Prep / Installation:

First install CUDA, most places will tell you that CUDA is incompatible with gcc-4.3 but this is not true if you make a few small changes to the configuration.mk file, please see this post about installing CUDA in ubuntu

Install the required packages (or use package manager):

sudo apt-get install python-numpy -y
sudo apt-get install build-essential python-dev python-setuptools libboost-python-dev -y

Download pycuda from: http://pypi.python.org/pypi/pycuda

Untar the achive:

tar xzvf pycuda-0.94rc.tar.gz

Configure, make and install:

cd pycuda-0.94rc
./configure.py –cuda-root=/usr/local/cuda –cudadrv-lib-dir=/usr/lib –boost-inc-dir=/usr/include –boost-lib-dir=/usr/lib –boost-python-libname=boost_python-mt –boost-thread-libname=boost_thread-mt
make -j 4
sudo python setup.py install

Problems

I had some problems with the compliation because it was complaining about pytools missing, this was resolved by removing and reinstalling python-setuptools:

sudo apt-get remove python-setuptools
sudo apt-get install python-setuptools

Running the example

You should now be able to run the hellogpu.py demo in the examples folder and use the download-examples-from-wiki.py to download further demos from the pycuda wiki.

Next
I plan to learn how to use PyCuda and aim to make a MD5 cracker as discussed previously (here and here)

The passwords were stored as MD5 hashes with a random 6 character alphanumeric salt. To create the MD5 hash of the password the salt was prefixed to the password and then the combination was hashed. Thanks to this method we can employ a simple bruteforce/dictionary attack on the passwords. I will start with the wordlists creation, then results I obtained to keep your interest, and finally show my python code.

Creating wordlists
I already has two reasnoble sized dictionaries that I use for different things like wordcube. I used john the ripper on my double sized dictionary to create lots of common permutations on words, such as captial first letter, and a number affixed to the end. To do this you run john with the following parameters, where dic.txt is the input dictionary and dic_plus_rules.txt is the output from john with all of the additions it has made.

john –wordlist=dic.txt –rules –stdout > dic_plus_rules.txt

I also download two wordlists from openwall, one which is a list of ~3100 common passwords, and one labelled ALL that has a large amount of words (~4 million) in various languages. Because of the highly compressible nature of text the files are available in small gzip files. ALL is 11.5Mb which unzips to 41.4Mb and password 12kb which unzips to 21.8kb. There are also more wordlists avaliable for different languages, but the ALL file includes these.

The size of all of the wordlists I used is shown below:

Results

Comical passwords

Here are some of the more amusingly bad passwords, the number in brackets shows the frequency of the password.

Crap passwords: 123456 (18), password (4), 1234567 (4), 123456789 (3) 12345678 (2), 12345 (2), abc123 (2), asdfgh (2), nintendo (2), 123123, abcd1234, abcdefg, qwerty
Self-describing passwords: catholic, cowboy, creator, doger, ginger, killer, maggot, player, princess, skater, smallcock, smooth, super, superman, superstar, tester, veggie, winner, wolverine
Some other passwords:bananas, cheese, cinnamon, hampster ,DRAGON, dribble1, poopie, poopoo

Python Program

# -*- coding: utf-8 -*-
#pymd5cracker.py
import hashlib, sys
from time import time

# Change to commandline swtiches when you have the time!
hash = ""
hash_file = "hash2.csv"
wordlist = "mass_rules.txt"; 

# Read the hash file entered
try:
	hashdocument = open(hash_file,"r")
except IOError:
	print "Invalid file."
	raw_input()
	sys.exit()
else:
	# Read the csv values seperated by colons into an array
	hashes=[]
	for line in hashdocument:
		line=line.replace("\n","")
		inp = line.split(":")
		if (line.count(":")<2):
			inp.append("")
		hashes.append(inp)
	hashdocument.close();

# Read wordlist in
try:
	wordlistfile = open(wordlist,"r")
except IOError:
	print "Invalid file."
	raw_input()
	sys.exit()
else:
	pass

tested=0
cracked=0
tic = time()
for line in wordlistfile:

	line = line.replace("\n","")
	tested+=1
	for i in range(0,len(hashes)):

		m = hashlib.md5()
		m.update(hashes[i][2]+line)
		word_hash = m.hexdigest()
		if word_hash==hashes[i][1]:
			toc = time()
			cracked+=1
			hashes[i].append(line)
			print hashes[i][0]," : ", line, "\t(",time()-tic,"s)"

	# Show progress evey 1000 passwords tested
	if tested%1000==0:
		print "Cracked: ",cracked," (",tested,") ", line

# Save the output of this program so we can use again
# with another program/dictionary adding the password
# to each line we have solved.
crackout = open("pycrackout.txt","w")
for i in hashes:
	s=""
	for j in i:
		if s!="":
			s+=":"
		s+=j
	s+="\n"
	crackout.write(s)
crackout.close()

print "Passwords found: ",cracked,"/",len(hashes)
print "Wordlist Words :", test
print "Hashes computed: ",len(hashes)*tested
print "Total time taken: ",time()-tic,'s'

Next

• Play with more dictionaries
• Speed up code:
  • Add multi-threading: My experience with multi-threading in python is that it doesn't work well for cpu intensive tasks, if you know otherwise please let me know.
  • Have a look at PyCUDA to see if I can use my graphics card to speed up the code significantly (another type of mutli-threading really...) without having to change language like in my previous post of CUDA MD5 cracking
• Remove hash once found to stop pointless checking
• Add command line switches to all it to be used like a real program

I wrote a quick python script which will accept an input text and create a random substitutional cipher and encrypt it. It then outputs the cipher alphabet and the encrypted text.

Source code:

# -*- coding: utf-8 -*-
import sys
from random import randint
from string import maketrans

if (len(sys.argv)>1):
	# Normal alphabet
	alphabet="abcdefghijklmnopqrstuvwxyz"

	# Randomly create a new cipherbet
	cipherbet=""
	left=alphabet
	for i in range(0,len(alphabet)):
		x=randint(0,len(left)-1)
		cipherbet+=left[x]
		left=left[:x]+left[x+1:]

	# Get input text to translate
	text=sys.argv[1].lower()

	trantab = maketrans(alphabet,cipherbet)
	text=text.translate(trantab)

	# Replace unused letters in cipherbet with _'s
	for i in cipherbet:
		if i not in text:
			cipherbet=cipherbet.replace(i,"_")

	# Print cipherbet (solution) and the text (cryptogram)
	print cipherbet
	print text

Example usage

python create_cipher.py “The Science gets done. And you make a neat gun. For the people who are still alive.”
b_lpievrm_acqxuj_fzdgwn_o_
dri zlmixli vidz puxi. bxp oug qbai b xibd vgx. euf dri jiujci nru bfi zdmcc bcmwi.

