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Once an MU is going to handoff to an adjacent MAP of the other WISP domain, the MU will be collaboratively issued a D-coin generated by the sMAP and the nMAPs. A D-coin is composed of seven components, as shown in Fig. 22.5, where Issuer is the current serving WISP, Receiver is the tMAP of the handoff target WISP, B represents the face value of this D-coin, Enc(PMK) refers to a pairwise master key (PMK) encrypted with the receiver s public key, SP is a hash chain newly generated by the MU as the spending proof described in Section 22.3.5, Exp is the expiration date, and Sig is the issuer s signature on the above six components. Among the seven components, a digital signature issued by the issuer plays a critical role in building up the trust relationship among WISPs. In the following, we will show how a sMAP collaborates with its nMAPs to locally issue a D-coin. Assume that every neighboring MAP periodically broadcasts its public key certi cate along with service set identi er (SSID). After deciding the handoff target MAPtMAP@WISPB, the MU can easily ensure the security of tMAP s public key by (1) validating the public key certi cate by checking the RB s signature on it and (2) ensuring that this public key certi cate has not been revoked by checking the CRL stored at the cMAP. After that, the D-coin issuing protocol can be performed as follows.

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UPC-A Barcode C# .NET Generation SDK. UPC-A , short for Universal Product Code version A, is a commonly used linear barcode, especially in America. It can only encode 10 characters, i.e., digit 0-9. In general, an UPC-A barcode will encode 11 data and 1 check digit.

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20 Sep 2006 ... It's an extension of UPC (Universal Product Code). ... A helper method is required to check the code's checksum. ... so that we can use it in order to calculate the EAN-13 code for a given ISBN later on (see later post). ... The first digit is part of the number system, a code to represent the country of origin. In the ...

Similarly, we can prove that Election( 1/2 j 1 ) and Election( 1/2 j ) succeed in electing a leader with probability at least 1 e 1/2 and 2e 2, respectively Therefore, Elec2 tion(Di) fails to elect a leader with probability at most 1 This completes the proof Let D i = Di Di Di be an infinite sequence, where denotes the concatenation of sequences For example, D 2 = 1 , 1 , 1 , 1 , 1 , 1 , Suppose that every station knows the 2 4 2 4 2 4 upper bound u of the number n of the station Since Election(D log u ) elects a leader with probability at least 1 from Lemma 32, t times iteration of Election(D log u ) fails 2 to elect a leader with probability 1/2t Also, the t times iteration runs in t log u time slots.

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Therefore, we have: Lemma 33 An oblivious protocol Election(D log u ) elects a leader in log f log u time slots with probability at least 1 1/f for any f 1 1033 Oblivious Leader Election for Scenario 3 Let V = v(1), v(2), be a nondecreasing sequence of positive integers such that 1 v1 v2 holds For such sequence V, let P(V) = Dv(1) Dv(2) Dv(3) be the infinite sequence of probabilities For example, if V = 1, 2, 3, , then P(V) = D1 D2 D3 = 1 1 1 1 1 1 , , , , , , We are going to evaluate the performance Election(P(V)) for various 2 2 4 2 4 8 sequences V For a sequence V = v(1), v(2), , let l(V) denote the minimum integer satisfying v[l(V)] log n .

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provide all of the data elements of the order Typically, merchants will have performed an authorization prior to making this call, so they can provide information such as address veri cation results and the card security results to the fraud-scoring service These services are typically set up to process orders in a real-time environment, but this does not mean merchants can t use them in a batch mode The service typically takes a matter of seconds to evaluate an order to determine the level of risk associated with it Once a fraud-scoring service is done, it will provide one of several data points back to the merchant Make sure to check what the service provider will return: there should be a pass or fail result, a score, and descriptors.

where (11.36) follows from Theorem 11.1.1 and (11.38) follows from Theorem 11.1.2. We give a slightly better approximation for the binary case.

In other words 1 v(1) v(2) < v[l(V)] log n v[l(V) + 1] v[l(V) + 2] 1 1/4 1 1 e 1 e 1/2 (1 2e 2) < 4 2 2.

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