Getting Apache Drill working on Windows 10

Apache Drill is a program that allows you to query many data sources from a single query using SQL. It supports a variety of NoSQL databases and file systems, including HBase, MongoDB, MapR-DB, HDFS, MapR-FS, Amazon S3, Azure Blob Storage, Google Cloud Storage, Swift, NAS and local files.

I basically want to be able to query CSV files, sqlite files, parquet files and other RDBMS using SQL with a unified interface from python. I am looking into Apache Drill, hopefully it will solve all my problems, I'll just be using it in local mode.

Download apache drill from here: https://drill.apache.org/download/. I did the direct file download, I got version 1.13.0.

To get drill working, you'll also need Oracle JDK v8 from here: http://www.oracle.com/technetwork/java/javase/downloads/jdk8-downloads-2133151.html. I got version jdk-8u171-windows-x64.exe

Use 7zip to unzip the tar.gz file to some directory. From the command prompt, cd to the directory, it should have the following contents:

bin
conf
git.properties
jars
KEYS
LICENSE
NOTICE
README.md
sample-data
winutils

cd to the bin folder and try to run

sqlline -u jdbc:drill:zk=local

The first error I got was: "WARN: JAVA_HOME not found in your environment. Please set the JAVA_HOME variable in your environment to match the location of your Java installation". This is pretty easy to fix, make sure java JDK is installed and you have an environment variable called JAVA_HOME (see here for how to create one) that has the value "C:\Program Files\Java\jdk1.8.0_171" (assuming that is where you installed it).

After doing that, the next error I got when running sqlline -u jdbc:drill:zk=local was: Error: Failure in connecting to Drill: org.apache.drill.exec.rpc.RpcException: Failure setting up ZK for client. (state=,code=0) plus a whole lot more java error stuff. Then no matter what I typed it simply said No current connection. To fix this one, you have to put quotes around the argument passed to sqlline like this:

sqlline -u "jdbc:drill:zk=local"

Note that to exit sqlline I found Ctrl d to be the easiest, this simulates an end of file character and the program stops gracefully. Now I can run commands like:

USE cp; 
SELECT  employee_id, first_name FROM `employee.json`; 

Connecting to sqlite databases from apache drill

Of course you can connect to sqlite databases directly using the sqlite3 console program, but I would like to connect to sqlite dbs, csvs, parquet files and other RDBMS all from the same query. To do that I need the JDBC drivers for sqlite, which are available here: https://bitbucket.org/xerial/sqlite-jdbc/downloads/. I got version sqlite-jdbc-3.21.0.jar. This has to be copied to the 3rdparty folder in the apache drill install directory: C:\Users\james\Downloads\apache-drill-1.13.0\jars\3rdparty on my system. Make sure you close and reopen sqlline/drill before trying to use the driver.

Unfortunately, we can't access sqlite dbs in the same way as files, we have to create a new storage plugin for each sqlite database we have. I downloaded the sample database chinook.db from here: http://www.sqlitetutorial.net/sqlite-sample-database/. Extract the database chinook.db somewhere and remember where it is. I got the list of tables by opening the db in sqlitestudio.

Now we need to open the drill web console by navigating to http://localhost:8047/storage/ (make sure sqlline or drill is running in the background). We need to create a new storage plugin, we'll call it chinook. Down the bottom of the page there is a text entry, put chinook in there and and press the create new storage plugin button. You'll be met with a text entry field that is expecting JSON. Enter the following, but replace the path that I have entered with the path to your database file:

{
  "type": "jdbc",
  "driver": "org.sqlite.JDBC",
  "url": "jdbc:sqlite:C:\\Users\\james\\Downloads\\chinook.db",
  "username": null,
  "password": null,
  "enabled": true
}

Now we can query the sqlite database using drill (you may need to close and restart sqlline/drill):

SELECT * FROM chinook.customers WHERE lastname like 'g%';
+-------------+------------+------------+---------------------------------------------------+--------------------------+
| CustomerId  | FirstName  |  LastName  |                      Company                      |                Address   |
+-------------+------------+------------+---------------------------------------------------+--------------------------+
| 1           | Luφs       | Gonτalves  | Embraer - Empresa Brasileira de Aeronßutica S.A.  | Av. Brigadeiro Faria Lim |
| 7           | Astrid     | Gruber     | null                                              | Rotenturmstra▀e 4, 1010  |
| 19          | Tim        | Goyer      | Apple Inc.                                        | 1 Infinite Loop          |
| 23          | John       | Gordon     | null                                              | 69 Salem Street          |
| 27          | Patrick    | Gray       | null                                              | 1033 N Park Ave          |
| 42          | Wyatt      | Girard     | null                                              | 9, Place Louis Barthou   |
| 56          | Diego      | GutiΘrrez  | null                                              | 307 Macacha Gⁿemes       |
+-------------+------------+------------+---------------------------------------------------+--------------------------+
7 rows selected (0.185 seconds)

Unfortunately, not all tables could be read. If I try to open the employees table, i get:

 SELECT * FROM chinook.employees;
Error: DATA_READ ERROR: Failure while attempting to read from database.

sql SELECT *
FROM employees
plugin sqlite
Fragment 0:0

[Error Id: ac42e2eb-a8ac-478e-b5bd-de663f7bc93e on DESKTOP:31010] (state=,code=0)

The sqlite JDBC driver doesn't seem to like the datetime fields, use something like sqlitestudio to change the datatype of the fields to strings, then everything will work. You'll just have to convert the string dates to actual date objects in code later on.

Using Apache drill from python

I think the easiest way of doing this is to use pydrill (run pip install pydrill to get it). This however uses the rest api, which is likely not the most efficient. Using the ODBC drivers and pyodbc will likely be better, but I have not tested it yet.

from pydrill.client import PyDrill

drill = PyDrill(host='localhost', port=8047)

if not drill.is_active():
    raise ImproperlyConfigured('Please run Drill first')

res= drill.query('SELECT * FROM chinook.customers LIMIT 5')

for result in res:
    print(result)

# pandas dataframe
df = res.to_dataframe()
print(df)

Generating Gray codes with python

Gray codes are often used in positioning systems, see the wiki article for more details. I want to generate gray codes with python, and I'd like to generate all possible gray codes for a specific number of positions and number of bits. The maximum number of positions for a gray code is 2^numbits i.e. a 4 bit gray code can have at most 16 positions. It can also have fewer. If you want the gray code to be circular i.e. the first and last code have hamming distance 1, then there needs to be an even number of positions. Odd is fine if you don't care if it is circular.

There are of course algorithms for generating gray codes, but a constraint solver works really well, and also allows you to generate all possible gray codes for a specific number of positions. The python code below is for generating gray codes using python-constraint:

from constraint import *
from itertools import product

numbits = 4
numpos = 16

# this generates a list of all binary vectors of numbits bits
vecs = list(product(*(((1,0),)*numbits)))

problem = Problem()
# we will have numpos variables, i.e. one var for each code,
# each var is an index into vecs, has 2^numbits possible vals
problem.addVariables(list(range(numpos)), list(range(len(vecs))))

# vecs holds our binary vectors. 
# we want True if hamming distance between vecs[a] and vecs[b] is 1
# False otherwise. This will be our constraint. 
def hammingDist1(a,b):
    hamming = sum([vecs[a][i]^vecs[b][i] for i in range(numbits)])
    if hamming == 1: return True
    return False

# add the important constraints
for i in range(numpos):
    j = (i+1)%numpos
    problem.addConstraint(hammingDist1,[i,j])

# make sure all the vectors are different    
problem.addConstraint(AllDifferentConstraint())

# print out the first solution we come across
sol = problem.getSolution()
for i,j in enumerate(sol):
    print(vecs[sol[i]])
    

The main thing to note is that we add constraints so that each code has a hamming distance of 1 with its neighbour, and also that all the codes are diferent.

An example run looks like this (for a 4-bit, 16 position code):

C:\Users\james\Desktop>python gray.py
(0, 0, 0, 0)
(0, 0, 0, 1)
(0, 0, 1, 1)
(0, 0, 1, 0)
(0, 1, 1, 0)
(0, 1, 1, 1)
(0, 1, 0, 1)
(1, 1, 0, 1)
(1, 0, 0, 1)
(1, 0, 0, 0)
(1, 0, 1, 0)
(1, 0, 1, 1)
(1, 1, 1, 1)
(1, 1, 1, 0)
(1, 1, 0, 0)
(0, 1, 0, 0)

Latex Dictionary Template

I recently needed a latex dictionary template, there are a couple around on the internet, but they are all under non-commercial licences so I couldn't use them. Fortunately it is pretty easy to make one yourself, I have provided one here. Consider it to be under the wtfpl licence.

A short example of what the definitions look like:

%Copyright 2017 James Lyons
%This work is free. You can redistribute it and/or modify it under the
%terms of the Do What The Fuck You Want To Public License, Version 2,
%as published by Sam Hocevar. See http://www.wtfpl.net/ for more details.
\documentclass[twoside]{book}
\usepackage[utf8]{inputenc}
\usepackage{multicol}
\usepackage{fancyhdr} 
\usepackage[bindingoffset=0.2in,margin=0.75in,paperwidth=5.25in,paperheight=8in]{geometry}

% set the headers so that first mark is always on left, second on right
% this is so that each word is put in the header correctly 
\pagestyle{fancy} 
\fancyhead[LO,LE]{\rightmark}
\fancyhead[RE,RO]{\leftmark}
% set the font
\renewcommand*\rmdefault{bch}
\newcommand{\comment}[1]{}
% remove chapter numbering
\usepackage{titlesec}
\titleformat{\chapter}
  {\Large\bfseries} % format
  {}                % label
  {0pt}             % sep
  {\huge}           % before-code
% remove section numbering
\setcounter{secnumdepth}{0}
% this is how we define the actual dictionary items. markboth is for 
% setting the page headers so that first and last words appear there.
\newcommand{\ditem}[1]{\item[#1] \markboth{\footnotesize \textbf{#1}}{\footnotesize \textbf{#1}}}
\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%% title page
\begin{center}

\thispagestyle{empty}
\vbox{\vspace{5cm}
{\huge The Dictionary Title}}
\vfill
{\large Author}
\vfill

\end{center}
\clearpage
%%%%%%%%%%%%%%%%%%%%%%%% table of contents
\tableofcontents

%%%%%%%%%%%%%%%%%%%%%%%% Preface
\chapter{Preface}

This is the preface.