YAML is a recursive acronym that stands for YAML Ain’t Markup Language and below is a simplified example taken from the YAML site. The layout is similar to python with indents representing the nested layers. Semicolons are used to seperate variable names and value pairs.

name   : Joe Bloggs
product:
    - sku         : BL394D
      quantity    : 4
      description : Basketball
      price       : 450.00
    - sku         : BL4438H
      quantity    : 1
      description : Super Hoop
      price       : 2392.00
tax  : 251.42
total: 4443.5

To use YAML in python you will need the PyYAML library avaliable here or in via your package manager in linux. (I.E. “sudo apt-get install http://pyyaml.org/wiki/PyYAML” in ubuntu/debian).

The code for then using YAML is very simple, first we import the yaml library, then we open our settings file we have created and then use yaml’s load feature to load our settings into a python dictionary, and then finally printing it out so we can see what it contains

import yaml
f=open('settings.cfg')
settings=yaml.load(f)
print settings

Using the example YAML file above saved to settings.cfg this python script will output the following when run:

{'product': [{'sku': 'BL394D', 'price': 450.0, 'description':
'Basketball', 'quantity': 4}, {'sku': 'BL4438H', 'price': 2392.0,
'description': 'Super Hoop', 'quantity': 1}], 'total': 4443.5, 'tax':
251.41999999999999, 'name': 'Joe Bloggs'}

We can now access all the information in the file very easily (I.E. settings['name'] will give us the name Joe Bloggs).

YAML supports much much more than what is discussed here and further information can be found on the YAML website.

Screen shot of PyMan in action

Enhancements over original code

• start menu
• sounds
• multiple ghosts
• not having to hold down a directional button constantly
• can press a direction before a turn and will turn at next opertunity (like real pacman)
• probably more stuff that I’ve forgotten about by now

Source
Source Files – The source and files, to run the game just run “python PyMan.py”

References
I believe the site where I got the original code from was learningpython.com (not 100% sure) and the following links to the tutorials:
Tutorial 1
Tutorial 2
Tutorial 3

Ardunio Dicemella Annotated Photo

It (Arduino Dicemella) has 14 digital input/output pins (of which 6 can be used as PWM outputs), 6 analog inputs, a 16 MHz crystal oscillator, a USB connection, a power jack, an ICSP header, and a reset button. It contains everything needed to support the microcontroller; simply connect it to a computer with a USB cable or power it with a AC-to-DC adapter or battery to get started.

They are very useful for people who know how to program but have little experience with hardware interaction.

Programming the arduino
This post will not contain in-depth detail on how to program the arduino, instead focussing briefly on setting up serial (over serial or usb cable) communications in order to talk to a python script. The arduino can be programmed via a IDE provided by the creators in a C-style hardware language.

Code example

int ledPin = 13;            // choose the pin for the LED
int inputPin = 2;          // choose the input pin (for a pushbutton)
int val = 0;                // variable for reading the pin status

void setup() {
  pinMode(ledPin, OUTPUT);      // declare LED as output
  pinMode(inputPin, INPUT);     // declare pushbutton as input
}

void loop(){
  val = digitalRead(inputPin);  // read input value
  if (val == HIGH) {            // check if the input is HIGH
    digitalWrite(ledPin, HIGH);  // turn LED ON
  } else {
    digitalWrite(ledPin, LOW); // turn LED OFF
  }
}

Arduino LED switch circuit off

Arduino LED switch circuit on

Now we add a few lines to enable the writing of information from our arduino over the serial connection. We first need to set up the transfer speed in our setup (Serial.begin(9600);). Then we can simply send messages over serial using Serial.print(“message\n”);. You can choose between print and println with the difference been that the latter automatically appends the newline char, so we would use the former to write multiple things to the same line. Below is our modified code:

Serial write example

int ledPin = 13;           // choose the pin for the LED
int inputPin = 2;         // choose the input pin (for a pushbutton)
int val = 0;               // variable for reading the pin status

void setup() {
  pinMode(ledPin, OUTPUT);      // declare LED as output
  pinMode(inputPin, INPUT);     // declare pushbutton as input
  Serial.begin(9600);
  Serial.print("Program Initiated\n");
}

void loop(){
  val = digitalRead(inputPin);  // read input value
  if (val == HIGH) {            // check if the input is HIGH
    digitalWrite(ledPin, HIGH);  // turn LED ON
    Serial.print("LED Activated\n");
  } else {
    digitalWrite(ledPin, LOW); // turn LED OFF
  }
}

We now add into this code the ability to receive information via serial. Below is the modified example which removes the action of the button and replaces it by activating the LED when ‘Y’ is sent via serial.

Serial read example

int ledPin = 13;  // choose the pin for the LED
int val = 0;      // variable for reading the pin status
char msg = '  ';   // variable to hold data from serial

void setup() {
  pinMode(ledPin, OUTPUT);      // declare LED as output
  Serial.begin(9600);
  Serial.print("Program Initiated\n");
}

void loop(){
        // While data is sent over serial assign it to the msg
	while (Serial.available()>0){
		msg=Serial.read();
	}

  // Turn LED on/off if we recieve 'Y'/'N' over serial
  if (msg=='Y') {
    digitalWrite(ledPin, HIGH);  // turn LED ON
    Serial.print("LED Activated\n");
    msg=' ';
  } else if (msg=='N') {
    digitalWrite(ledPin, LOW); // turn LED OFF
  }
}

Interaction with python

First we import the serial library to python in order to communicate with the arduino (this includes talking over usb).

import serial

We then attempt to connect to our arduino on /dev/ttyUSB0, using try and except to catch an exception if we are unable to find the arduino on USB0. The 9600 corresponds to the baud rate (speed of communication) that we are using with the arduino and should be the same as set in the program on the arduino otherwise your communication may appear garbled.

try:
	arduino = serial.Serial('/dev/ttyUSB0', 9600)
except:
	print "Failed to connect on /dev/ttyUSB0"

The address will be /dev/ttyUSB# where # is replaced by a number for arduinos connected via usb and /dev/ttyS# where # is replaced by a number for arduinos connected via serial. If you are not sure of the location of your arduino, it can be found in the arduino IDE or you can write some python to scroll through possible locations until a response is found

locations=['/dev/ttyUSB0','/dev/ttyUSB1','/dev/ttyUSB2','/dev/ttyUSB3',
'/dev/ttyS0','/dev/ttyS1','/dev/ttyS2','/dev/ttyS3']

for device in locations:
	try:
		arduino = serial.Serial(device, 9600)
	except:
		print "Failed to connect on",device

You may need to be careful as other devices can be connected. For example if I try to connect to /dev/ttyS0 I will connect to the wacom tablet on my laptop.

