Approximate List-Decoding and

Uniform Hardness Amplification

Russell Impagliazzo (UCSD)Ragesh Jaiswal (UCSD)

Valentine Kabanets (SFU)

Hardness Amplification

Given a hard function we can get an even harder function

f FHard function Harder function

Hardness

f

{0, 1}n

s

{0, 1}n

• A function f is called δ-hard for circuits of size s (Algorithm with running time t), if any circuit of size s (Algorithm with running time t) makes mistake in predicting the function on at least δ fraction of the inputs

δ.2n

XOR Lemma

f

f f f

XOR

0/1

0/1

{0, 1}n{0, 1}nk

k

fk

fk:{0, 1}nk {0, 1}fk(x1,…, xk) = f(x1) … f(xk)

XOR Lemma: If f is δ-hard for size s circuits, then fk is (1/2 - ε)-hard for size s’ circuits (ε = e-Ω(δk), s’ =

s·poly(δ, ε))

XOR Lemma Proof: Ideal case

C

A

C (which computes f for at least (1 - δ) fraction of inputs)

(which computes fk for at least (½ + ε) fraction of inputs)

whp

XOR Lemma Proof

C

A

C (which computes f for at least (1 - δ) fraction of inputs)

Advice (|Advice|=poly(1/ε))

C1 Cl

One of them computes f for at least (1 - δ) fraction of inputs

l = 2|Advice| = 2poly(1/ε)

(which computes fk for at least (½ + ε) fraction of inputs)

whp

A “lesser” nonuniform reduction

Optimal List Size Question: What is the reduction in the

list size we should target? A good combinatorial answer using

error correcting codes

C

A

C1 Cl

whp

XOR-based Code [T03]Think of a binary message msg on M=2n bits as a truth-table

of a Boolean function f.The code of msg is of length Mk where code(x1,…,xk) = f(x1)

… f(xk)

msg

x (|x| = n)

x = (x1, …, xk)

f(x1) … f(xk)

code

f(x)

List Decoder

m XOR Encoding c w Decoding

m1,…,ml

Decoder•Local•Approximate•List

≈ (1/2 + )

≈ (1 - δ)

Information theoretically l should be O(1/2)

channel

The List Size

•The proof of Yao’s XOR Lemma yields an approximate local list-decoding algorithm for the XOR-code defined above

• But the list size is 2poly(1/) rather than the optimal poly(1/)

• Goal: Match the information theoretic bound on list-decoding i.e. get advice of length log(1/)

The Main Result

The Main Result

C ((½ + ε)-computes fk)

A

C ((1 - δ)-computes f)

• ε = poly(1/k), δ = O(k-0.1)• Running time of A and size of C is at most poly(|C|, 1/ε)

whp

Advice(|Advice| = log(1/ε))

The Main Result

C ((½ + ε)-computes fk)

A

C ((1 - δ)-computes f)

• ε = poly(1/k), δ = O(k-0.1)• Running time of A and size of C is at most poly(|C|, 1/ε)

w.p. poly(ε)

The Main Result

We get a list size of poly(1/ε) … which is optimal but…

ε is large: ε = poly(1/k)

C((½ + ε)-computes fk)

A

C ((1 - δ)-computes f)

A’

C1Cl

At least one of them (1 - ρ)-computes f

l = poly(1/ε)

Advice(|Advice| = log(1/ε))

whpw.p. poly(ε)

Advice efficient XOR Lemma

Uniform Hardness Amplification

Uniform Hardness Amplification

What we want

f hard wrt BPP g harder wrt BPP

What we get

f hard wrt BPP/log g harder wrt BPPAdvice efficient XOR Lemma

Uniform Hardness Amplification

What we can do:

f Є NP: hard wrt BPP f’ Є NP: hard wrt BPP/log[BDCGL92]

g Є ?? harder wrt BPP

Advice efficient XOR Lemma

• g not necessarily Є NP but g Є PNP||

• PNP||: poly-time TM which can make polynomially many parallel Oracle queries to an NP oracle

h Є PNP||: hard wrt BPP

Simple average-case reduction

g Є PNP||: harder wrt BPP

1/nc ½ - 1/nd

Trevisan gives a weaker reduction (from 1/nc to (1/2 – 1/(log n)α) hardness) but within NP.

