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  • 0 posts edited
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May
7
comment How is HMAC(message,key) more secure than Hash(key1+message+key2)
Thanks, changed. Also added a note stressing the importance of padding.
May
7
revised How is HMAC(message,key) more secure than Hash(key1+message+key2)
Stress the importance of padding; edited body
May
6
comment How is HMAC(message,key) more secure than Hash(key1+message+key2)
Correct; that applies to the version without the $10^t$ padding. Yasuda notes: "We note that it is the lack of appropriate filling between the message M and the last key K, rather than the usage of a single key, that contributes to this key recovery attack."
May
6
comment How is HMAC(message,key) more secure than Hash(key1+message+key2)
Being the envelope MAC was the point! A single key is enough to be secure, though two independent keys are also fine.
May
6
comment How is HMAC(message,key) more secure than Hash(key1+message+key2)
OK, changed it to a less confusing formula, that matches Yasuda's paper.
May
6
comment How is HMAC(message,key) more secure than Hash(key1+message+key2)
$1$, followed by enough $0$s to pad out to block length. I think it should be intelligible now?
May
6
revised How is HMAC(message,key) more secure than Hash(key1+message+key2)
Use Yasuda's notation
May
6
comment How is HMAC(message,key) more secure than Hash(key1+message+key2)
Right, I got the message padding wrong. Fixed now.
May
6
revised How is HMAC(message,key) more secure than Hash(key1+message+key2)
added 1 character in body
May
6
answered How is HMAC(message,key) more secure than Hash(key1+message+key2)
May
4
revised Factoring two RSA moduli $N_i=p_i\cdot q_i$ knowing that $p_2=p_1+2$?
deleted 4 characters in body
May
4
revised Factoring two RSA moduli $N_i=p_i\cdot q_i$ knowing that $p_2=p_1+2$?
edited body
May
4
answered Factoring two RSA moduli $N_i=p_i\cdot q_i$ knowing that $p_2=p_1+2$?
Apr
20
awarded  Enlightened
Apr
20
awarded  Nice Answer
Mar
29
awarded  Revival
Mar
28
answered What aspect of elliptic curve encryption paradigms makes them especially susceptible to quantum based attack algorithms?
Dec
17
comment N way collision of hashes
The exact work factor is worked out in Suzuki et al: $(n!)^{1/n}\cdot T^{1-1/s}$.
Nov
19
comment Do data-dependent rotations have any advantage over fixed rotations?
Probably. Analysis is still more difficult on data-dependent rotations, and so it may be wise to assume the attacker can force them to have no difference. This is especially the case in hash functions, where the attacker controls almost everything; here's an example. When assuming no differences in the rotations, fixed rotations start to seem like the better choice.
Nov
19
answered Do data-dependent rotations have any advantage over fixed rotations?