H H Sample Clauses

H H. In the binary case ( A and B both have dimension two), the above two conditions are equivalent and sufficient for the possibility of quantum key agree- ment: all entangled binary states can be purified. The same even holds if one Xxxxxxx space is of dimension 2 and the other one of dimension 3. However, for larger dimensions there are examples showing that these conditions are not equivalent: There are entangled states whose partial transpose has no negative eigenvalue, hence cannot be purified [17]. Such states are called bound entangled, in contrast to free entangled states, which can be purified. Moreover, it is be- lieved that there even exist entangled states which cannot be purified although they have negative partial transposition [9].‌

H H. – Commitment. From Lemma 2, all honest parties that complete AVSS-Sh would agree on the same h and c. According to the collision-resistance of hash function, the adversary cannot find a Cj = C such that h = (Cj) = (C ) with all but negligible probability, so there is a fixed C except with negligible probability. Moreover, C is computationally binding conditioned on DLog assumption, so all honest parties agree on the same polynomial A∗(x) committed to C , which fixes a unique key∗, and they also receive the same cipher c. So there exists a unique m∗ = c key∗, which can be fixed once some honest party outputs in AVSS-Sh. Now we prove that m∗ can be reconstructed when all honest parties activate AVSS-Rec. Any honest party outputs in the AVSS-Sh subprotocol must receive 2f + 1 Ready messages from distinct parties, at least f + 1 of which are from honest parties. Thus, at least one honest party has received 2f + 1 Echo messages from distinct parties. This ensures that at least f + 1 honest parties get the same commitment C and a valid quorum proof Π. Due the unforgeability of signatures in Π, that means at least f +1 honest parties did store valid shares of A∗(x) and B∗(x) along with the corresponding commitment C except with negligible probability. So after all honest parties start AVSS-Rec, there are at least f + 1 honest parties would broadcast KeyRec messages with valid shares of A∗(x) and B∗(x). These messages can be received by all parties and can be verified by at least f + 1 honest parties who record C . With overwhelming probability, at least f + 1 parties can interpolate A∗(x) to compute A∗(0) as key and broadcast it, and all parties can receive at least f + 1 same key∗ and then output the same m∗ = c key∗ as they obtain the same ciphertext c from AVSS-Sh.

H H. If the set of bases is large enough, then for all z there is a basis with posi- tive intrinsic information, hence the mean is also positive. Clearly, this result is stronger if the set of bases is small. Nothing is proven about the achievable size of such sets of bases, but it is conceivable that max dim A, dim B bases are always sufficient.

H H. The optimal power allocation policy to (P1-full) is thus given by

H H. In the binary case ( A and B both have dimension two), the above two conditions are equivalent and su cient for the possibility of quantum key agree- ment: all entangled binary states can be xxxx xx. The same even holds if one Xxxxxxx space is of dimension 2 and the other one of dimension 3. However, for larger dimensions there are examples showing that these conditions are not equivalent: There are entangled states whose partial transpose has no negative eigenvalue, hence cannot be xxxx xx [17]. Such states are called bound entangled, in contrast to free entangled states, which can be xxxx xx. Moreover, it is be- lieved that there even exist entangled states which cannot be xxxx xx although they have negative partial transposition [9].

H H. So we have shown on the one hand that the agents not in ε1 at tj are in ε2 at t1. On the other hand, the agents in ε1 at tj remain in δ1 at t1 from (5.5), and therefore remain in ε2 at t1 because δ1 ε2 . Hence, at time t1, ε2 (x(tj)) has at least two agents. Let V2 and V2∗ be a partition of the node set V such that i ∈ V2 if xi(t1) ∈ Hε2 and i ∈ V2∗ otherwise. Note that by (5.5) (5.5) k ∈ V1 =⇒ xk(tj) ∈ Hε1 =⇒ xk(t1) ∈ Hδ1 ⊂ Hε2 =⇒ k ∈ V2, so V1 ⊂ V2. In particular ck2 , the center node of G([τk2 , τk2 + Tj]), is in V2 because it is in V1. Then we can apply the same argument to conclude that there are a t2 ∈ [t1, tj + k2T ] and an i in V2∗ such that xi(t2) ∈ Hε3 and therefore, Hε3 has at least three agents at t2. Repeating this argument n − 1 times leads to the result that there is a tn−1 ∈ [tj, tj + kn−1T ] ⊂ [tj, tj + T¯] such that Hεn has n agents at tn−1. Hence, V1(x(tn−1)) ≤ V1(x(tj)) − εn = V1(x(tj)) − η(V1(x(tj))), and (5.4) follows.

H H i,j for j = 1...

H H. Example 1 (cont’d). We have shown in Section 3.2 that the resulting quan- tum state is entangled if and only if the intrinsic information of the cor- responding classical situation (with respect to the standard bases) is non- zero. Here, we show that such a correspondence also holds on the pro- tocol level. First of all, it is clear for the quantum state that QPA is possible whenever the state is entangled because both A and B have dimension two. On the other hand, the same is also true for the corre- sponding classic√al situation, i.e., secret-key agreement is possible whenever D/(1 − D) < 2 (1 − δ)δ holds, i.e., if the intrinsic information is positive. This is shown in Appendix C. There we describe the required protocol, more precisely, the advantage-distillation phase (called repeat-code protocol [20]), in which Xxxxx and Xxx use their advantage given by the authenticity of the public-discussion channel for generating new random variables for which the legitimate partners have an advantage over Eve in terms of the (Xxxxxxx) information about each other’s new random variables. For a further discus- sion of this example, see also [15]. ♦ Example 2 (cont’d). The quantum state ρAB in this example is bound entan- gled, meaning that the entanglement cannot be used for QPA. Interestingly, but not surprisingly given the discussion above, the corresponding classical distribution has the property that I(X; Y Z) > 0, but nevertheless, all the known classical advantage-distillation protocols [20], [22] fail for this distri- bution! It seems that S(X; Y ||Z) = 0 holds (although it is not clear how this fact could be rigorously proven, except by proving Conjecture 1 directly). ♦

H H have proposed a fallback. However, a Byzantine aggregator may be successful in preventing the fallback by forming a cluster of one honest observation max with fc Byzantine observations with value max +d. Had the protocol switched to the fallback protocol, the smallest value of the median produced by Algorithm 3 would have been min. Therefore, even in this case, the upper bound for the error would be min max 1+d. The arguments with respect to Byzantine observations forming a cluster with min would be symmetric and do not result in any change in the error upper bound. Theorem 5 (Least error upper bound). A protocol for agreeing on a , where 3ft+1 nodes participate out of which ft of these nodes could be Byzantine, and a median of values from non- deterministically chosen3 2ft + 1 nodes is proposed as the S, would have an error upper bound of at least Hmin −Hmax .

H H. FIR2.4 FIR1.9 FIR2.7 Hydro pole landscaped garden outdoor patio & cafe 15 cars charging station 185 Scale: 1:250 Design: LM, MM Drawn: MM well