Timeline for Clarification on how rainbow tables work
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 27 at 16:42 | vote | accept | user1068636 | ||
| Nov 29, 2012 at 14:21 | vote | accept | user1068636 | ||
| Feb 22, 2020 at 22:56 | |||||
| Nov 29, 2012 at 2:30 | history | edited | user1068636 | CC BY-SA 3.0 |
added 1775 characters in body
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| Nov 27, 2012 at 2:08 | history | edited | Thomas Pornin |
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| Nov 27, 2012 at 2:06 | answer | added | Thomas Pornin | timeline score: 11 | |
| Nov 26, 2012 at 14:54 | comment | added | user1068636 | Hi Henning - No I did not mix up the plaintext and password. The key is all zeros as I said and the password is "0x56". The rainbow table is constructed using all-zero key. You have to pretend as though you do not know "0x56" is the password. The only thing we know about the password is it is 2-hex characters and was encrypted with a key that is all zeros. We also know the AES of the password is 1bb171d2afef6e7f1ccb895ee9b35fe0. | |
| Nov 26, 2012 at 7:48 | history | edited | Henning Klevjer |
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| Nov 26, 2012 at 7:05 | comment | added | Henning Klevjer |
First of all, AES is an encryption algorithm, not a hash function. What you seem to want to do is to calculate a rainbow table for everything encrypted with the key 0x56. Can you clarify this? Also, in the assertion The password "56" is a two hex characters that was encrypted with a key of all zeros. you mix key and plaintext.
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| Nov 26, 2012 at 6:07 | history | asked | user1068636 | CC BY-SA 3.0 |