Once you have connected to your arduino successfully you can write information to it using write and read information sent from it using read (you will need to import time to use the sleep function). If your arduino does not send any messages via serial then attempting to readline will result in your program hanging until it receives a message.

try:
	arduino.write('Y')
	time.sleep(1)
	print arduino.readline()
except:
	print "Failed to send!"

So the python code should now look like the following and we should be able to control the LED over serial.

import serial
import time

locations=['/dev/ttyUSB0','/dev/ttyUSB1','/dev/ttyUSB2','/dev/ttyUSB3',
'/dev/ttyS0','/dev/ttyS1','/dev/ttyS2','/dev/ttyS3']  

for device in locations:
	try:
		print "Trying...",device
		arduino = serial.Serial(device, 9600)
		break
	except:
		print "Failed to connect on",device   

try:
    arduino.write('Y')
    time.sleep(1)
    print arduino.readline()
except:
    print "Failed to send!"

The above will send the character ‘Y’ (Y for Yes please turn on the LED) to the arduino wait for 1 second and then read from the arduino which will have hopefully posted a response to our ‘Y’. Using the program on this should turn the LED on, and report LED Activated back via serial to our python program. This should be enough for people to get started with ardunios and communicating with them in python.

References

• Arduino – The arduino website with everything you are likely to need (programming examples and reference guide, and hardware information)
• Arduino tutorial – a basic and easy to understand tutorial on programming the arduino
• Python port of arduino-serial.c – By John Wiseman from which I based my program.
• original arduino-serial.c – by Tod E. Kurt.
• Sparkfun – Here is a good place to purchase ardunio and other electronics parts. Try coolcomponents if your from the uk like me
• Dealextreme – Hong Kong based retailer that sells a lot of cheap DIY electronics and also has worldwide free delivery with no min spend (crazy). Does take about two weeks to arrive though (uk).

Below is a table of the frequency of letters in the English language:

And shown graphically:

English Letter Frequency

Using the following code we can use frequency analysis to find the solution to ciphertext created using the Caesar shift demonstrated previously (see Caesar shift and Caesar shift using makestrans).

from Numeric import *
from string import maketrans

def translator(text,alphabet,key):
	trantab = maketrans(alphabet,key)
	return text.translate(trantab)

def caesar_decode(ciphertext,s):
	alpha="abcdefghijklmnopqrstuvwxyz"
	return translator(ciphertext,alpha,alpha[-s:]+alpha[:-s])

class frequency_analysis:
	def __init__(self, ciphertext):
		self.cor=[0.64297,0.11746,0.21902,0.33483,1.00000,0.17541,
		0.15864,0.47977,0.54842,0.01205,0.06078,0.31688,0.18942,
		0.53133,0.59101,0.15187,0.00748,0.47134,0.49811,0.71296,
		0.21713,0.07700,0.18580,0.01181,0.15541,0.00583]
		self.ciphertext=ciphertext.lower()
		self.freq()
		self.min_error()
		self.key=self.minimum[0]
		self.solution=caesar_decode(self.ciphertext,self.minimum[0])

	def freq(self):
		self.arr=zeros(26,Float64)
		for l in self.ciphertext:
			x=ord(l)
			if (x>=97 and x<=122):
				self.arr[x-97]+=1.0
		self.arr/=max(self.arr)

	def error(self):
		e=0
		for i in range(0,len(self.arr)):
			e+=abs(self.arr[i]-self.cor[i])**2
		return e

	def min_error(self):
		self.minimum=[0,10000]
		for rot in range(0,25):
			e=self.error()
			print rot,e
			if e<self.minimum[1]:
				self.minimum[1]=e
				self.minimum[0]=rot
			x=self.arr[-1]
			del self.cor[-1]
			self.cor.insert(0,x)

ciphertext="ymjwj fwj ybt ydujx tk jshwduynts: tsj ymfy bnqq "+\
"uwjajsy dtzw xnxyjw kwtr wjfinsl dtzw infwd fsi tsj ymfy bnqq "+\
"uwjajsy dtzw ltajwsrjsy. ymnx nx f ajwd nrutwyfsy qjxxts yt "+\
"wjrjgjwjxujhnfqqd ktw fyyfhpx zxnsl kwjvzjshd fsfqdxnx bmnhm "+\
"wjvznwj qtsljw ufxxflj tk yjcy ns twijw yt fhmnjaj gjyyjw wjxzqyx."
FA=frequency_analysis(ciphertext)
print FA.solution

This code will calculate the error in statistical frequency for each letter squared to generate an error for each possible rotation. Using a sufficiently long piece of ciphertext this code should accurately reveal the Caesar rotation use. The table below shows the error for each rotation:

Rotation	Error
0	4.11797847386
1	3.05305477067
2	3.70059678828
3	3.66330931218
4	3.5078619579
5	0.361318100755
6	3.17289666386
7	3.66072641654
8	3.39769855873
9	1.74854802027
10	2.92550921273
11	2.67524757297
12	2.86847189573
13	3.06980318397
14	2.56886153328
15	2.17180117031
16	2.24503724763
17	2.95579718798
18	1.74002183444
19	1.83328601011
20	1.74779021766
21	2.71332097813
22	1.5409364067
23	1.83209213494
24	1.54904808883

The lowest error is for 5 rotations (correctly so) with an error of 0.361318100755, the next lowest error is 22 rotations with an error of 1.5409364067. This is ~4.3x difference, which gives a very large degree of confidence to our solution and below is the deciphered text.

there are two types of encryption: one that will prevent your sister from reading your diary and one that will prevent your government. this is a very important lesson to remeberespecially for attacks using frequency analysis which require longer passage of text in order to achieve better results.

