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SSL Review part 1

Most of us use and rely on SSL everyday. The mathematical workings of the RSA [Rivest, Shamir, Adleman] algorithm are not overly complex but mapping everything back to what happens in reality requires detailed understanding. Skipping over the need for SSL (for confidential and authenticated exchange of a symmetric key over and insecure medium) I will review the mathematical workings then how they are applied in real world examples.

There are also details in previous posts – RSA1, RSA2

Mathematics 

Step
Components
1. public key – e
standard practice to choose: 65537 
2. random primes p,q
Let’s use:

579810099525248565010050509754571001027,

6989752565699505597485398979958574481969
3. key modulus – n
 n = pq:

4052729130775091849638047446256554071699019514021047339267026030072286291982163
4. φ(n) = (p – 1)(q – 1)
 φ(n):

4052729130775091849638047446256554071691449951355822585104530580582573146499168
5. find e that is co-prime with φ(n)
 already using a prime,which will be co- prime.. – e = 65537 (in binary 10000000000000001)
6. (ed) mod φ(n) = 1d = e–1 mod φ(n) –  Modular multiplicative inverseMore than one answer
using Extended Euclidean algorithm:

944402082567056818708092537028397604145319798848072425038015030084640082599681,

4997131213342148668346139983284951675836769750203895010142545610667213229098849,

..+ 2φ(n), +3φ(n))
7. private key – d
 d = e–1 mod φ(n):

944402082567056818708092537028397604145319798848072425038015030084640082599681

With a public key (e), a key modulus (n) and a private key (d) we can apply the RSA algorithm.

Message (mess) = 911

RSA encrypt -> mess ^ e mod n  = 911 ^65537 mod 4052729130775091849638047446256554071699019514021047339267026030072286291982163

RSA encrypted message (ciph) = 3095021178047041558314072884014000324030086129008597834642883051983162360819331

RSA decrypt -> ciph ^ d mod n = 3095021178047041558314072884014000324030086129008597834642883051983162360819331 ^ 944402082567056818708092537028397604145319798848072425038015030084640082599681 mod

4052729130775091849638047446256554071699019514021047339267026030072286291982163

= 911

How is that secure?

When Alice encrypts using Bob’s public key (e) along with the key modulus (n) the output is a protected cipher.

An eavesdropper does not know the private key so decryption is very difficult:

Attacker must solve:

(unknown val, x) ^ e mod n = ciph

x ^65537 mod 4052729130775091849638047446256554071699019514021047339267026030072286291982163 = 3095021178047041558314072884014000324030086129008597834642883051983162360819331

OR, easier – try to determine the private key:

The attacker knows e and n (which = pq). When we created the private key (step 6 above) we conducted:  e–1 mod φ(n) – Modular multiplicative inverse which is relatively fast for us to calculate.

The attacked does not know  e–1 mod φ(n) though. φ(n) = (p – 1)(q – 1). The attacker knows that n is a composite prime = pq (where p and q are both primes).

So… if the attacker can solve p * q = n (where they know n) then RSA is insecure.

Thankfully the process of Integer factorization is so much harder than the process of creating p,q,n, φ(n), e and d that online business and confidentiality can be maintained to acceptable levels.

Threats to RSA

It would be extremely valuable to malicious individuals/groups and  (more importantly) intelligence organizations make large integer factorization efficient enough to break RSA.

However the theoretical aspects of RSA are not generally recognized as the main source of vulnerability

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downloading a youtube playlist and converting to mp3 (osx/linux(ubuntu))

Downloading youtube audio and converting mp3 is very simple now thanks to these guys: https://rg3.github.io/youtube-dl/

Installing:

# on Ubuntu
sudo apt-get install youtube-dl avconv
curl https://raw.githubusercontent.com/SecurityShift/tools/master/scripts/random/GetYoutubePlaylist.sh -o GetYoutubePlaylist.sh
chmod 755 GetYoutubeAudio.sh

https://raw.githubusercontent.com/SecurityShift/tools/master/backup_scripts/google_drive_backup.py

