SHA-0 是如何被破解的? - 少量的哈希冲突有什么意义?

发布于 2024-11-19 11:01:00 字数 779 浏览 4 评论 0原文

我想了解 SHA0 哈希函数是如何被破坏的。据我所知,利用生日问题/鸽子保持原则,发现了哈希冲突。 http://www.mail-archive.com/cryptography%40metzdowd.com /msg02554.html 包含示例消息。

我无法找到/理解什么: 这是否意味着存在一种及时的数学方法来始终产生哈希冲突?

我最终能否找到给定 m1 的 m2 ,使得 m1 != m2, sha(m1) == sha(m2) 或者仅可能在可能消息的子集中?改写:我的密码是否有可能出现另一条冲突消息?

找到 2 个随机长消息(例如上面的链接中具有相同哈希值的消息)有何意义?为什么他们必须筛选长随机消息以查找冲突,而不是计算出冲突像“棕色的狗跳过狐狸”这样的实用信息?

几个哈希冲突的例子似乎并不像为任何消息生成冲突的及时方法那么重要,但所有帖子都讨论了前者。

感谢您的帮助/您的时间!我读过很多帖子/文章,但无法解决我的困惑。我怀疑我对 MD5 等其他损坏的哈希函数也有同样的问题。

编辑:

论文(解释查找冲突的改进方法)中引用的回答

I wanted to understand how the SHA0 hash function was broken. I understand that utilising the birthday problem/pigeon-hold principle, hash collision(s) were found. http://www.mail-archive.com/cryptography%40metzdowd.com/msg02554.html contains an example message.

What I’m having trouble finding/understanding:
Does this mean there is a timely, mathematical way to ALWAYS produce a hash collision?

Can I eventually find a m2 for a given m1 such that m1 != m2, sha(m1) == sha(m2) or is it only possible on a subset of possible messages? Rephrased: Are the chances of my password having another message for a collision guaranteed?

What is the significance of finding 2 random long messages such as in the link above that have the same hash value? Why did they have to sift through long random messages for a collision instead of figuring a collision for a practical message like “The brown dog jumped over the fox” ?

A couple examples of hash collisions don’t seem as important as a timely method to generate a collision for any message, but all the posts talk about the former.

Thanks for any help/your time! I've read alot of posts/articles, but can't work my brain around my confusion. I suspect I have the same questions for other broken hash functions like MD5.

EDIT:

The paper (explaining improved method for finding collisions) referenced in the answer

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梦回梦里 2024-11-26 11:01:00

来自维基百科

2005年2月,小云的攻击
王、Yiqun Lisa Yin、于洪波
宣布可以找到
2^39 中 SHA-0 的冲突
操作。

以目前可用的计算能力,这种复杂性完全不足以满足加密目的。它保证在非常合理的时间内发现任何消息的冲突。

From Wikipedia:

In February 2005, an attack by Xiaoyun
Wang, Yiqun Lisa Yin, and Hongbo Yu
was announced which could find
collisions in SHA-0 in 2^39
operations.

That kind of complexity is completely insufficient for cryptographic purposes with the computing power currently available. It guarantees the discovery of a collision for any message in a very reasonable amount of time.

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