xcthulhu

5 ms·

1.
by xcthulhu 10y ago
Well, a couple months back I attended a talk on a Haskell implementation of the Noise protocol. The programmer admitted that he was a cryptography novice, and in fact a Haskell novice. As a result the code he wrote is needlessly abstract -
2.
by xcthulhu 10y ago
> Why do we do this, still? Is it not irresponsible? One reason is security - if cryptography isn't done in constant time then there is the potential for timing attacks. This is why BitCoinJ ditched BouncyCastle for their own JNI w
3.
by xcthulhu 12y ago
The Universal Approximation Theorem[1] asserts that you only ever need one hidden layer, which at least asserts that "an (approximate) simplification exists". But I can't say off the top of my head how you'd collapse an
4.
by xcthulhu 13y ago
Okay, for one: 1. Dynamically Typed Languages (LISP, Erlang, Smalltalk, Python, Ruby, and Javascript) are all easier than Haskell, JAVA, C++ or any other typed language for working with ADTs. Whether the language is OO or not isn't re
5.
by xcthulhu 13y ago
Poincaré was a genius but math has moved forward since the 19th century. Here's what I'd argue is the most important take away from set theory: for a given framework of mathematics, there are conjectures one can make that can be n
6.
by xcthulhu 15y ago
Isn't Microsoft's schizophrenic relationship with FOSS old news? The same company that claims "Linux is a cancer" also employs K.Y. Srinivasan, a topic linux kernel contributor [1], and Simon Peyton Jones of Haskell fame. Microsoft's compe
7.
by xcthulhu 16y ago
> Bernard Linsky seems to have written a chapter on Chwistek and type theory in The Golden Age of Polish Philosophy. You can read the first page [3] but I haven't been able to find the entire thing online. Try gigapedia.com ;-D
8.
by xcthulhu 16y ago
>No one uses ramified type theory these days, at least not that I am aware In the 1920s Frank Ramsey proved that the theory of ramified types + "The Axiom of Reducibility" is equivalent to the theory of simple types: http://en.wikipedi
9.
by xcthulhu 16y ago
Check out the axioms of Peano Arithmetic: http://en.wikipedia.org/wiki/Peano_axioms#Arithmetic If we use Dedekind's recursive definition of addition, and the definitions 1=S(0) and 2=S(S(0)), then: S(0) + S(0) = S(S(0)+0) = S(