RECUADRO 8 EVOLUCIÓN RECIENTE DEL ACCESO DE LAS PYMES ESPAÑOLAS A LA FINANCIACIÓN EXTERNA
5 COMERCIO EXTERIOR DE BIENES Cifras acumuladas de los últimos doce meses
We prove the following theorem about asynchronous extractor:
Theorem 4.3.49. Assume we have explicit construction of the optimal seeded ex- tractor in Definition 4.3.41. There exist constants c1, c2 > 0 such that if k > log p
and n ≤ poly(p) then as long as t > p0.1, for sufficiently large p, there exists a
(t, p/c1 − c2(log log klog C + 1)t, + 2−k
Ω(1)
) asynchronous extractor that runs in at most
1.1 log C
log log k + 2 rounds in the full-information model. Here is the error of the extractor
in Assumption 4.3.16.
Remark 4.3.50. The parameter p0.1 can be replaced by pδ for any constant δ > 0
and the constant 1.1 can be replaced by 1 + α for any constant α > 0.
To prove this theorem we need the following lemma, which is similar to
Lemma 4.3.27for the synchronous extractor.
Lemma 4.3.51. Let Cl be the number of independent (n, k) sources needed for IExt
or SRIExt in round l of the asynchronous extractor. Define Badl to be the number of
honest players in Al that don’t have a valid somewhere random string for l ≥ 2 and
let Bad1 = 0. Assume we use the optimal seeded extractor inDefinition 4.3.41 in the
1. ∀l, Badl ≤ 2t and Cl = o(log n)
2. If Cl> log k, then Cl+1 ≤ log kc0 Cl
3. If Cl≤ log k, then Cl+1 is a constant C ≤ c0
Proof. Similar as in Lemma 4.3.27, we prove 1 by induction. For l = 1, the number of independent (n, k) sources needed for IExt is C1 = O(log nlog k) = o(log n) by Theo-
rem 3.5.9. Also Bad1 = 0 < 2t.
Let b be the constant promised to exist by Lemma 4.3.39. Now assume for round l, Cl = o(log n) and Badl ≤ 2t. Note bCl = o(log n) = o(log t) as n =
O(poly(p)) and t > p0.1. Let U
l denote the subset of honest players in Al that have
a valid somewhere random source for l ≥ 2, and the subset of honest players in Al
for l = 1, then |Ul| ≥ 3bt − t − Badl ≥ 3(b − 1)t. Thus |Ul|
|Al| ≥
b−1
b , which means the
honest players consist of a large fraction of the players.
Consider Protocol 4.3.45for round l. An AND-disperser G = (N, M, bD, α, β) is constructed with M = |Al| = 3bt, D = Cl, α = b−1b . By Lemma 4.3.39, N ≤
M dbD = M dbCl for some constant d, and there exists a subset V ⊂ [N ] with |V | =
βN > µbDN s.t. Γ(V ) ⊂ Ul. V is the set of “good” tuples that consist of all honest
players (note |V | > 0 as D = o(log N )).
Let W = {v ∈ [N ], |Γ(v)∩Ul| < D} and γ = |W |/N , then β > µbD = 30 > 3γ.
W is the set of “bad” tuples that consist of less tha D = Cl honest players.
Now consider the extractor Ext : [N1] × [D1] → [M1]. As M1 = N and we
of [M1]. Let S1 = {v ∈ [N1],
|Γ(v)∩V |
D1 ≤ β − 0}, S2 = {v ∈ [N1],
|Γ(v)∩W |
D1 ≥ γ + 0}
and S0 = [N1]/(S1 ∪ S2). S0 is the set of players in B that have roughly the right
fraction of “good” and “bad” tuples in the neighbors. As Ext is a (K, 0) extractor,
byProposition 4.3.44|S1| ≤ K and |S2| ≤ K. Therefore |S0| ≥ N1− 2K = |B| − 2K.
For each honest player uj ∈ S0, |Γ(uj) ∩ V | > (β1− 0)D1 and |Γ(uj) ∩ W | <
(γ + 0)D1. Thus uj has at least (1 − γ − 0)D1 + 1 > (1 − 20)D1 + 1 = D2
neighbors in [M1]/W . Each of these neighbors corresponds to a tuple of size bD that
consists of at least D honest players. For each of these tuples, uj will eventually
receive D strings. Thus uj will receive D strings from D2 tuples eventually. Also,
as D2 = (1 − 20)D1 + 1 > (1 − (β − 0))D1 + 1, at least one of these D2 tuples
must be in V and thus consists of all honest players. The strings sent from this tuple and received by uj are all from honest players. Therefore if uj finishes computing
the D2 sjqs, then at least one of them is close to uniform and thus uj obtains a valid
somewhere random string yj. We have at least |S0| − t = |B| − t − 2K such honest
players, therefore eventually every player uj in B will finish computing yj or receive
|B| − t − 2K “complete” and abort. Thus the protocol for any round will eventually end.