Techniques

Techniques

Advice efficient Direct Product Theorem

A Sampling Lemma Learning without Advice

Self-generated advice Fault tolerant learning using faulty advice

Direct Product Theorem

f

f f f

concatenation

0/1

{0, 1}k

{0, 1}n

{0, 1}nk

k

fk:{0, 1}nk {0, 1}k

fk(x1,…, xk) = f(x1) | … | f(xk)

fk

• Direct Product Theorem: If f is δ–hard for size s circuits, then fk is (1 - ε)-hard for size s’ circuits (ε = e-Ω(δk), s’ = s·poly(δ, ε))

• Goldreich-Levin Theorem: XOR Lemma and Direct Product Theorem are saying the same thing

XOR Lemma from Direct Product Theorem

CDP (poly(ε)-computes fk)

A2

C ((1 - δ)-computes f)

•ε = poly(1/k), δ = O(k-0.1)

•Using Goldreich-Levin Theorem

C ((½ + ε)-computes fk)

A1

whp

w.p. poly(ε)

LEARN from [IW97]

LEARN [IW97]

CDP (-computes fk)

Advice: n/2 pairs of (x, f(x)) for independent uniform x’s

C ((1 - δ)-computes f)

whp

•ε = e-Ω(δk)

Goal

LEARN [IW97]

CDP (-computes fk)

Advice: n/2 pairs of (x, f(x)) for independent uniform x’s

C ((1 - δ)-computes f)

whp

•ε = e-Ω(δk)

LEARN’

w.p. poly() No advice!!!

•ε = poly(1/k), δ = O(k-0.1)

• We want to eliminate the advice (or the (x, f(x)) pairs). In exchange we are ready to make some compromise on the success probability of the randomized algorithm

Self-generated advice

Imperfect samples We want to use the circuit CDP to

generate n/ pairs (x, f(x)) for independent uniform x’s

We will settle for n/ pairs (x,bx) The distribution on x’s is statistically close

to uniform and for most x’s we have bx= f(x).

Then run a fault-tolerant version of LEARN on CDP and the generated pairs (x,bx)

How to generate imperfect samples

A Sampling Lemma

nk

2nk

x1 x2xkx3

•D is a Uniform Distribution

A Sampling Lemma

nk

G

x1 x2xkx3

•|G| >= 2nk

•Stat-Dist(D, U) <= ((log 1/)/k)1/2

Getting Imperfect Samples

G: subset of inputs on which CDP(x) = fk(x) |G| >= 2nk

Pick a random k-tuple x, then pick a random subtuple x’ of size k1/2 With probability x lands in the “good” set

G Conditioned on this, the Sampling Lemma

says that x’ is close to being uniformly distributed

If k1/2 > the number of samples required by LEARN,then done!

Else…

Direct Product Amplification

CDP CDP’ which poly(ε)-computes fk’

where (k’)1/2 > n/ε2

??

CDP CDP’ such that for at least poly(ε) fraction of k’-tuples, x CDP’(x) and fk’(x) agree on most bits

Putting Everything Together

CDP for fk CDP’ for fk’

DP Amplification

Sampling

Fault tolerant LEARN

pairs (x,bx)

circuit C (1-)-computes f

with probability > poly()

Repeat poly(1/) times to get a list containing a good circuit for f, w.h.p.

Open Questions

Open Questions

Advice efficient XOR Lemma for smaller For ε > exp(-kα) we get a quasi-polynomial list size

Can we get an advice efficient hardness amplification result using a monotone combination function m (instead of )? Some results: [Buresh-Oppenheim, Kabanets,

Santhanam] use monotone list-decodable codes to re-prove Trevisan’s results for amplification within NP

Thank You