On Uniform Amplification of Hardness in NP

Luca TrevisanSTOC 05

Paper Review

Present by Hai Xu

Uniform Algorithm

Averaged over random inputs Averaged over the internal coin tosses of the

algorithm

Amplification of Hardness

Starting from a problem that is know (or assumed) to be hard on average in a weak sense, we can define a related new problem which is hard on average in the strongest possible sense

Yao’s XOR Lemma

From a Boolean function f: {0,1}n→{0,1}, we define a new function f k(x1,…,xk) = f(x1) … f(xk).

Yao’s XOR Lemma says if every circuit of size ≤ S makes at least a δ faction of errors in computing f(x) for a random x, then every circuit of size ≤ S•poly(δε/k) makes at least a 1/2 – ε fraction of errors in computing f k(), where ε is constant and roughly Ω((1-δ)k)

Amplification of Hardness in NP

Want to prove: if L is a language in NP such that every efficient algorithm (or small family of circuits) errs on at least a 1/poly(n) fraction of inputs of length n, then there is a language L’ also in NP such that every efficient algorithm (or small circuit) errs on a 1/2−1/n(1) fraction of inputs

Yao’s XOR Lemma cannot prove this directly

Previous Work

O’Donnell proved that for every balanced Boolean function f: {0,1}n→{0,1} and parameters ε, δ (>0), there is an integer k = poly(1/ε, 1/δ) and a monotone function g: {0,1}k→{0,1}, such that if every circuit of size S makes at least a δ fraction of errors in computing f(x) given x, then every circuit of size S •poly(ε, δ) makes at least a 1/2 − ε fraction of errors in computing fg,k = g(f(x1), . . , f(xk)) given (x1, . . , xk), then there is a circuit of size poly(1/ε, 1/δ)•S that makes at most a δ fraction of errors in computing f(x)

Balanced Problems

This proof only works for balanced decision problems, i.e., for a random instance of a given length n, there is a probability 1/2 that the answer is YES and a probability 1/2 that the answer is NO

Some Improvement

For balanced problems, Dr. O’Donnell proved the amplification of hardness from 1-1/poly(n) to 1/2 +1/n1/2- ε

For general problems, he proved the amplification of hardness from 1-1/poly(n) to 1/2 +1/n1/3- ε

For balanced problems, Healy et al improved the amplification range from 1-1/poly(n) to 1/2 +1/poly(n)

Limitation of Previous Work

All above proofs are all based on balanced problems

Dr. Trevisan’s Previous Contribution

In FOCS 03, Dr. Trevisan proved:

With every language L in NP, there is a probabilistic

poly-time algorithm that accept probability

≥ 3/4+1/(log n)α with input length n. Then we can

extend this range to 1-1/p(n). He improved the amplification range from

1-1/(log n)α to 3/4 +1/(log n)α, where α > 0 and α=const

Major Contribution of This Paper

The uniform analysis of amplification of hardness using the majority function

Proved lemma that with every language L in NP, there is a probabilistic poly-time algorithm that accept probability≥ 1/2+1/(log n)α with input length n. Then for every language in NP and polynomial p, there is a probabilistic poly-time algorithm that succeeds with probability 1-1/p(n) on input with length n, where α > 0 and α=const

Majority Function

If L NP with input length n and ∈f: {0,1}n→{0,1} to the Boolean function, for an odd integer k, we define:

,1 1

,

( ,..., ) : majority ( ),..., ( )

() is still a balanced function

maj kk k

maj k

f x x f x f x

f

Proof Main Idea I

For every problem in NP, there is an efficient algorithm that solves the problem on a 1- ε fraction of inputs with length n, then for every problem in NP, there is an efficient algorithm that solve the problem on a 1-1/p(n) fraction with input length n with a small amount of non-uniformity

Proof Main Idea II

Based on the balanced function f(), a new function , is introduced. If an efficient algorithm can solve F on a 1- ε fraction of inputs, then there will be an efficient algorithm that can solve f on a fraction of inputs (ε is a positive constant). To simplify the proof procedure, t is set to 1/5 in this paper.

1 ( )

: 0,1 0,1O tn

F

1 1 tn

Proof Main Idea III

For every search problem in NP and every p(), there is an efficient algorithm that solves the search problem on a 1-1/p(n) fraction with input length n and a small amount of non-uniformity

Eliminating the non-uniformity

Detailed Proof of II

We want to prove:– L in NP and L’ is poly bounded by a computable

function l(n)– If circuit C’ solves L’ on α≥1-ε with input length

l(n)– Then another circuit C can solve L on

with input length n

1 51 2 n

Parameter Settings

t = 1/5 a>δ Then we further define:

2 7 and k k c

7 8 6 8 20 8 7: 2a

0 015

7 /8 ,1 1 12/ 7

1

1

1 maj k

i i i i i i i ii

f fn

k f f n n k

Proof

Solve δi and ki recursively, we obtain:

Let r to be the largest index such that δr<α. Then we could know

We also could know that the length of fr is:

11 2 7

5 7 8

1 7

5 8

1,

i

ii ik n

n

8/ 7 2r a

21 2 7 7 7 1 2 161 1 ... 1 8 15 7 8 8 8 5 7 35

1 2...

r

rn k k k n n n

Proof cont’d

Based on majority function, we can let:

g() is the majority function and With recursively apply a lemma, we are able

to obtain circuit C0 agrees f on at least fraction of inputs

1 1( ,..., ) ( ( ),..., ( ))r K Kf x x g f x f x16/35K n

0 15

11 1

n

Proof cont’d

Now we need to create another circuit Cr with input length nK and

Then Cr should agree fraction of inputs

Furthermore, we could construct a circuit C which agrees with f at least faction of inputs

Finally, we conclude that C agrees with L on at least fraction of inputs

16/35K n

1 1 1n K K n

1/51 1 n

1/5 1/51 1 1 1 2n n n

Conclusion

With the proof, we are able to convert a small error into a larger number of error

Generalized the amplification of hardness problem in NP

Input length is an important factor to decide the success probability

Acknowlegement

Thanks for the great help from Fenghui Zhang and Jie Meng

Homework

Mathematically describe one of the major contribution of this paper, and what is the improvement than previous work?

Due on May 3, 2006