7 research outputs found

    On an authentication scheme based on the Root Problem in the braid group

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    Lal and Chaturvedi proposed two authentication schemes based on the difficulty of the Root Problem in the braid group. We point out that the first scheme is not really as secure as the Root Problem, and describe an efficient way to crack it. The attack works for any group.Comment: This paper has been withdrawn by the author. One of the claims is incorrect as written. We are working on correcting and generalizing it. This will be published in another pape

    A Strong Blind Signature Scheme over Braid Groups

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    The rapid development of quantum computing makes public key cryptosystems not based on commutative algebraic systems hot topic. Because of the non-commutativity property, the braid group with braid index more than two becomes a new candidate for constructing cryptographic protocols. A strong blind signature scheme is proposed based on the difficulty of the one-more matching conjugacy problem in the braid groups, in which the signer can not relate the signature of the blinded message to that of the original message. The usage of random factor ensures that the blind signatures of the same message are different and avoids the weakness of simultaneous conjugating. The scheme can resist the adaptively chosen-message attack under the random oracle model

    Security Analysis and Design of Proxy Signature Schemes over Braid Groups

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    The braid groups have attracted much attention as a new platform of constructing cryptosystems. This paper firstly analyzes the security vulnerabilities of existing proxy signature schemes over braid groups and presents feasible attacks. Then a new proxy signature scheme is proposed based on the difficulty of the conjugacy search problem and the multiple conjugacy search problem. Security analysis shows that the proposed scheme satisfies the security requirements of proxy signature

    Groups With Two Generators Having Unsolvable Word Problem And Presentations of Mihailova Subgroups

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    A presentation of a group with two generators having unsolvable word problem and an explicit countable presentation of Mihailova subgroup of F_2×F_2 with finite number of generators are given. Where Mihailova subgroup of F_2×F_2 enjoys the unsolvable subgroup membership problem.One then can use the presentation to create entities\u27 private key in a public key cryptsystem

    Double shielded Public Key Cryptosystems

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    By introducing extra shields on Shpilrain and Ushakov\u27s Ko-Lee-like protocol based on the decomposition problem of group elements we propose two new key exchange schemes and then a number of public key cryptographic protocols. We show that these protocols are free of known attacks. Particularly,if the entities taking part in our protocols create their private keys composed by the generators of the Mihailova subgroups of Bn, we show that the safety of our protocols are very highly guarantied by the insolvability of subgroup membership problem of the Mihailova subgroups

    On an authentication scheme based on the root problem in the braid group, lanl.arXiv.org ePrint Archive, September 2005, Online available at http:// arxiv.org/ps/cs.CR/0509059

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    Abstract. Lal and Chaturvedi proposed two authentication schemes based on the difficulty of the Root Problem in the braid group. We point out that the first scheme is not really as secure as the Root Problem, and describe an efficient way to crack it. The attack works for any group. 1. The first authentication scheme The basic definitions are given in [2]. We only describe the scheme itself. We work in the braid group Bn where n is even. Let LBn = 〈σ1,..., σn/2−1 〉 and UBn = 〈σn/2+1,..., σn〉. In the sequel, multiplication of elements of Bn means concatenation and reduction to left canonical form. Key Generation. Alice chooses integers r, s ≥ 2, a ∈ LBn, and b ∈ UBn. The public key is (X = a r b s, r, s), and the secret key is (a, b). Authentication. Bob chooses c ∈ UBn and d ∈ LBn, and sends Alice the challenge Y = c r d s. Alice responds with Z = a r Y b s. Bob verifie
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