使用指定的散列函数解密电子邮件地址散列 [英] Decrypt email address hashed using specified hashing function

问题描述

所以我试图解决这个问题。我应该获得哈希函数散列的电子邮件地址。

的秘密电子邮件地址散列如下：搜索结果，
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

首先，我们生成一系列字符串前缀，长度为
，增加2。例如，如果我们的秘密电子邮件地址是
[e-mail]，我们将生成：


hellow
hellowor
...
[e-mail]

然后，对于每一个前缀s，我们计算出以下散列J：md5（md5（e）
+ s + md5（s））[其中+是字符串连接运算符，e是您的电子邮件地址]。最后，我们将所有哈希字符串J连接到
，形成上面的长哈希值！

例如，对于[e-mail]，我们将计算： >

md5（md5（[e-mail]'）+'he'+ md5（'he'））+

md5（md5（[e-mail] '）+'hell'+ md5（'hell'））+

md5（md5（[e-mail]'）+'hellow'+ md5（'hellow'））+

...

为了简单起见，您可以假设我们的电子邮件地址只有
包含字母数字字符和这4个字符：_。 @ +

我对这类问题并不熟悉。如果你能给我一个关于我应该做什么的概念（一个游戏计划），那会很有帮助。假设我对哈希（或相关的数据结构）知之甚少是安全的。这个问题对我来说是学习这些东西的好机会。

解决方案

只要e是已知的，就可以很容易地伪造一个碰撞，那很可能是原始的。

md5有128位输出...所以把你的长字符串分割成单独的md5十六进制字符串

现在如果已知e，那么您需要猜测的第一个散列值是2个字符......详尽搜索将总共为[a-z0-9 _。@ +] ^ 2 = 40 ^ 2 = 1600个可能的候选项。 。

一旦遇到碰撞，你就会知道这个秘密的前两个字母......

对于下一个散列，所有你必须猜测的是自从第一步已经知道的前两个字符之后的下一个2个字符...

为所有哈希重复按照你的大字符串J的出现顺序，因为你只需要猜测每个散列的2个字符，并且可以重用发现的前一个散列的冲突的信息，这相当简单...

就复杂性而言，

需要计算e的md5散列值只有一次，你有2个concatinations和2 md5的计算每个猜测...每个哈希你有一个1600的ex1ustive搜索候选库，所以平均你可以说你需要800个猜测每个哈希在J

请记住这不是解密......它仍然是搜索碰撞......但很可能它是原始数据，因为md5很可能会几乎可以肯定地说）是可以避免碰撞的可能的投入......

So I am trying to solve this problem. I am supposed to obtain the email address hashed by the hash function.

The secret email address is hashed below:

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

First, we generated a series of string prefixes with lengths increasing by 2. For example, if our secret email address was [e-mail], we would generate:
he
hell
hellow
hellowor
... [e-mail]

Then, for every prefix s, we computed the following hash J: md5(md5(e) + s + md5(s)) [where + is the string concatenation operator and e is your email address]. Finally, we concatenated all hash strings J to form the long hash above!

For example, for [e-mail], we would compute:

md5(md5('[e-mail]') + 'he' + md5('he')) +
md5(md5('[e-mail]') + 'hell' + md5('hell')) +
md5(md5('[e-mail]') + 'hellow' + md5('hellow')) +
...

For the sake of simplicity, you can assume that our email address only contains alphanumeric characters and these 4 characters: _.@+

I am not familiar with this kind of problem. If you could give me a rough idea of what I am supposed to do (a game plan), that would be of great help. It is safe to assume that I do not know much about hashing (or the related data structures). This problem would be a great opportunity for me to learn these things.

解决方案

as long as e is known, you can quite easily forge a collision, that is likely to be the original ...

md5 has a 128 bit output ... so devide your long string there into seperate md5 hex strings

now if e is known, all you need to guess for the first hash are 2 characters ... exhaustive search will be a total of [a-z0-9_.@+]^2 = 40^2 = 1600 possible candidates ...

once you have a collision, you "know" the first 2 chars of the secret ...

for the next hash, all you have to guess are the next 2 chars of s since the first two are already known from step 1 ...

repeat for all hashes in order of appearance in your large string J ... since you only have to guess 2 chars per hash and can reuse information from the found collision of the previous hash, this is fairly easy ...

in terms of complexity, the md5 hash of e needs to be calculated only once, and you have 2 concatinations and 2 md5 calculations per guess ... per hash you have an exaustive search candidate pool of 1600 so as and average you can say that you will need 800 guesses per hash in J

keep in mind that this is no decrytion ... it is still a search for a collision ... but most likely it will be the original data, since md5 will most likely (one could almost be tempted to say certainly) be collision free for the possible inputs ...

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