Future
The frequency analysis presented here can be used along with some other techniques in order to crack the viginere cipher.

maketrans takes an input alphabet and an output alphabet and can then be used on a string. This can be used to greatly simplify the two ciphers I produced previously. Example below:

input_alphabet="abcde"
output_alphabet="12345"
trantab = maketrans(input_alphabet,output_alphabet)
text="translate abcdefg"
print text.translate(trantab)
# This will output:
# tr1nsl1t5 12345fg

This nows means my code can be rewritten in just a fraction of what it was before:

from random import shuffle
from string import maketrans

alphabet="abcdefghijklmnopqrstuvwxyz" + \
	 "1234567890ABCDEFGHIJKLMNOPQRSTUVWXYZ"+ \
         ":.;,?!@#$%&()+=-*/_<> []{}`~^"

def translator(text,alphabet,key):
	trantab = maketrans(alphabet,key)
	return text.translate(trantab)

def caesar_encode(plaintext,s,alphabet):
	return translator(plaintext,alphabet,alphabet[s:]+alphabet[:s])

def caesar_decode(ciphertext,s,alphabet):
	return translator(ciphertext,alphabet,alphabet[-s:]+alphabet[:-s])

def substitution_encode(plaintext,alphabet):
	randarray=range(0,len(alphabet))
	shuffle(randarray)
	key=""
	for i in range(0,len(alphabet)):
		key+=alphabet[randarray[i]]

	return translator(plaintext,alphabet,key),key

def substitution_decode(ciphertext,key,alphabet):
	return translator(ciphertext,key,alphabet)

# Example useage
plaintext="The wheels on the bus go round and round. round" + \
" and round. round and round. The wheels on the bus go round and"+ \
" round, all through the town!"

print
print "SUBSTITUTION"
ciphertext,key=substitution_encode(plaintext,alphabet)
print "Key: ", key
print "Plaintext:", plaintext
print "Cipertext:", ciphertext
print "Decoded  :", substitution_decode(ciphertext,key,alphabet)

print
print "CAESAR SHIFT"
ciphertext=caesar_encode(plaintext,5,alphabet)
print "Key: ", 5
print "Plaintext:", plaintext
print "Cipertext:", ciphertext
print "Decoded  :", caesar_decode(ciphertext,5,alphabet)

This will output the following

SUBSTITUTION
Key: &fywQ.%!lmx_sRGu:{<(5jqAXvMFgk]SIY[4iK+3P8pcV$2da@DT/ZnOJ*E>-r},BH16zUb#L?N`e7C ~9t)oW^=;h0
Plaintext: The wheels on the bus go round and round. round and round.round and round. The wheels on the bus go round and round, all through the town!
Cipertext: O!Q)q!QQ_<)GR)(!Q)f5<)%G){G5Rw)&Rw){G5Rw,){G5Rw)&Rw){G5Rw,{G5Rw)&Rw){G5Rw,)O!Q)q!QQ_<)GR)(!Q)f5<)%G){G5Rw)&Rw){G5RwH)&__)(!{G5%!)(!Q)(GqR6
Decoded : The wheels on the bus go round and round. round and round.round and round. The wheels on the bus go round and round, all through the town!

CAESAR SHIFT
Key: 5
Plaintext: The wheels on the bus go round and round. round and round.round and round. The wheels on the bus go round and round, all through the town!
Cipertext: Ymj`2mjjqx`ts`ymj`gzx`lt`wtzsi`fsi`wtzsi@`wtzsi`fsi`wtzsi@wtzsi`fsi`wtzsi@`Ymj`2mjjqx`ts`ymj`gzx`lt`wtzsi`fsi`wtzsi$`fqq`ymwtzlm`ymj`yt2s&
Decoded : The wheels on the bus go round and round. round and round.round and round. The wheels on the bus go round and round, all through the town!

Conclusion
maketrans is awesome

Subsitution Cipher
Because a Caesar shift only rotates the alphabet there are only 25 possible unique solutions, this leaves the cipher quite vulnerable to brute force. If rather than just rotating the alphabet and keeping it ‘linear’ we can shuffle it to create a substitution cipher. This improves the number of possible solutions to a shocking
(26! – 1) = 4.03291461126605635584e26 (unfortunatly, the substitution cipher is alot weaker than it seems as it is vunerable to several different cryptanalysis attacks).

Substitution Cipher Diagram

We start with the simple Caesar shift function with the following changes:

• A new array is introduced filled with the numbers 0 – the length of the alphabet. This array is shuffled using shuffle from the random module. The alphabet substitution dictionary is created using this array to decide which letters go where (this is probably clearer in the code than in my explanation).
• With the Caesar shift we only needed to know the number of rotations in order to decrypt the text, now we need a full list of the letter substitutions. This is stored as the key and will be needed in order to decode our substitution cipher.
• We also now store the alphabet outside of the function so that it can be used in a decode function. This function was added along with some example usage to make the full process more understandable

from random import shuffle

alphabet="abcdefghijklmnopqrstuvwxyz"

def substitution(alphabet,plaintext):

	# Create array to use to randomise alphabet position
	randarray=range(0,len(alphabet))
	shuffle(randarray)

	key=""

	#Create our substitution dictionary
	dic={}
	for i in range(0,len(alphabet)):
		key+=alphabet[randarray[i]]
		dic[alphabet[i]]=alphabet[randarray[i]]

	#Convert each letter of plaintext to the corrsponding
	#encrypted letter in our dictionary creating the cryptext
	ciphertext=""
	for l in plaintext:
		if l in dic:
			l=dic[l]
		ciphertext+=l
	for i in alphabet:
		print i,
	print
	for i in key:
		print i,
	print
	return ciphertext,key

# This function decodes the ciphertext using the key and creating
# the reverse of the dictionary created in substitution to retrieve
# the plaintext again
def decode(alphabet,ciphertext,key):

	dic={}
	for i in range(0,len(key)):
		dic[key[i]]=alphabet[i]

	plaintext=""
	for l in ciphertext:
		if l in dic:
			l=dic[l]
		plaintext+=l

	return plaintext

# Example useage
plaintext="the cat sat on the mat"
ciphertext,key=substitution(plaintext)
print "Key: ", key
print "Plaintext:", plaintext
print "Cipertext:", ciphertext
print "Decoded  :", decode(ciphertext,key)

Running this will output the following (This will be different on each run due to the use of random to generate the key).

