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network profile
| bio | website | |
|---|---|---|
| location | Windsor, Canada | |
| age | 39 | |
| visits | member for | 1 year |
| seen | Oct 29 '12 at 1:32 | |
| stats | profile views | 1 |
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Oct 30 |
awarded | Student |
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Oct 29 |
awarded | Scholar |
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Oct 29 |
accepted | Why, or when, to use an Initialization Vector? |
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Oct 28 |
revised |
Why, or when, to use an Initialization Vector? added 155 characters in body |
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Oct 28 |
comment |
Why, or when, to use an Initialization Vector? Is there any attack to be had in my example; where i leave the IV always at zero? |
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Oct 28 |
awarded | Editor |
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Oct 28 |
comment |
Why, or when, to use an Initialization Vector? @PaŭloEbermann In that case it would be whatever the "default" is; which is documented to be all zero's. But if AES doesn't support an IV, then i guess my IV is "undefined". |
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Oct 28 |
revised |
Why, or when, to use an Initialization Vector? added 983 characters in body |
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Oct 28 |
asked | Why, or when, to use an Initialization Vector? |
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May 13 |
comment |
Is it fair to assume that SHA1 collisions won't occur on a set of <100k strings @owlstead "Is it fair to assume that SHA1 collisions won't occur on a set of <100k strings" Yes, it is fair to assume that SHA1 collisions won't occur on a set of <100k strings. |
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May 12 |
awarded | Teacher |
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May 12 |
comment |
Academic papers on GPU password cracking? @Ninefingers i saw the title, and wondered what the answer was - only to find the question closed. The question is clear, and it is useful. You closed it, so you win; i just disagree. |
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May 12 |
answered | Does a public key absolutely need to be used to initiate an encrypted session? |
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May 12 |
comment |
Academic papers on GPU password cracking? @Ninefingers It's not "please list". It's "are there any": a true/false answer. Either there are or there are not. For example, "Has anyone ever found two messages that collide in SHA1"; i'm not asking for a comprehensive list of all messages - i'm asking if it has ever happened. |
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May 12 |
comment |
Time Capsule cryptography? The downside of having a time capsule that you don't want opened until 20 years from now: is that you must spend 18 years encrypting the puzzle. Considering Moore's law it will then take 2 years for the public to decrypt it. Total: 20 years. |
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May 12 |
awarded | Supporter |
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May 12 |
answered | Is it fair to assume that SHA1 collisions won't occur on a set of <100k strings |