\chapter{The Dictionary}
% we'd like the text size to be a bit smaller for the definitions
\footnotesize
% we'd like two columns of definitions
\begin{multicols}{2}

\begin{description}
\section{A}
\ditem{Abandon} \textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abandoned} \textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abbey}\textit{-v.} definition 1 \textit{-n.} definition 2
\ditem{Abide} \textit{-v.} definition 1 \textit{-n.} definition 2
\end{description}

\end{multicols}
 
\end{document}

Creating crosswords with python-constraint [part 2]

In the previous post I was generating the patterns of white and black squares. In this post I'd like to actually fill the patterns with words so that everything matches up. This sort of problem is very well suited to a constraint satisfaction solver, as it can try many possibilities until something is found that works. The trick is specifying the constraints in an efficient manner. My first solution was to implement it the same way as the magic word squares, but this was really inefficient and there is a much better way.

I put the code for this on github, it's a bit messy but works: https://github.com/jameslyons/crossword_gen.

The trick is you don't need to represent all the different squares in the crossword as separate variables and put the constraints on those, your actual variables will be indexes into the word list, and the constraints will be whether e.g. character 4 of word 1 is the same as character 2 of word 7. These constraints will obviously be different for every crossword and will have to be generated each time, but this isn't too hard.

The inputs from the previous step are hex strings e.g.

3EF9C51EBFAFAAF8FE37577BFB9D5CEFEF75763F8FAAFAFEBC51CFBEg

This can be converted to a binary string by replacing e.g. 3 with '0011', same as decoding hex anywhere. The function I use is down the bottom of the previous post. This gives a binary string that can be represented as the white and black squares of the crossword.

Example run

The code takes as argument the hex representation of the pattern. It first prints a representation of the crossword with 'X's where the letters go, then once it gets solved it outputs the solved crossword, along with the list of words that got used. The dictionary I am using has some weird words in it, so you need to pay some attention to that if you want nice crosswords (I recommend using something like the 3of6game.txt from the 12dicts wordlist, it is specifically for games). The final thing that gets printed is the crossword represented as a string, this is passed to some typesetting code I have that produces nice pdfs of the crosswords.

>pypy cw_fill.py BEFB4516BFAFAAF8FE27573BFBDD5DEFEE75723F8FAAFAFEB4516FBEh
building word lists...
X   X X X X X   X X X X X   X
X   X       X   X       X   X
X   X   X X X X X X X   X   X
X X X X   X   X   X   X X X X
X       X X X X X X X       X
    X X X   X   X   X X X
X X X   X X X X X X X   X X X
X   X X X   X   X   X X X   X
X X X   X X X X X X X   X X X
    X X X   X   X   X X X
X       X X X X X X X       X
X X X X   X   X   X   X X X X
X   X   X X X X X X X   X   X
X   X       X   X       X   X
X   X X X X X   X X X X X   X
starting...
count: 0
s . c r u m b . r e m a n . b
w . r . . . r . i . . . o . r
a . a . p h o e b u s . s . i
m o w n . e . l . r . g y m s
i . . . c r i m i n e . . . k
. . p o o . n . n . v e g . .
a v o . i n f a u n a . i l k
s . s u n . l . t . d i g . i
p h i . a l a n i n e . o f t
. . t a g . m . l . r a t . .
i . . . e y e l e s s . . . o
m o t s . e . o . p . w a t t
i . o . s t o p g a p . m . t
d . r . . . a . e . . . i . a
e . c l a c k . l u l l s . r

['swami', 'asp', 'imide', 'craw', 'posit', 'torc', 'coinage', 'her', 'yet', 'bro',
 'inflame', 'oak', 'elm', 'lop', 'rib', 'inutile', 'gel', 'urn', 'spa', 'evaders',
 'nosy', 'gigot', 'amis', 'brisk', 'kit', 'ottar', 'crumb', 'reman', 'phoebus',
 'mown', 'gyms', 'crimine', 'poo', 'veg', 'avo', 'infauna', 'ilk', 'sun', 'dig',
 'phi', 'alanine', 'oft', 'tag', 'rat', 'eyeless', 'mots', 'watt', 'stopgap',
 'clack', 'lulls']
solution time = 1.05939997405
s.crumb.reman.bw.r...r.i...o.ra.a.phoebus.s.imown.e.l.r.gymsi...crimine...k..p
oo.n.n.veg..avo.infauna.ilks.sun.l.t.dig.iphi.alanine.oft..tag.m.l.rat..i...ey
eless...omots.e.o.p.watti.o.stopgap.m.td.r...a.e...i.ae.clack.lulls.r

The code

This takes the hex string output from the pattern generation step as an argument. It uses a wordlist called 'fewerwords.txt'. Each time it is run it should give different results because the word indexes are shuffled each time. Sometimes it will take a long time to find a solution, I suggest just killing it and running it again. Most of the time this will get a new starting point that allows it to be solved very quickly, though some patterns are just harder to solve than others.

from __future__ import print_function
from constraint import *
from timeit import default_timer as timer
from itertools import combinations
from math import sqrt
from test_bin2hex import hex2bin
import sys

''' to run from pypy
pypy cw_fill.py 3EF9C51EBFAFAAF8FE37577BFB9D5CEFEF75763F8FAAFAFEBC51CFBEg
'''
# this program will take a crossword pattern and try to fill it with words

if len(sys.argv) < 2: print('supply pattern as argument')
else:
    pattern = hex2bin(sys.argv[1]) 
    pattern = [int(i) for i in pattern]

s = int(sqrt(len(pattern)))

print('building word lists...')
# extract the word positions from the pattern
wordlocs = []
hwordlocs = []
vwordlocs = []
for i in range(s):
    hword =  []
    vword = []
    for j in range(s):
        if pattern[i*s + j] == 0: print(' ',end=' ')
        else: print('X',end=' ')
        if pattern[i*s + j] == 1:
            hword.append(i*s + j)
        else:
            if len(hword) > 1: hwordlocs.append(hword)
            hword = []         
        
        if pattern[j*s + i] == 1:
            vword.append(j*s + i)
        else:
            if len(vword) > 1: vwordlocs.append(vword)
            vword = []
        
        if j == s-1:
            if len(vword) > 1 and vword not in vwordlocs: vwordlocs.append(vword)
            if len(hword) > 1 and hword not in hwordlocs: hwordlocs.append(hword)
            
    print('')
wordlocs.extend(vwordlocs)
wordlocs.extend(hwordlocs)

# now that we have the words, add constraints so each word is in the dictionary
maxL = 10

wordlist = []
f = open("fewwords.txt")
for line in f:
    w = line.split()[0].lower()
    wordlist.append(w)

wordlen = []
olen = 0
for c,word in enumerate(wordlist):
    if olen < len(word):
        wordlen.append(c)
    olen = len(word)

problem = Problem()
from random import shuffle

for c,w in enumerate(wordlocs):
    L = len(w)-3
    r = list(range(wordlen[L],wordlen[L+1]))
    shuffle(r)
    problem.addVariables([c], list(r))

problem.addConstraint(AllDifferentConstraint())

def eq(a,b):
    def temp(c1,c2):
        word1 = wordlist[c1]
        word2 = wordlist[c2]
        if word1[a]==word2[b]: return True
        return False
    return temp
        
for c1,word1 in enumerate(wordlocs):
    for c2,word2 in enumerate(wordlocs):
        if word1==word2: continue
        intsc = list(set(word1).intersection(set(word2)))
        if len(intsc) > 0:
            a = word1.index(intsc[0])
            b = word2.index(intsc[0])
            problem.addConstraint(eq(a,b),[c1,c2])    
        
print('starting...')
        
def getwords(solution):
    listq = []
    for locs in wordlocs:
        word = ''.join([solution[loc] for loc in locs])
        listq.append(word)        
    return listq
              
# pretty print the output
start = timer()
b = ''
for count,a in enumerate(problem.getSolutionIter()):  
    print('count: '+str(count))
    sol = ['.' for i in range(s*s)]
    for key,val in a.iteritems():
        for c,char in enumerate(wordlist[val]):
            sol[wordlocs[key][c]] = char      
    b=sol
    for i in range(s):
        for j in range(s): 
            print(sol[i*s+j],end=' ')
        print('')
    print('')  
    # get the words we have used
    zz = getwords(sol)
    print(zz)
    break
end = timer()
print("solution time = " + str(end-start))
print(''.join([str(b[i]) for i in range(s*s)]))

Creating Crosswords with python-constraint [part 1]

Intro

I put the code for this on github, it's a bit messy but works: https://github.com/jameslyons/crossword_gen.

In previous posts I have been playing with python-constraint, and building crosswords is something that interests me for some reason. I decided to do it in a two step process: step 1 generate the patterns of white and black squares, step 2 fill the patterns with letters. This post will deal with step 1 only, the next post will deal with step 2.

Generating the patterns is well suited to using a constraint solver since you can just specify the things you'd like to be true of the final result, then ask the constraint solver to do it for you. In this case we have e.g. 15 by 15 binary variables, we choose the constraints so that the final pattern looks like a crossword puzzle.

What do pretty patterns look like

I am interested in automatically generating patterns that look like nice crossword patterns e.g.

Some of the things I'd like to be true of the patterns I generate:

1. I'd like 4-fold symmetry, because I think it looks good.
2. No 2x2 blocks of white squares.
3. No 2x2 blocks of black squares.
4. No length 1 words.
5. No length 2 words.
6. No words longer than MAXLEN.
7. No sequences of black squares longer than 3.
8. White squares must be fully connected.

Some of these conditions are pretty easy to implement with python-constraint, the fully connected constraint I can just apply to the final solutions and discard them at that point. By fully connected I mean you can get from one square to every other square in the pattern, so you can't get disjoint islands.

