neizod

            DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
                    Version 2, December 2004

 Copyright (C) 2012 Nattawut Phetmak <http://about.me/neizod>

 Everyone is permitted to copy and distribute verbatim or modified
 copies of this license document, and changing it is allowed as long
 as the name is changed.

            DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE

neizod

array[1..3] of int: divisors = [2, 3, 7];

var 1..100: answer;

constraint forall(d in divisors)( answer mod d == 0 );

solve satisfy;

output [ "The ultimate answer is \(answer).\n" ];

neizod

neizod

\NeedsTeXFormat{LaTeX2e}
\ProvidesClass{independent-study}

\DeclareOption*{\PassOptionsToClass{\CurrentOption}{book}}
\ProcessOptions\relax
\LoadClass[11pt,a4paper,oneside]{book}


% thai book
\usepackage[english,thai]{babel}

neizod

#!/usr/bin/env Rscript

library(rgl)
library(sp)


calc_radian <- function(i, tick, uplim) {
    2 * pi * (i / tick) * (uplim / 360)
}

neizod

random.int <- function(n) { sample.int(n, 1) }


random.coupon <- function(...) {
  count <- 0
  have.coupon <- logical(...)
  while (!all(have.coupon)) {
    have.coupon[random.int(...)] <- TRUE
    count <- count + 1
  }

neizod

Keybase proof

I hereby claim:

• I am neizod on github.
• I am neizod (https://keybase.io/neizod) on keybase.
• I have a public key whose fingerprint is 48AB A6A9 A002 DCD8 6E24 DFB0 0E27 2831 8C9B 4947

To claim this, I am signing this object:

neizod

Break the NSA Encryption

Challenge on Gist from this Facebook Post provide an interesting question on cracking the NSA encryption. Doesn't this hook you enough? Let's roll. :D

Note, you can also test the validity of logic in this file with the command:

$ python3 -m doctest break_nsa.rst --verbose

neizod

#!/usr/bin/env python3

from collections import deque


class Node(object):
    def __init__(self):
        self.value = None
        self.left = None
        self.right = None

neizod

type Vector = (Double, Double)


dotV :: Vector -> Vector -> Double
dotV (x, y) (u, v) = x * u + y * v


addV :: Vector -> Vector -> Vector
addV (x, y) (u, v) = (x + u, y + v)