Schneier on Security

Clive Robinson • March 30, 2011 7:51 AM

Mind you what does it say about the organisation that took this man on as an “IT worker”?

Or is his real role in “IT” lower than say a data entry clerk?

I look forward to the Defence Experts article when it comes out (he’s actually a very good writer and an “inventor” of security devices in his own right that the UK Gove decided to nick)

Clive Robinson • March 30, 2011 11:23 AM

@ phred14,

“Next thing someone’s going to suggest some sort of cypher usi a deck of cards and a Solitaire-like game.”

Only if they are showing a bit of bias 😉

http://www.ciphergoth.org/crypto/solitaire/

Clive Robinson • March 31, 2011 5:50 AM

@ Mark,

“How does the alphabet length affect a subsitution cypher?”

The simple answer is it’s very dependant on many other things.

Thus you need to first consider the cipher “block size” or alphabet size in more general terms (ie not in substitution / code book modes).

For instance in a stream cipher it is effectivly bit by bit encryption therefore the alphabet size is 2 (0,1). However used properly stream ciphers can be very strong (the OTP is effectivly a stream cipher with a user selected alphabet size) That is the strength comes not from the alphabet but from the stream generator producing an unknown binary alphabet for each bit.

Then consider a large code book where messages are enciphered not on individual input alphabet charecters but on collections of charecters that go to form largish blocks. An example would be using AES 256 in Code Book mode with the input alphabet being Baudot (32bit) 51 charecters per block.

Further then consider how you actually break a simple substitution cipher, you do it by frequency counts of individual chars in the cipher text or binims (2 chars) trinims (3 chars) etc and compare to the frequency table of the assumed input language and alphabet.

If you break the frequency count corelation between the input alphabet and cipher text alphabet you make the job considerably more difficult.

Shannon and others before him recognised this but Shannon actually gave a mathmatical way to measure the “redundancy” in ciphertext alphabets and gave a new meaning to the word “entropy” and produced a measure for a given alphabet and language called the “unicity” distance. Thus any message bellow this length (about 26 chars in english slightly less in a single case alphabet) is in theory not uniquely decodable from the cipher text. If I remember correctly the unicity distance for DES in ordinary Code Book was just over two blocks…

So one way to make a substitution cipher less amenable to the standard attack methods is to some how break up the statistics of the individual chars and their relationships when in pairs tripples etc.

There are a couple of ways to do this the first is to simply transpose the input message in blocks greater than the language unicity distance in a method that uses a uniformly increasing steped size (that is similar to selecting the wire order on a rotor wheel). There are known fairly easy pencil and paper ways to do this based on “code words” (look up WWII “poem codes” to see how it was done for double transposition).

The second thing that can be done is to change the frequency appearence of plain text chars to “flatten the statistics”. Again there are fairly easy ways to do this with pencil and paper (see the “straddeling checkerboard” in for instance the Foote version of the Russian “nilihist code”).

Variations of these were used in what was considered to be a very strong pencil and paper cipher in the 1950’s one such version was the VIC Cipher which also included a lagged Fibonacci generator for generating an adition key stream. It remained unbroken even to the NSA untill it’s basic structure was revealed by a defector.

Thus it becomes clear that as Shannon noted a repeated application of transposition then substitution could produce a very strong cipher system (which is still what most block ciphers actually do).

But the required “complexity” of the system is very much dependent on how much traffic is to be sent.

The usuall way to attack unknown substitution or transposition ciphers is by the “indepth method” where messages assumed encoded under the same or related keys are stacked up and subjected to various anylitical techniques to reveal any latent structure and statistics and by repeated application un roll the layers one by one like pealing an onion with only your bare hands.

So the less traffic sent in the system and under any one key the better…

Finaly one weak system in many ciphers is “standard openings” and “standard closings” a simple way to hide these is to use “Russian Coupling” where you split the cipher text up into blocks and transpose them such that the message begining and ending is somewhere in the middle.

So yes it is possible to make semi-secure pencil and paper ciphers, but usually they suffer from the human ability to make unintentional mistakes, so the systems are in general weak to allow for human failings…

Clive Robinson • April 1, 2011 4:19 AM

@ Robert in San Diego,

“I know someone out there has released an alphabet die, so the biggest problem with OTP generation now is sourcing, and disposing of, the carbon paper you’d need to create the two sheets without using a copier.”

An “alphabet die” I’d like one as a “desk toy”, but you don’t need it.

All you realy need to generate a rough and ready OTP is two dice of the same size but different colours (or two dice of the same colour but ink in the spots on one) and a simple six by six grid.

You can make either a “letter” or “number” OTP with the grid.

For a “letter OTP” you fill the alphabet in five letters at a time in the grid rows (ie in the 1 to 5 columns and leave the sixth blank) in the first five rows. In the sixth row you put the Z in the sixth column.

To use first decide which die is for the columns and which for the rows (and stick with it for the entire time you make the OTP).

For each letter throw both dice and look up the intersect square, if it contains a letter write it down, if it’s blank (and it should be on average for 10 in 36 throws) just throw both dice again untill you get a valid letter.

For a “Number OTP” just put 1-5 in the odd rows, and 6-0 in the even rows giving you thirty filled spaces and six blank spaces.

Providing you are only making a small amount of KeyMat two dice are ok.

You will also find it’s a lot quicker if you can use a 1lb (454g) glass “jam jar” or other transparent container with a lid put both dice inside. Then shake it sufficiently hard with your non writting hand so the dice hit the lid and bottom three or four times and let them fall to the bottom. With a little practice you can get one letter or number every 4 or 5 seconds and can keep it up at that rate to write down a thousand letter pad on A4 paper (ie ten 5 letter groups every third line).

I do it slightly differently, I use a very old PC and a dot matrix printer and two part stationary, I wrote a small program years ago in Apple Pascal (later converted to turbo Pascal then Turbo C) where you just type in the letters or numbers one by one and it prints them out in a nice format (ie six five letter groups in three line boxes fifteen rows to a page with a serial number at the top etc) on two part fan fold stationary that is also punched for putting in a ring bound folder.

Surprisingly you can turn out about 15 pages an hour and a couple of days days work gives me a couple of hundred pages which is all the out station emergancy KeyMat I need for a year or so these days.

Importantly the printer “ribbon” goes back in the safe with the disk, jar and folder of printed fanfold or it goes out the back to the BBQ pit where it gets reduced to less than ashes.

One important thing to note with OTP’s in use, you need a piece of glass slightly bigger than your “pad sheet” to write on otherwise you could leave a tell tale impression (why do you think all ‘signals’ officers/ yeomen / asorted REMFs have glass tops to their tables, it’s not there just to hold down the photos or standing odrers 😉