# on OSX
brew install youtube-dl libav
curl https://raw.githubusercontent.com/SecurityShift/tools/master/scripts/random/GetYoutubePlaylist.sh -o GetYoutubePlaylist.sh
chmod 755 GetYoutubeAudio.s

Contents:



Usage: sh GetPlaylist.sh [youtube playlist URL] [output_directory]


Example:


sh ./GetYoutubePlaylist.sh http://www.youtube.com/playlist?list=PL702CAF4AD2AED35B ~/Music


To get the youtube playlist URL view the playlist by clicking on its name, no videos will be playing, then click on share and the url will be highlighted:


youtube-playlist
click on share and the URL will appear


		


	


	

		
	


	






	



	


		
			

				
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					Functional Programming - Scala Random				

			


			
Virtual Enigma Machine Implementation

		


			

		


		
	





	


		


			
After watching a video on numberphile thought it would be interesting to implement a virtual version on the Enigma cipher machine.


Step 1: Install Scala and Eclipse IDE for Scala.


http://www.scala-lang.org/downloads, http://scala-ide.org/


Step 2: Review the numberphile video and how the Enigma machine works




Step 3: Implement each function of the machine


    

  1. Rotors

      

    • Rotor choices, rotors have different circuitry (0-24) – static value for each message
      • 

    • Initial rotor combination (0-25) – static value for each message
      • 

    • Rotor increment – Faster rotor, middle rotor, slow rotor
      • 

    

    2. 

  2. Plug board

      

    • 10 wires
      • 

    • swapping 2 letters
      • 

    • static value for each message
      • 

    

    3. 



Step 4: Write an interface between the functions and a basic interface (text?)


Step 5: Test!


Conclusion: Using Scala (trying anyway) the implementation is not very complex. What is more difficult is replicating the original machines where the output character could not equal the input character. The limitation Turing used to crack the machines. Might see if I can implement that later then test out the cracking method. Not sure how easy it would be to break the current implementation…


CODE (EnigmaMachine.scala, EnigmaMachine.class):


 


 


// Virtual reconstruction of what the enigma machines did
// The replication is of the improved version where the input could result in the same output

object EnigmaMachine {
//////////////// ENCRYPTION/DECRYPTION FUNCTIONS \\\\\\\\\\\\\\\\\\\\\
    //to replicate the simple machine circuit chars are converter to 0 - 25
    def convertMessageToSimpleInt(rawInput: String):List[Int] = {
        val userMessage = rawInput.toUpperCase
      (for (x <- userMessage) yield if (x != 32) (x.toInt - 65) else 26).toList
    }
    
    //for all integers in the list, convert
    def convertMessageToString(input: List[Int]): String = {
      (for (x <- input) yield if (x == 26) " " else (x + 65).toChar).mkString
    } 
    
    def enigmaGo(rotorChoices: List[Int], rotorPositions: List[Int], inputMessage: List[Int], encrypt: Boolean): List[Int] = {    
      // incrementing the rotors is dont based on the number of previous chars.. instead of the previous position
      def incrementRots(nth: Int): List[Int] = {
        val rot0 = (rotorPositions(0) + nth) % 26 //fast rotor
        val rot1 = (rotorPositions(1) + (nth / 26)) % 26 //medium rotor
        val rot2 = (rotorPositions(2) + ((nth / 26) /26)) % 26 //slow rotor
        List(rot0, rot1, rot2)
      }
      //applies the rotor encryption then the scramble board
      def transmuteElement(element: Int, rotPosApp: List[Int]): Int = {
        val rotorTotal = rotorList(rotorChoices(0))(rotPosApp(0)) +
              rotorList(rotorChoices(1))(rotPosApp(1)) +
              rotorList(rotorChoices(2))(rotPosApp(2)) 
        
      def scramble(inputVal:Int):Int = plugMap1.find(x => (x._1 == inputVal) || (x._2 == inputVal)) match {
            case Some(y) => if (y._1 == inputVal) y._2 else y._1 
          case None => inputVal
      }
        if (element == 26) element
        else if (encrypt) (scramble(element) + rotorTotal) % 26 else { 
            //decrypt is a bit messy atm :( 
            if (element - (rotorTotal % 26) < 0) scramble(element - (rotorTotal % 26) + 26) else scramble(element - (rotorTotal % 26))
          }     
      }
      