The implementation

Examples of what the program outputs:

3EF9C51EBFAFAAF8FE37577BFB9D5CEFEF75763F8FAAFAFEBC51CFBEg
X X . . . . . X . . . . . X X
. . . X X X . X . X X X . . .
. X . X . . . . . . . X . X .
. . . . X . X . X . X . . . .
. X X X . . . . . . . X X X .
. X . . . X . X . X . . . X .
. . . X . . . . . . . X . . .
X X . . . X . X . X . . . X X
. . . X . . . . . . . X . . .
. X . . . X . X . X . . . X .
. X X X . . . . . . . X X X .
. . . . X . X . X . X . . . .
. X . X . . . . . . . X . X .
. . . X X X . X . X X X . . .
X X . . . . . X . . . . . X X

The hex string is the binary to hex encoding of the pattern, 'X's are zeros, '.'s are ones. The hex string will be passed to step 2 which will fill in the words using a wordlist.

It's pretty easy to generate a few million example patterns with this program (if you are willing to wait a day), which is what I suggest doing. Once you have a heap of patterns you can further filter them by whatever criteria you want e.g. no more than X 3-letter words, no more than X% black squares etc. Once you have the filtered list, you can try filling them with words.

from __future__ import print_function
from constraint import *
from itertools import combinations
from math import sqrt,floor

# this program will create crossword patterns, but not fill it with words
# i.e. just patterns of zeros and ones for the black and white squares

# tests if a solution is fully connected i.e. can every point be reached from
# any random point. True if fullyconnected, false if disjoint parts.
# Could easily be made more efficient, but this works fine.
def isfullyconnected(solution):
    L = int(sqrt(len(solution)))
    # get list of all non-zero points
    plist = []
    for i in range(L):
        for j in range(L):
            if solution[i*L + j]: plist.append((i,j))
    
    # start top right corner, try to get to every other point
    reached = {plist[0]:1}
    rlen = len(reached)
    oldrlen = 0
    while rlen != oldrlen:
      temp = list(reached.keys())
      for i,j in temp:
        if (i+1,j) in plist: reached[(i+1,j)]=1
        if (i-1,j) in plist: reached[(i-1,j)]=1
        if (i,j+1) in plist: reached[(i,j+1)]=1
        if (i,j-1) in plist: reached[(i,j-1)]=1
      oldrlen = rlen
      rlen = len(reached)
    # if rlen is not eq number of all points, we couldn't reach some points
    if rlen < len(plist): return False
    return True    
    
from random import randint
    
def getPattern(sidelen=15,maxlen=9):
    problem = Problem()
    s = sidelen
    maxWordLen = maxlen
    problem.addVariables(list(range(s*s)), [0,1])
  
    # ensure there are no 2x2 blocks of ones or zeros
    def no4x4block(*x):
        if sum(x)==0: return False
        if sum(x)==len(x): return False
        else: return True

    for i in range(s-1):
        for j in range(s-1):
            problem.addConstraint(no4x4block,[i*s+j,(i*s+j)+1,(i*s+j)+s,(i*s+j)+s+1])
    
    #ensure matrix is symmetric  
    def eq(a,b):  
        return a==b

    for i in range(s):
        for j in range(s):
            problem.addConstraint(eq,[i*s+j,(s-i-1)*s+j])
            problem.addConstraint(eq,[i*s+j,i*s+(s-j-1)])
            problem.addConstraint(eq,[i*s+j,(s-i-1)*s+(s-j-1)])
       
    #ensure no words longer than maxWordLen
    def maxlen(*x):
        if sum(x) == len(x): return False
        else: return True

    seq = list(range(maxWordLen+1))
    for i in range(s - maxWordLen):
        for j in range(s):
            problem.addConstraint(maxlen,[(k+i)+j*s for k in seq])
            problem.addConstraint(maxlen,[(k+i)*s+j for k in seq])
    
    # no two letter words
    def no2letter(*x):
        if x == (0,1,1,0): return False
        else: return True
    
    seq = list(range(4))
    for i in range(s - 3):
        for j in range(s):
            problem.addConstraint(no2letter,[(k+i)+j*s for k in seq])
            problem.addConstraint(no2letter,[(k+i)*s+j for k in seq])

    # no two letter words around the edge either
    def no2letterEdge(*x):
        if x == (1,1,0): return False
        else: return True
    
    seq = list(range(4))
    for i in range(s):
        problem.addConstraint(no2letterEdge,[i,i+s,i+2*s])
        problem.addConstraint(no2letterEdge,[i*s,i*s+1,i*s+2])
        problem.addConstraint(no2letterEdge,[i*s+s-1,i*s+s-2,i*s+s-3])
        problem.addConstraint(no2letterEdge,[i+s*(s-1),i+s*(s-2),i+s*(s-3)])
    
    # no one letter squares
    def no1letter(*x):    
        if sum(x) == 0: return True
        if sum(x[1:]) == 0: return False
        return True
    
    seq = list(range(4))
    for i in range(s):
      for j in range(s):
        vals = [i*s+j]
        if i-1 >= 0: vals.append((i-1)*s+j) 
        if i+1 < s: vals.append((i+1)*s+j) 
        if j-1 >= 0: vals.append(i*s+(j-1)) 
        if j+1 < s: vals.append(i*s+(j+1)) 
        problem.addConstraint(no1letter,vals)
 
    # no more than three 0's in a row
    def nothreeblank(*x):
        if sum(x) == 0: return False
        return True
    
    seq = list(range(4))
    for i in range(s - 3):
        for j in range(s):
            problem.addConstraint(nothreeblank,[(k+i)+j*s for k in seq])
            problem.addConstraint(nothreeblank,[(k+i)*s+j for k in seq])
   
    for sol in problem.getSolutionIter():  
        if not isfullyconnected(sol): continue
        yield(sol)
    
from test_bin2hex import bin2hex
import sys

print(len(sys.argv))
if len(sys.argv)<2: sidelen = 11
else: sidelen = int(sys.argv[1])
print('using sidelen',sidelen)

f = open('pattern'+str(sidelen)+'.txt','w')
for count,sol in enumerate(getPattern(sidelen=sidelen,maxlen=8)):
    f.write(bin2hex(''.join([str(sol[a]) for a in range(sidelen*sidelen)]))+'\n')
    if count % 100 == 0: 
      print('count: '+str(count))
      print(bin2hex(''.join([str(sol[a]) for a in range(sidelen*sidelen)])))
      for i in range(sidelen):
        for j in range(sidelen): 
            if sol[i*sidelen+j]==1: print('.',end=' ')
            else: print('X',end=' ')
        print('')
      print('')
    #break

The hex string is generated using the following python script (test_bin2hex.py). The extra letters are so we can encode binary strings that aren't necessarily divisible by 4.

bin = ['0000','0001','0010','0011','0100','0101','0110','0111','1000','1001','1010','1011','1100','1101','1110','1111',
       '0','1','00','01','10','11','000','001','010','011','100','101','110','111']
hex = "0123456789ABCDEFghijklmnopqrst"

def bin2hex(binstr):
    map = dict(zip(bin,hex))
    ret = ''
    for i in range(0,len(binstr),4):
        sub = binstr[i:i+4]
        ret += map[sub]
    return ret
    
def hex2bin(hexstr):
    map = dict(zip(hex,bin))
    ret = ''
    for i in hexstr:
        ret += map[i]
    return ret

Magic Word Squares with python-constraint

This post will deal with the seemingly boring topic of magic word squares - I define these to be squares of letters where each row is a dictionary word, and each column is also a dictionary word. In my wordlist, there are 3,641 4-letter words (I talk about how I generated the wordlist below). How many possible 4 by 4 word squares do you think there are? Would you believe 90.6 million? I was rather surprised to discover there were so many. What about 8 by 8 word squares (given there are 12,317 8 letter words in the wordlist)? would you believe 0? It turns out it gets much harder to find word squares as you increase the side length.

I'll display some of the word squares and other stats before I display the python code I used to generate all these things further down.

Some example word squares

a z o n o w t o n	f u z z a l o e g a r s s n i t	f r i z z l a r e e o d o r s p i n o t s i s s y
z y g o t e y e l p e d g l u i n g o p i a t e t e n t e r e d g e r s		e s c a p e s s e a l a n t c a s e t t e a l e r t e r p a t t e r n e n t e r a l s t e r n l y

How many word squares are there?

I counted how many word squares there are for each side length. I was suprised how many there are for 4x4 squares, but then there are no 8x8 squares or higher. The No. of words is how many words of that particular length occur in the wordlist.

Side Length	No. of words	No. of squares
3	1,014	2,607,858
4	3,641	90,669,454
5	6,683	10,991,077
6	9,559	68,518
7	11,960	98
8	12,317	0
9	11,181	0
10	8,760	didn't check
11	6,296	didn't check
12	4,082	didn't check
13	2,493	didn't check
14	1,326	didn't check
15	720	didn't check

Building the wordlist

I identified two wordlists which I might use: the scrabble dictionary, and googles billion word corpus. Unfortunately, the scrabble dictionary has so many weird words I've never heard of before, and google's dataset has a heap of misspellings and gibberish. To get my wordlist I found the intersection of the two, with the hope being that the result will be just reasonably common proper English words.

