jameslyons

for i = 1:10
    print(i)
end

jameslyons

import urllib
from xml.etree.ElementTree import parse

URL = '''http://jameslyons0.blogspot.com.au/feeds/posts/default?max-results=9999'''

xml = parse(urllib.urlopen(URL)).getroot()
info = {}
for post in xml.getchildren():
    if post.tag=='{http://www.w3.org/2005/Atom}entry': 
        headers,title,link=[],None,None    

jameslyons

# we need 2 helper mappings, from letters to ints and the inverse
L2I = dict(zip("ABCDEFGHIJKLMNOPQRSTUVWXYZ",range(26)))
I2L = dict(zip(range(26),"ABCDEFGHIJKLMNOPQRSTUVWXYZ"))

key = 3
plaintext = "DEFEND THE EAST WALL OF THE CASTLE"

# encipher
ciphertext = ""
for c in plaintext.upper():

jameslyons

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

int main(int argc, char *argv[]){

    char plaintext[] = "DEFEND THE EAST WALL OF THE CASTLE";
    int L = strlen(plaintext);
    char *ciphertext = malloc(sizeof(char)*(L+1));

jameslyons

// enciphering
plaintext = "DEFEND THE EAST WALL OF THE CASTLE";  
var key = 3;
ciphertext = "";    var re = /[a-z]/;
for(i=0; i<plaintext.length; i++){ 
   if(re.test(plaintext.charAt(i))) ciphertext += String.fromCharCode((plaintext.charCodeAt(i) - 97 + key)%26 + 97); 
   else ciphertext += plaintext.charAt(i); 
} 

// deciphering

jameslyons

# return the autocorrelation of x
function ac(x)
    N = length(x)
    # zero pad x 
    x = [x,zeros(N-1)]
    R = irfft(abs(rfft([2,3,1,0,0])).^2, 2*N-1)
    R[1:N]
end

x = [2,3,1] 

jameslyons

function levinson(R,L)
    a = zeros(L,L)
    P = zeros(1,L)

    # for m = 1
    a[1,1] = -R[2]/R[1]
    P[1] = R[1]*(1-a[1,1]^2)

    # for m = 2,3,4,..L
    for m = 2:L

jameslyons

function arcov(x,L)
    N = length(x)
    phi = zeros(L+1,L+1)
    for k = 0:L, i=0:L, n=L:N-1
        phi[k+1,i+1] += x[n+1-k]*x[n+1-i]
    end
    phi = phi./(N-L)

    a = phi[2:end,2:end]\-phi[2:end,1]
    P = phi[1,1] + phi[1,2:end]*a

jameslyons

function armcov(x,L)
    N = length(x)
    phi = zeros(L+1,L+1)
    for k = 0:L, i=0:L, n=0:N-L-1
        phi[k+1,i+1] += x[n+1+k]*x[n+1+i] + x[n+1+L-k]*x[n+1+L-i]
    end
    phi = phi./(2*(N-L))

    a = phi[2:end,2:end]\-phi[2:end,1]
    P = phi[1,1] + phi[1,2:end]*a

jameslyons

• Update HISTORY.rst
• Update version number in my_project/__init__.py
• Update version number in setup.py
• Run the tests:

python setup.py test
tox

• Commit the changes: