulo 7 at least! P ( 1 ) = 1, P ( 2 ) = 4, and P ( 3 ) = 4 n packet message. Why? ( 0 , P ( 0 )) = ( 5 , P ( 5 )) ( mod 5 ) Send Packets: ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 4 ) , ( 4 , 7 ) , ( 5 , 2 ) , ( 6 , 0 ) Optimal. Notice that packets contain “x-values”. Bad reception! Questions for Review Polynomials. Send: ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 4 ) , ( 4 , 7 ) , ( 5 , 2 ) , ( 6 , 0 ) Recieve: (1,1) (3,4), (6,0) You want to encode a secret consisting of 1,4,4. Reconstruct? How big should modulus be? Format: ( i , R ( i )) . ◮ ..give Secret Sharing. Larger than 144 and prime! Lagrange or linear equations. ◮ ..give Erasure Codes. You want to send a message consisting of packets 1,4,2,3,0 P ( 1 ) = a 2 + a 1 + a 0 ≡ 1 ( mod 7 ) through a noisy channel that loses 3 packets. Error Correction: P ( 2 ) = 4 a 2 + 2 a 1 + a 0 ≡ 4 ( mod 7 ) How big should modulus be? Noisy Channel: corrupts k packets. (rather than loss.) Larger than 8 and prime! P ( 6 ) = 2 a 2 + 3 a 1 + a 0 ≡ 0 ( mod 7 ) Additional Challenge: Finding which packets are corrupt. Send n packets b -bit packets, with k errors. Modulus should be larger than n + k and also larger than 2 b . Channeling my inner linear algebra genius ... P ( x ) = 2 x 2 + 4 x + 2 Message? P ( 1 ) = 1 , P ( 2 ) = 4 , P ( 3 ) = 4.

Error Correction The Scheme. Properties: proof. P ( x ) : degree n − 1 polynomial. Problem: Communicate n packets m 1 ,..., m n Send P ( 1 ) ,..., P ( n + 2 k ) on noisy channel that corrupts ≤ k packets. Receive R ( 1 ) ,..., R ( n + 2 k ) Satellite 3 packet message. Send 5. At most k i ’s where P ( i ) � = R ( i ) . Reed-Solomon Code: Properties: 1 2 3 1 2 1. Make a polynomial, P ( x ) of degree n − 1, (1) P ( i ) = R ( i ) for at least n + k points i , that encodes message. A B C D E (2) P ( x ) is unique degree n − 1 polynomial ◮ P ( 1 ) = m 1 ,..., P ( n ) = m n . that contains ≥ n + k received points. Corrupts 1 packets. ◮ Comment: could encode with packets as coefficients. Proof: 3 1 2 1 2 (1) Sure. Only k corruptions. 2. Send P ( 1 ) ,..., P ( n + 2 k ) . A B’ C D E (2) Degree n − 1 polynomial Q ( x ) consistent with n + k points. Q ( x ) agrees with R ( i ) , n + k times. After noisy channel: Recieve values R ( 1 ) ,..., R ( n + 2 k ) . P ( x ) agrees with R ( i ) , n + k times. Properties: Total points contained by both: 2 n + 2 k . P Pigeons. GPS device (1) P ( i ) = R ( i ) for at least n + k points i , : n + 2 k . Total points to choose from H Holes. (2) P ( x ) is unique degree n − 1 polynomial Points contained by both : ≥ n . ≥ P − H Collisions. that contains ≥ n + k received points. = ⇒ Q ( i ) = P ( i ) at n points. = ⇒ Q ( x ) = P ( x ) . Example. Slow solution. Example. Received R ( 1 ) = 3 , R ( 2 ) = 1 , R ( 3 ) = 6 , R ( 4 ) = 0 , R ( 5 ) = 3 Brute Force: For each subset of n + k points Find P ( x ) = p 2 x 2 + p 1 x + p 0 that contains n + k = 3 + 1 points. Fit degree n − 1 polynomial, Q ( x ) , to n of them. Message: 3 , 0 , 6. All equations.. Check if consistent with n + k of the total points. Reed Solomon Code: P ( x ) = x 2 + x + 1 ( mod 7 ) has If yes, output Q ( x ) . P ( 1 ) = 3 , P ( 2 ) = 0 , P ( 3 ) = 6 modulo 7. p 2 + p 1 + p 0 ≡ 3 ( mod 7 ) ◮ For subset of n + k pts where R ( i ) = P ( i ) , Send: P ( 1 ) = 3 , P ( 2 ) = 0 , P ( 3 ) = 6 , P ( 4 ) = 0 , P ( 5 ) = 3 . method will reconstruct P ( x ) ! 4 p 2 + 2 p 1 + p 0 ≡ 1 ( mod 7 ) (Aside: Message in plain text!) 2 p 2 + 3 p 1 + p 0 ≡ 6 ( mod 7 ) ◮ For any subset of n + k pts, Receive R ( 1 ) = 3 , R ( 2 ) = 1 , R ( 3 ) = 6 , R ( 4 ) = 0 , R ( 5 ) = 3 . 2 p 2 + 4 p 1 + p 0 ≡ 0 ( mod 7 ) 1. there is unique degree n − 1 polynomial Q ( x ) that fits n of 1 p 2 + 5 p 1 + p 0 ≡ 3 ( mod 7 ) P ( i ) = R ( i ) for n + k = 3 + 1 = 4 points. them 2. and where Q ( x ) is consistent with n + k points = ⇒ P ( x ) = Q ( x ) . Assume point 1 is wrong and solve..no consistent solution! Reconstructs P ( x ) and only P ( x ) !! Assume point 2 is wrong and solve...consistent solution!

