Code 48 : Gronsfeld Cipher

Gronsfeld Cipher

Background

The cipher has been named after Johann Franz Graf Gronsfeld-Bronkhorst (†1719). He was the imperial field marshal in the Bavarian national uprising of 1705/06.

Principle

The cipher is identical to the Vigenère cipher with the exception that only numbers can be used as keys. Instead of 26, there are only 10 possible Caesar ciphers that can be used.

Fig. 1: Key table for the Gronsfeld cipher.¹

The message "Geheimnis" has been encoded according to the key "326" which results in "JGNHKSQKY".

Security

The security is also almost identical to that of the Vigenère cipher, but the number of possible mappings is significantly smaller in the Gronsfeld cipher.
With a key length of 3, there are 26x26x26 possible mappings for the Vigenère cipher but only 10x10x10 possible mappings for the Gronsfeld cipher.

Weblinks

http://en.wikipedia.org/wiki/Bifid_cipher

References

¹ Kryptographiespielplatz

The Gronsfeld Cipher

Author: R. Morelli

This page describes a method for attacking a Gronsfeld cipher. It is based on the approach described in F. Pratt, Secret and Urgent, NY: Bobbs-Merrill, 1939.
The Gronsfeld cipher is a variation of the Vigenere cipher in which a key number is used instead of a keyword, e.g., 14965. Usually the key does not contain repeated digits.
Here's a message written in a Gronsfeld Cipher.

	cjifk qywtj ioipo wovlh ncxlo peosg gxrkx 
	baiiq caguy rxrlq klcoy vewql nhsut oiddg 
	qdrap dnfwk owpgw gzlsk xlt

For this problem, I've simplified things as follows: we allow only the digits between 0-5 (a-d) to be used in the key. The method for attacking a Gronsfeld cipher involves the following steps:

Step 1. Write the first line of the message, and then write under each of its letter, the letters that precede it in the alphabet. Since we know that this version of Gronsfeld uses only numbers between 0-5, (a-f), we need 6 rows. I've numbered the rows and columns so that we can refer to them.

   0  1  2  3  4  5  6  7
0  c  j  i  f  k  q  y  w  tj ioipo wovlh ncxlo peosg  gxrkx  (Message)
1  b  i  h  e  j  p  x  v  si hnhon vnukg mbwkn odnrf  fwqjw
2  a  h  g  d  i  o  w  u  rh gmgnm umtjf lavjm ncmqe  evpiv
3  z  g  f  c  h  n  v  t  qg flfml tlsie kzuil mblpd  duohu
4  y  f  e  b  g  m  u  s  pf ekelk skrhd jythk lakoc  ctngt
5  x  e  d  a  f  l  t  r  oe djdkj rjqgc ixsgj kzjnb  bsmfs

Step 2. Construct all reasonable trigrams using combinations of letters from the first three columns -- i.e., columns 0-2 -- taking 1 letter from each column. For example, we can get the trigram 'ahe' by picking from rows 2,2,3. We would say that the number code for 'ahe' is 223. Since this represents the first word of the message, the trigrams formed should be possible ways to start a word or phrase. In this case, 'ahe' could be the start of 'ahead.' Actually, it's not a very likely trigram, since it repeats the number 2. Make a table of the trigrams, their number codes (which represent a portion of the possible key number) and their frequencies, from Table XII in Pratt.

Trigram    Code         Frequency (Table XII in Pratt)

aid        215          24 ******** 
age        234          20 ********
aff        243          9
ahe        224          2
agi        230          3
agg        232          3
big        114          4
chi        010          22 ******** repeated numbers
che        024          27 ********
cei        050 052      13 
bed        155          2
bee        154          32 ********
bei        150          19 ********
bef        153          8
beg        152          5

Step 3. Pick the most reasonable looking trigrams from the list in step 2. In this case we've picked the following entries:

aid     215     24 ******** 
age     234     20 ********
bee     154     32 ********
bei     150     19 ********
che     024     27 ********

They are all relatively frequent trigrams. They could be used as the prefix of the first word. None of them involves a repeated digit in its number code, which rules out 'chi.'

Step 4. For each of the likely trigrams, apply the number formulas to each succeeding trigram in the message. For example, if we apply 024, to the letters in columns 1,2,3 we get the trigram, 'jgb'; if we apply it to the letters in columns 2,3,4 we get 'idg,' and so on. A partial table has been constructed below. Impossible trigrams are marked with (*). Filling in the rows for 'aid' and 'age' is left as an exercise.

Column     1     2     3     4     5

aid 215
age 234
bee 154    idb   hag   efm*  jlu*  pts
bei 150    idf   hak   efq*  jly*  ptw
che 024    jgb*  idg   fim   kou   qws*

Step 5. Note that in the table above, some of the trigrams for 'bee' and 'bei' are reasonable looking, but they don't combine well with the assumption that 'bee' or 'bei' form the first three letters of the message. For example, we can get 'bee--pts' by combining 'bee' with the trigram that starts in column 5, the first column that has a possible trigram, since 'efm' and 'jlu' are impossible. Similarly, we can get 'bei--ptw' by combining 'bei' and 'ptw', which also starts in column 5. Neither of these strings ('bee--pts' or 'bei--ptw') look very promising as the start of the clear message. On the other hand, combining 'che' as the prefix with the trigram that begins at column 4 ('kou'), gives the following partial string: 'che-kou.' That looks pretty promising. So let's work on it.

Step 6. Now, working with our partial solution, that begins, che-kou, replace the blank with each of the 6 letters from column 3 of the table in step 1. This gives us all possible trigrams for columns 2-3-4 that are consistent with che and kou. This list consists of:

efk, eek, edk, eck, ebk, eak

We want to eliminate 'efk,' 'edk,' and 'ebk' from this list, leaving Ôeek,Õ ÔeckÕ and Ôeak.Õ If we make these substitutions we get the following candidates for partial solutions:

      Candidate   Number Code  Comment	
	
      cheekou     0241024      Possibly cheek our or cheek out
      checkou     0243024      Possible check out or check our
      cheakou     0245024      Not very likely

Notice that a cycle is beginning to appear that goes 024-024 and we now have two candidates 02410241 and 02430243. If we replace the 7th letter for each of these candidates we get:

     02410241 = cheekouw     Impossible
     02430243 = checkout     ********* Solution!!!! ***********

Step 7. To complete the decipherment, apply the key number 0243 to the rest of the cipher text. You can use CryptoTool to complete this if you use the keyword 'aced.'

For Further Study and Enjoyment

CryptoToolJ. Try using CryptoToolJ break the message given at the top of the page. Even though CryptoTool does not have a Gronsfeld Analyzer, it should be able to analyze it with the Vigenere Analyzer.

CryptoToolJ. Try using CryptoToolJ to create and analyze your own Gronsfeld cryptograms.

The Vigenere Cipher. The Gronsfeld Cipher is a simple variant of the Vigenere Cipher. 

The Gronsfeld cipher is essentially a Vigenere cipher, but uses numbers instead of letters. So, a Gronsfield key of 0123 is the same as a Vigenere key of ABCD. This online version lets you encode and decode messages with a keyed alphabet as well, to allow for maximum flexibility.

Encrypt 
Decrypt : 
http://rumkin.com/tools/cipher/gronsfeld.php