      //Accumulator hold the input and output lists, when input list is empty, return output.. also tail recursive
      def enigmaAccu(remInput: List[Int], rotPos: List[Int], accumOutput: List[Int]): List[Int] = {
              if (remInput.isEmpty) accumOutput
            else enigmaAccu(remInput.tail, incrementRots(accumOutput.length), accumOutput:+(transmuteElement(remInput.head, rotPos)))
      }
      
      enigmaAccu(inputMessage, rotorPositions, Nil) //start the process
    }
////////////////////   \\\\\\\\\\\\\\\\\\\\\\\\\
    
//////////////// UI STUFF \\\\\\\\\\\\\\\\\\\\\
    def useMachine(rotorChoices: List[Int], rotorStartVals: List[Int], activity: String) = {
            spamLine
            val inputMessage = convertMessageToSimpleInt(askInput("Please enter a message to be " + activity + ", use only a-z: "))
            val resultingMessage = enigmaGo(rotorChoices, rotorStartVals, inputMessage, if (activity == "encrypted") true else false)
            spamLine
            println("Your message has been " + activity + ": ")
        println(convertMessageToString(resultingMessage))
            spamLine
            exit(0)
    }

    def getRotorChoices(s1:String, p: String => Boolean):List[Int] = {
        val slot0 = getUserInput(s1 + "fast slot: ", "Entry Invalid.", p).mkString
        val slot1 = getUserInput(s1 + "medium slot: ", "Entry Invalid.", p).mkString
        val slot2 = getUserInput(s1 + "slow slot: ", "Entry Invalid.", p).mkString
        List(slot0.toInt,slot1.toInt,slot2.toInt)
    }
    
    def spamLine = println("##############################################")
    
    def askInput(statement: String) = { 
      val input = readLine(statement)
      if (input == "") "99" else input
    }
    
    def askAgain(statement: String, f: => String) = {
        println(statement)
        f
    } 
    
    def inputVals(f1: => String, f2: => String) = Stream.cons( f1, Stream.continually(f2))
    
    def getUserInput(s1: String, s2: String, p:String => Boolean) = inputVals(askInput(s1), askAgain(s2, askInput(s1))).find(p)
////////////////////   \\\\\\\\\\\\\\\\\\\\\\\\\
    
//////////////// MAIN \\\\\\\\\\\\\\\\\\\\\ 
        def main(args: Array[String]) {
            println("### WELCOME TO THE ENIGMA MACHINE! ###" )
            // GET INITIAL MACHINE SETTINGS FROM USER 
            // ROTOR PLACEMENT CONFIG
            val rotorChoices = getRotorChoices("Choose rotor (0 - 4) for ", x => (x.toInt >= 0 && x.toInt < 5)) 
            // ROTOR START VALS 
            val rotorStartVals =  getRotorChoices("Choose rotor START VAL (0 - 25) for ", x => (x.toInt >= 0 && x.toInt < 25))
            // PLUG MAP
            println("There is only 1 plug map option at present:") 
            println(plugMap1.toString.drop(4))
            //ENCRYPT OF DECRYPT - ONLY RELEVENT FOR GUI TEXT.. bleh not quite need to reverse circuit manually
            val userChoice = getUserInput("Press 1 to encrypt, 2 to decrypt: ", "Entry Invalid.", x => (x == "2" || x == "1"))
            userChoice.mkString match {
              case "1" => useMachine(rotorChoices, rotorStartVals, "encrypted") 
              case "2" => useMachine(rotorChoices, rotorStartVals, "decrypted")
              case default => exit(0)
            }
    }
////////////////////   \\\\\\\\\\\\\\\\\\\\\\\\\

