Timeline for Block cipher with key longer than block size
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 30, 2014 at 15:38 | comment | added | sashank | but the relation is not established correctly. the factor does not depend on cardinality of block size but cardinality on the set of permutations possible which is the crucial part to be specific | |
| Jan 30, 2014 at 10:33 | history | edited | CodesInChaos | CC BY-SA 3.0 |
added 251 characters in body
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| Jan 30, 2014 at 10:26 | comment | added | CodesInChaos | @sashank Apart from using "size" instead of "cardinality" this answer looks correct to me. It only argues that there are multiple keys which encrypt one particular $x$ to the same value. It doesn't argue that there are equivalent keys, which encrypt all $x$ to the same value. This fits the question, which is about that particular property, not about equivalent keys, like the question you linked. | |
| Jan 30, 2014 at 10:19 | history | rollback | CodesInChaos |
Rollback to Revision 1
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| Jan 30, 2014 at 9:07 | history | edited | sashank | CC BY-SA 3.0 |
corrected the math
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| Jul 2, 2013 at 10:27 | comment | added | user7139 | Yes, I see what you mean. Looking at my comment question now it looks as though I had $E(k_0,x)=x$ in mind, which should be avoidable in theory. | |
| Jul 1, 2013 at 20:24 | comment | added | archie | Every block cipher has to be a bijective mapping over the input/output space $X$ - i.e. every input maps to and from one output using the encryption/decryption functions (this is entirely natural when you think about what a block cipher needs to do: encrypt every possible input value and decrypt that output value back to the input value. When $|K| > |X|$ you still have the same bijective mapping, but the selection of which mapping is used (and the mechanics of how it is effected) is based on the key supplied. | |
| Jul 1, 2013 at 6:05 | comment | added | user7139 | Would it not be possible to generate a set of permutations where no element in any permutation has the same place twice? This would avoid the "collisions" at the cost of having fewer keys. | |
| Jun 30, 2013 at 22:56 | vote | accept | CommunityBot | ||
| Jun 30, 2013 at 21:06 | history | answered | archie | CC BY-SA 3.0 |