Key: miylbsowutgdkfvjepqhazrncx
Plaintext: the cat sat on the mat
Cipertext: hwb ymh qmh vf hwb kmh
Decoded : the cat sat on the mat

Improvement by using additional characters
Adding additional characters into the substitution will it more difficult to solve. For example if we change our alphabet from:

alphabet=”abcdefghijklmnopqrstuvwxyz”

If we include capital letters, numbers from 0 -9 and special characters:

alphabet=”ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890″+ \
“:.;,?!@#$%&()+=-*/_<> []{}`~^”+ \
“abcdefghijklmnopqrstuvwxyz”

We increase the number of possible solutions from (26!-1) to (63!-1) which is 1.98260831540444006412e87. This also has the added benefit of making the encrypted text alot cooler and harder to guess at by eye (unfortunately still very easy to see what character represents space).

Key: MXjAarLzWqFePI7E botO5f1kym29RZd3Sh8JTiQGKVDp6YBCsU4nucNgwlx0vH
Plaintext: this plaintext will be much more of a challenge to decode compared to the Caesar shift cipher
Cipertext: tzWoHEeMWIta1tHfWeeHXaHPOjzHP7baH7rHMHjzMeeaILaHt7HAaj7AaHj7PEMbaAHt7HtzaHjMaoMbHozWrtHjWEzab
Decoded : this plaintext will be much more of a challenge to decode compared to the caesar shift cipher

Future
The substitution cipher is a lot more secure than Caesar shift cipher but unfortunately is very insecure towards frequency analysis. In future posts I will address using frequency analysis and methods to prevent this type of attack as well as improving on this cipher by creating multiple-dicitionary based ciphers to create Vigenère style ciphers.

I imagine most people reading this will enjoy the simple challenge of solving some encrypted text. I have used this code to make some ciphertext, try and decode it! (extra points for knowing where it is from):

^’VtuBbtv3vut1u-w.G^’Vt&vnu-tZBtZnuIvwvtwn
-qbtuB6GN3vutbqBS-qt.BSt&wB~vtV.tqv1wbG}u5t
~nDDv5tVvG}u5tbBwvtVvtbBt@nvIvZG}u5tbqwv6tv
3vw.t@nvIvtnubBt1tUnwvG}Ztbqv.t&Swuv5tnbtqS
wbt&vI1SZvt^t61ZtZBtq1@@.tUBwt.BSm)B6tbqvZv
t@BnubZtBUt51b1tV1~vt1t&v1SbnUSDtDnuvG}u5t6
v’wvtBSbtBUt&vb1GQv’wvtwvDv1Znu-tButbnVvG

(note: newlines were placed to make it fit and do not represent a character)

Update: Made code a little cleaner by moving dictionary outside of functions.

This post aims to detail the creation (in python) of one of the simplest forms of encryption; the simple Caesar shift (or shift cipher). The Caesar shift takes the normal alphabet and maps it to a an identical alphabet with a rotation. The cipher will be written in such a way that it can be easily expanded on to create more complex encryption schemes with little modification.

Caesar shift alphabet diagram

The above image shows a diagrammatic representation of a Caesar shift of 3 (alphabet transposed onto a rotation of itself with a displacement of 3). Below shows the entire alphabet in plaintext and in ciphertext, followed by a simple sentence in plaintext and in cipher text.

Below is the code to convert plaintext into ciphertext along with an example of the usage:

def caesar(plaintext,shift):

	alphabet=["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"]

	#Create our substitution dictionary
	dic={}
	for i in range(0,len(alphabet)):
		dic[alphabet[i]]=alphabet[(i+shift)%len(alphabet)]

	#Convert each letter of plaintext to the corrsponding
	#encrypted letter in our dictionary creating the cryptext
	ciphertext=""
	for l in plaintext.lower():
		if l in dic:
			l=dic[l]
		ciphertext+=l

	return ciphertext

#Example useage
plaintext="the cat sat on the mat"
print "Plaintext:", plaintext
print "Cipertext:",caesar(plaintext,3)
#This will result in:
#Plaintext: the cat sat on the mat
#Cipertext: wkh fdw vdw rq wkh pdw

Explanation
Here we have written a function that is given the plaintext and an arbitary rotation (including negative) and returns a ciphertext. The function creates a dictionary, mapping each letter to another letter that is ‘shift’ letters away from it in the alphabet (the modulus (%) is used so to wrap the alphabet back to the start if it falls off the end). It then converts each letter of the plain text into the corresponding encrypted letter using the dictionary.

Future
In the next post I will discuss methods to improve the Caesar shift and how to turn it into a full substitution cipher (where the alphabet is shuffled rather than rotated). The Caesar shift is very susceptible to brute forcing and frequency analysis which I shall explain and create a program to defeat these encryptions in a future post.

Example WordCube image for the 12th December 2009 from www.stealthcopter.com/wordcube/2009/12/12

Below is a function I wrote to check if an input was a valid anagram (or partial anagram, as it isn’t essential to use every letter). The function works by cycling over each letter of word we are testing (word), and checks if the letter is valid (checked against chkword). If the letter is valid then it removes the letter from the original word and moves to the next letter until we run out of letters (returns True) or if the letter is invalid (returns False).

def anagramchk(word,chkword):
	for letter in word:
		if letter in chkword:
			chkword=chkword.replace(letter, '', 1)
		else:
			return False
	return True

f=open('english', 'r')
word=raw_input('Input letters (starting with mandatory letter) :')
minlen=4
count=0
for l in f:
	l=l.strip()
	if len(l)<=len(word) and len(l)>=minlen and word[0] in l and anagramchk(l,word):
		if len(l)==len(word):
			print l,'  <-- Long word'
		else:
			print l
		count+=1
f.close()
print count

This will output a list of the possible words, along with a total. The results can be seen for the WordCube in the example above here (To prevent spoiling it if you’d like to have a go at it yourself).

As always I’d be interested to see if anyone knows any faster methods or any other general improvements or comments.

The dictionary file can be found here (not perfect):
here

It would be computationally expensive to calculate all the common factors of two numbers and then compare to find the highest factor in common. Instead (as usual) there is a mathematical theory that can be used to speed up this process, in this case it’s the Euclidean algorithm.

The Euclidean algorithm subtracts the smaller number from the larger number, then using this new number and the smaller number repeats this process until the numbers are equal and hence the subtraction goes to zero. This is the highest common factor.

Example:
Find the highest common factor of 252 and 105

252-105=147
147-105=42
105-42=63
63-42=21
42-21=21
21-21=0

Below is the python code to preform this

def hcf(no1,no2):
	while no1!=no2:
		if no1>no2:
			no1-=no2
		elif no2>no1:
			no2-=no1
	return no1

This method is quite quick even and scales well for larger numbers. If you know of a faster method or can see any improvements to make then please let me know.