In general.. Ditty... Where oh where can my bad packets be? E ( 1 )( p n − 1 + ··· p 0 ) ≡ R ( 1 ) E ( 1 ) ( mod p ) P ( x ) = p n − 1 x n − 1 + ··· p 0 and receive R ( 1 ) ,... R ( m = n + 2 k ) . 0 × E ( 2 )( p n − 1 2 n − 1 + ··· p 0 ) ≡ R ( 2 ) E ( 2 ) ( mod p ) Oh where, Oh where . has my little dog gone? . p n − 1 + ··· p 0 ≡ R ( 1 ) ( mod p ) . Oh where, oh where can he be p n − 1 2 n − 1 + ··· p 0 E ( m )( p n − 1 ( m ) n − 1 + ··· p 0 ) ≡ R ( 2 ) ( mod p ) ≡ R ( n + 2 k ) E ( m ) ( mod p ) With his ears cut short · And his tail cut long p n − 1 i n − 1 + ··· p 0 Idea: Multiply equation i by 0 if and only if P ( i ) � = R ( i ) . ≡ R ( i ) ( mod p ) Oh where, oh where can he be? All equations satisfied!!!!! · p n − 1 ( m ) n − 1 + ··· p 0 But which equations should we multiply by 0? Where oh where...?? ≡ R ( m ) ( mod p ) Oh where, Oh where We will use a polynomial!!! That we don’t know. But can find! have my packets gone.. wrong? Error!! .... Where??? Errors at points e 1 ,..., e k . (In diagram above, e 1 = 2.) Oh where, oh where do they not fit. Could be anywhere!!! ...so try everywhere. � n + 2 k � Runtime: possibilitities. Error locator polynomial: E ( x ) = ( x − e 1 )( x − e 2 ) ... ( x − e k ) . With the polynomial well put k Something like ( n / k ) k ...Exponential in k !. But the channel a bit wrong E ( i ) = 0 if and only if e j = i for some j Where, oh where do we look? Multiply equations by E ( · ) . (Above E(x) = (x-2).) How do we find where the bad packets are efficiently?!?!?! All equations satisfied!! Example. ..turn their heads each day, Finding Q ( x ) and E ( x ) ? E ( 1 )( p n − 1 + ··· p 0 ) ≡ R ( 1 ) E ( 1 ) ( mod p ) Received R ( 1 ) = 3 , R ( 2 ) = 1 , R ( 3 ) = 6 , R ( 4 ) = 0 , R ( 5 ) = 3 . Find P ( x ) = p 2 x 2 + p 1 x + p 0 that contains n + k = 3 + 1 points. . . ◮ E ( x ) has degree k ... E ( i )( p n − 1 i n − 1 + ··· p 0 ) Plugin points... ≡ R ( i ) E ( i ) ( mod p ) ( 1 − 2 )( p 2 + p 1 + p 0 ) ≡ ( 3 )( 1 − 2 ) ( mod 7 ) . ( 1 − e )( p 2 + p 1 + p 0 ) ≡ ( 3 )( 1 − e ) ( mod 7 ) E ( x ) = x k + b k − 1 x k − 1 ··· b 0 . . . ( 2 − e )( 4 p 2 + 2 p 1 + p 0 ) ( 2 − 2 )( 4 p 2 + 2 p 1 + p 0 ) ≡ ≡ ( 1 )( 2 − e ) ( mod 7 ) ( 1 )( 2 − 2 ) ( mod 7 ) E ( m )( p n − 1 ( n + 2 k ) n − 1 + ··· p 0 ) ≡ R ( m ) E ( m ) ( mod p ) = ⇒ k (unknown) coefficients. Leading coefficient is 1. ( 3 − 2 )( 2 p 2 + 3 p 1 + p 0 ) ≡ ( 6 )( 3 − 2 ) ( mod 7 ) ( 3 − e )( 2 p 2 + 3 p 1 + p 0 ) ≡ ( 3 )( 3 − e ) ( mod 7 ) ◮ Q ( x ) = P ( x ) E ( x ) has degree n + k − 1 ... ( 4 − 2 )( 2 p 2 + 4 p 1 + p 0 ) ( 4 − e )( 2 p 2 + 4 p 1 + p 0 ) ≡ ≡ ( 0 )( 4 − 2 ) ( mod 7 ) ( 0 )( 4 − e ) ( mod 7 ) ...so satisfied, I’m on my way. ( 1 − 2 )( 4 p 2 + 5 p 1 + p 0 ) ≡ ( 3 )( 5 − 2 ) ( mod 7 ) ( 5 − e )( 4 p 2 + 5 p 1 + p 0 ) ≡ ( 3 )( 5 − e ) ( mod 7 ) m = n + 2 k satisfied equations, n + k unknowns. But nonlinear! Q ( x ) = a n + k − 1 x n + k − 1 + a n + k − 2 x n + k − 2 + ··· a 0 Let Q ( x ) = E ( x ) P ( x ) = a n + k − 1 x n + k − 1 + ··· a 0 . Error locator polynomial: ( x − 2 ) . = ⇒ n + k (unknown) coefficients. Equations: Multiply equation i by ( i − 2 ) . All equations satisfied! Q ( i ) = R ( i ) E ( i ) . Number of unknown coefficients: n + 2 k . But don’t know error locator polynomial! Do know form: ( x − e ) . and linear in a i and coefficients of E ( x ) ! 4 unknowns ( p 0 , p 1 , p 2 and e ), 5 nonlinear equations.