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A Quadratic Lower Bound for Three-Query Linear Locally Decodable Codes Over Any Field

A Quadratic Lower Bound for Three-Query Linear Locally Decodable Codes Over Any Field. David Woodruff IBM Almaden. Linear Locally Decodable Codes.

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A Quadratic Lower Bound for Three-Query Linear Locally Decodable Codes Over Any Field

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  1. A Quadratic Lower Bound for Three-Query Linear Locally Decodable Codes Over Any Field David Woodruff IBM Almaden

  2. Linear Locally Decodable Codes A (q, , )-linearlocally decodable code (LDC) C: Fn! Fm is a linear map with a (possibly adaptive) decoder A such that 8x 2 Fn, 8 i 2 [n] and 8 y for which (y, C(x)) < m, 1. Pr [A(i, y) = xi] ¸ 1/|F| +  2. A queries at most q positions of y

  3. Applications • LDCs • Local Decoding of Large Files • Private Information Retrieval • Complexity Theory • Linearity important for various applications • Succinct representation / efficient encoding • Streaming • Matrix rigidity • … • Infinite fields, e.g., F = R, important in these applications

  4. Previous Bounds • Assume ,  are constants • q = 1: LDCs do not exist [KT] • q = 2:m = exp(n) for any field F [DS, KdW, Hadamard] • q = 3:for general |F|, only trivial m = (n) Upper bounds have m super-polynomial in n, but sub-exponential [Y, E]

  5. Our Result For any constants , , and any field F, we show for q = 3, any linear LDC C: Fn! Fm satisfies m = (n2) First non-trivial lower-bound for general F -Result holds, e.g., if F is the field of real or complex numbers

  6. Proof Overview 1. Make the decoder non-adaptive 2. Find a special set in the recovery graphs 3. The projection 4. Recursive projection

  7. Making the Decoder Non-Adaptive Lemma: Any (3, , )-LDC can be made into a (3, ’, ’)-LDC with a non-adaptive decoder -If,  are constant, then so are ’, ’ - Works for any field F Combining this with previous work, seems possible to get - (n2 / (|F| log2 n)) bound for q = 3 [KdW] - (n3/2) bound for q = 3 for any F [KT] In contrast, we will get (n2) for any F

  8. Proof of Lemma Lemma: Any (3, , )-LDC can be made into a (3, ’, ’)-LDC with a non-adaptive decoder -If,  are constant, then so are ’, ’ - Works for any field F Proof: Since LDC is linear, there are vectors v1, …, vm2 Fn for which for all x: C(x) = <v1, x>, <v2, x>, …, <vm, x> For each standard unit vector ei, there must be a matching Mi of (m) disjoint triples {vi1, vi2, vi3} which span ei The new decoder chooses such a triple uniformly at random

  9. Proof Overview 1. Make the decoder non-adaptive 2. Find a special set in the recovery graphs 3. The projection 4. Recursive projection

  10. The Recovery Hypergraphs • Vertices of G are v1, …, vm • Hyperedges of G are 3-edges which occur in some Mi • the 3-edge is then labeled ei • (mn) hyperedges • Lemma: there is a non-empty sub-hypergraph G’ µ G with minimum degree ¯n for a constant ¯ > 0 • Proof: iteratively remove minimum degree vertex until minimum degree larger than original average degree / 3

  11. Finding a Special Set • Choose any v 2 G’, and consider N(v), its set of neighbors in G’ • Since v has degree >¯n and occurs at most once in each Mi, from {v} [ N(v), we can span (n) different ei • Hence, |N(v)| = (n)

  12. The Picture v … e3 en e1 N(v) … e9 e2 e1 e101 N(N(v)) We know m ¸ |N(N(v))|, so let’s lower bound |N(N(v))|

  13. Proof Overview 1. Make the decoder non-adaptive 2. Find a special set in the recovery graphs 3. The projection 4. Recursive projection