We start off by writing a function to check if a given word is an anagram of another word. This will be used to check a list of dictionary words against a given word in order to find its anagrams. The function accepts two string inputs, the word we are testing (word), and the given anagram to check it against (chkword).

The function loops over every letter in the word and if it is not present in chkword then we know that it cannot be an anagram of the chkword. Otherwise it is valid and that letter is removed from the chkword so that each letter may only be used once. For example checking “netting” with “testing” would fail on the second n.

def anagramchk(word,chkword):
	for letter in word:
		if letter in chkword:
			chkword=chkword.replace(letter, '', 1)
		else:
			return 0
	return 1

Now we need to get the anagram from the user. The following using raw_input to read what the user types in the console and assign it to a variable, with the optional argument used to print text so the user knows they need to do something.

wordin=raw_input('Enter anagram:\n')

We can now join both these together and loop over our dictionary file. We strip each line to remove any extra space and gumph at the end of the lines. If the length of a line is less than 4 then it is ignored,this can be changed to whatever minimum you like, so if you want it to be the same length as the input you could replace it with len(line)==len(wordin). We then use the function already written earlier to check if the word is a valid anagram and print the word if it is. Then we finish by closing the file.

f=open('english.txt', 'r')
for line in f:
	line=line.strip()
	if len(line)>=4:
		if anagramchk(line,wordin):
			print line
f.close()

Hopefully this has been helpful to someone. As ever if you have any suggestions or improvements then please leave a comment. Below is the complete version of the code

def anagramchk(word,chkword):
	for letter in word:
		if letter in chkword:
			chkword=chkword.replace(letter, '', 1)
		else:
			return 0
	return 1

wordin=raw_input('Enter anagram:')

f=open('english.txt', 'r')
for line in f:
	line=line.strip()
	if len(line)>=4:
		if anagramchk(line,wordin):
			print line
f.close()

The dictionary file I used was one I found in /usr/share/dict/british-english (most linux distro’s should have this or similar) it can also be found here

def factors(n):
	fact=[1,n]
	check=2
	rootn=sqrt(n)
	while check<rootn:
		if n%check==0:
			fact.append(check)
			fact.append(n/check)
		check+=1
	if rootn==check:
		fact.append(check)
        fact.sort()
	return fact

As always any help in speeding up my function would be appreciated. I think it could be improved using a sieve method or by checking if its divisible by 2 first then using check+=2 starting at 3 to improve the speed of the loop by a factor of 2.

Below is a the commented version to help explain what’s going on if it’s not clear.

def factors(n):
	# 1 and n are automatically factors of n
	fact=[1,n]
	# starting at 2 as we have already dealt with 1
	check=2
	# calculate the square root of n and use this as the
	# limit when checking if a number is divisible as
	# factors above sqrt(n) will already be calculated as
	# the inverse of a lower factor IE. finding factors of
	# 100 only need go up to 10 (sqrt(100)=10) as factors
	# such as 25 can be found when 5 is found to be a
	# factor 100/5=25
	rootn=sqrt(n)
	while check<rootn:
		if n%check==0:
			fact.append(check)
			fact.append(n/check)
		check+=1
	# this line checks the sqrt of the number to see if
	# it is a factor putting it here prevents it appearing
	# twice in the above while loop and putting it outside
	# the loop should save some time.
	if rootn==check:
		fact.append(check)
	# return an array of factors sorted into numerial order.
        fact.sort()
	return fact

After a bit of work I’ve written up the queens example as an evolutionary algorithm in python. Here goes:

The queens problem
The following is the quote from wikipedia regarding the problem.

The eight queens puzzle is the problem of putting eight chess queens on an 8×8 chessboard such that none of them is able to capture any other using the standard chess queen’s moves. The queens must be placed in such a way that no two queens would be able to attack each other. Thus, a solution requires that no two queens share the same row, column, or diagonal.

The above image is one of the many solutions to the problem. Hint: if you are looking for many solutions once you have found one, all of its 4 rotations (if unique) are solutions too.

Check out the wikipedia article for more information

Code
This is the current code with medium amount of comments, I have also provided a heavily commented version for people with less python experience (scroll to bottom).

# -*- coding: utf-8 -*-
##################################################################
##################################################################
# QUEENS EXAMPLE
# - Program Type: Demonstration of Evolutionary alogorithms
# - Authour: Matthew Rollings
# - Date: 14/09/09
##################################################################
##################################################################

# Import randint from the random module
from random import randint

# Creates a array I just use for convenience
board=[1,2,3,4,5,6,7,8]

##################################################################
# FUNCTIONS

# Function to check amount of attacking queens
def check(array):

 # Collisions represent number of attacking queens, starts at zero obv.
 collisions=0
 for i in range(1,9):
  if i not in array:
   print "DUPLICATE NUMBER ERROR - KILLING STUPID GENE" # Never happen 
   return 0

 # Total Collisions
 # For each queen in the array
 for i in range(0, 8):
  col=0
  # For every other queen in the array
  for j in range(0, 8):
   # avoids checking self vs self
   if i!=j:
    # if queen i is on a diagonal from queen j, then the
    # difference in magnitude x-cord will equal the difference in
    # the magnitude of the y-coord
    if abs(board[i]-board[j])==abs(array[j]-array[i]):
     # Sets a variable to tell the next part that a collision was detected
     col=1
  # If a collision was detected add one to collisions
  if col==1:
   collisions+=1
 # Return 8-colllisions, so that 0 is bad (8 attacking queens) and 8 is good (no attacking queens)
 return 8-collisions

# The reproduce function, takes two arguements, the two parents
def reproduce(array1, array2):

 # FIRST BABY (mother first then father)
 # Takes first half of mothers gene
 baby=array1[0:4]
 # Can't just add two halves as I will get duplicate numbers
 for i in array2:
  # add fathers genes going in order of numbers left
  if i not in baby:
   baby.append(i)
 # Add a little variation by giving a percentage chance of mutation
 if randint(0,100)>20:
  baby=mutate(baby)
 # Add the baby to the population and add its corresponding fitness
 population.append(baby)
 fitness.append(check(baby))

 # SECOND BABY (arrays just swapped around, Father first then mother)
 baby=array2[0:4]
 for i in array1:
  if i not in baby:
   baby.append(i)
 if randint(0,100)>20:
  baby=mutate(baby)
 population.append(baby)
 fitness.append(check(baby)) 

