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

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

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

neizod

#!/usr/bin/env python3

from numpy import matrix, zeros


class RandomWalk(object):

    limpow = 10**9
    offset = [(0,-1), (0,1), (-1,0), (1,0)]

neizod

#!/usr/bin/env python3

import sys
from itertools import count


class Point(object):
    def __init__(self, pp):
        self.x, self.y = pp

neizod

#include <iostream>
using namespace std;

int arr[] = {2,5,3,9,1,0,7,4,8,6};

void show(int i=-1) {
    for (int k=0; k<10; k++) {
        if (i-1 == k) cout << "\x1b[37;1m";
        if (i == k) cout << "\x1b[31;1m";
        cout << arr[k];

neizod

<h3>encrypted sentence</h3>
<script src="http://code.jquery.com/jquery-2.1.0.min.js"></script>

<input name="text" size="40" />
<button id="mk">make</button>

<hr />
<h3>decrypted sentence</h3>
<div id="box"></div>
<hr />