on uniform amplification of hardness in np luca trevisan stoc 05 paper review present by hai xu
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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)
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
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