The code

I used python-constraint to solve the squares. A basic (very slow) implementation is shown below

from constraint import *
problem = Problem()
s = 6
problem.addVariables(range(s*s), list("abcdefghijklmnopqrstuvwxyz"))

# build the wordlist from the wordlist text file
wordlist = {}
f = open("fewer_words.txt")
for line in f:
    w = line.split()[0].lower()
    wordlist[w] = 1

# we need a function: true if word is in wordlist, false if not
# this is characterwise checking, so a sequence of chars is provided, 
# we need to join them into a string before checking if they are in the list
def inlist(*word): 
    w = ''.join(word)
    if w in wordlist: return True
    return False
   
# add row and col constraints
for i in range(0,s): problem.addConstraint(inlist,range(s*i,s*i+s))
for i in range(0,s): problem.addConstraint(inlist,range(i,s*s,s))

# pretty print the output
a = problem.getSolution()
for i in range(s):
    for j in range(s): print a[i*s+j],
    print('')

It can be sped up a lot by adding way more constraints on what bigrams, trigrams etc are possible. I apologise for the scruffy code, I was more concerned with getting it to work efficiently than making it look nice

from constraint import *
from timeit import default_timer as timer
from itertools import combinations

problem = Problem()
s = 3
problem.addVariables(list(range(s*s)), list("abcdefghijklmnopqrstuvwxyz"))

wordlist = {}
f = open("fewer_words.txt")
for line in f:
    w = line.split()[0].lower()
    if len(w) != s: continue 
    wordlist[w] = 1

# build a list of monograms in each position
mono = []
for j in range(s): mono.append({})
for i in wordlist.keys():
    for j in range(s):
        mono[j][i[j]] = 1
    
# build a list of bigrams at each pair of positions
bgpos = list(combinations(list(range(s)),2))
bg = {}
for j in bgpos: bg[j] = {}
for i in wordlist.keys():
    for j in bgpos:
        bg[j][i[j[0]]+i[j[1]]] = 1
        
tripos = list(combinations(list(range(s)),3))
tri = {}
for j in tripos: tri[j] = {}
for i in wordlist.keys():
    for j in tripos:
        tri[j][i[j[0]]+i[j[1]]+i[j[2]]] = 1

qpos = list(combinations(list(range(s)),4))
q = {}
for j in qpos: q[j] = {}
for i in wordlist.keys():
    for j in qpos:
        q[j][i[j[0]]+i[j[1]]+i[j[2]]+i[j[3]]] = 1

fpos = list(combinations(list(range(s)),5))
f = {}
for j in fpos: f[j] = {}
for i in wordlist.keys():
    for j in fpos:
        f[j][i[j[0]]+i[j[1]]+i[j[2]]+i[j[3]]+i[j[4]]] = 1

def opposite():
    return lambda x,y: x==y
    
# we need a function: true if word is in wordlist, false if not
def inlist(*word): 
    w = ''.join(word)
    if w in wordlist: return True
    return False
   
def inmono(i):
    return lambda x: x in mono[i]
    
def inbg(i,j):
    return lambda *x: ''.join(x) in bg[(i,j)]   
   
def intri(i,j,k):
    return lambda *x: ''.join(x) in tri[(i,j,k)]   

def inq(i,j,k,l):
    return lambda *x: ''.join(x) in q[(i,j,k,l)]   

def inf(i,j,k,l,m):
    return lambda *x: ''.join(x) in f[(i,j,k,l,m)]   
    
# add row and col constraints
for i in range(0,s): problem.addConstraint(inlist,list(range(s*i,s*i+s)))
for i in range(0,s): problem.addConstraint(inlist,list(range(i,s*s,s)))

# add row monogram constraints
for j in range(s):
    for i in range(s):
        problem.addConstraint(inmono(i),[j*s +i])
        problem.addConstraint(inmono(i),[i*s +j])

#print(bg.keys())
for j in range(s):
    for a,b in bgpos:
        problem.addConstraint(inbg(a,b),[j*s +a,j*s +b])
        problem.addConstraint(inbg(a,b),[a*s +j,b*s +j])

for j in range(s):
    for a,b,c in tripos:
        problem.addConstraint(intri(a,b,c),[j*s +a,j*s +b,j*s +c])
        problem.addConstraint(intri(a,b,c),[a*s +j,b*s +j,c*s +j])

for j in range(s):
    for a,b,c,d in qpos:
        problem.addConstraint(inq(a,b,c,d),[j*s +a,j*s +b,j*s +c,j*s +d])
        problem.addConstraint(inq(a,b,c,d),[a*s +j,b*s +j,c*s +j,d*s +j])

for j in range(s):
    for a,b,c,d,e in fpos:
        problem.addConstraint(inf(a,b,c,d,e),[j*s +a,j*s +b,j*s +c,j*s +d,j*s +e])
        problem.addConstraint(inf(a,b,c,d,e),[a*s +j,b*s +j,c*s +j,d*s +j,e*s +j])

# pretty print the output
start = timer()
for count,a in enumerate(problem.getSolutionIter()):
  if count % 1000 == 0: print 'count:',count   
  for i in range(s):
    for j in range(s): print a[i*s+j],
    print('')
  print('')  
end = timer()

print("solution time = " + str(end-start))
print count

for i in range(s):
    for j in range(s): print a[i*s+j],
    print ''
print ''  

Constraint Programming with python-constraint

I have recently been looking at constraint programming and I thought I would post some of my code as I go. This is using the python-constraint package.

Doing things with python-constraint is pretty easy. To constrain a variable to have a certain value, we can use InSetConstraint. To make sure all the variables in a set are different, we use the AllDifferentConstraint. These are enough to solve sudoku.

Sudoku

I don't think Sudoku needs any introduction. The constraints are fairly easy to specify: each row must be all different, each col must be all different and each 3x3 block must be all different. Also we have to ensure that the given starting state is used, for this we use equality constraints. We use 1 variable for each number in the sudoku square, so 81 variables all up.

from constraint import *
problem = Problem()
problem.addVariables(range(9*9), range(1,10))

# specify the given starting sudoku problem here, dots are blank squares
sudoku = '\
.42986.1.\
....2...5\
......82.\
4.1..793.\
.683.254.\
.758..1.2\
.39......\
7...4....\
.1.67325.'

# Make sure our solution matches the specified starting thing
for i,char in enumerate(sudoku):
    if char != '.':  problem.addConstraint(InSetConstraint([int(char)]),[i])
    
# add row and col constraints
for i in range(9): 
    problem.addConstraint(AllDifferentConstraint(),range(9*i,9*i+9))
for i in range(9): 
    problem.addConstraint(AllDifferentConstraint(),range(i,81,9))
# add block constraints
for i in [0,3,6,27,30,33,54,57,60]: 
    ind = []
    for j in range(3):
        for k in range(3):
            ind.append(i + j*9 + k)
    problem.addConstraint(AllDifferentConstraint(),ind)

# pretty print the output
a = problem.getSolution()
for i in range(9):
    for j in range(9): print a[i*9+j],
    print ''

The Golomb Ruler

A bit of info on Golomb Rulers: here. If you know how to improve the code below, leave a comment. Golomb rulers are described by their order, which is the number of marks on their edge. The length of a Golomb ruler is the distance between the outer two marks and represents the longest distance it can measure.

There is no requirement that a Golomb ruler measures all distances up to their length. However, if a ruler does measure all distances, it is classified as a perfect Golomb ruler. There are no perfect Golomb rulers above order 4.

Finally, a Golomb ruler is described as optimal if no shorter ruler of the same order exists. The code below first looks for perfect rulers, then increments length each time until an optimal ruler is found. It starts to slow down by order = 7, it takes a few minutes.

This problem is harder than the sudoku problem, we have to use a set of auxiliary variables to represent the pairwise differences, then apply the AllDifferentConstraint to the auxiliary variables.

from constraint import *
from itertools import combinations

order = 7 # the number of marks on the Golomb ruler
diffs = [] # the set of pairwise differences between marks
for i in combinations(range(order),2): diffs.append(i)
length = len(diffs)

def lessthan(a,b): return a < b
def diffeq(a,b,res): return (b-a)==res     

while True:
    print 'trying length: '+str(length)
    problem = Problem()
    problem.addVariables(range(order), range(0,length+1))
    problem.addVariables(diffs,    range(1,length+1))

    # make sure the first mark is 0 and last is length
    problem.addConstraint(InSetConstraint([0]),[0])
    problem.addConstraint(InSetConstraint([length]),[order-1])

    # ensure that the marks are in increasing order
    # make sure all possible pairwise relations are constrained
    for i in diffs: problem.addConstraint(lessthan,[i[0],i[1]])

    # ensure that the differences reflect the marks
    for i in diffs: problem.addConstraint(diffeq,[i[0],i[1],i])
    
    # all the differences must be different    
    problem.addConstraint(AllDifferentConstraint(),diffs)
   
    # pretty print the output
    solutions = problem.getSolutions()
    for ruler in solutions:
        for mark in range(order): print ruler[mark],
        print ''
    if len(solutions) > 0: break # if we find some solutions, quit
    length = length + 1

example output for order = 7, note the shortest ruler is length 25, though a perfect ruler would be length 21:

trying length: 21
trying length: 22
trying length: 23
trying length: 24
trying length: 25
0 4 9 15 22 23 25
0 3 4 12 18 23 25
0 2 3 10 16 21 25
0 2 5 14 18 24 25
0 2 6 9 14 24 25
0 2 7 13 21 22 25
0 2 7 15 21 24 25
0 1 4 10 18 23 25
0 1 7 11 20 23 25
0 1 11 16 19 23 25

Magic Squares

Magic Squares seem to be very common examples when doing constraint programming, so I may as well include the code for them. Info on Magic Squares can be found here. This code is quite slow, I thought it would be much faster. This may be because python-constraint is just kind of slow, or maybe the problem is harder than I thought.

from constraint import *

problem = Problem()
size = 4
problem.addVariables(range(size*size), range(1,size*size+1))
total = size*(size*size + 1)/2

# make sure rows add to total
for i in range(size): 
    problem.addConstraint(ExactSumConstraint(total),range(i*size,i*size+size))
for i in range(size): 
    problem.addConstraint(ExactSumConstraint(total),range(i,size*size,size))
# add the diagonal constraints
diag1,diag2 = [],[]
for i in range(size):
    diag1.append(i*size + i)
    diag2.append(i*size + (size-i-1))

problem.addConstraint(ExactSumConstraint(total),diag1)
problem.addConstraint(ExactSumConstraint(total),diag2)
problem.addConstraint(AllDifferentConstraint())
   
# pretty print the output
solution = problem.getSolution()
for i in range(size):
    for j in range(size):
        print solution[i*size + j],
    print ''

Terraforming Venus

I thought I would look at some of the figures involved in terraforming Venus, discussions of which can be found all over the place e.g. here. The numbers from this page come from here. Venus is the planet next to earth as you go towards the Sun. It has a year of 224.7 Earth days, and a day length of 243 Earth days (longer than the year!). It also rotates in the opposite direction to most other planets, likely due to some large collision early in its life.

I'll quote a little bit from wikipedia: Venus has an extremely dense atmosphere composed of 96.5% carbon dioxide, 3.5% nitrogen, and traces of other gases, most notably sulfur dioxide. The mass of its atmosphere is 93 times that of Earth's, whereas the pressure at its surface is about 92 times that at Earth's—a pressure equivalent to that at a depth of nearly 1 kilometre under Earth's oceans. The density at the surface is 65 kg/m3, 6.5% that of water or 50 times as dense as Earth's atmosphere at 20 °C at sea level. The CO2-rich atmosphere generates the strongest greenhouse effect in the Solar System, creating surface temperatures of at least 735 K (462 °C).

