如何解密SHA-512散列数据 [英] How to decrypt SHA-512 hashed data

问题描述

更新：

SHA-512是单向的，因此我不需要尝试破解密码，因为它更简单重置密码。如果有人知道从SHA-512哈希文本中获取原始密码，请告诉我。否则，我正在从这个问题继续。谢谢你所有的答案。

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原问题：

我读过很多文章，指出SHA-512哈希不能取消哈希。但是，这里有各种SHA-1 +算法的源代码： https：/ /tls.mbed.org/sha-512-source-code 。

我想知道是否有可能逆转这种编码，方式来解密SHA-512散列文本。 Linux使用SHA-512哈希来加密密码。作为系统管理员，我宁愿仅根据需要解密或取消散列此信息，而不是猜测密码是正确还是不正确，并查看散列是否匹配。创建新的密码可能会导致大量额外的时间和金钱。如果您对发布此信息感到不适，并希望私下讨论，请随时索取我的联系信息。

谢谢！

解决方案

为什么你不相信你读过的东西？

加密哈希函数不能颠倒。

想法实验：您有200个字节传递给SHA512，超出64个字节。有些东西已经丢失。你如何重拾失去的东西？

以类似的方式，如果你有一个整数，比如说123，mod是10，那么结果就是3.现在反过来 - 哦，它可能是3,13,23,33,123,9343453，* 3。

Update:

SHA-512 is one-way so I will need to not attempt to crack passwords because it is easier to simply reset a password. If someone is aware of getting the original password from a SHA-512 hashed text, please let me know. Otherwise, I am moving on from this question. Thank you for all of your answers.

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Original Question:

I have read a lot of articles that state that SHA-512 hashing cannot be unhashed. However, there is a source code for the various SHA-1 + algorithms here: https://tls.mbed.org/sha-512-source-code.

I would like to know if it is possible to reverse this coding, in a way, to decrypt SHA-512 hashed text. Linux encrypts their passwords with SHA-512 hashing. As a systems administrator, I would prefer to simply decrypt or unhash this information as needed, rather than guessing whether a password is correct or incorrect and see if the hash matches. Creating new passwords can cause a lot of extra time and money. If you do not feel comfortable publishing this information and would like to discuss it privately, feel free to request my contact information.

Thank you!

解决方案

Why do you not believe what you have read?

Cryptographic hash functions can not be reversed.

Thought experiment: You have 200 bytes you pass to SHA512, out come 64 bytes. Something has been lost. How do you regain what is lost?

In a similar manner if you have an integer, say 123, and mod by 10 the result would be 3. Now reverse that–oh it could have been and of 3, 13, 23, 33, 123, 9343453, *3.

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