32,416 reputation
562124
bio website bolet.org/~pornin
location Quebec City, Canada
age 39
visits member for 3 years, 3 months
seen 15 hours ago

Cryptographer, programmer in several languages (C, Java, several assemblies, Pascal, Forth...). I also have a life.


15h
comment AES-GCM Disadvantage
One may add a performance issue: the not-too-slow implementations of GCM use tables that exhaust L1 caches on small architectures (like the 32-bit Mips found in cheap home routers). It has been reported that on such systems, EAX mode outperforms GCM.
Oct
21
awarded  Nice Answer
Oct
17
answered MD5 update functions
Oct
8
awarded  Enlightened
Oct
8
awarded  Nice Answer
Oct
7
awarded  Nice Answer
Sep
30
awarded  Explainer
Sep
18
comment What is the length of an RSA signature?
That trick can be extended; e.g. you can skip a complete byte (8 bits) if the verifier is ready to compute 256 RSA verifications, while it is trying to guess the missing bits. Maybe more importantly, in ISO/IEC 9796-2, part of the signed data can be embedded "for free" in the signature, so while the signature value has size $n$ bits (or $n-1$ with the trick you describe), the overhead implied by the presence of the signature can be substantially smaller, depending on the situation.
Sep
7
awarded  Good Answer
Sep
5
awarded  Nice Answer
Aug
26
awarded  Enlightened
Aug
26
awarded  Nice Answer
Aug
25
comment Efficiently map $2^n$ unique 64-bit vectors to $2^n$ $n$-bit vectors where $n < 64$?
If all possible inputs are known a priori then you can put 64 million entries into exactly 64 million buckets with no collision -- but the mapping will have to know the 64 million entries. And the gist of my answer is that this cannot, in general, be done without using enough ROM or RAM to actually encode the 64 million possible values one way or another.
Aug
23
answered Efficiently map $2^n$ unique 64-bit vectors to $2^n$ $n$-bit vectors where $n < 64$?
Aug
18
awarded  Enlightened
Aug
18
awarded  Nice Answer
Aug
14
answered Why can't you decrypt an encrypted message with just the public key?
Aug
13
awarded  Nice Answer
Aug
13
answered How is SHA1 different from MD5?
Aug
12
comment linear computations over bilinear pairings
Two elements in a group are opposite to each other if their product is 1 (or their sum is 0, if you use an additive notation for the group).