//////////////// CONSTANTS \\\\\\\\\\\\\\\\\\\\\
    val rotorList = List(List(6, 17, 4, 3, 23, 22, 21, 5, 0, 19, 20, 11, 16, 9, 13, 24, 18, 1, 8, 2, 7, 10, 12, 14, 25, 15)
    ,List(2, 7, 23, 12, 13, 15, 18, 1, 8, 11, 0, 3, 10, 5, 24, 4, 17, 19, 6, 9, 20, 14, 25, 21, 16, 22)
    ,List(23, 1, 19, 18, 0, 10, 3, 8, 9, 21, 2, 12, 13, 14, 24, 6, 20, 16, 11, 17, 15, 7, 4, 25, 22, 5)
    ,List(21, 22, 16, 2, 18, 20, 15, 9, 5, 1, 17, 10, 3, 6, 0, 13, 11, 23, 4, 12, 19, 8, 7, 25, 24, 14)
    ,List(25, 3, 11, 7, 5, 15, 24, 10, 21, 6, 12, 17, 8, 23, 2, 18, 19, 14, 4, 0, 20, 9, 22, 1, 16, 13))
    
    val plugMap1 = List((3,7),(11,16),(18,24),(17,9),(21,19),(14,13),(0,12),(1,25),(22,20),(4,6))
////////////////////   \\\\\\\\\\\\\\\\\\\\\\\\\
}



		


	


	

		
		


			

		


		
	


	






	



	


		
			

				
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MapReduce Programming Model and Hadoop

		


			

		


		
	





	


		


			
After hearing the praises of Hadoop I had a brief look it and the Map/Reduce paradigm. Most info was read from Wikipedia and references in wiki articles. In particular, this paper from Jeffrey Dean and Sanjay Ghemawat.


Hadoop is an opensource software framework that implements MapReduce and Google File System. The aim is to enable easy and robust deployment of highly parallel data set processing programs. It is important to note that the MapReduce model is applicable to embarrassingly parallel problems. Processing can occur on data that is stored in a database or filesystem.


Map/Reduce refers to the steps in the model:


Map: A master node takes an input problem and divides it into sub-problems, passing them to worker nodes. Worker nodes can further divide sub-problems.  For example, in the problem of counting word occurrence in a document the Map function will output a key/value pair every time it sees a a specified work – ie: (“searchterm”, 1).


Reduce: The reduce function takes the list of word(key)/values and sums the occurrences:


(foo, (1)


(bar, 1)


(foo, 1)


(bar, 1)


(foo, 1)


The output of reduce in this case could be: (foo, 3).


The MapReduce model becomes powerful when considering giant datasets can be processed in large clusters very efficiently using this model.


Hadoop Run-time Takes care of:


    

  • Parallelalization
    • 

  • Partitioning of input data
    • 

  • Scheduling programs execution across machines
    • 

  • Handling machine failures
    • 

  • Managing inter-machine communication
    • 



Hadoop aims to enable developers with little distributed programming experience to utilize compute resources such as EC2.


With the emergence of ‘big data’ and the apparent value that can be extrapolated from massive databases/datastores, many organisations have found the limits of traditional relational databases. Hadoop has such a big buzz because it can pass the processing boundaries of relational database software and enable the extrapolation of value. The video below is a decent explanation of this point by data scientists at SalesForce.com







		


	


	

		
	


	






	



	


		
			

				
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Parsing HTML with Python

		


			

		


		
	





	


		


			
Came across an interesting scenario this week where I decided to try and use Python for parsing HTML. Turns out to quite straight forward and something I could imagine using more often in the future.


We were trying to grab the a dynamic link from this page at twitch http://www.twitch.tv/directory/StarCraft%20II:%20Wings%20of%20Liberty.


The link we were after was the most viewed channel at any particular time:


The most viewed channel is always in this location/element


First attempt was to load up that page in an <iframe> allowing our own JavaScript to grab the relevant URL then forward. Twitch.tv however forces forwarding when iframes are detected, presumably for proprietary reasons. So javascript was not really an option.


After experimenting with Python’s lxml http://lxml.de/ it was really straight forward and effective. It is apparently quite efficient using C core (http://stackoverflow.com/a/6494811/692180) too. The module I used is documented quite well here: http://lxml.de/lxmlhtml.html. With a little bit of trial an error a brief python script successfully grabs the relevant information.


Then using PHP’s popen() method I can simply call the python script and use the return value as a php variable for a header redirect.