# Mutate the array given to the function
def mutate(array1):
 #Chooses two random places (WARNING CAN BE SAME POSITION)
 a=randint(0,7)
 b=randint(0,7)
 # Swaps the two over (temporary var must be used (c))
 c=array1[a]
 array1[a]=array1[b]
 array1[b]=c
 return array1

# Prints the population and corresponding fitness to the screen
# Only used for debugging
def printpop():
 # Prints the group
 for i in range(0, len(population)):
  print population[i],fitness[i]

##################################################################
# VARIABLES
# The size of the population (how many genes, more = faster convergance but more cpu + memory, fewer = opp.)
popsize=40
# The starting variation of the group because I havn't created a better method
# for starting off. I just add arrays of 1,2,3,4,5,6,7,8 to the population and
# give them the number of mutations between the following two values to make
# psudo random starting group
variation=[4,14]
# How many of the genes to kill each time in percentage, each dead gene will be replaced by a new child 
die=0.40
# Kill limit is calculated from the above percentage
kill_limit=die*popsize

##################################################################
# MAIN

population=[]
fitness=[]
for i in range(0,popsize):
 population.append([1,2,3,4,5,6,7,8])
 a=0
 while a<randint(variation[0],variation[1]):
  a+=1
  population[i]=mutate(population[i])
 fitness.append(check(population[i]))

maxi=0
generations=1

while maxi!=8:

 # Picks top # in group to be parents, kills rest
 killed=0
 # starting at the lowest fitness (0 and increasing untill kill limit reached)
 x=0
 while killed<kill_limit:
  for i in range(0,popsize):
   # Try is here to catch any errors
   try:
    if fitness[i]==x:
     # This bit removes the crappy gene from the population and from fitness
     population.pop(i)
     fitness.pop(i)
     # increases the kill counter
     killed+=1
    if killed==kill_limit:
     break
   except:
    break
  # increments fitness
  x+=1

 babies=0
 cpop=len(population)-1 #current population
 while babies<killed:
  # produces two babies from two random parents (should prob give fittest parents preference)
  reproduce(population[randint(0,cpop)],population[randint(0,cpop)])
  babies+=2 

 generations+=1

 # Looks for highest fitness in the group and checks if any have won!
 maxi=0
 for i in range(0,popsize):
  if fitness[i]>maxi:
   maxi=fitness[i]
   if maxi==8:
    print population[i]
    break

print "Took",generations,"generations"

Quick Analysis
I did a very quick analysis keeping the kill off percentage at 0.4 and varying the population size to see the improvement on generations required before a solution is found. I’ve also added a link (scroll to bottom) to the .ods file I used to calculate this (and an excel .xls file for the windows chaps). Below is the graph I calculated for different population sizes and it looks like we would expect, decreasing as the population increases. I only did 10 averages for each point and there were some anomalies, which are to be expected with this method of programming as it essentially relies on random numbers so it is a bit noisey:

Links
Note the following might be more desirable to copy and pasting the above as they use tabs rather than spaces.
queens.py – The normal version
queens_commented.py – The heavily commented version.

evolutionary.ods – Analysis in ods format (open office)
evolutionary.xls – Analysis in excel format

If you have any corrections or suggestions please contact me.

IOError: [Errno 2] No such file or directory: 'test2/file.txt'

I imagine this is deliberate as you don’t want to be accidentally creating folders. So to make a single folder we import the os module, and use the mkdir command

import os
os.mkdir('newfolder')

This is good, but if we try and make a subdirectories using the single command then it will fail as this command will only try and create the last directory in the string you pass. IE: if you pass dir1/dir2/dir3 it will assume dir1 and dir2 already exist and try to make dir3, which will fail with:

OSError: [Errno 2] No such file or directory: 'dir1/dir2/dir3'

So in order for this to work we would need to make it recursivly create subdirectories, but thankfully someone already went to the trouble for us with ‘supermakedir’ or makedirs.

import os
os.makedirs('dir1/dir2/dir3')

So this is probably useful enough for most people, but I needed to create folders in order for a file to get dropped in it, however I was only given the full path for the file not seperated into filename and directory path.

So I made the following function that accepts an input such as ‘dir1/dir2/dir3/filename.txt’ and cuts it into the folder path using os.path.dirname(filename), and then checks if the folder exists using os.path.exists(folder), and then makes the folders if they don’t already exist.

def mkdirnotex(filename):
	folder=os.path.dirname(filename)
	if not os.path.exists(folder):
		os.makedirs(folder)

First I’ll explain the maths so you can visualise what’s going on. As we should know _pi_ is the ratio of circle’s radius to its circumference, which is conveniently the same as the ratio of a circle’s area to the square of its radius (wiki…)

So what we are going to be doing is picking lots of random coordinates in an x-y grid and calculating if they are within the circle or the square.

We will assign the radius to be 1, because that makes it easy to work with. By default a random number in python ( random() ) will return a floating point number between 0 and 1. To test if a point is within a circle we simply use Pythagoras.

So if the sqrt(a**2+b**2)<=1 then the point lies inside the circle’s radius. In the diagram above we see that point A lies within the circle, and point B lies outside the circle.

We can really don’t need to use the whole circle as it has symmetry, so we can just take a quartre, which makes the generating of random numbers easier as you only need to use a random number for x and y between 0 and 1, rather than -1 and 1. It will look like the diagram below.

Now for a confusing bit of maths. We are calculating the ratio of the area of a circle to the area of a square.

# Area of circle
A=pi*r**2
# where r = 1
A = pi
# Area of square
A = l ** 2
# in this case (see diagram) our square's length is twice the radius
l=2*r
A=(1+1)**2 = 4

#Therefore our ratio will be pi : 4.
# Which means we must multiply our result by four to get pi.

Final version (efficient for using)

from random import *
from math import sqrt
inside=0
n=1000
for i in range(0,n):
	x=random()
	y=random()
	if sqrt(x*x+y*y)<=1:
		inside+=1
pi=4*inside/n
print pi

Below we can see the values it creates

n	calc	error
1	4.00000000	0.73686317
10	3.60000000	0.45840735
100	3.24000000	0.09840735
1000	3.06400000	-0.07759265
10000	3.16160000	0.02000735
100000	3.14140000	-0.00019265
1000000	3.14293600	0.00134335
10000000	3.14117920	-0.00041345

So we can see that the program quickly solves pi to about two decimal places, but it is a terribly inefficient method and will struggle to get much more accuracy than this.

