blocks|key|507970|text|语句"X哈希函数已被破坏“意味着哈希函数算法中存在一个缺陷，使得冲突的生成速度比通过强暴产生的速度要快。看看这篇文章是布鲁斯·施耐尔写的+-他说，SHA-1碰撞现在可以更快地产生，仅此而已。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|507971|因此，是的，它们对强制执行都同样不安全，但这不是"X散列函数已被破坏“语句的意义所在。|507972|entityMap|0|LINK|mutability|MUTABLE|url|http://www.schneier.com/blog/archives/2005/02/sha1_broken.html^0|1I|E|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]]|C|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]]]

The statement "X hash function has been broken" means that there's a defect in the hash function algorithm such that a collision can be generated faster than via bruteforcing. Look at <a href="http://www.schneier.com/blog/archives/2005/02/sha1_broken.html" rel="noreferrer">this post by Bruce Schneier</a> - he says that a SHA-1 collision can now be generated much faster, that's all.

So yes, they are all equally insecure to bruteforcing, but that's not what "X hash function has been broken" statement is about.

blocks|key|2061333|text|现代密码哈希算法用什么方法来防止碰撞攻击？|type|blockquote|depth|inlineStyleRanges|entityRanges|data|2061334|密码学家巴特·普莱尔博士论文是对这一课题的一个很好的概述：密码散列函数的分析与设计。|unstyled|offset|length|2061335|您可能特别感兴趣的是可证明安全的加密散列函数的概念，它存在但目前被认为不实用。|2061336|entityMap|0|LINK|mutability|MUTABLE|url|http://homes.esat.kuleuven.be/~preneel/phd_preneel_feb1993.pdf|1|http://en.wikipedia.org/wiki/Provably_secure_cryptographic_hash_function^0|0|T|C|0|0|A|C|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@$E|U|F|V|1|W]]|A|$]]|$1|G|3|H|5|D|7|X|8|@]|9|@$E|Y|F|Z|1|10]]|A|$]]|$1|I|3|-4|5|D|7|11|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]|Q|$5|L|M|N|A|$O|R]]]]

<blockquote>
 What measures do modern cryptographic
 hashing algorithms use to prevent
 collision attacks?
</blockquote>

The PHD thesis of cryptographer Bart Preneel is a good overview of this subject: <a href="http://homes.esat.kuleuven.be/~preneel/phd_preneel_feb1993.pdf" rel="nofollow noreferrer">Analysis and Design of Cryptographic Hash Functions</a>. 

You might be especially interested in the concept of <a href="http://en.wikipedia.org/wiki/Provably_secure_cryptographic_hash_function" rel="nofollow noreferrer">provably secure cryptographic hash functions</a>, which exist but are currently not considered practical.

blocks|key|1705729|text|MD5不是抗碰撞性。维基百科的定义是：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1705730|抗碰撞是密码散列函数的一个性质:如果很难找到到相同输出的两个哈希输入，即两个输入a和b使得H(a)+=+H(b)，则哈希函数具有抗冲突性。|blockquote|style|BOLD|1705731|MD5生成一个128位散列，现在可以在几秒钟内中断。|1705732|现代密码哈希算法用于防止碰撞攻击的措施是什么？|1705733|防止这种情况的一种方法是增加散列结果的大小(以位为单位)。这使得密码专家有时间调查漏洞。我已经很久没有做this...but了，是的，简而言之:-)|1705734|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Collision_resistance^0|5|4|0|0|0|3|0|0|0|N|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@$A|11|B|12|G|H]]|9|@]|C|$]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|14|8|@$A|15|B|16|G|H]]|9|@]|C|$]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|18|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

MD5 is not <a href="http://en.wikipedia.org/wiki/Collision_resistance" rel="nofollow noreferrer">collision resistant</a>. The definition from Wikipedia states:

<blockquote>
 Collision resistance is a property of cryptographic hash functions: a
 hash function is collision resistant
 if it is hard to find two inputs that
 hash to the same output; that is, two
 inputs a and b such that H(a) =
 H(b).
</blockquote>

MD5 generates a 128 bit hash that can now be broken within seconds now.

<blockquote>
 What measures do modern cryptographic hashing algorithms use
 to prevent collision attacks?
</blockquote>

One way to prevent it is to increase the size of the hashing results (output size in bits). That allows time for cryptographers to investigate vulnerabilities. It's been ages since I done this...but yeah, in a nutshell :-)

blocks|key|2061268|text|我只有部分答案。=)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2061269|根据我目前的理解，从数据d获得某个散列h(d)的机会对于所有散列结果都是相等的。|blockquote|2061270|这不是真的。散列的目的是每次对某些数据获得相同的散列h(d)。因此，该算法可能没有任何产生哈希的“机会”。哈希算法所面临的困境是，从d计算h(d)应该是容易的，而从h(d)计算有效的d应该是困难的。|2061271|如果从h(d)计算d的速度比蛮力快，则算法被认为是“非密码安全的”。|2061272|不难看出，并不是所有的哈希算法都必须同样不安全，因为它使用返回d的前几位的散列函数。无论您占用多少位，查找冲突都是微不足道的。|2061273|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$]]|$1|E|3|F|5|6|7|O|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|P|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Q|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|R|8|@]|9|@]|A|$]]]|L|$]]

I only have a partial answer. =)

<blockquote>
 From my current understanding, the chance of getting a certain hash h(d) from data d is equal for all hashing results. 
</blockquote>

This is not true. The point of hashing is to get the same hash h(d) for some data d every time. The algorithm therefore may not have any "chances" of producing a hash. The dilemma a hash algorithm is facing is, that calculation of h(d) from d should be easy, but the calculation of a valid d from h(d) should be hard.

An algorithm is considered "not cryptographically secure", if calculating d from h(d) can be done faster than brute-force.

Its easy to see that not all hash algorithms have to be equally insecure by taking the hash function that return the first several bits of d. Regardless of the number of bits you take, its trivial to find collisions.

blocks|key|508039|text|是的，“密码破译”就是这个意思。恐怕我不是专家，所以我不能回答第二个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|508040|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Yes, "cryptographically broken" means precisely that. I'm afraid I'm not an expert, so I can't answer the second question.

I've noticed things such as

<blockquote>
 MD5 has been cracked for collisions and is no longer cryptographically secure; use SHA-1 instead. SHA-1 has been cracked for collisions and is no longer cryptographically secure; use SHA-2 instead.
</blockquote>

From my current understanding, the chance of getting a certain hash <code>h(d)</code> from data <code>d</code> is equal for all hashing results. This implies, then, that the only strengthening mechanism for a hashing algorithm is to return longer hashes.

This also implies that all hashes (when not taking hash result length into account) are equally insecure to brute forcing, and that cryptographically broken only refers to quicker attacks other than brute force searching.

Is this true? What measures do modern cryptographic hashing algorithms use to prevent collision attacks?

Strength of hashing algorithms

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我注意到了一些事情，比如MD5已因碰撞而被破解，不再具有密码安全性；请改用SHA-1。沙-1因碰撞而被破解，不再具有密码安全性；改用SHA-2。根据我目前的理解，从数据h(d)获得某个散列d的机会对于所有散列结果都是相等的。这就意味着，哈希算法的唯一强化机制是返回更长的散列。这也意味着所有散列(当不考虑哈希结果长度时)对于蛮力强制同样不安全，密码中断只指比蛮力搜索更快的攻击。这是真的吗？现代密码哈

问哈希算法的强度
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