Terraforming Venus would involve removing most of the sulfuric acid and carbon dioxide, needless to say this is a monumental task that we don't currently have the technology for. Some of the truly staggering figures:

Total Mass of Venus Atmosphere	= 4.80E+20 kg
Percent Atmosphere CO2	= 96.50%
Total Mass of CO2	= 4.63E+20 kg

For comparison:

Total Mass of Earth's Atmosphere	= 5.10E+18 kg
Mass Ratio Venus to Earth	= 94.118

Utilizing the Bosch reaction (CO2 + 2H2 -> C + 2H2O) , combining hydrogen with carbon dioxide to make carbon graphite and water could be used to remove Venus' carbon dioxide. This reaction requires the introduction of iron, cobalt or nickel as a catalyst and requires a temperature level of 530-730 degrees Celsius. Since the reaction releases some heat, it could be self sustaining if you could supply enough hydrogen. A problem with this is that that the production of elemental carbon tends to foul the catalyst's surface, which is detrimental to the reaction's efficiency [ref].

Molecular Weight of CO2	= 44
Molecular Weight of 2*H2	= 4
Total Initial Molecular Weight	= 48
Molecular Weight of C (graphite)	= 12
Molecular Weight of 2*H2O	= 36
Total Final Molecular Weight	= 48

You could use water from e.g. comets to get hydrogen for the reaction. You would have to use Electrolysis to split the water and initiate the reaction. Alternatively you could reuse the water output from the Bosch reaction , recover the Hydrogen and continue. But this way you wouldn't have oceans on Venus at the end.

If you were to provide hydrogen for the reaction (4E19 kg), this massive reaction would create an ocean nearly as large as 1/4 of the Earth’s ocean:

Ratio of 2*H2O to CO2 (36 / 44)	= 0.818
Resultant Mass of H2O	= 3.79E+20 kg
Density of H2O	= 1.000 g / cm^3
	= 1,000,000.000 g / M^3
	= 1,000.000 kg / M^3
Volume of Resultant H2O	= 3.79E+17 M^3
	= 3.79E+08 kM^3
Area of Venus Surface	= 4.602E+08 kM^2
Average Depth of H2O	= 0.824 kM
Total Volume of Earth's Oceans	= 1.300E+09 kM^3
Average Depth of Earth's Oceans	= 3.682 kM

It would also result in the deposition of a layer of graphite with an average thickness over the entire surface of Venus roughly equal to a 40 story building:

Ratio of graphite to CO2 (12 / 44)	= 0.273
Resultant Mass of graphite	= 1.26E+20 kg
Density of C (graphite)	= 2.230 g / cm^3
	= 2,230,000.000 g / M^3
	= 2,230.000 kg / M^3
Volume of Resultant C (graphite)	= 5.66E+16 M^3
	= 5.66E+07 kM^3
Area of Venus Surface	= 4.602E+08 kM^2
Average Depth of C (graphite)	= 0.123 kM

So after converting all the carbon dioxide to graphite, you would be left with ~120 meter thick blanket of carbon over the planet. Though this may not be terrible, carbon makes a pretty good base for soils. There would still be a great deal of atmospheric pressure due to all the oxygen though, and a spark would result in a mighty conflagration that would undo all your work. A better idea may be to trap the carbon dioxide as carbonate rocks, but for this you would need large quantities of calcium and/or magnesium (About 8×10^20 kg of calcium or 5×10^20 kg of magnesium would be required). This would have the advantage of removing a lot of the oxygen, which would lower the atmospheric pressure. Unfortunately Magnesium carbonate begins to decompose at 350 degrees C, so Calcium might be a better bet. You would also need to remove all the sulphuric acid since it would dissolve all your carbonate rocks, using up the sulphuric acid but producing more carbon dioxide.

Of course you have to power all this as well, electrolysis doesn't come cheap. Solar panels can work well on Venus because the sunlight is 4 times stronger. Perhaps some genetically engineered plant...

CODZ ciphers, testing RC4 in python

Some of the CODZ ciphers consist of binary data, and it is not clear what sort of cipher has created them. Notably 2,3,7 and 10 are base64 converted binary data. Ciphers 4,5,11, and 14 are hex data. I will be dealing with 2,3,7 and 10 in this post.

Cipher 2 decrypts to: dd3ed3b56ff21a7751ebbbaf6ffd1086f190a6b29b49c2a47e656bcc91768346df0321db23505874cb9f8a71d978490e and 3a411e6479e27ee31d601d486c7c807dbd20d85273ad599f90a7126caae9406be1434fdbbeeefa45d9c6cbf5ad4fb4f7 depending on wether you read the base64 data forwards or backwards. Either way you get exactly 384 bits of binary, which is suspiciously similar to the SHA384 hash length. I tried running john the ripper on the two hashes for a day or so, but nothing turned up. Either it is not SHA384 or we're gonna need some better word lists.

Ciphers 3,7 and 10 are longer. I have tested their entropy after breaking them up into different length segments, and the entropy is pretty much always equal to random, this means they are highly unlikely to have been generated by a classical cipher. More likely something modern like RC4. I have tried decrypting the 3 ciphers as RC4 using a wordlist for the keys, but no luck.

Code for testing entropy:

from base64 import decodestring
from bitarray import bitarray
from binascii import b2a_hex
from math import log

ctext = 'iW9cXmzOU7ZuZBtW40b3ng...' # I cut the rest of the cipher off to save space
data = decodestring(ctext)

a = bitarray(endian='little')
a.frombytes(data)

for bitlen in range(1,16):
  code = {}
  for i in range(2**bitlen):
    string = format(i,'0'+str(bitlen)+'b')
    code[i] = bitarray(string)
  num = a.decode(code)
  freq = {}
  for i in num:
      if i in freq: freq[i] += 1.
      else: freq[i] = 1.
  N = len(num)
  en = 0
  for i in freq.values():
    p = i/N
    en -= p*log(p)
  print bitlen,' entropy: ',en,' if random: ',-log(1./2**bitlen)

For decrypting RC4, we keep the key corresponding to the decrypt with the lowest entropy, if the wrong key is used something high entropy will result, but if it decrypts to text we should get something easily identifiable. Python code for decrypting RC4 with a word list:

data = base64.b64decode(ctext)

def rc4(data,key):
    S = range(256)
    j = 0
    out = []

    #KSA Phase
    for i in range(256):
        j = (j + S[i] + ord( key[i % len(key)] )) & 0xFF
        S[i] , S[j] = S[j] , S[i]

    #PRGA Phase
    i = j = 0
    for char in data:
        i = ( i + 1 ) & 0xFF
        j = ( j + S[i] ) & 0xFF
        S[i] , S[j] = S[j] , S[i]
        out.append(chr(ord(char) ^ S[(S[i] + S[j]) & 0xFF]))
    return out
    
# find a good word list and use it here
keylist = open("C:\Users\james\Desktop\cipher_stuff\simplesub_word\\count_1w.txt")

besten = 10e10
bestkey = ""
bestout = ""
for count,key in enumerate(keylist):
  key = key.split()[0].strip()
  for i in range(3):
    if i == 0: key = key[0].upper() + key[1:].lower()
    elif i == 1: key = key.upper()
    else: key = key.lower()
    out = rc4(data,key)
    # compute entropy of the decoded text
    freq = {}
    for i in out:
        if i in freq: freq[i] += 1.
        else: freq[i] = 1.
    N = len(out)
    en = 0
    for i in freq.values():
        p = i/N
        en -= p*log(p)
    if en < besten:
        besten = en
        bestkey = key
        bestout = ''.join(out[:20])
  if count % 500 == 0: print count, key, bestkey,besten,bestout
print count, key, bestkey,besten,bestout

CODZ 8. sha_paper

I havn't actually played the game, I just like breaking codes.

Apparently there are some new CODZ ciphers, the transcripts can be seen here: http://www.callofdutyzombies.com/topic/183529-all-revelations-ciphers-texture-files/#comment-1759652.

This post will look at the following cipher:

bx re yh zy bf lm kt ut yg se tb sx ky co jh km aq we tx wx
cy ji ut vt kn vc gx aw ij av qn lg ef fj uq bd kn sv cx fn
je wr rk kn cg aw xq vn zf li fh vz wt ta ia ij zf eh uf tj 
qm yg hl yq cx ij vw ig de qz tg nj rs er vk tm sa yv tw hr 
hs lt vy kr qc tv gh hb jn yb qh er ut gk et cs wv jl rh xo 
wr ex hr xt zi kc xs qs fd wd cm ku ah fh fj lf ui ly sh vf 
au xm hx qw dl gi cx vb dh wt xm kv un ej kt kt ye cq jd ef 
eh zv xt he uz tg cl jw nr tw ur vo jt jo ru iq iy rz ey ho 
gd nq yn bq ul ai fh bu ji ho nw qg yg vj if yv zu id jc gh 
ke xr qf cq ra it gw dl fc gq ti iu qu ny vr gy sj rh iu hi 
wr mv ym zi lk re vk xu ry uq gs ve gd yn bq ch ky er qh jr 
ho ya el ky zj ei hz cb if dk

There are 460 characters total, 25 distinct characters, IC is 0.0418. Counts:

a: 12, c: 18, b: 11, e: 23, d: 12, g: 20,
f: 19, i: 24, h: 29, k: 21, j: 23, m: 9,
l: 14, o:  7, n: 15, q: 23, s: 12, r: 24,
u: 20, t: 26, w: 18, v: 23, y: 25, x: 19,
z: 13

Since this cipher uses pairs of characters, and there are 25 distinct characters, it is highly likely to be foursquare, bifid, playfair or some other digraphic substitution cipher based on a 5 by 5 grid. Usually I and J are combined, but it looks like in this cipher P is missing, so It was combined with something else.

Other ciphers that use a 25 letter alphabet include BAZERIES, BIFID, CADENUS, CHECKERBOARD, CM BIFID, FOURSQUARE, PHILLIPS, PHILLIPS-C, PHILLIPS-RC, PLAYFAIR, SERIATED PLAYFAIR, TRI-SQUARE, TWIN BIFID and TWO-SQUARE. Unfortunately this is quite a lot, and trying them all will be time consuming.