Now if no player aborts then all honest players obtain valid somewhere random strings. Otherwise consider the first honest player who aborts, he aborts because he receives |B| − t − 2K “complete”. Let the number of faulty players in B be tB, then
at least |B| − t − 2K − tB “complete” are from honest players that don’t abort. Thus
at least |B| − t − 2K − tB honest players obtain valid somewhere random strings. As
Now consider the number of independent (n, k) sources needed for a player in round l + 1. The somewhere random string yj obtained by player uj ∈ Al+1 at the
end of round l is of size D2 × k, where D2 = (1 − 20)D1 ≤ D1. By Lemma 4.3.42,
D1 = O(KM12 0
polylogN1) = 2O(D)polylog(t). Therefore by Theorem 3.5.10 the number
of independent (n, k) sources needed in round l + 1 is
Cl+1 = O( log D2 log k ) = O( D log k + log log t log k ) ≤ c1( Cl log k + 1) for some constant c1 > 0 as k > log p > log t.
Now it’s clear that as long as Cl = ω(1), Cl+1 < Cl. As C1 = o(log n) we have
Cl = o(log n) for all l. Thus claim 1 holds.
Let c0 = 2c1. If Cl ≤ log k, then Cl+1 ≤ c1(log kCl + 1) ≤ c0 is a constant.
Therefore claim 3 holds.
If Cl> log k, then log kCl > 1 and thus
Cl+1≤ c1( Cl log k + 1) ≤ 2c1( Cl log k) ≤ c0 log kCl. Thus claim 2 holds.
Proof of Theorem 4.3.49. Run Protocol 4.3.47 with round number R to be chosen later. Let Cl be the number of independent (n, k) sources needed in round l. By
Lemma 4.3.51, there exists a constant c0 > 0 such that if Cl ≥ log k, then Cl+1 ≤ c0
CR+1 will be a constant C ≤ c0. In round R + 1 we runProtocol 4.3.34on AR+1where
each ui1 ∈ AR+1 waits to receive C − 1 strings from the other players and compute
zi1 = SRIExt(xi1, xi2, ..., xiC, yi1). By the same analysis in Theorem 4.3.35, at least
|AR|/C − BadR− t ≥ |AR|/c0− 3t honest players end up with private random bits.
By the recursion Cl+1 ≤ log kc0 Cl we get
Cl0 ≤ c0 log k l0−1 C1 = c0 log k l0−1 C
To ensure Cl0 ≤ log k it suffices to have
c0 log k l0−1 C ≤ log k. we get l0 ≥
log C − log log k log log k − log c0
+ 1 = log C − log c0 log log k − log c0
Thus it suffices to take
R =
log C − log c0
log log k − log c0
≤ (1 + o(1)) log C log log k + 1.
Now |AR+1| = p−3bRt, therefore the number of honest players that get private
random bits is at least |AR+1|/c0− 3t = p/c0− 3(bR/c0+ 1)t. In each round when we
apply the extractor, the error will increase by 2−kΩ(1), thus the total error is at most + R2−kΩ(1).
Note that R ≤ (1 + o(1))log log klog C + 1. Thus for p large enough R < 1.1 log Clog log k + 1. Together with the fact C = o(log n) and k > log p = Ω(log n) this implies that the error R2−kΩ(1) = 2−kΩ(1).
Choose c1 = c0, c2 = 3(1.1b/c0+ 1), we have thatProtocol 4.3.47is a (t, p/c1−
c2(log log klog C + 1)t, + 2−k
Ω(1)
) asynchronous extractor that runs in at most 1.1 log Clog log k + 2 rounds in the full-information model.
If we use the extractor in Theorem 4.3.43, then we get the following theorem:
Theorem 4.3.52. There exists a constant c1 > 0 such that if log log log nlog log k = o(1)
and n = poly(p), then as long as t > p0.1, for sufficiently large p there exists a
(t, p−c1(log log klog C +1)t, +2−k
Ω(1)
) asynchronous extractor that runs in at most 1.1 log Clog log k+2 rounds in the full-information model. Here is the error of the extractor in Assump- tion 4.3.16.
Remark 4.3.53. The parameter p0.1 can be replaced by pδ for any constant δ > 0 and the constant 1.1 can be replaced by 1 + α for any constant α > 0.
