Decryption Sample Clauses

Decryption. If there has been proper receipt pursuant to Section 2.1 the receiving party shall attempt to decrypt the Transaction. If the Transaction decryption is unsuccessful, the receiving party shall send the applicable error message to the sending party. The sending party shall attempt to correct the error and promptly retransmit the Transaction or notify the receiving party in an attempt to solve the problem. If the Transaction can not be authenticated an applicable error message will be sent to the sending party however, if the sending party’s identity can not be ascertained, then the transmission will not be deemed a Transaction.

Decryption. After receiving a ciphertext (c1, c2, c3), anyone who has a valid message-signature pair (s,σ) can extract m = c3 / e(σ,c1)e(H(s), c2).

Decryption. Group member can decrypt the message sent from Eve by using his/her private key as the following. For Xxxxx: M = M′ ⊕ H{SK1A Z1(g2) (g3)( g5) Z2 SK2A } For Bob: M = M′ ⊕ H{SK1B Z1(g1) (g3)( g5) Z2 SK2B } For Xxxxxxx: M = M′ ⊕ H{SK1C Z1(g1) (g2)( g5) Z2 SK2C }

Decryption. In decryption process, user B decrypts the ciphertext message Ci = [C1, C2, …, Ci] using its own private key (d, n) to get the original corresponding plaintext blocks Mi as Mi = Ci (mod n). After that, user B concatenates all decrypted ciphertext blocks Mi to get the message M. ECC is a public key encryption technique based on elliptic curve theory which is the set of the point that satisfies the specific mathematical equation: y2= x3+ ax + b (Stallings, 2006; Yu, 2012). Before talking about the ECC encryption/decryption, we will discuss in a brief elliptic curve equation. A finite field or Galois Field (GF) is a field that contains a finite number of elements which called its order, such as the fields of prime order GF(p) (also denoted Zp or Fp), which is a field of prime number of order p (size of field). The finite field with elliptic curve defined as follow: Let a and b ϵ Zp, where Zp = {0, 1,… p - 1 } and p > 3 be a prime, such that 4a³ + 27b² != 0 (mod p). A non-singular elliptic curve y² = x3 + ax + b over the finite field GF (p) is the set Ep (a, b) of solutions (x,y) ϵ Zp × Zp to the congruence y² = x³ + ax + b (mod p), where a and b ϵ Zp are constants such that 4a³ + 27b² € 0 (mod p). Together with a special point Ҩ called the point at infinity or zero points (Xxxxx et al., 2015; Xxxxxxxxx, 2006; Yu, 2012). Let P (Xp, Yp) and Q (XQ, YQ) be two point on an elliptic curve y² = x³ + ax + b (mod p), and let G be the base point on Ep (a, b) whose order be n, that is nG = G + G + … + G (n times) = Ҩ, R (XR, YR) = P + Q is computed as follows: XR = ( €²- Xp – XQ) (mod p), YR = ( €( Xp – XR) – Yp) (mod p), Where €= (YQ – Yp)/( XQ – Xp) (mod p), if P € Q and €= (2X ² + a )/( 2YP) (mod p), if y y2 = x3 + ax + b Figure 2-1. Elliptic Curve. Before encryption and decryption process in elliptic curve algorithm, the plaintext message m will be encoded to be sent as an elliptic curve point Pm ϵ Ep (a, b). This point Pm then will be encrypted as a cipher text and then subsequently decrypted. In the key generation process, user B has the elliptic curve Ep (a, b) defined over a finite field GF (p), and a base point G ϵ Ep (a, b) whose order is n, that nG = Ҩ. User B selects a private key nB randomly on the interval [ 1, n – 1 ] and computes his/her public key PB = nB G.

Decryption. The recipient IDi set a vector a1 = (0,…0,1,0,…0) with n elements, and only the ith element is 1. Then A is a n × n matrix The recipient IDi can solve the following system of equations (x1,x2,…,xn) × A = ( 1 1 … 1). With (x1,x2,…,xn) they can get To decrypt the ciphertext, the recipient IDi needs to compute e(Xxxx, r QV1) , which with knowledge of the private key Si it can do via: e(Ppub, r QV1) = e(Ppub, r(x1Qi + x1 QV2 + … + xn QVn)) = e(Ppub, rx1Qi) e(Ppub, r(x2 QV2+ … + xn QVn)) = e(rP, x1sQi) e(Ppub, x2rQV2+ … + xn rQVn) = e(U1, x1Si) e(Ppub, x2U2 + … + xn Un) Then the recipient can compute K = V ⊕ H2(e(U1, x1Si) e(Ppub, ∑n i=2 xiUi ))

Decryption. After receiving a ciphertext (c1, c2, c3), anyone with a valid message-signature pair (s, σ) can extract c3 m = . e(σ, c1)e(H(s), c2) The correctness of the proposed ASBB scheme follows from a direct verifica- tion. Define by (R1, A1) (R2, A2)= (R1R2, A1A2) and by σ1 σ2 = σ1σ2. For security, we have the following claims in which Claim 2 follows from the definition of and , and the security proof of Claim 3 can be found in the full version of the paper [23].

Decryption. Acronis cannot decrypt your files if Reseller has elected to encrypt. However, You have selected the location of your backup and understand that local laws where the selected data centers are located may be different than the laws of the country in which you reside. Acronis will comply with the local laws of the jurisdiction You reside and also the jurisdiction where the data center housing Your data is located. As a result, You acknowledge that Acronis or Acronis affiliates may use servers and other equipment to provide the Acronis Backup Cloud Product and Platform that are located in the United States or in other countries where litigants, law enforcement, courts, and other agencies of the government may have the right to access data stored within their jurisdictions upon terms and conditions provided by local law.

Decryption. Bob can decrypt the message sent from Xxxxx by using his private key; SK1B, SK2B and an identity of Xxxxx, XXX as the following. M = M′ ⊕ H{ SK1B Z1 (g1) Z2SK2B }

Decryption. Given , the plaintext can be retrieved using the private key:

Decryption. The recipient optionally can verify the encrypted DSRC key with the public key KS-PUB of the originator and decrypts the DSRC key with his private key KR-PRIV. Signature verification KS-PUB Signature OK / not OK RSA decryption KR-PRIV encrypted Key DSRC Key