Security Analysis Sample Clauses

Security Analysis. 4.2.1. Session Key Security Xx uses the session key to encrypt the information sending over Internet. Therefore, if the session key is secure, it means that the communication in the cloud meeting is also security. The proposed solution has the Diffie–Xxxxxxx problem. Even if attackers capture Tai (x) or Tbi (x), they still can not generate authentication information. Moreover, we consider random value ai and bi, so it is difficult for attackers to compute ski and SK = h(Sn, ski). Therefore, the session key is security in PL-GAKA.
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Security Analysis. In this section, we will analyze the security of our proposed scheme. The main assumption for guarantee of security lies in:
Security Analysis. An inherent problem with EKA is the length of the blocks it communicates during the commitment phase. As they are only 64 bits long, their hashes can be easily brute-forced by a capable adversary. One of the ways of overcoming this problem could be to use key strengthening. The main idea being that sensors hash each of their 64 bit blocks ’2n’ times before transmitting them. An attacker while brute forcing has to generate a candidate block (64 bits long) and hash it ’2n’ times before knowing weather the candidate is the actual block. Therefore an additive increase in the number of hash computations for the sender results in a multiplicative increase in the hashing requirements for the adversary. This results in a 2n fold increase in the processing required for brute- forcing a 64 bit block which is analogous to brute forcing a 64 + n bit block. We recommend that sensors choose the value of ’n’ as high as possible. However, key strengthening is an expensive proposition. The need to hash each of the 20 blocks 2n times increases the costs further. For example if we choose ’n’ to be 16 (which makes the block as secure as a 80 bit block) the total number of hash operations required to computed would be 131072!. The choice of this value actually used therefore, may depend upon the capabilities of the sensor and the amount of energy available at them, and hostility of the environment in which the subject carrying the sensors is. In a home environment for example, no key strengthening may be required, while in a shopping mall the maximum possible value of ’n’ would be necessary. The use of key strengthening is however a stop-gap solution, we are currently working on generating longer blocks during the commitment phase so that brute-forcing them becomes impractical. Assuming that the brute forcing of blocks is infeasible, the commitment and de-commitment phases makes it very difficult for adversaries to know the key being agreed upon. There are 4 reasons for this - a) The blocks are not exchanged 0.9985 0.998 Averag Entropy 0.9975 0.997 0.9965 . Σ
Security Analysis. This section provides the security analysis of the proposed scheme focused on no requirement of global time synchronization, providing forward secrecy provision, and secure against password guessing attack, replay attack and user identity guessing attack.
Security Analysis. A A A A A In our scheme, an adversary is able to manipulate locally- stored temporary states, reject to provide the correct proof, and refuse to make a revocation. The goal of is to get profit by tempering or canceling state transitions. We focus on the most important attack, dependency attack, where first submits commitments about TX1 with user B colluding with , then makes another latency-first transaction TX2 with user U . tries to get profit by tempering or canceling TX2 by not providing the correct proof about TX1. A
Security Analysis. A-TGDH satisfies our stated security goals with the following assumptions. Since key confirmation is essential for achieving perfect forward secrecy [4], we assume that it has been implemented as described in Section V-B. Also, we assume that there exists only a passive adversary E that monitors the flow of blinded key messages. We further assume that E cannot solve the Xxxxxx-Xxxxxxx problem [6] (i.e., given only , p, x mod p, and y mod p, it is infeasible for E to compute xy mod p) and the discrete logarithm problem (i.e., given only , p, and x mod p, it is infeasible for E to compute x). The following proof is based on [3], [15].
Security Analysis. The proposed protocol has the security properties of i i =a0 + IDi(a1 + IDi(a2 + IDi(. . . ))) key freshness, key confidentiality and mutual authen- .... {IDn, ..}pus , sign sig0 = τ pr0 (ID0,Y,U )
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Security Analysis. In this section, we analyze the security characteristics of ≥ 2g · qes+1 n · q ins qsend+1 · qH3 the PCAKA scheme under the above ”Security Model”. C This is the opposite of the difficulty of the ECCDH problem. Thus we obtain the conclusion that any PPT adversary can not fake a legal login information or the corresponding response information with a non-negligible probability.
Security Analysis. E In the proposed protocol, the agreed secret K and EK are dynamically generated in every session, furthermore, there is no decryption oracle for any fixed EK to help the adversary in the running of the protocol. Hence, we only require the one-time symmetric-key encryption scheme to be passively secure as defined in Definition 2. While, for the simplicity of security analysis, we specifically require that is the “one time pad” with pseudorandom bit stream scheme, which is an efficient and passively secure one-time encryption scheme [18].
Security Analysis. We now examine our CL-AK protocol in relation to the security attributes defined in Section 2, and informal ar- guments are provided to support our claims: • Forward secrecy. – Full forward secrecy: The compromise of both party’s long-term partial private key, i.e. xU , SU of UP and xR, SR of RP, gives no informa- tion about any previously established session keys. Suppose Eve knows all the long-term partial private keys xID and SID (in whice ID ∈ {U, R}), to extract a past session key, he must compute abP ∈ G1 from P , aP and bP . Without the knowledge of a and b, this is exactly an instance of the computational Xxxxxx- Xxxxxxx(CDH) problem in G1 that Xxx is not able to solve. – Perfect forward secrecy: Xxxxxxx at a moment the master key s known only to the GTA is com- promised. Since the established session keys are computed with the ephemeral private keys of UP and RP, Xxx still have to solve the CDH problem in G1 to reveal the session key. This means that our protocol has perfect forward se- crecy(PFS). q • Known-key secrecy. Each run of the protocol be- tween UP and RP shall produce a unique session key which depends on every particular ephemeral private key a and b ∈ Z∗ of UP and RP. Even if the adver- xxxx Xxx has learned some other session keys, he can not compute the keying point abP ∈ G1 from them aP and bP , because when he has no access to a and b, he faces the computational Xxxxxx-Xxxxxxx prob- lem which is believed to have no polynomial time Table 1: Computational and bandwidth efficiency com- parisons Item ↓ / Protocol → AP’s Ours Pairing 1 0 Point multiplication 3 2 algorithm to compute. Hence our protocol has the property of known-key security.
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