I am trying 4 variants of the above the cipher, the original cipher: bkreyh..., the reversed cipher: kdfibc..., forward but with each pair reversed: xberhy..., and each pair forwards, but the pairs from last to first: dkifcb... .

BAZERIES is a substitution + transposition, which means its IC would be that of normal text. Since the IC of this text is ~0.04, I am confident in saying it is not BAZERIES . It also can't be CADENUS because the message length is not a multiple of 25. CHECKERBOARD uses a 5x5 key square, but can at maximum use 20 characters in the ciphertext.

Possibilities after thinking a bit: BIFID, CM BIFID, TWIN BIFID, FOURSQUARE, PHILLIPS, PLAYFAIR, SERIATED PLAYFAIR, TRI-SQUARE, and TWO-SQUARE. The next piece of information is that none of the letter pairs in the ciphertext contain a doubled letter e.g. AA, BB, CC etc. This is highly unlikely to occur with most ciphers, but it is a peculiarity of PLAYFAIR. So this puts playfair near the top of the list.

I have tried the playfair cryptanalysis program over here on all four variants and it hasn't worked. I did a test sentence of the same length which it cracked in just a few iterations, so I am fairly confident in saying it is not a vanilla playfair, or at least if it is then something else is going on as well (such as some sort of route cipher).

For bifid, there is no obvious period identifiable (see here for how to compute it), and some quick simulated annealing didn't turn anything up.

Regarding the possibility of routes, 460 has a lot of factors: 2x230,4x115,5x92,10x45,20x23. This means lots of routes around the rectangles when the characters are lined up right. Exploring this would add a lot of extra time to the cracking effort.

The quest continues...

CODZ cipher 13. kin_paper_torn straddle checkerboard?

I haven't actually played the game, I just like breaking codes.

Apparently there are some new CODZ ciphers, the transcripts can be seen here: http://www.callofdutyzombies.com/topic/183529-all-revelations-ciphers-texture-files/#comment-1759652.

This post will look at the following cipher:

@CB?>?>C@F?A?>CCCG??@C>?@C@C>???G=?C?=C@CGBC@CB?G@A
?=CB?C?G<?>CB@C=CB?<FG<?>C>C>CA?B?AGA?F?C?@CA?G=?B?>C@B?

TLDR

EDIT: I'm beginning to think that this is not the final decryption, that they have used a slightly modified method which results in some extra characters. There is too much decrypted for me to think that it is totally wrong, but I believe there is an extra step I'm missing somewhere in the middle. </EDIT>

The ciphertext is a straddle checkerboard. Converting the 10 symbols to digits, and reversing the string, we get (using @=0, C=1, B=2, ?=3, >=4, F=5, A=6, G=7, = is 8, <=9, Note that this mapping is arbitrary, and any other mapping would work just as well):

320143238736103135367632361414143975932181021439731321 83607321012710183138733341010341033711143635014343210

See the link for a description of the straddle checkerboard. Using this key:

  0123456789
  C.D.JYPEKW     
1 NXV.T.QFUL 
3 ZSOGIARVBH

The message decrypts to OCTOBERNSAREFORTTTHEYFOUNDTHESOURCEONVENUSBEGINNINGEXTRACTION. with spaces: OCTOBER NSA REPORT T THEY FOUND THE SOURCE ON VENUS BEGINNING EXTRACTION. There are likely a couple characters wrong still, or the slight mistakes are intentional.

My approach

There are 10 different characters, 107 in total. The counts for each character are as follows: A: 6, @: 11, C: 24, B: 9, G: 8, F: 3, =: 5, <: 3, ?: 28, >: 10. It so happens that 10 characters can be converted to digits, so we can look at existing ciphers that use digits e.g. the straddle checkerboard, GRANDPRE, MONOME-DINOME, MORBIT, NIHILIST SUBSTITUTION, POLLUX or TRIDIGITAL. There are probably also more.

Converted to digits it is: 0123434105363411173301430101433378313810172101237063812313 7934120181239579341414163236763531301637832341023. If you reverse it, you get 320143238736103135367632361414143975932181021439731321 83607321012710183138733341010341033711143635014343210. I'll be talking about the non reversed ciphertext, but I'll be applying all the steps to both just in case. Using the above ciphertext and this calculator, we can compute the most likely cipher algorithms, and it has concluded that it is most likely a nihilist substitution. Unfortunately, it can't be nihilist because it has an odd number of characters, and nihilist requires even. It can't be grandpre for the same reason. CryptoCrack (from here) can't break it as GRANDPRE, MORBIT, NIHILIST SUBSTITUTION, POLLUX or TRIDIGITAL.

That really only leaves straddle checkerboard as the last candidate from the ciphers I can think of. Assuming it is a straddle checkerboard, we first have to find to location of the blanks in the key. If we can find those locations, we can decrypt in with any letters and we will have a plain substitution cipher. There are 10 choose 2 = 45 ways of putting the first 2 blanks, and 20 choose 2 = 190 ways of putting the second. This means there are 45*190 = 8550 possible configurations of putting the blanks. Some of these will be invalid though, e.g. the key:

   0 1 2 3 4 5 6 7 8 9
   f k m   c p d   y e
3: h b i g q r o s a z
7: l u t j n w v x    

can't have 38, 39, 78 or 79 in the ciphertext. If these numbers do exist in the ciphertext, then this is not a possible set of blanks. After trying to decrypt our ciphertext with all 8550 keys, only 3340 are actually valid blank positions. In addition to this, most of the blank positions result in identical outputs to other blank positions, so we can discard all the duplicate outputs, leaving only 45 substitution ciphers that we have to try.

After trying to decrypt the resulting 45 substitution ciphers from the forward cipher and another 45 from the reversed cipher, one of the decrypts, namely SALSYFWIRVWFESWLLLZFDHSOIBLZFRSOWAFSIKFIORYFTUIIUITFJLWVALUSI, can be decrypted to OCTOBERNSAREFORTTTHEYWOUNDTHESOURCEONMENUSBEGINNINGEXTRACTION. There are likely a couple characters wrong still, or the slight mistakes are intentional.

This is a bit of python code that will spit out the possible substitution ciphers for a given straddle checkerboard ciphertext (Please excuse the rough code, I wrote everything pretty quick):

I have actually since discovered there is a much shorter way of getting to the final candidates, but I'll leave this code here. To rank the final candidates, use index of coincidence, the correct one is almost always the one with the IC closest to 0.07.

import random
import sys
from itertools import combinations

ctext = '6909746723099383772753870703607230943837727093872638757438333832743772974928387272384175943874720383270'

''' decrypt a straddle checkerboard cipher ctext given a key
 key should look like e.g. 'fkm.cpd.yehbigqrosazlutjnwvx..' 
 it should be 30 chars in length, and have 4 dots. 
 2 dots in the first 10 chars, and 2 in the last 20.
 - if it returns 0 in the first result, the key was invalid
def scdecrypt(ctext,key):
    dotpos = []
    for i,k in enumerate(key): 
        if k == '.': dotpos.append(i)
    output = ""        
    if dotpos[0] > 9: return 0,output
    if dotpos[1] > 9: return 0,output
    if dotpos[2] < 10: return 0,output
    if dotpos[3] < 10: return 0,output
    flag = 0
    for cc in ctext:
        c = int(cc)
        if key[c] != '.' and flag == 0: 
            output += key[c]
        elif flag == 1: 
            if key[10+c] == '.': return 0,""
            output += key[10+c]
            flag = 0
        elif flag == 2: 
            if key[20+c] == '.': return 0,""        
            output += key[20+c]
            flag = 0
        elif c == dotpos[0]: flag = 1     
        elif c == dotpos[1]: flag = 2
        else: 
            return 0,output
    return 1,output

''' see if word and ct are possibly the same substitution cipher 
returns 0 in the first part of the result if they are not '''        
def canstart(ct,word):
    key = {}
    if len(word) > len(ct): return 0,key
    for i,char in enumerate(word):
        if ct[i] not in key: 
            if char not in key.values():           
                key[ct[i]] = char
            else: return 0,key
        elif key[ct[i]] == char: pass
        else: return 0,key
    return 1,key

# get a list of all the possible blank spot permutations
dp = []
a = combinations(range(10),2)
for i in a:
    b = combinations(range(20),2)
    for j in b:
        dotpos = list(i)
        dotpos.append(10+j[0])
        dotpos.append(10+j[1])
        dp.append(dotpos)

# we have all possible blank spot positions, try decrypting with each and see if we can discard any
count = 0
texts = []
for d in dp:
    key = list('ABCDEFGHIJKLMNOPQRSTUVWXYZ')
    for pos in d: key.insert(pos,'.')
    
    bool,ptext = scdecrypt(ctext,key)
    if bool == 0: continue
    texts.append(ptext)
    count += 1

# now we have a heap of valid decrypts, discard duplicates
lst = []
for i in range(len(texts)):
    cs = False
    for j in range(len(lst)):
        if canstart(lst[j],texts[i])[0]: cs = True
    if not cs: lst.append(texts[i])

# print the unique possible decrypts, they must be solved as substitution ciphers
for i,j in enumerate(range(len(lst))):
    print lst[j]

The CODZ Mob of the dead ADFGX cipher

The ADFGX cipher from CODZ has been unsolved for quite a while, even with many people trying to break it. I thought I would document some of my progress on this page. Other Discussions:

• https://www.reddit.com/r/codes/comments/5542it/looking_to_hire_or_work_with_a_cryptographer_to/
• http://www.callofdutyzombies.com/topic/154218-adfgx-cipher/
• https://www.reddit.com/r/CODZombies/comments/4vxy7f/adfgx_key_word_mega_thread/
• https://www.reddit.com/r/CODZombies/comments/3wfhht/adfgx_cipher_still_unsolved_and_treyarch_employee/
• https://www.reddit.com/r/CODZombies/comments/3d94uz/mob_of_the_dead_cipher/
• http://www.callofdutyzombies.com/topic/182194-mob-of-the-dead-loading-screen-code/
• http://s13.zetaboards.com/Crypto/topic/7047345/1/

There are of course other discussions around, but I'm not sure how much useful information is in them. Apparently the developers of the game have said that this was one of the early ciphers they made, and it may be too difficult to break. But we will have a go anyway.