Resources to check out:
This blog post – Solves pi via taylor series expansion
Super pi – Program that calculate pi often used for benchmarking

Method 1

def ispalindrome(num):
	n=str(num)
	while len(n)>1:
		print n
		if n[0]!=n[-1]:
			return 0
		n=n[1:-1]
	return 1

I thought that it would be faster avoiding a string conversion, and to somehow use the modulus (modulo) function. However when I came to write it, I found it quite difficult to code, and I’m sure there must be a better way.

Method 2

def ispalindrome2(num):
	l=1
	while num/10**l>=1.0:
		l+=1
	r=0
	d=[]
	for i in range(1,l+1):
		p=num%10**i-r
		r+=p
		p=p/10**(i-1)
		d.append(p)

	for i in range(0,l/2):
		if d[i]!=d[-i-1]:
			return 0
	return 1

Doing the speed tests show that Method 1 is over 5.2 times faster than Method 2.

Method 1
0.355437994003 Seconds elapsed
Method 2
1.85815691948 Seconds elapsed

Update: Mike of mikemeat sent me his method using slices (Method 3) and I adapted it slight (Method 4), then shortly after I realised it could be even more efficient by not needing to differentiate between even and odd strings (Method 5).

Method 3

def ispalindrome3(x):
    z = str(x)
    if len(z)%2 == 0 and z[:len(z)/2]==z[-len(z)/2:][::-1]:
       return 1
    if len(z)%2 != 0 and z[:(len(z)- 1)/2]==z[(-len(z) + 1)/2:][::-1]:
       return 1
    else:
	return 0

Method 4

def ispalindrome4(x):
	z = str(x)
	if z[:len( z)/2]==z[len( z)/2+len( z)%2:][::-1]:
		return 1
	return 0

Method 5

def ispalindrome5(x):
	z = str(x)
	l=len(z)/2
	if z[:l]==z[-l:][::-1]:
		return 1
	return 0

Method 1
0.357168912888 Seconds elapsed
Method 2
1.83943104744 Seconds elapsed
Method 3
0.179126977921 Seconds elapsed
Method 4
0.179482936859 Seconds elapsed
Method 5
0.149376153946 Seconds elapsed

Method 1:

def digitsum(x):
	total=0
	for letter in str(x):
		total+=int(letter)
	return total

I thought that this could be improved using ord, which converts a letter into its decimal ascii number. Numbers ’0′, ’1′, ’2′ … ’9′ correspond to the ascii values of 48 – 57 and then took the moduli of this with 48 to give the integer value. I later realised that this was completely nonsensical and should have just subtracted 48, but I decided to include it for the purposes of the speed test.

Method 2:

def digitsum2(x):
	total=0
	for letter in str(x):
		total+=ord(letter)%48
	return total

Method 3:

def digitsum3(x):
	total=0
	for letter in str(x):
		total+=ord(letter)-48
	return total

Speed Test:
The test uses a long number and one million repetitions for each method.

from time import time

# .. functions go here

# Nice long number to sum
x=981234153134415646571899783156122451653

tic = time()
for i in range(0,1000000):
	digitsum(x)
print time() - tic, 'Seconds elapsed'

tic = time()
for i in range(0,1000000):
	digitsum2(x)
print time() - tic, 'Seconds elapsed'

tic = time()
for i in range(0,1000000):
	digitsum3(x)
print time() - tic, 'Seconds elapsed'

Results:

#Method 1
29.3496568203 Seconds elapsed
#Method 2
12.185685873 Seconds elapsed
#Method 3
9.59367895126 Seconds elapsed

So we can see that the first method is much slower, avoiding the integer conversion by using ord speeds it up the function by ~60% and that using subtraction rather than modulus (a division based operation) saves a further ~20% on top of this.

word=raw_input('Crossword Solver \nuse * as a wildcard: ')
f=open('dic.txt', 'r')
for line in f:
	line=line.strip()
        if len(line)==len(word):
		good=1
		pos=0
		for letter in word:
			if not letter=='*':
				if not letter==line[pos]:
					good=0
			pos+=1
		if good==1:
			print line
f.close()

Example usage:

Crossword Solver
use * as a wildcard: *arn*val
carnival

The dictionary file I used is 608.2Kb with 80,368 english words and avaliable here

# os module required
import os

# retrieves information for the harddrive where root is mounted
# in windows replace this with "C:\" or the relevant drive letter
disk = os.statvfs("/")

# Information is recieved in numbers of blocks free
# so we need to multiply by the block size to get the space free in bytes
capacity = disk.f_bsize * disk.f_blocks
available = disk.f_bsize * disk.f_bavail
used = disk.f_bsize * (disk.f_blocks - disk.f_bavail) 

# print information in bytes
print used, available, capacity

# print information in Kilobytes
print used/1024, available/1024, capacity/1024

# print information in Megabytes
print used/1.048576e6, available/1.048576e6, capacity/1.048576e6

# print information in Gigabytes
print used/1.073741824e9, available/1.073741824e9, capacity/1.073741824e9

You can argue about if they should be KiB or KB if you want, but i take them as 1024 bytes in a kilobyte

However sometimes this is not necessary if you only require one prime number, or if memory limitations mean a sieve would be inconceivable. The following is the code I wrote to check if a number is prime or not in python. It tests upwards checking if the number is perfectly divisble, and as it does this it lowers the maximum number need to reach to stop as we know that if a number is not divisible by 2 then it will not be uniquely divisible (a repatition of a divislbe may exist) by anything greater than N/2

def isprime(number):
	if number<=1:
		return 0
	check=2
	maxneeded=number
	while check<maxneeded+1:
		maxneeded=number/check
		if number%check==0:
			return 0
		check+=1
	return 1

I hope that this made sense, and is useful to someone. If anyone has any more efficient methods I would be happy to hear from you.

Check out my profile to see which problems I’ve completed. If you’ve completed any I haven’t please get in contact and give me some tips.

Update:
Mike sent a suggestion that I could speed up the program by ignoring all factors of 2, it could also propbably be sped up by looking at 3,4,5 .. etc until certain point.

def isprime(number):
	if number<=1 or number%2==0:
		return 0
	check=3
	maxneeded=number
	while check<maxneeded+1:
		maxneeded=number/check
		if number%check==0:
			return 0
		check+=2
	return 1

So lets test the speed difference by doing 1,000 checks of the number 982,451,653 (known to be prime from this usefulsite)

Original Code: 9.99917411804 Seconds elapsed
Improved Code: 5.2977039814 Seconds elapsed

That’s approximately a factor of two speed increase, but this makes me think that a combination of the sieve method with this one may lead to an even further increase.