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The Goldreich-Levin Theorem: List-decoding the Hadamard code

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Presentation on theme: "The Goldreich-Levin Theorem: List-decoding the Hadamard code"— Presentation transcript:

1 The Goldreich-Levin Theorem: List-decoding the Hadamard code
Amnon Aaronsohn ECC Course, TAU

2 Outline Motivation Probability review Theorem and proof

3 Decoding Fix an (n, k, d) code C, and suppose there is an unknown message sk We are given a vector yn which is equal to the codeword C(s) with at most m of the places corrupted Suppose we want to find possible values sk for the original message so that dH(C(s),y)m If m<d/2 then there's a unique solution If d/2<m<d there could be multiple solutions

4 Hadamard Codes [2n, n, 2n-1]2 linear code
The encoding for a message xFn is given by all 2n scalar products <x,y> for yFn (Note: all string related math here is mod 2.) Why is the relative distance 1/2? We will see a probabilistic algorithm that provides list decoding for Hadamard codes when up to 1/2-e of the bits are corrupted

5 Basic probability theory review
Random variables (discrete) Expected value (m) E(X) = Sxp(x) Variance (s2) Var(X) = E[(X-E(X))2] = E[X2]-E[X]2

6 Binary random variables
Pr(X=1)=p, Pr(X=0)=1-p Often used as indicator variables E(X)=… Var(X) = p(1-p) ≤ 1/4

7 Majority votes Consider a probabilistic algorithm that returns a binary value (0 or 1), with probability > 1/2 of returning the correct result We can amplify the probability of getting the correct answer by calling the algorithm multiple times and deciding by the majority vote In order for this to work well there should be some independence between the algorithm’s results in each invocation

8 Independence Events A1,...,An are independent if
Pr[A1,...,An] = Pr[A1]...Pr[An] Likewise, random variables X1,...,Xn are independent if for each possible assignment x1,...,xn: Pr[X1=x1,...,Xn=xn] = Pr[X1=x1]...Pr[Xn=xn]

9 Pairwise independence
A set of r.v.'s (or events) is pairwise independent if each pair of the set is independent Does one type of independence imply the other?

10 Example: xors of random bits
Let X1,…,Xk be independent binary r.v.’s with p=1/2 For each non-empty subset of indexes J define XJ = iJ xi (= SiJ xi) The XJs are (1) uniformly distributed (2) not mutually independent (3) pairwise independent Can be trivially extended to random vectors

11 Chernoff bound Reminder: we want to improve the accuracy of an algorithm by calling it multiple times and deciding by majority vote The probability of not getting a simultaneous occurance of the majority of n independent events, each having probability p≥1/2+e, has the upper bound Pr(error) ≤ exp{-2ne2}

12 Chebyshev inequality For any r.v. X with expected value μ and variance s2: Pr(|X-m|≥a) ≤ s2/a2 Can be used to get an upper bound for the probability of not getting a majority of n pairwise independent events with p≥1/2+e: Pr(error) ≤ 1/(4ne2)

13 Back to the decoding problem
Message space {0,1}n Think of codewords as binary functions: c=Had(s)  x c(x)=<s,x> Input: function f:{0,1}n{0,1}, representing a codeword with noise Output: a list L of possible messages s.t. for each sL, f agrees with Had(s) at p fraction of the function inputs: Prx[f(x)=<s,x>] = p Time complexity in terms of calls to f

14 No error case: p = 1 Unique decoding
In this case we can recover the ith bit of the message by computing f(ei) where ei is the string with 1 at the ith position and 0 everywhere else.

15 Low error case: p = 3/4+e Unique decoding
Why not simply use f(ei) as before? Probabilistic algorithm: Estimate-Had(x): For j = 1…k (k to be fixed) Choose rj{0,1}n randomly aj f(rj+x) - f(rj) Return majority(a1,…,ak) Now set the ith bit of the solution to Estimate-Had(ei)

16 Analysis Consider this part:
Choose rj{0,1}n randomly aj f(rj+x) - f(rj) If both f(rj+x) and f(rj) are correct then aj = f(rj+x) - f(rj) = <s, rj+x> - <s, rj> = <s,x> Using a union bound we get Pr[aj <s,x>] ≤ 2(1-p) = 1/2-2e

17 Analysis (contd.) Since we take a majority vote of a1,…,ak we can use the fact that they’re independent to get a Chernoff bound of at most e-(ke2) on the probability of error The probability of getting some bit wrong is Pr[Estimate-Had(ei) is wrong for some i] ≤ ne-(ke2) Taking k = O(logn/e2) gives an O(nlogn/e2) algorithm with arbitrarily small error Note that the error probability is doubled, so doesn’t work with p<3/4

18 General case: p = 1/2+e List decoding
The Goldreich-Levin theorem gives a probabilistic algorithm for this problem. Specifically: Input: Function f() as before Output: List L of strings such that each possible solution s appears with high probability: Prx[f(x)=<s,x>] ≥ 1/2+e  Pr[sL] ≥1/2 Run time: Poly(n/e)

19 The algorithm (almost)
Suppose that we somehow know the values of Had(s) in m places. Specifically, we are given the strings r1,…,rm and the values b1,…,bm where bj = <s,rj>, for an unknown s We can then try to compute the value of Had(s) in any x: Estimate-With-Guess(x , r1,…,rm , b1,…,bm): For J {1,...,m} (Jf) aJ f(x+SjJ rj) - SjJ bj Return majority of all aJ Now get the bits of s by calling Estimate-With-Guess with ei as before

20 Analysis The idea here is that due to linearity we can get the correct values in more places than we are given For any J {1,...,m} define rJ=SjJ rj. Then <s, rJ>=<s, SjJrj>=SjJ<s, rj >=SjJ bj If the rjs are uniformly random so are the rJs The probability of getting aJ wrong is therefore the probability of getting f(x+rJ) wrong, which is bounded by 1/2-e

21 But! The rJs are not independent, so Chernoff bound can’t be used
However, they are pairwise independent so we can use Chebyshev Pr[EWG(x , r1,…,rm , b1,…,bm) <s,x>] ≤ 1/(2me2) when the ris are independent and chosen uniformly and for each i, bi=<s,ri> We can recover all bits with an error of at most n/(2me2). Taking 2m = O(n/e2) gives an O(n2/e2) algorithm with arbitrarily small error

22 Completing the algorithm
We don’t actually have the correct values for the bis But if m is small we can try all 2m combinations – for each solution one of them must be correct! The final algorithm: 1. Choose r1,…,rm randomly 2. For each (b1,…,bm){0,1}m: 2.1 For i=1,..,n aiEWG(ei , r1,…,rm , b1,…,bm) 2.2 Output (a1,…,an) Complexity: O(n3/e4)

23 Finally Now that we can generate a list where every possible solution appears with probability  1/2, we can re-run the algorithm a constant number of times to get an arbitrary small probability to miss a given solution

24 Summary We saw a list decoding algorithm for Hadamard code, enumerating with high probability all strings with distance arbitrarily close to 1/2 to a given string Sample f() at uniformly distributed points so that the adversary won’t be able to affect result Generate points in a linear subspace spanned by a small number of points, for which we can try all combinations Results in pairwise independent trials, so we can apply Chebyshev inequality

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