EDIT: later thoughts

This paragraph has been added later, but I have decided that most of the analysis on this page is useless because I can't crack length 34 substitution ciphers using all the knowledge and code I have (which is mostly over here). I have made a couple of my own length 34 substitution ciphers, and I can't get anywhere even close to cracking them (If it were e.g. 100 characters it would be totally possible). So I might think about how to solve that problem. But until then, there is little point in the analysis below, because even without the ADFGX keyword steps it is impossible to solve (for me at the moment).

Some Observations

An ADFGX cipher with 68 characters means the plaintext has only 34 characters. Unfortunately this is close to the unicity distance for substitution ciphers, so it may be unbreakable from that perspective unless some extra information can be found. Most substitution cipher solvers can't break text this short as there are incorrect solutions that will score higher (using ngram scorers) than the true decryption. The ciphertext is as follows:

FFGXGD
GFFAGF
GGDDGF
FFXXFF
FDGFFG
FDGFFG
FDGGFF
FGFGAA
FXFXDX
XFXDGF
FAGGFF
    AF

There are two components to an ADFGX key: a keyword e.g. 'GERMAN' and a substitution key for the Polybius square. Much of the discussion on various forums is about finding the ADFGX keyword, however I think this is unneccessary. With a length 6 keyword there are only 6! = 720 possible keys, and only 48 possibilities if we assume the last two columns are the first 2 characters of the keyword (this is a reasonable assumption because they are dangling at the end like that).

This means we can decrypt the cipher using each of the possible 48 keyword permutations (this gives us 48 Polybius square ciphers), then use an arbitrary Polybius square key e.g. "ABCDEFGHIKLMNOPQRSTUVWXYZ" to get a 48 ciphers which are just a substitution cipher. From here, ideally, it would just take a good substitution cipher solver applied to each of the 48 candidates to determine which can be broken resulting in English text. It turns out that, due to symmetry of the polybius square, there are actually only 24 possible substitution ciphers, which makes our job easier. Unfortunately 34 characters is just too short for most substitution cipher crackers, or alternatively one of the previous assumptions is wrong.

Of course I am not the first to think of this, and many people have tried this exact procedure to no avail.

The Repeated Rows

There are two identical rows in the ciphertext: FDGFFG FDGFFG. This means, regardless of the keyword, there will be 6 letters in the plaintext that consists of 3 letters repeated e.g. THETHE. How often does this sort of thing occur in English? Quite often actually. I have a corpus of around 50 million English sentences from news websites and wikipedia which can be used to determine the frequency of this sort of occurrence.

It so happens that in every million english sentences, you would expect around 34,000 to have a repeated triple like this, consisting of around 1100 distinct sequences (i.e. many of the 34,000 are repeats of previously seen sequences). In total after 50 million sentences 6164 distinct 6-character sequences were found, occurring 1.89 million times in total. The most common are: THATHA (235167), ANDAND (176355), INGING (122567), THETHE (89908), REAREA (67514), NTONTO (42511), ETHETH (36691), SSESSE (31542), NDINDI (31259), ASSASS (29754), followed by many other rarer ones.

To utilise this information, it may be possible to fix the letters in the substitution key so that the repeated 3-letter combo is e.g. fixed to THATHA. This might make it easier for an n-gram scorer to identify the rest of the letters if some are known. I haven't tried this yet.

A Dictionary attack on the Polybius bit

Apparently past ciphers in the game have used keywords to generate keys for ciphers. This means we may be able to use a dictionary attack to generate polybius keys for decrypting the adfgx. This means we don't have to crack the substitution component, which would make things easier.

After doing this, the best decryption I could get from the 48 keyword permutations was VOGANHEMOWIGHHEHHENHEKANZEWESINHAC with a 4-gram score of -166. So this means either the polybius key was not a dictionary word (my dictionary has about 500k words in it), the adfgx keword is not length 6, or something other assumption we have made is wrong.

Installing RNNLIB on Ubuntu 14.04

I have previously looked at CURRENNT, but now I'll go through how to use RNNLIB, which is a different rnn library. RNNLIB doesn't use cuda, but it can do multi-dimensional rnns and also can use the CTC loss function which can be handy. Let's get right into it. Download RNNLIB from sourceforge here.

You'll have to install the netcdf library:

sudo apt-get install libnetcdf-dev

You should also install boost: sudo apt-get install libboost-all-dev. I have libboost1.55 installed already.

There is an installation guide over here if you want some additional pointers. First run configure and make:

configure
make

Fixing the errors

The first error I got was:

Mdrnn.hpp:239:6:   required from here
Helpers.hpp:311:41: error: ‘range_min_size’ was not declared in this scope, and no declarations were found by argument-dependent lookup at the point of instantiation [-fpermissive]
  size_t size = range_min_size(r1, r2, r3);

To fix this one you need to put templates code above the code that uses them. Moving the 4 templates for range_min_size to the top of the file should fix all the issues in Helpers.hpp. I cut lines 532-547 and pasted the lines to line 163 of Helpers.hpp.

Running make again I got: Container.hpp:157:24: error: ‘resize’ was not declared in this scope, and no declarations were found by argument-dependent lookup at the point of instantiation. To fix this one change on line 157 of containers.hpp "resize" to "this->resize."

That was all I needed to do to get this version installed.

Running the CURRENNT rnn library

Previously I have made posts about installing Currennt and writing data files for Currennt. This post will be about actually running it assuming it is installed and the data files have been written.

The first thing you'll need to specify is the config file, which I called config.cfg. Its contents look like this:

max_epochs_no_best   = 20
max_epochs           = 100
learning_rate        = 1e-5
network              = network.jsn 
train                = true
train_file           = full_train_currennt.nc
val_file             = full_test_currennt.nc
weights_dist         = normal
weights_normal_sigma = 0.1
weights_normal_mean  = 0
stochastic           = true
validate_every       = 1
parallel_sequences   = 20
input_noise_sigma    = 0
shuffle_fractions    = true
shuffle_sequences    = false

Many of the options are described on the currennt wiki. The network.jsn is a file that I will describe down below and contains the network configuration e,g, number of layers, layer types etc. The files full_train_currennt.nc and full_test_currennt.nc were created using the MATLAB script in a previous post. If you set parallel sequences too high you'll run out of memory on your GPU, but higher means it will train faster.

The network configuration

Info on the network configuration can be found over here. The network I used looks like this:

{
    "layers": [
        {
            "size": 20,
            "name": "input",
            "type": "input"
        },
        {
            "size": 200,
            "name": "level_1",
            "bias": 1.0,
            "type": "blstm"
        },
        {
            "size": 200,
            "name": "level_2",
            "bias": 1.0,
            "type": "blstm"
        },
        {
            "size": 4,
            "name": "output",
            "bias": 1.0,
            "type": "softmax"
        },
        {
            "size": 4,
            "name": "postoutput",
            "type": "multiclass_classification"
        }
    ]
}

actually running stuff

Now to run things, just do:

/path/to/currennt/build/current config.cfg

and everything should work, just wait for it to finish training. This will output a file called trained_network.jsn which we will need during testing. For testing, you will need another HDF5 file just like the training one, except with the test sequences in it. For testing we have another config file which I called ff_config:

network          = trained_network.jsn
cuda             = 0
ff_output_format = csv
ff_output_file   = test_results
ff_input_file    = full_test_currennt.nc

Notice that cuda is off, I found it ran faster in feed forward mode without cuda. To do testing run:

/path/to/currennt/build/current ff_config.cfg

Now all you need to do is parse the output file test_results.csv if you need the individual predictions, it is in comma separated variable format so it is not hard to work out.

Writing HDF5 files for Currennt

In a previous post I explained how to install currennt. This post will deal with writing the data files. The next post will be about actually running things. Currennt uses data written to HDF5 files, so whatever format your data is in you have to write it to a HDF5 file then use currennt to process it. This post will use matlab to generate the HDF5 files, but e.g. python could easily do it as well.

Recurrent neural networks predict things about sequences, so for this example we'll be using protein sequences as the example. Predicting things like phonemes for speech files is exactly the same. For each amino acid in a protein, we'll be predicting the secondary structure, which can be C, E, H or X. The training information will be a L by 20 matrix, this means the protein is of length L and each amino acid has 20 features (we'll use PSSM features, of which there are 20, extracted using PSI-BLAST.). I will just be using 400 training proteins and 100 test proteins, it is often beneficial to try things out on small datasets because you get the results quicker and can figure out problems quicker.

HDF5 files

Currennt needs the data written to HDF5 files, but it needs a heap of extra parameters written to the files as well as the data, this section will specify what you need to write to them, then I'll put some actual MATLAB code in the next section.

The data we'll be using is protein data, there will be 400 training proteins, each of length L (the sequences can be anywhere from about 20 to 2000 in length). There are 20 features per amino acid, this means each protein is represented by a matrix of size 20 by L, and there are 400 matrices.

These parameters are for classification, if you want something else like regression you may have to use different dimensions and variables e.g. numLabels only makes sense for classification. HDF5 files have dimensions and variables, and they need to be specified separately. The dimensions that currennt needs specified in the HDF5 files are as follows:

numTimesteps: the total number of amino acids if you lined 
      up all the proteins end to end
inputPattSize: the number of features per amino acid, in this case 20.
numSeqs: the number of proteins
numLabels: the number of classes we want to predict, in this case 4
maxLabelLength: the length of the longest string used to hold 
      the class label name
maxTargStringLength: to be honest I'm not sure about this one, I 
      just set it to be 5000 and things seem to work
maxSeqTagLength: this is another one I'm not sure about. I set 
      it to 800 and it seems to work.

Now we have specified the dimensions, now we want to specify the variables i.e. the actual data we will be using for features and labels. They are specified like so:

seqTags,          size: maxSeqTagLength x numSeqs
numTargetClasses: size: 1
inputs,           size: inputPattSize x numTimesteps
seqLengths,       size: numSeqs
targetClasses,    size: numTimesteps
labels,           size: maxLabelLength x numLabels

You then just have to write the relevant data to each of the variables and run currennt. Some MATLAB code that does all this stuff is provided below. Note that this matlab file will just create one file for e.g. training. If you want a separate dataset for e.g. testing (strongly recommended) then you will need to run the code a second time changing the file names and protein dataset.

Installing Currennt on ubuntu 14.04

This post is all about how to install Currennt on ubuntu 14.04. The next post will deal with how to write the data files that Currennt needs as well as actually running currennt. Currennt is a recurrent neural net library that is sped up using cuda. Installing it is not as simple as it could be.

