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Bernstein Bound on WCS is Tight: Repairing Luykx-Preneel Optimal Forgeries

机译:WCS的Bernstein束手无策:修复Luykx-Preneel最佳伪造品

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In Eurocrypt 2018, Luykx and Preneel described hash-key-recovery and forgery attacks against polynomial hash based Wegman-Carter-Shoup (WCS) authenticators. Their attacks require 2~(n/2) message-tag pairs and recover hash-key with probability about 1.34 × 2~(-n) where n is the bit-size of the hash-key. Bernstein in Eurocrypt 2005 had provided an upper bound (known as Bernstein bound) of the maximum forgery advantages. The bound says that all adversaries making O(2~(n/2)) queries of WCS can have maximum forgery advantage O(2~(-n)). So, Luykx and Preneel essentially analyze WCS in a range of query complexities where WCS is known to be perfectly secure. Here we revisit the bound and found that WCS remains secure against all adversaries making q √n × 2~(n/2) queries. So it would be meaningful to analyze adversaries with beyond birthday bound complexities. In this paper, we show that the Bernstein bound is tight by describing two attacks (one in the "chosen-plaintext model" and other in the "known-plaintext model") which recover the hash-key (hence forges) with probability at least 1/2 based on √n × 2~(n/2) message-tag pairs. We also extend the forgery adversary to the Galois Counter Mode (or GCM). More precisely, we recover the hash-key of GCM with probability at least 1/2 based on only √n/ℓ × 2~(n/2) encryption queries, where ℓ is the number of blocks present in encryption queries.
机译:在Eurocrypt 2018中,Luykx和Preneel描述了针对基于多项式哈希的Wegman-Carter-Shoup(WCS)身份验证器的哈希密钥恢复和伪造攻击。他们的攻击需要2〜(n / 2)个消息标签对,并以大约1.34×2〜(-n)的概率恢复哈希密钥,其中n是哈希密钥的比特大小。 Bernstein在Eurocrypt 2005中提供了最大的伪造优势上限(称为Bernstein界限)。边界表示,所有对WCS进行O(2〜(n / 2))个查询的对手都可以具有最大的伪造优势O(2〜(-n))。因此,Luykx和Preneel从本质上分析了WCS完全安全的一系列查询复杂性中的WCS。在这里,我们重新审视了边界,发现WCS对于所有进行q <<√n×2〜(n / 2)个查询的对手都是安全的。因此,分析具有超出生日限制的复杂性的对手将是有意义的。在本文中,我们通过描述两种攻击(在“选择纯文本模型”中使用一种,在“已知纯文本模型”中使用另一种)来恢复哈希密钥(因此伪造),从而证明伯恩斯坦边界是紧的。基于√n×2〜(n / 2)个消息标签对至少为1/2。我们还将伪造对手扩展到Galois Counter Mode(或GCM)。更准确地说,仅基于√n/ℓ×2〜(n / 2)加密查询,我们以至少1/2的概率恢复GCM的哈希密钥,其中ℓ是加密查询中存在的块数。

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