Başlığıma hayır, sorumun vücuduna evet. Aslında bu hemen genelleştirir
gelmez her polinom uzunlukta oyuna değil düşmanlar kodunu kullanın.
Kullanacağımı unutmayın C düşmanlar için değil, A,
Teorem 2 gösterimi
ile eşleşecek şekilde .
Hemen hemen tüm kehanetler için O, bir P / poli var
oracle algoritması varC öyle ki Prx[R(x,CO(x))] ihmal edilemez.
Hemen hemen tüm kehanetler için O,
en fazla d + n d boyutunda bir dizi devre olacak şekilde pozitif bir tamsayı d
vardır .
Prx∈{0,1}n[R(x,CO(x))] sonsuz-sık sık 1/(nd).
Sayılabilir katkı maddesi ile, sıfır olmayan bir oracles kümesi için pozitif bir tamsayı d vardır. O, en fazla d + n d boyutunda bir dizi devre vardır, öyle ki
Prx∈{0,1}n[R(x,CO(x))] sonsuz-sık sık 1/(nd).
J böyle bir reklam olsun ve z
, n'yi girdi olarak alan ve sözlükbilimsel olarak en az oracle devresini en fazla j + n olarak çıkaran (zorunlu olarak verimli olmayan) oracle algoritması olsun.
j
en üst düzeye çıkarmak Prx∈{0,1}n[R(x,CO(x))]. Borel-Cantelli'nin kontraseptifiyle ,1/(n2)<ProbO[1/(nj)<Prx∈{0,1}n[R(x,(zO)O(x))]]sonsuz sayıda için
Böyle n için,
1/(n2+j)=1/((n2)⋅(nj))=(1/(n2))⋅(1/(nj))<ProbO,x∈{0,1}n[R(x,(zO)O(x))]
.
İzin VermekA be the oracle-algorithm that takes 2 inputs, one of which is n, and does as follows:
Choose a random n-bit string x. Attempt to
[parse the other input as an oracle-circuit and run that oracle-circuit on the n-bit string].
If that succeeds and the oracle-circuit's output y satisfies R(x,y), then output 1, else output 0.
(Note that A is not just the adversary.)
For infinitely many n, 1/(n2+j)<ProbO[AO(n,zO(n))].
Let p be as in Theorem 2, and set f=2⋅p⋅(j+nj)⋅n(2+j)⋅2.
By Theorem 2, there exists an oracle-function S such that with P as in that theorem,
if 1/(n2+j)<ProbO[AO(n,zO(n))] then
1/(2⋅(n2+j))=(1/(n2+j))−(1/(2⋅(n2+j)))=(1/(n2+j))−1/(2⋅2⋅(n(2+j)⋅2))−−−−−−−−−−−−√
=(1/(n2+j))−(p⋅(j+nj))/(2⋅2⋅p⋅(j+nj)⋅(n(2+j)⋅2))−−−−−−−−−−−−−−−−−−−−−−−−−−√=(1/(n2+j))−(p⋅(j+nj))/(2⋅f)−−−−−−−−−−−−√
<ProbO[AO(n,zO(n))]−(p⋅(j+nj))/(2⋅f)−−−−−−−−−−−−√≤ProbO[AP(n,zO(n))].
For n such that 1/(n2+j)<ProbO[AO(n,zO(n))]:
In particular, there exists [an oracle-circuit C of size at most j+nj] and
[an assignment of length at most f] such that with that input and presampling,
A's probability of outputting 1 is greater than 1/(2⋅(n2+j)).
Oracle-circuits of size at most j+nj can be represented with poly(n) bits, so for p is bounded
above by a polynomial in n, which means f is also bounded above by a polynomial in n.
By construction of A, that means there are oracle-circuits of size at most j+nj and a
polynomial-length assignment such that when run with that presampling, the circuits' probability of finding a solution is greater than 1/(2⋅(n2+j)). Since such circuits cannot make queries longer than j+nj bits, presampled inputs longer than that can be ignored, so such presampling can be efficiently-and-perfectly simulated with a random oracle and poly(n) hard-coded bits. That means there are polynomial-size oracle circuits such that with a standard random oracle, the circuits' probability of finding a solution is greater than 1/(2⋅(n2+j)). Such a random oracle can in turn be efficiently-and-perfectly simulated with just ordinary random bits, so there are polynomial-size probabilistic non-oracle circuits whose probability of finding a solution is greater than 1/(2⋅(n2+j)). In turn, by hard-coding optical randomness, there are polynomial-size deterministic (non-oracle) circuits whose probability (over the choice of x) of finding a solution is greater than 1/(2⋅(n2+j)).
As shown earlier in this answer, there are infinitely many n such that 1/(n2+j)<ProbO[AO(n,zO(n))], so there is a polynomial such that
the sequence whose n-th entry is the lexicographically least
[circuit C of size bounded above by that polynomial] which maximizes Prx∈{0,1}n[R(x,C(x))]
is a P/poly algorithm whose probability (over the choice of x) of finding a solution is non-negligible.
Therefore the implication's in my question's body always hold.
To get the same implication for other polynomial-length games, just
change this proof's A to make it have the input oracle-circuits play the game.