First step: grap the source code from here: currennt on sourceforge. I am using current-0.2-rc1.zip.

You'll first need to install cuda:

sudo apt-get update
sudo apt-get install cuda

You'll also have to install the netcdf library:

sudo apt-get install libnetcdf-dev

You should also install boost: sudo apt-get install libboost-all-dev. I have libboost1.55 installed already.

Now you can start the installation. Extract the Currennt zip file to a directory and cd to that directory. Now, create a folder called build and cd into that, then run make.

mkdir build
cd build
cmake ..
make

When I did this I got the error:nvcc fatal : Value 'compute_13' is not defined for option 'gpu-architecture'. This can be fixed by editing the CMakeLists.txt, change compute_13 to compute_30, and rerun cmake and make.

The next make got a little bit further, but I got the error: error: namespace "thrust" has no member "transform_reduce". This one can be fixed by adding the line: #include <thrust/transform_reduce.h>to the top of the file /currennt_lib/src/layers/Layer.hpp .

That was all I had to do this time, previously I have had a heap of different propblems when installing currennt on older computers, but this time seemed relatively painless.

Numerically Checking Neural Net Gradients

It is highly recommended by everyone that implements neural networks that you should numerically check your gradient calculations. This is because it is very easy to introduce small errors into the back-propagation equations that will make it seem like the network is working, but it doesn't do quite as well as it should. A simple numerical check can let you know that all your numbers are right.

This tutorial will go through the process of how to numerically compute the derivatives for a simple 1 hidden layer neural network. The exact same process is used for checking the derivatives of e.g. LSTMs, which are much more complicated.

The main thing to remember is the error measure you are using for your neural network. If you are using Mean Squared Error then the error \(E = \dfrac{1}{N}\sum^N_{n=1} \sum^K_{k=1} (y_k - x_k)^2 \) is what we want to minimise. In the previous equation, \(y_k\) is the desired label and \(x_k\) is the network output. \(N\) is the minibatch size, if you are just putting in a single input at a time then \(N\) is one. This single number \(E\) tells us how good our network is at predicting the output, the lower the value of \(E\) the better. Our aim is to adjust the weights so that \(E\) gets smaller.

For a neural net with 1 hidden layer we have 2 sets of weights and 2 sets of biases. The output is computed as follows:

\( a = \sigma (i*W_1 + b_1) \)
\( x = \sigma ( a*W_2 + b_2) \)

\(i\) is our input vector, \(a\) is the hidden layer activation, and \(x\) is the network output.

The way to numerically check the gradient is to pick one of the weights e.g. element (1,1) of \(W_1\), and to add and subtract a small number to/from it e.g. 0.0001. This way we have \(W_1^+\) and \(W_1^-\) (note that only element (1,1) is changed, all the other weights stay the same for now). We now have to compute \(x^+\) and \(x^-\) using the new slightly modified weights. Then we can compute \(E^+\) and \(E^-\) using \(x^+\) and \(x^-\). The gradient of E is then \((E^+ - E^-)/(2*0.0001)\).

Now that we have the derivative of \(E\) with respect to weight (1,1) of \(W_1\), we have to do it for all the other weights as well. This follows the exact same procedure, we just add/subtract a small number from a different weight. The final matrix of derivatives should exactly match the gradient as calculated by back-propagation. This python code: nn_2layer.py implements both back-propagation and the numerical check for a simple single hidden layer nn.

Caffe Tutorial - Part 1 - basic feedforward nets

Caffe is a deep learning framework often used for image recognition projects, but I've found that the tutorials aren't that great, or at least I have trouble finding solutions to the problems I have. So I've decided to write a sequence of tutorials to lay out information that I can revise in the future and that others might use as well.

The first tutorial (this one) will be a basic single hidden layer feed-forward neural net trained on the fisher iris dataset. I'll use MATLAB to write the feature files that caffe uses, for this tutorial we'll use layer type HDF5Data because HDF5 files are easy to write.

Also try this lenet tutorial as it covers some things that this page does not.

Writing the Data Files

This is a short matlab script for writing the train and test files, each file must contain both the features and labels. Other languages can be used e.g. python has a library for writing hdf5 files as well. Note that fisheriris is built in to matlab. When we load it the features are 150x4, when it gets written to the hdf5 files it has to be transposed (4x150). The labels come as a cell array of strings, we need to convert them to integers starting at 0. We randomly choose 100 samples to be in the train set and 50 to be in the test set.

We use innerProduct layers as these are fully connected feedforward layers, and they are initalised using the 'xavier' algorithm which is explained in this other blog post.

We now have 2 files: iris_train.hdf5 containing the train features and labels, and iris_test.hdf5 containing the test features and labels.

Specifying the Network

At this point I recommend reading a bit about layers, nets and blobs, you can also look at the types of layers available. Our network will be a single hidden layer with 20 nodes. The iris dataset has 3 classes, so our output layer will have 3 nodes. We use the rectified linear unit (ReLU) activation, and softmax outputs.

We are doing training and testing in this network, this is why there are 2 data layers, one with phase: TRAIN and one with phase: TEST. One annoying bit is that we can't directly specify the location of the HDF5 file, we have to specify the location of a txt file that contains as its first line the location of the HDF% file. So "./files/train_file_location.txt" that appears in the network specification actually has as its only contents:

./files/iris_train.hdf5

which is the location the matlab script saved the hdf5 files to.

At the end of this script, we have SoftmaxWithLoss as well as softmax layers. This is because we want to see the accuracy, and the loss can't be used with the accuracy. We pass the softmax outputs to the Accuracy layer so we can see how well it performs. Note that accuracy is only computed during phase: TEST.

Specifying the Solver

The solver is pretty standard.

Running Everything

Now you probably want to run it, just type the following into a terminal while being in the correct directory:

/path/to/install/caffe-master/build/tools/caffe train --solver=./iris_caffe_solver.prototxt

A look at the second Feynman cipher - fractionated morse [part 4]

The Feynman ciphers are a set of 3 ciphers given to Richard Feynman, the first of which has been solved, but the second 2 remain unsolved. This is part 4 of a discussion about the second cipher.

I have talked about this cipher three times previously in Part 1, Part 2 and Part 3. In part 1 I was playing around with vigenere type ciphers, in part 2 I decided to step back a little bit a figure out what the ciphertext is not i.e. identify the cipher algorithms it can't be. Part 3 looked at the 3x3 Hill cipher and some of its variants.

So far I have been working my way through the cipher types I think it may be. In part 2 I identified fractionated morse as a candidate cipher. It turns out breaking fractionated morse is not too difficult, the procedure is similar to breaking standard substitution ciphers. I will start with a short description of Fractionated morse, then we will look at how to break it.

Description of Fractionated Morse

There are many descriptions of the algorithm around, the ACA has a good description, I will reproduce a part of G. worley's description. It assumes you know morse code.

If Alice wanted to send Bob the
message "I love you." she would tap over the telegraph:

Pt:  I love you.
Et:  ..xx.-..x---x...-x.xx-.--x---x..-xx.-.-.-

After encoding the plaintext in Morse Code, the encoded text is broken into
n-graphs, usually trigraphs because there are 26 Morse Code trigraphs (you never
have three dividers in a row, so you don't count that one).  Here's a sample
trigraph mapping (using a keyword):

Keyword:  LOOKINGGLASS

L O K I N G A S B C D E F H J M P Q R T U V W X Y Z
. . . . . . . . . - - - - - - - - - x x x x x x x x
. . . - - - x x x . . . - - - x x x . . . - - - x x
. - x . - x . - x . - x . - x . - x . - x . - x . -

So, breaking our message into trigraphs we get:

     . x . - x . . - - - . x - .
     . . . - . - x . x - . x . -
     x - x - . x x - - x - . - x
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~
Ct:  K T K H R G B D P J O Y D G

Cracking this is much harder because the dividers are obfuscated.  The only way
to attack this is to do a statistical analysis for common Morse Code trigraphs.

The cipher assumes 'x' is placed between each letter and 'xx' is placed between each word.

Cryptanalysis of Fractionated Morse

Lets look at the following Fractionated Morse table:

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
. . . . . . . . . - - - - - - - - - x x x x x x x x
. . . - - - x x x . . . - - - x x x . . . - - - x x
. - x . - x . - x . - x . - x . - x . - x . - x . -

If you look at the table above, you can see that certain ciphertext letter combinations are not possible e.g. RS, RT, ... RZ cannot occur because we can't have xxx in the plaintext Morse. Also IS, IT, ... IZ can't happen for the same reason. Other impossible combinations: CY, CZ, FY, FZ, OY, OZ etc. We also can't have e.g. ABD because there are no Morse letters that go '.....-.-.'. You should be able to come up with many other impossible combinations. We can also use frequency information, since THE occurs very often in English, we expect 'xx-x....x.xx'='ZSCI' to be quite common in the ciphertext (with this particular key, if you look at the counts provided down below, we see ZSCI is actually the most common quadgram). This means certain keys will be very unlikely if they lead to impossible transitions, others more likely if they lead to likely transitions.

The trick to breaking Fractionated Morse is that finding the key can be done completely independently of the English->Morse translation. Knowing the statistics of a particular Fractionated Morse key means we can just find the key that translates to it, without ever checking to see if the putative plain text even decodes properly. The Idea is to start with a random key, then continuously swap characters in the key trying to make the ciphertext look like it has been enciphered with the key "ABCDEFGHIJKLMNOPQRSTUVWXYZ". Once this is done, we can just decrypt it. To do this we need quadgram statistics for a lot of English text that has been enciphered with the key "ABCDEFGHIJKLMNOPQRSTUVWXYZ". I did this for 40 million sentences, or around 1 billion characters. The results are linked below.

Code for breaking Fractionated Morse ciphers is here, with the required statistics here. I have run the code on many possible routes through the second Feynman cipher, and nothing resulted in any good looking plaintexts. I would expect it to work, since breaking length 100 Fractionated Morse ciphertexts is not too difficult using the code above, and F2 is 261 characters. This leads me to believe that F2 is not a Fractionated Morse cipher.

From here I'll try to write an efficient running key solver for vigenere, beaufort, variant and porta rules and see how that goes.