Links to source:


get_link.py – Python script


get_link.php – PHP calling python and using return value :


 


error_reporting(E_ALL);
$handle = popen('python -i ./get_link.py 2>&1', 'r');
$read = fread($handle, 2096);
pclose($handle);
header('Location: '.$read);



 


Python source:


import urllib
from lxml import html

def getElement():
        f = urllib.urlopen("http://www.twitch.tv/directory/StarCraft%20II:%20Wings%20of%20Liberty")
        # Read from the object, storing the page's contents in 's'.
        s = f.read()
        f.close()# Read from the object, storing the page's contents in 's'.
        doc = html.document_fromstring(s)
        doc = doc.get_element_by_id('directory_channels')
        #rVal = trimDoc.text_content()
        doc = doc.find_class('thumb')
        rVal = html.tostring(doc[0]).split()[2]
        return makeUrl(rVal)

def makeUrl(thumbString):
        return "http://twitch.tv" +  thumbString[6:-2]

if __name__ == "__main__":
        print getElement()



 


		


	


	

		
	


	






	



	


		
			

				
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Pascals Triangle – python

		


			

		


		
	





	


		


			
Wrote a quick script that attempts to complete pascals triangle recursively, thanks Feng.

** Note, after getting some feedback made a number of changes, thanks Fry**

see source: pascals.py


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#########################################################

# Python Script: Pascals triangle #

# Enter How many rows you want... #

# Mark Culhane 2011, straightIT.com #

#########################################################



#Initialize

tri =[[1],[1]]

n = 0



def pascals():

if len(tri) rowNum = len(tri) - 1

colCnt = len(tri[rowNum])



if colCnt == rowNum:

tri[rowNum].extend([1])

tri.append([1])

else:

p1 = tri[len(tri)-2][len(tri[len(tri)-1])-1]

p2 = tri[len(tri)-2][len(tri[len(tri)-1])]

tri[rowNum].extend([p1+p2])

pascals()

return



def displaySol():

i = 0

for x in tri:

if i != n:

print(tri[i])

i = i + 1



if __name__ == "__main__":

print "Pascals Triangle"

n = long(raw_input("How many rows do you want?:"))

print "Result: "

pascals()

displaySol()


		


	


	

		
	


	






	



	


		
			

				
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C Programming language

		


			

		


		
	





	


		


			
Over the summer break I wanted to get a better grounding in the C programming language. I was looking for a decent text book (preferably free) which was suitable for someone new to C but with an understanding of other languages and programming in general. After perusing a few I settled on:


http://publications.gbdirect.co.uk/c_book/


Working through the book was moderately interesting. Getting bogged down in memory allocation and array stepping did not really seem worthwhile though. Considering the C++ does not require in depth knowledge of these concepts meant that spending time on them was for interest sake only with little chance of use in the future.  Below are some links to some of the simple exercises I did for reference:


http://70.40.214.44/sourcecode/CLearning/


Some definite reasons for doing more learning in the C language are:



C


		


	


	

		
	


	






	



	


		
			

				
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Getting iPhone working on Linux (without jailbreaking)

		


			

		


		
	





	


		


			
Steps to Using iTunes and iPhone on Ubuntu (using windows virtual machine)




    

  1. Install git, with the following command: sudo apt-get install git-core
    2. 

  2. Install iPhone connectivity the driver with the following command: git clone git://github.com/dgiagio/ipheth.git
    3. 

  3. Install VitualBox PUEL (http://www.virtualbox.org/wiki/Linux_Downloads) , ensure it is PUEL, no OSE Edition
    4. 

  4. Install Samba and share Music folder (instructions: https://help.ubuntu.com/9.04/serverguide/C/samba-fileserver.html)
    5. 

  5. Open Sun Virtual Box, create a virtual machine (ensure in vm settings USB is enabled and iPhone filter selected).
    6. 

  6. Install iTunes on the vm and access your songs through shared folder.
    7. 



All up takes about 2 hrs max and is the easiest way to get your iPhone working on Ubuntu without jail breaking.


Continue reading “Getting iPhone working on Linux (without jailbreaking)”