A Quadratic Lower Bound for Three-Query Linear Locally

Decodable Codes Over Any Field

David Woodruff

IBM Almaden

Linear Locally Decodable Codes

A (q, , )-linear locally 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

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

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]

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

Proof Overview

1. Make the decoder non-adaptive

2. Find a special set in the recovery graphs

3. The projection

4. Recursive projection

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

Proof of LemmaLemma: 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, …, vm 2 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

Proof Overview

1. Make the decoder non-adaptive

2. Find a special set in the recovery graphs

3. The projection

4. Recursive projection

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

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)

The Picture

v

N(v)e1

…e3 en

N(N(v))

e2

e9e1e101

We know m ¸ |N(N(v))|, so let’s lower bound |N(N(v))|

…

Proof Overview

1. Make the decoder non-adaptive

2. Find a special set in the recovery graphs

3. The projection

4. Recursive projection

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

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

The New Picture

v

Se1

…e3 en

N(S)

e2

e9e1e101

a1 a2a3 a4

a5 a6

b1 b2b3 b4 b5 b6

b7

Apply linear map L

0 0 0 00 0

c1c2

c3 c4 c5 c6 c7

…3-edges

preserved by L

Reduction to Two Queries

e2

e9e1e101

0 0 0 0 0 0

c1 c2c3 c4 c5 c6 c7

…

• 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

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’, ei 2 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)

Proof Overview

1. Make the decoder non-adaptive

2. Find a special set in the recovery graphs

3. The projection

4. Recursive projection

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 ei to 0

e1…e3 en

e2

e9e1e101

…

0

000000

• Each component has rank at most 1

connectedcomponents

The Connected Components• We get a collection of rank 1 components:

…

• With good probability, (n2) edges from S to N(S) have their labels projected to 0

• Let ci be 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

C1 C2… Cs

Can’t have a bunch of components with a constant

number of vertices

Can’t have a bunch of components with a constant

number of vertices

Projecting one vertex in each of the n/100 largest components to zero, we project a much larger set

of vertices to 0

Projecting one vertex in each of the n/100 largest components to zero, we project a much larger set

of vertices to 0

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)

An (n2/log log n) Lower BoundSuppose |N(S)| = o(n2/log log n):

– Σi=1s ci = o(n2/ log log n)

– Σi=1s e(Ci) = (n2), and so Σi=1

s 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=1 s 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