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. 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. – 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 H the global state Ψ factorizes, i.e., Ψ = ψAB ψE, where ψAB A B and ψE E. In this case Xxxxx and Xxx are independent of Eve: Xxx cannot obtain any information on Xxxxx’s and Xxx’s states by measuring her system. After a measurement, Xxxxx and Xxx obtain a classical distribution PXY . In accordance with Xxxxxxxx’x principle that all information is ultimately physical, the classical scenario arises from a physical process, namely the measurements performed. Thus the quantum state Ψ , and not the distribution PXY Z, is the true primitive. Note that only if also Xxx performs a measurement, PXY Z is at all defined. It is clear however that it might be advantageous (if technologically possible)for the adversary not to do any measurements before the public discus- sion. Because of this, staying in the quantum regime can simplify the analysis. When Xxxxx and Xxx share many independent systems2 ρAB, there are basi- cally two possibilities for generating a secret key. Either they first measure their systems and then run a classical protocol (process classical information) secure against all measurements Eve could possibly perform (i.e., against all possible distributions PXY Z that can result after Eve’s measurement). Or they first run a quantum protocol (i.e., process the information in the quantum domain) and then perform their measurements. The idea of quantum protocols is to process the systems in state ρAB and to produce fewer systems in a pure state (i.e., to 1 We consider pure states, since it is natural to assume that Xxx controls all the environment outside Xxxxx and Xxx’s systems.
H H. ∈ H We claim that the off-diagonal entry z3 of Zj 2 is non-zero. Indeed, if z3 = 0, then Zj = diag(z1, z2) with z1, z2 = 1 and Aj is the product of the elliptic curves corresponding to z1 and z2, contradicting the fact that Aj is simple (Theorem I5.2). The claim and Corollary 5.19 together now give an upper bound on log(1/z3), which is polynomial in log ∆ by Lemma 3.6. We now have 4 1/4 1/2 uj = O(max{ √3π ∆0, 9 ∆1 ∆0 }), 0 h′ = O˜(∆1/2∆1/2), and by Theorem 9.1 also log D = O˜(∆3/2∆5/2). p = O˜(∆3/2∆5/2) and also rj = O˜(∆3/2∆5/2). We find that p is dominated by our bounds on log D, hence we have √— 0 Finally, we can bound the running time. Under the assumption that K is given as K = Q( a + b√∆0), where ∆0 is a positive fundamen- tal discriminant and a, b are positive integers such that a < ∆0, we can factor (a2 — b2∆0)∆2 and hence find the ring of integers in step 1 in ˜ time O(∆). As shown in Section 3.3, step 2 takes time O(∆ time O˜(D) = O˜(∆3/2∆5/2). 1/2 ). Step 3 takes is 2h′ = O˜(∆1/2∆1/2) by Lemmas 8.1 and 3.5. In particular, steps 4 For every j, step 4a takes time polynomial in log ∆ by Lemma 3.6 and Theorem 5.18. The same holds for steps 4b and 4c and each sum- mand of step 5. The number of iterations or summands of these steps j ˜ 1 0 and 5 take time O˜(∆1/2∆1/2). of this step together take time O˜(∆7/2∆11/2). We now come to the most costly step. By Theorem 7.16, it takes time O(r2) to do a single iteration of step 6a. In particular, all iterations ˜ of this step together take time O˜(∆2∆3). Finally, by Theorem 10.5, step The j-th iteration of step 6b takes time O(r) and hence all iterations of time is needed for the final two steps. 0 7a takes time O˜(h′) times O˜(p), which is O˜(∆2∆3). The same amount The output consists of h′ + 1 rational coefficients, each of which has a bit size of O˜(∆3/2∆5/2), hence the size of the output is O˜(∆2∆3). ^ ^ ˜ ˜ This proves the main theorem, except when using the polynomials HK,n (n = 2, 3) of Section 2.2. With the naive method of evaluat- ing HK,n that we described in Algorithm 11.1, it takes O(h1) times as much time to evaluate HK,n from the Igusa invariants as it does to eval- the rest of the algorithm. uate HK,n. This O˜(∆5/2∆7/2) is still dominated by the running time of running time of Algorithm 11.1 is O˜(∆2∆3), which can still be improved It also follows that, if one uses Xxxxxx’x [20] quasi-linear method of evaluating theta constants as mentioned in Section 7.4, the heuristic if better bounds on the...
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H H. Corollary 3 Let Ψ A B E and ρAB = TrHE (PΨ). Then the following statements are equivalent:
H H. Due to the strong deviation from the original least-squares fitting employed by Xxxxxx et al. (2006), it was necessary to validate the parameters obtained from the GDV dataset. For this pur- pose, we pooled the GDV data for all districts in the state of Baden-Württemberg and compared the obtained model parameters against those value published by Heneka & Ruck (2008). The re- sults presented in Tab. 21 show good agreement of the individual parameters across the different sources. As the wind sources are not directly comparable, the parameters shown in brackets were rescaled according to 4. Regarding these values, only 2, which represents the wind range from beginning to total destruction, shows a significant difference of approximately 15% as compared to the original values. While we report only those results that relate to the best performing model setup, results related to the Baden-Württemberg calibration applied to entire Germany (similarly to Heneka & Xxxxxxx 2011) are included in the Electronic Supplementary Material for special interest. Summary & Discussion According to wikipedia (8.10.2014), "A library is an organised collection of sources of information and similar resources, made accessible to a defined community for reference or borrowing." With the aim of providing such a collection within the RAMSES Deliverable 1.2, we have assembled impact functions in three complementary parts, which are or will be published in scientific journals as explicitly demanded in WP1.
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. FIR2.4 FIR1.9 FIR2.7 Hydro pole exisitng septic system 180 180 185 landscaped garden outdoor patio & cafe FIR1.1 FIR1.5 190 15 cars 175 190 190 charging station 185 S ite P la n Scale: 1:250 Design: LM, MM Drawn: MM well Issued: for application Date: May 3, 2019