Proof. [Proof Sketch] Run Protocol 4.3.47 with round number R to be chosen later. Now basically repeat the proof of Lemma 4.3.51 and Theorem 4.3.49. Let Badl be
the number of honest players in Al that don’t have a valid independent somewhere
random source for l ≥ 2, and define Bad1 = 0. Let Cl be the number of independent
(n, k) sources needed for a player in round l. Again by induction we can show that ∀l, Badl ≤ 2t. What is different now is the relation between Cl+1 and Cl.
Consider the (K, 0) extractor graph Ext : [N1] × [D1] → [M1] constructed in
the protocol. Similar as inLemma 4.3.42, we have
D1 = max{
M1
K2 0
, 2O((log log N1)3log(1/0))}.
Note M1
K2 0
= 2O(D) and O((log log N
1)3log(1/0)) = O(D(log log t)3). Thus
D1 = 2O(D(log log t)
3)
≤ 2O(Cl(log log p)3) = 2O(Cl(log log n)3).
The last equality follows because n = poly(p). Thus
Cl+1 = O( log D1 log k ) ≤ O( Cl(log log n)3 log k ) ≤ c0(log log n)3 log k Cl for some constant c0 > 0.
Therefore
Cl ≤
cl−10 (log log n)3(l−1) logl−1k C.
If log log log nlog log k = o(1), then log k > c0(log log n)3 and Cl will eventually decrease
to 1. Let l0 be the first round where Cl0 ≤ 1 and R = l0 − 1, then by the end of
round R all but BadR+1 of the honest players in AR+1 obtain private random bits by
computing zj = BasicExt(xj, yj).
cl0−1 0 (log log n)3(l0−1) logl0−1k C ≤ 1 We get l0 ≥ log C
log log k − 3 log log log n − log c0
+ 1
Thus it suffices to take
R =
log C
log log k − 3 log log log n − log c0
≤ (1 + o(1)) log C log log k + 1.
The equality follows because log log log nlog log k = o(1).
If AR+2 6= φ, then in round R + 1, we run Protocol 4.3.46 with A = AR+1
and B = AR+2. As |AR+1| ≥ 10t and BadR+1 ≤ 2t, every player in AR+2 will
eventually receive |AR+1| − BadR+1 − t ≥ 7t strings. Among these strings at most
BadR+1+ t ≤ 3t are not 0 close to uniform and independent of each other. Thus the
concatenated string sj is 0 close to have min entropy rate at least 47 and independent
of xj. Therefore by Theorem 3.5.7 zj = Raz(sj, xj) is 2−k
Ω(1)
close to uniform and independent of the transcript so far.
Now there are at most 3bRt honest players in the first R rounds. In round R + 1 there can be at most BadR+1≤ 2t honest players that don’t get private random
bits. Therefore the number of honest players that get private random bits is at least p − t − 3bRt − 2t = p − (3bR + 3)t. In each round when we apply the extractor, the error increases by 2−kΩ(1), thus the total error is at most + R2−kΩ(1).
Note that R ≤ (1 + o(1))log log klog C + 1. Thus for p large enough R < 1.1 log Clog log k + 1. Together with the fact C = o(log n) and log log log n = o(log log k) this implies that the error R2−kΩ(1) = 2−kΩ(1).
Choose c1 = 3(1.1b + 1), we haveProtocol 4.3.47 is a (t, p − c1(log log klog C + 1)t, +
2−kΩ(1)) asynchronous extractor that runs in at most 1.1 log Clog log k + 2 rounds in the full- information model.
Using the extractor in Theorem 3.5.9 where C = O(log nlog k), the above theorem gives the following theorem about (n, k) sources, and corollaries which tolerate a linear fraction of faulty players.
Theorem 4.3.54. There exists a constant c > 0 such that if log log log nlog log k = o(1) and n = poly(p), then as long as t > p0.1, for sufficiently large p there exists a (t, p −
clog log nlog log kt, 2−kΩ(1)) asynchronous extractor that runs in O(log log n/ log log k) rounds in the full-information model.
Corollary 4.3.55. There exists a constant c > 0 such that for every constant δ > 0 and p large enough, there is a (t < pc, p − ct, 2−kΩ(1)) asynchronous extractor that runs in at most 2 rounds for k = nδ in the full-information model.
Corollary 4.3.56. There exists a constant c > 0 such that if n = poly(p), then for every constant δ > 0 and p large enough, there is a (p0.1 < t < δpc, p − ct/δ, 2−kΩ(1)) asynchronous extractor that runs in at most 1/δ + 1 rounds for k = 2logδn in the
Remark 4.3.57. The asynchronous network extractor tolerates a linear fraction of faulty players and guarantees a linear fraction of honest players end up with private random bits, even for min-entropy roughly as small as k = 2logδn.
In the case k = nδ, we don’t need t > p0.1 and we can deal with n ≥ poly(p), since C is a constant. Moreover if t = Θ(p), then the protocol runs in one round.