  14. Setup • We have a minimum degree ¯n hypergraph G’ whose hyperedges are sets in Mi • We have found a set N(v) with |N(v)| > ¯ n • By definition of G’,N(v) is incident to (n2) hyperedges. • Let S ½ N(v) be an arbitrary subset of ¯ n/2 linearly independent vertices • Let E be a set of n-¯ n/2 standard unit vectors for which E [ S is a basis of Fn

  15. The Projection • By definition of G’,N(v) is incident to (n2) hyperedges. • Let S ½ N(v) be an arbitrary subset of ¯ n/2 linearly independent vertices • Let E be a set of n-¯ n/2 standard unit vectors for which E [ S is a basis of Fn Define a linear map L: L(s) = 0 for all s 2 S L(e) = e for all e 2 E

  16. The New Picture v … e3 en e1 0 a5 a6 S a3 a4 0 a1 a2 0 0 0 0 … 3-edges preserved by L e9 e2 e1 e101 N(S) c3 c4 c5 c6 b7 c7 c2 b3 b4 b5 b6 b1 c1 b2 Apply linear map L

  17. Reduction to Two Queries 0 0 0 0 0 0 … e9 e2 e1 e101 c3 c4 c5 c6 c7 c1 c2 • Each vertex in S has degree > ¯n, so at least ¯ n/2 • of 3-edges incident to it are preserved by L • We get (n2)2-edges on a set of |N(S)| vertices

  18. Isoperimetric Inequality • [Bollobas, GKST, DS]: Given r vectors in Fn for which for each i 2 [n], there is a matching Mi’ of 2-edges for which for each {a, b} 2 Mi’, ei2 span(a, b), then: • r log r = (Σi=1n |Mi’|) • In our setting, • r = |N(S)| • Σi=1n |Mi’| = (n2) • Hence, m ¸ |N(S)| = (n2/log n)

  19. Proof Overview 1. Make the decoder non-adaptive 2. Find a special set in the recovery graphs 3. The projection 4. Recursive projection

  20. Boosting the Lower Bound • If |N(S)| = (n2), then done. So suppose |N(S)| = o(n2) • Let’s also project a random constant fraction of eito 0 0 … e3 en e1 0 0 0 0 0 0 connected components … e9 e2 e1 e101 • Each component has rank at most 1

  21. The Connected Components • We get a collection of rank 1 components: … Can’t have a bunch of components with a constant number of vertices … C1 C2 Cs • With good probability,(n2) edges from S to N(S) have their labels projected to 0 • Let cibe the # of vertices in Ci, let e(Ci) be the # of edges • Σi=1s ci· |N(S)| = o(n2) • Σi=1s e(Ci) = (n2) • But, each Ci obeys isoperimetric inequality! • e(Ci) · ci log ci Projecting one vertex in each of the n/100 largest components to zero, we project a much larger set of vertices to 0

  22. Wrapping Up • To get (n2), repeatedly project large connected components to 0, then a new fraction of standard unit vectors adjacent to these components, obtaining new components, etc. • Gradually enlarge the set of vertices that is projected to 0 while preserving a large fraction of standard unit vectors • Summary: we show for any constants , , and any field F, any 3-query linear LDC C: Fn! Fm satisfies m = (n2)

  23. An (n2/log log n) Lower Bound Suppose |N(S)| = o(n2/log log n): • Σi=1s ci = o(n2/ log log n) • Σi=1s e(Ci) = (n2), and soΣi=1s ci log ci = (n2) Order the Ci so that c1¸ c2¸ … ¸ cs Lemma:Σi=1t ci = (n log n / log log n) for t = n/100 Proof: If not, then cj for j ¸ t, is o(log n / log log n). But then: (n2) = Σi=1s e(Ci) · o(n log2 n / log log n) + Σj=(s+1)t cj log cj But Σj=(s+1)t cj log cj = o(Σi=1s ci log log n), a contradiction. If we project one vertex in each of C1, …, Ct to 0, we project a set of (n log n / log log n) vertices to 0. New set incident to (n2 log n / log log n)3-edges, and we can lower bound neighborhood

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